Power Machines N5 2019 combined

T1390(E)(A11)T NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 11 April 2019 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated steam tables (Appendix to BOE 173) Students need drawing instruments. Non-programmable calculators may be used. This question paper consists of 5 pages and a formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Start each question on a NEW page. 5. Use only BLUE or BLACK ink. 6. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- QUESTION 1 1.1 There are two laws of thermodynamics, namely Joule's law (the first law) and Clausius' theory (the second law). State Clausius's theory. (2)  1.2 What is the function of a governor in a mechanical mechanism? (2) 1.3 What are the TWO basic products of combustion resulting from a chemical reaction between a hydrocarbon and oxygen? (2) 1.4 Name TWO particulars of manufacturers' information that should be engraved or printed on the boiler. (2) 1.5 Name the TWO classifications of steam turbines depending on how steam expansion takes place. (2) [10] QUESTION 2 A certain gas with a volume of 0,5 m3 is heated isothermally from an initial pressure of 220 kPa and a temperature of 30 °C to a pressure of 1,5 MPa. The gas is then further expanded at a constant volume to a temperature of 20 °C. The gas characteristic constant is 0,296 kJ/kg. K and the specific heat capacity at constant volume is 0,709 kJ/kg.K. Calculate the following: 2.1 The volume at the end of the isothermal process  (3) 2.2 The final pressure at the end of the constant volume process (3) 2.3 The specific heat capacity at constant pressure in kJ/kg.K (2) 2.4 The quantity of heat transferred during the isothermal process and the heat received (absorbed) during the expansion process in kJ/kg (4) 2.5 The total change in internal energy for the entire process in kJ/kg (4) 2.6 The total change in entropy for the entire process in kJ/kg  (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- QUESTION 3 Steam with a mass of 2 kg at 2 500 kPa and 250 °C is expanded to 250 kPa. The steam is 9% wet. The index of expansion . The specific heat capacity of superheated steam is 2 kJ/kg.°C. Calculate the following: 3.1 The total enthalpy of superheated steam in kJ (4) 3.2 The total volume of superheated steam in m3 (4) 3.3 The total enthalpy of wet steam in kJ (4)  3.4 The total volume of wet steam in m3 (4) 3.5 The total change in internal energy in kJ (4) [20] QUESTION 4 4.1 A quantity of 3 200 kg of steam per hour flows into a condenser at a pressure of 11 kN/m2 and is 9% wet. The air leakage is 1,5 kg per 1 000 kg steam. The temperature of the air pump and the condensate is 41,6 °C. Cooling water has a mass flow rate of 12 kg/s and it enters at a temperature of 20 °C. Take the gas characteristic constant as 0,296 kJ/kg.K and the specific heat capacity of water as 4,187 kJ/kg.°C.  Calculate the following: 4.1.1 The outlet (final) temperature of the cooling water (6) 4.1.2 The capacity of the air pump in m3/minute (5) 4.2 The composition of a gaseous fuel has the following structural analysis according to volume: H2 = 37% CH4 = 29% CO = 18% CO2 = 1% N2 = 15% Air composition consists of 21% oxygen and 79% nitrogen Calculate the following: 4.2.1 The volume of air required to completely burn 1 m3 of the fuel  (5) 4.2.2 The total percentage analysis, by volume, of ALL the products of combustion (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- QUESTION 5 In a steam turbine test rig, a recorded quantity of 2 kg of steam per hour enters an impulse steam turbine at a velocity of 900 m/s and is discharged through a nozzle inclined at 20° to plane of rotation of the turbine. The steam enters a series of turbine runner blades that are rotating at an average tangential velocity of 350 m/s, with each runner blade having an outlet angle of 32°. Take the relative velocity coefficient for the turbine system as the steam flows over the runner blades as 0,75. 5.1 Construct a velocity diagram in the ANSWER BOOK on landscape to a scale of 1 cm = 50 m/s. Indicate all the angles and the velocity values in m/s on the velocity diagram.  NO MARKS WILL BE AWARDED FOR THE VELOCITY DIAGRAM IF THE INSTRUCTIONS ARE NOT FOLLOWED ACCORDINGLY. (10) 5.2 Use the diagram to determine the following: 5.2.1 The angle at which the steam leaves the turbine 5.2.2 The inlet angle of the moving blade (2 × 1) (2)  5.3 Use the diagram to calculate the power developed by the turbine. (3) [15] QUESTION 6 Air enters a single stage, single-acting reciprocating air compressor at a rate of 15 m3 per minute at a temperature of 42 °C and a pressure of 100 kPa. The air leaves the compressor at a pressure of 700 kPa. The compressor runs at a speed of 350 r/min and a clearance volume of 5% of the swept volume. Take the index of compression as n = 1,3 and the mechanical efficiency as 85% of the reciprocating compressor. Calculate the following: 6.1 The cylinder volume of the compressor in m3 (9) 6.2 The delivery temperature of the compressor  (2) 6.3 The power required to drive the compressor in kW (4) [15] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

(8190035) -1- POWER MACHINES N5 FORMULA SHEET 1. Q  W  U n 1 n 1 T2  P2  n V  17.     1  2. U  mC T T1  P1   V2  3. Q  mC T  1  1 T2  P2   V  18.     1  4. Q  mC p T T1  P1   V2  19. h  h f  h fg V2 5. Q  P1V1 ln V1 20. h  hg  C p T  T V  21. h  h f  h fg  hg 6. S  m C ln 2  R ln 2   T1 V1  n  1  hsup  1 941  22. Vsup    7. W  P1V n  P  V2 8. W  P1V1 ln  Vm V1 23. Vg P1V1  P2V2 9. W  M n 1 24.  M m P1V1  P2V2 25. U = H - PV 10. W  1 1 2 26. gZ1  U1  P1V1  C1  Q  11. R  C p  C 2 1 Cp gZ 2  U 2  P2V2  C2 2  W 12.   2 C  m s (h2  h1 ) 27.   P1V1 P2V2 13.  m  CV T1 T2  14. PV  mRT m s (h2  h1) 28. EE   m  2 257 15. P1V1  P2V2 101,325 P1V1n  P2V2n 29. p  ( Bm  M m ) 16. 760 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- 100  8  30. m C  8 H 2  S  O2  23  3   y y 31. Cx H y   x  O2  xCO2  H 2O  4 2  O  32. H .C.V .  (CVC .C )  CVH 2  H 2  2   (CVs .S )  8  33. L.C.V .  H .C.V .  h fg (9H 2 ) (mw  me ) C p T 34. H .C.V .  mf  n 1   n 1   n   P2  n   n   P2  n  35. W  P1Ve      1  mRT1      1  n  1   P1    n  1   P1        1  1 Ve Vc  P2  n  36. c  .100  1     1  1     (rp ) n Vs Vs  P1      V 37.   .100 Vs  38. Fc  m(C fe  C fi )  39. P  mU [Cwi  (Cwe )] 2U [Cwi  (Cwe )] 40.  .100 Cai 2 41. U = DN  42. m V  AC 43. (m  M ) g  m 2h  44. Vs  D2 L 4 45. θ1  tc  twi Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- 46. θ2  tc  two θ1  θ2 47. Log.temp.diff.  θ n 1 θ2 P  48. Piso  P1 V1 n  2   P1   n 1    P1 V1    1  n P2 n 49. Pact  n 1  P    1   Piso 50. Niso   100 Pact Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE NATIONAL CERTIFICATE POWER MACHINES N5 11 APRIL 2019 This marking guideline consists of 15 pages. CHIEF MARKER: _____D SAROTI_________ SIGNATURE: _______________ INTERNAL MODERATOR: N NOMANDLA_ SIGNATURE: MCM: M DU PREEZ SIGNATURE: DATE: 13 APRIL 2019 Kindly note that this symbol √ mean ½ Mark and this symbol  mean 1 Mark Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2 POWER MACHINES N5 QUESTION 1 1.1 Clausius's theory states that heat can flow from a hot substance (higher temperature) to a cold substance (lower temperature) without the aid of external work (naturally) and without naturally reversing the flow process.  (2) OR Clausius theorem states that a system exchanging heat with external reservoirs and undergoing a cyclic process, is one that ultimately returns a system to its original state.  (2) 1.2  To control or maintain the desired speed of an engine or a mechanical mechanism.  OR  To regulate or keep the speed of an engine within required limits or specifications.  (Any 1 x 2) (2) 1.3  Carbon dioxide or CO2   Water vapour (steam) or H2O (2) 1.4  The name of a manufacturer  The factory number  SABS standards ( SABS Stamp)   The year of construction  The design pressure in Pascal  Complete guide to inspection  Tags for boilers  Valve regulations for boilers (Any 2 x 1) (2) 1.5  Impulse turbine  Reaction turbine (2) [10] Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -3- POWER MACHINES N5 Kindly note that this symbol √ mean ½ Mark and this symbol  mean 1 Mark QUESTION 2 2.1 √ √   (3) ALTERNATIVE METHOD 𝑃1 𝑉1 = 𝑚𝑅𝑇1 𝑃1 𝑉1 𝑚= 𝑅𝑇1 220𝑥0,5 𝑚 = 0,296𝑥303√ 𝑚 = 1,226𝑘𝑔√ 𝑃 𝑚 = 𝑅𝑇1 √ √ ½ x 6 =3 Marks 1 𝑚𝑅𝑇2 𝑉2 = √ 𝑃2 1.226𝑥0,296𝑥303 𝑉2 = √ 1500 𝑉2 = 0,0733𝑚3 √ (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- POWER MACHINES N5 2.2 P2V2 P3V3 from ,   mR T2 T3 P2 P3  ...V2  V3 √ T2 T3 P2T3 P3  √ T2 1500  293   303  P3  1450, 495 kPa (3)  2.3 R  C p  Cv C p  R  C v√ √ √ √ ½ x 4 =2 Marks C p  0, 296  0,709 C p  1,005 kJ / kg.K√ (2) 2.4 isothermal process : V2 Qisothermal  PV 1 1 ln  mC v  T2  T1  V1 √ 0, 073 √   220  0, 5  ln 0 0, 5 Qisothermal   211, 656 kJ  cons tan t volume process : Qvol  P  V  mcV  T3  T2  √ √  0  0, 709 (303  293) √ Qvol  7, 09 kJ / kg (4)  2.5 for isothermal process :  U isothermal  mC v  T2  T1   0, 709(293  293) √  U  0 kJ / kg  for cons tan t volume :  U volume  mC v  T3  T2   0, 709  303  293  √   U volume  7, 09 kJ / kg   U total   U volume   U isothermal  0  7, 09 √  7, 09 kJ / kg√ (4) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- POWER MACHINES N5 Kindly note that this symbol √ mean ½ Mark and this symbol  mean 1 Mark 2.6  V T   T P   S total   R ln 2  CV ln 2    C P ln 3  R ln 2   V1 T1   T2 P3   √  0, 073  √  293    √  303  √  1500     0, 296 ln    0, 709 ln     1, 005 ln    0, 296 ln     0, 5   293     293   1450, 495   √ √    0, 5695    0, 033728  0, 00993  4 Marks   S total   0, 546 kJ / kg  Alternative method :  V T   V T   S total   R ln 2  CV ln 2    R ln 3  CV ln 3   V1 T1   V2 T2   √  0, 073  √  293    √  0, 073  √  303     0, 296 ln    0, 709 ln      0, 296 ln    0, 709 ln     0, 5   293     0, 073   293   √ √    0, 56955    0, 02379  4 Marks   S total   0, 546 kJ / kg  (4) [20] QUESTION 3 3.1 At P1  2500 kPa : t s  223, 9 o C √ t ss  250 o C √ hg  2801kJ /kg  1 Mark h1  m  hg  c p  t ss  t s   √  2  2801  2  250  223, 9   √  2  2853, 2 √  5706, 4 kJ / kg √ (4) 3.2 At P1  2500 kPa : h1  2853, 2 kJ / kg  n  1  hsu  1941  v1    n  P  1, 3  1  2853, 2  1941     1, 3  2500  m3  0, 0842  2 kg  4 Marks kg  0,1684 m 3  (4) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- POWER MACHINES N5 Kindly note that this symbol √ mean ½ Mark and this symbol  mean 1 Mark 3.3 At P2  250 kPa : h f  535 kJ / kg  h fg  2181kJ / kg  h2  m  h f  xh fg  √  2  535  0, 91  2181   2  2519, 71  4 Marks  5039, 42 kJ / kg √ (4) 3.4 At P2  250 kPa : V g  0, 7184 m 3 / kg  V2  xV g  0, 91  0, 7184  m3  0, 6537  2 kg  kg 4 Marks  1, 307 m 3 / kg  (4) 3.5 U 1  h1  p1V1 √ √  5706 , 4   2500  0,1684   5285 , 4 kJ √ U 2  h2  p 2V2 √ √  5039 , 42   250  1, 307   4712 , 67 kJ √  U  U 1  U 2   5039 , 42  4712 , 67 √  572 ,85 kJ ( gained ) √ 4 Marks OR U  U 2  U 1  4712 , 67  5039 , 42 √   572 ,85 √  572 ,85 kJ (lost ) √ Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- POWER MACHINES N5 Kindly note that this symbol √ mean ½ Mark and this symbol  mean 1 Mark Alternative Method U 1  h1  p1V1 √ √  2853, 2   2500  0, 0842   2642 , 7 kJ / kg √ U 2  h2  p2V2 √ √  2519 , 71   250  0, 6537   2356 , 285 kJ / kg √  U  m U 1  U 2   2  2642 , 7  2356 , 285  √  572 ,83kJ 4 Marks  572 ,83 kJ ( gained )√ OR U  m ( U 2  U 1 )  2 ( 2356 , 285  2642 , 7 )√   572 ,83 kJ  572 , 83 kJ (lost )√ (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- POWER MACHINES N5 Kindly note that this symbol √ mean ½ Mark and this symbol  mean 1 Mark QUESTION 4 4.1 4.1.1 P  11 kPa From the steam table at : P  11 kPa h f  200 kJ / kg  h fg  2388 kJ / kg  tc  41, 6 o C hc  174 kJ / kg  heat emitted by steam  heat absorbed by wate m s  h f  xh fg  hc   mc p  T 3200  200   0, 91  2388   174   12  4,187  (T2  20)  3600  1954, 738 T2   20  50, 244 6 Marks  T2  58, 905 o C  (6) 4.1.2 From the steam table at t c  41, 6 oC Ps  8 kPa √ PT  Pa  Ps √ Pa  11  8 √ Pa  3 kPa √ 1, 5 kg air ma   3200 kg steam √ 1000 kg steam ma  4, 8 kg air / hr √ PaVa  mRTa √ mRTa Va  √ Pa 4, 8  0, 296  314, 6 Va  √ 60  3 √ ½ x 10 =5 Marks Va  2, 483 m 3 / min √ (5) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -9- POWER MACHINES N5 Kindly note that this symbol √ mean ½ Mark and this symbol  mean 1 Mark 4.2.1 Symbol % Equation O2 CO2 H 2O N2 √ 1√ 1  0,37 1  0,37 H2 37 H2  O2  H 2 O 2 0 0 2  0,37  0,185 √ √ √ 2  0, 29 1  0,29 2  0, 29 CH 4 CH 4  2O 2  CO 2  2 H 2 O 0 29  0,58  0,29  0,58 √ 1√ 1  0,18 1  0,18 CO 18 CO  O 2  CO 2 2 0 0 2  0,18  0,09 CO2 1 0 0,01 0 0 0,15 0  N2 15 N2 0 0 79 4,701  100 √ Total 100 0,855 0,48 0,95 3,366 (4) 100 Va   0,855 21 √ m 3 air V a  4,071 3 √ m fuel (1) Alternative solutions √ √ √ √ √ √ √ √ √ √ Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -10- POWER MACHINES N5 ALTERNATIVE METHOD (5) 4. 2. 2 Pr oducts : √ Percentage : V H 2O  0, 48 m H 2 O / m fuel 3 3 0, 48 VH 2 O   10 % √ 4,795 √ 0,95 VCO 2  0,95 m 3 CO 2 / m 3 fuel VCO 2   19 ,8 % 4,795 √ √ VN 2  3,366  70 , 2 % √ V N 2  3,366 m 3 N 2 / m 3 fuel 4,795 √ ½ x 8=4 Marks √ Total  4,795 m 3 products / m 3 fuel Total  100 % √ ALTERNATIVE METHOD √ √ √ √ √ √ √ √ ½ x 8=4 Marks √ √ (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -11- POWER MACHINES N5 NB If candidates did not follow instructions, allocate √½ mark for each correct position (√½ x10) =5 Marks QUESTION 5 (Mark an answer in the tolerance range of ±2,5%) 5.1 229,9125m/s to 240,875m/s  843,375m/s to 886,625m/s 307,125m/s to 322,875m/s (10) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -12- POWER MACHINES N5 5.2 5.2.1 From the velocity diagram :   850 (1) 5.2.2 From the velocity diagram :   32 0 (1) 5.3 5.3.1 2 𝑚= = 5,556𝑥10−4 √ 60𝑥60 𝑃𝑡𝑢𝑟𝑏𝑖𝑛𝑒 = 𝑚Ϫ𝐶𝑤 √ √ 𝑃𝑡𝑢𝑟𝑏𝑖𝑛𝑒 = 5,556𝑥10−4 𝑥865𝑥350 𝑃𝑡𝑢𝑟𝑏𝑖𝑛𝑒 = 168,2𝑊 √ NB ALLOCATE FULL MARKS FOR THE TURBINE POWER IF IT’S IN THE RANGE OF ±2,5% 𝑃𝑡𝑢𝑟𝑏𝑖𝑛𝑒 = 163,994𝑊 𝑡𝑜 172,405𝑊  (3) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -13- POWER MACHINES N5 QUESTION 6 6.1 Vs  V1  V3 1 √ and Vc  V3 √  Vs  V1  0, 05 Vs √ V1  1, 05 Vs  2 √  Vs √ 5 V3  100 V3  0, 05 Vs  3 √ 1  P n V4,  V3  4   P3  1  700  1,3√ V4   0, 05 VS     100  V4  0, 2233 Vs  4 √ 15 m 3 min √ Ve   min 350 r Ve  0, 0428 m 3√ Ve  V1  V4 0, 0428  1, 05 Vs  0, 2233 Vs  0, 8267Vs√ 0, 0 428 Vs  √ 0, 8267 Vs  0, 05177 m 3√ V1  1, 05  0, 05177 9 Marks  V1  0, 0543 m 3√ Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -14- POWER MACHINES N5 Alternative method : Sin ce, V1  SVs  CV also , V1  SVe  V4   1   CV   P  n  from , vol  1     1   100% 2   SVs   P1       √  √ 1   0, 05 SVs   700  1,3    1     1   100%  SV   100     s    1  0, 05  4, 46761  1  100%   82, 66195%  0 S Ve also , vol  0  100% SVs 0 0 SVe SVs   vol 15   0,8266195 0  SVs  18,146197 m 3 / min  0 also , SVs  SVs  N 18,146197 SVs   350  0, 05185 m 3 thus , V1  SVs  CV √   0, 05185   0, 05185  0, 05   V1  0, 054 m 3 / min √ √ (9) n 1 6.2 T2 P  n  2 T1  P1  n 1 P  n T2  T1  2  √  P1  1,3 1  700  1,3   315    √  100   T2  493, 552 K  (2) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -15- POWER MACHINES N5 n 1 6.3    P2  n P1 V e   1  o n Powercomp  n 1  P   1   √ √ 1,31   1, 3100 15   700  1,3    1, 3 1  60  100  1     61, 40688 kW  Powercomp from , mech   100% Powermotor 61, 40688 thus , Powermotor   0,85  Powermotor  72, 243kW  (4) [15] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

T1390(E)(A2)T NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 2 August 2018 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam tables (BOE 173) Superheated steam tables (Appendix to BOE 173) Nonprogrammable calculators and drawing instruments may be used. This question paper consists of 6 pages and a formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1390(E)(A2)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1390(E)(A2)T QUESTION 1 1.1 1.1.1 State the function of an air pump in a steam condenser. 1.1.2 Name the TWO most common air pumps. (2 × 2) (4) 1.2 Name TWO types of turbines used in a power station to generate electricity. (2) 1.3 State the toxic gas which is produced from incomplete combustion. (1) 1.4 State THREE industries where compressed air is used. (3) [10] QUESTION 2 A gas of 1 kg mass is expanded polytropically from an initial pressure of 1 300 kPa and a temperature of 250 °C until it occupies a volume of 2 m 3 and has a temperature of 37 °C. Take the gas constant R = 0,287 kJ/kg K. Calculate the following: 2.1 The final pressure after expansion (3) 2.2 The index of expansion (5) 2.3 The initial volume (3) 2.4 The work done during expansion (4) [15] QUESTION 3 The following information were recorded during the testing of a steam boiler. Mass of steam produced = 6 600 kg/hr Mass of coal used = 600 kg/hr Pressure of a steam boiler = 2 000 kPa Temperature of feedwater entering the boiler = 95,2 °C Calorific value of fuel = 32 MJ/kg Thermal efficiency of boiler = 80% Calculate the following using the steam tables: 3.1 The enthalpy of steam produced by the boiler (6) 3.2 The dryness fraction of steam produced (5) 3.3 The quantity of heat absorbed by steam in the boiler (4) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1390(E)(A2)T QUESTION 4 A test for dryness fraction was conducted using a combined separating and throttling calorimeter. The following information was recorded during the test: The pressure of steam in main line = 2500 kPa Mass of water collected in separating calorimeter = 0,2 kg Mass of steam condensed after throttling = 1,8 kg Pressure after throttling = 100 kPa Temperature after throttling = 125°C Specific heat capacity of superheated steam = 2,5 kJ/kg.K Calculate the following: 4.1 The dryness fraction of steam in the separating calorimeter. (3) 4.2 The dryness fraction of steam in the throttling calorimeter. (8) 4.3 The dryness fraction of the steam in the main line. (4) [15] QUESTION 5 5.1 During a test with a bomb calorimeter the following observations were recorded: Mass of fuel burned = 0,515 g Mass of water in the calorimeter = 2,42 kg Water equivalent of the calorimeter = 390 g Initial temperature of water in the calorimeter = 13 °C Final temperature of water in the calorimeter = 14,9 °C Quantity of hydrogen in the fuel = 15 % Latent heat of moisture = 2453 kJ/kg Specific heat capacity of water = 4,187 kJ/kg K Calculate the following: 5.1.1. The highest calorific value of the fuel in MJ/kg (4) 5.1.2 The lowest calorific value of the fuel in MJ/kg (3) 5.2 The Orstat apparatus gave the following analysis of the dry products from an engine’s exhaust gas by volume: CO2 = 8,7% O2 = 5,5% CO = 6,3% Remainder N2 = 79,5% Calculate the percentage analysis according to the mass using the table method. (8) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1390(E)(A2)T QUESTION 6 A single-stage, single-acting reciprocating air compressor has a piston diameter of 150 mm, a stroke length of 160 mm and runs at a speed of 540 r/min. Air enters at an initial pressure of 98 kPa and a temperature of 15 °C. The clearance volume is 6% of the stroke volume and the delivery pressure is 600 kPa. Compression and expansion is according to the law PV1,32 = C. Take R as 0,287 kJ/kg K. Calculate the following: 6.1 The effective stroke volume at inlet in m3/s (10) 6.2 The volumetric efficiency at inlet (2) 6.3 The power required (3) [15] QUESTION 7 A single-stage impulse turbine discharges gas axially. The gas leaves the nozzle at a velocity of 500 m/s and at an angle of 22°. The inlet angle of the moving blade is 28° and the outlet angle of the moving blade is 25°.The coefficient of friction across the blade is 0,9. 7.1 Construct a velocity diagram in the ANSWER BOOK (landscape) and enter ALL the values (m/s) onto the diagram. Use a length of 5 cm for the blade velocity and calculate the scale. NB: NO marks will be given if the values are not indicated on the diagram and if the diagram is not constructed to scale. (10) 7.2 Determine the following from the velocity diagram: 7.2.1 The blade velocity in m/s (1) 7.2.2 The inlet velocity of the moving blade in m/s (1) 7.2.3 The discharge velocity from the turbine in m/s (1) 7.2.4 The power developed per kg of gas per second in kW (2) [15] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) T1390(E)(A2)T POWER MACHINES N5 FORMULA SHEET 1. Q  W  U n 1 n 1 T2  P2  n V  17.     1  2. U  mC T T1  P1   V2  3. Q  mC T  1  1 T2  P2   V  18.     1  4. Q  mC p T T1  P1   V2  19. h  h f  h fg V 5. Q  P1V1 ln 2 V1 20. h  hg  C p T  T V  21. h  h f  h fg  hg 6. S  m C ln 2  R ln 2   T1 V1  n  1  hsup  1 941  7. W  P1V 22. Vsup    n  P  V2 8. W  P1V1 ln Vm V1 23.  Vg P1V1  P2V2 9. W  M n 1 24.  M m P1V1  P2V2 25. U = H - PV 10. W  1 1 2 26. gZ1  U1  P1V1  C1  Q  11. R  C p  C 2 1 2 Cp gZ 2  U 2  P2V2  C2  W 12.   2 C  m s (h2  h1 ) 27.   P1V1 P2V2 m  CV 13.  T1 T2  14. PV  mRT m s (h2  h1) 28. EE   m  2 257 15. P1V1  P2V2 101,325 16. P1V1n  P2V2n 29. p  ( Bm  M m ) 760 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1390(E)(A2)T 100  8  30. m C  8 H 2  S  O2  23  3   y y 31. Cx H y   x  O2  xCO2  H 2O  4 2  O  32. H .C.V .  (CVC .C )  CVH 2  H 2  2   (CVs .S )  8  33. L.C.V .  H .C.V .  h fg (9H 2 ) (mw  me ) C p T 34. H .C.V .  mf  n 1   n 1   n   P2  n   n   P  2 n   35. W  P1Ve      1  mRT1      1  n  1   P1    n  1   P1        1  1 Ve Vc  P2  n  36. c  .100  1     1  1     (rp ) n Vs Vs  P1      V 37.   .100 Vs  38. Fc  m(C fe  C fi )  39. P  mU [Cwi  (Cwe )] 2U [Cwi  (Cwe )] 40.  .100 Cai 2 41. U = DN  42. m V  AC 43. (m  M ) g  m 2h  44. Vs  D2 L 4 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1390(E)(A2)T 45. θ1  tc  twi 46. θ2  tc  two θ1  θ2 47. Log.temp.diff.  θ n 1 θ2 P  48. Piso  P1 V1 n  2   P1   n 1  n  P  n 49. Pact  P1 V1    1  2 n 1  P    1   Piso 50. Niso   100 Pact Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE NATIONAL CERTIFICATE POWER MACHINES N5 2 AUGUST 2018 This marking guideline consists of 9 pages. CHIEF MARKER: D SAROTI____________ SIGNATURE: INTERNAL MODERATOR: N NOMANDLA__ SIGNATURE: MARKER: MATUMBA L ____________ SIGNATURE: MARKER: MALOPE M.S____________ SIGNATURE: MARKER: SABU L____________ SIGNATURE: MCM: M DU PREEZ SIGNATURE: DATE: 4 AUGUST 2018 CONCESSION 1. Reduce marks for QUESTION 7 by 15 marks 2. The total 100 to be reduced by 15 marks to 85 for all candidates 3. Mark all candidates out of a total of 85 marks 4. Convert the mark achieved out of 85 per candidate to percentage 5. Record the percentage achieved on the mark sheet Downloaded from hlayiso.com

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 QUESTION 1 1.1 1.1.1 Function of an air pump is to remove unwanted airin a steam condenser to maintain the vacuum in the condenser 1.1.2  The Edwards pump or Reciprocating pump   The Hamilton pump or Rotary air pump or Centrifugal pump  (2 x 2) (4) 1.2  The impulse turbine  The reaction turbine (2) 1.3 Carbon monoxide (1) 1.4  Mining industry  Manufacturing industry  Motor industry  Medical Industry  Food Industry  Construction  Aeronautics ( There are so many ALTERNATIVES please accommodate other options given by candidates ) (3) [10] QUESTION 2 Given: T1 = 250°C + 273 = 523 K T2 = 273 + 37°C = 310 K V2 = 2 m3 R = 0,287 kJ/kg K m = 1 kg 2.1 P2V2  mRT 2 mRT 2 P2   T2 1  0, 287  310   2  44 , 485 kPa  (3) Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 ACCOMMODATE ALTERNATIVE METHODS AND GIVE FULL MARKS IF CANDIDATES OBTAINS THE CORRECT ANSWER 2.2 n 1 T2  P2  n   T1  P1  n 1  310  44 ,485  n   523  1300   310 n  1 44 , 485 log  log 523 n 1300 310 n 1 log   523 n 44 , 485 log 1300 n 1  0,155 n n  1  0,155 n  n (1  0,155 )  1 1 n 0,845 n  1,18  ALLOCATE FULL MARKS IF n = 1,2 (5) OR 2.2 ALTERNATIVE METHOD 𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛 𝑉1𝑛 𝑃2 = 𝑉2𝑛 𝑃1 𝑉1 𝑃2 𝑛 𝐼𝑛 ( ) = 𝐼𝑛( ) 𝑉2 𝑃1 𝑃2 𝑉1 𝑛 = 𝐼𝑛( )/𝐼𝑛 ( ) 𝑃1 𝑉2 44 485 0,1155 𝑛 = 𝐼𝑛( )/𝐼𝑛 ( ) 1300 2 𝑛 = 1,183 Downloaded from hlayiso.com (5) Copyright reserved

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 2.3 P1V1  mRT 1 mRT 1 V1   P1 1  0, 287  523   1300  0,115 m 3  (3) 2.4 P1V1  P2V2 Workdone  n  1  (1300  0,115 )  ( 44 , 485  2)   1,18  1 149 ,5  88 ,97   0,18  336 , 28 kJ  ( 4 Marks) ALTERNATIVE METHOD 2.4 mR (T1  T2 ) Workdone  n 1 1  0, 287 (523  310 )  1,18  1 61 .131  0,18  339 ,617 kJ ( 4 Marks ) NB the value for WORK DONE varies depending on the number of decimal places used by the candidate for n and V1( examiner did not specify so students would use different number of decimal places hence accommodate the following ranges for Work done Work Done Ranges from ( 302, 65kJ for n =1.2 to 339,614kJ for V1=115,462x10-3 m3 ) (4) [15] Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 QUESTION 3 3.1  m s ( h h )     2 1 100 m   CV   6600 ( h2 399 ) 0,80  100 from S .T 600  32000 0,80  32000  at 95 , 2  C ( h2  399 )  h1  399 kJ / kg  11 ( h2  399 )  2327 , 27 h2  2327 , 27  399 h2  2726 , 27 kJ /kg  (6) 3.2 Dryness factor h2  h f  xh fg  At 2000 kPa from S .T h2  h f x  hf  908 kJ / kg h fg  h fg  1889 kJ / kg  2726 , 27  908  1889   0,963  (5) 3.3 Heat absorbed by boiler ms  Q  h2 h1  mf   6600  ( 2726 , 27  399 ) 600  25599 , 97 kJ /kg  (4) [15] Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 QUESTION 4 Given: P = 2500 kPa Pss = 100 kPa Mw = 0,2 kg tss = 125°C Ms = 1,8 kg Cpss = 2,5 kJ/kg.K 4.1 Separating calorimete r Ms 1  Ms   mw 1,8   1,8  0,2  0 .9  (3) 4.2         (8) 4.3 dryness fraction of steam in the main line x  x1  x2    0,9  0,967   0,87  (4) [15] Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 QUESTION 5 5.1 5.1.1 Higher Calorific value ( H .C .V ) Heat emitted  Heat absorbed by water Mf  HCV  (M w  M e )C w  t   2,42  0,390   4,187  14 ,9  13  HCV  0,000515   43406 kJ /kg  43 , 41 MJ /kg  (4) 5.1.2 L.C .V .  H .C .V .  h fg  (9 H 2 )  43406  ( 2453  9  0,15 )   43406  3311 , 55   40094 kJ /kg  40 ,1MJ /kg  (3) 5.2 Gas (a) (b) (a)×(b) % Mass symbol %volume Molecular M Mass CO2 8,7 12+32= 44 3,828 3,828  100  12,93 % 29 ,612 O2 5,5 2×16 = 32 1,76 1,76  100  5,94 % 29 ,612 CO 6,3 12+16= 28 1,764 1,764  100  5,96 % 29 ,612 N2 79,5 14×2 = 28 22,26 22 ,26  100  75 ,17 % 29 ,612 TOTAL: 100 22,612 100% (8) [15] Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 QUESTION 6 Given: P1 = 98 kPa T1 = 15+273=288 K N = 540 r/min D =150 mm P2 = 600 kPa R = 0,287 kJ/kg.K L = 160 mm Vc =0,06 Vs 6.1 d 2 Stroke volume (V s )   L 4    0,15 2   0,16 4  0,002827 m 3 / stroke  Clearance volume (Vc )  0,06V s  0,06  0,002827  0,0001696 m 3 / stroke  P3V3n  P4V4n 1  P3  n V4    V3   P4  1  600  1, 32  0,0001696     98   0,0006692 m 3 / stroke  Cylinder volume (V1 )  V s  Vc  0,002827  0,0001696  0,0 02997 m 3 / stroke  Effective volume (Ve1 ) at inlet  V1  V4  2,997 10  3  6,692 10  4  0,002328 m 3 / stroke  N Effective volume / s  Ve1  60 540   0,002328  60  0,021 m / s 3  (10) Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE -10- T1390(E)(A1)T POWER MACHINES N5 6.2 Volumetric efficiency V1  V4    100 Vs 2,997 10  3  6,692 10  4   100  2,827 10  3  82 ,34 %  (2) n  1 6.3     P2  n 1  n N Power required  P1 Ve1   n  1  P1   60    1, 32  1   1,32  600  1, 32 540   98 0,002328   1  0,32  98   60   540  0,941 0,552  60  4,67 kW  (3) [15] QUESTION 7 DO NOT MARK [0] TOTAL: 85 Copyright reserved Downloaded from hlayiso.com

T1790(E)(A11)T NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 11 August 2017 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam tables (BOE 173), Superheated Steam Tables (Appendix to BOE 173) Candidates need drawing instruments. Nonprogrammable calculators may be used. This question paper consists of 5 pages and a formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1790(E)(A11)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1790(E)(A11)T QUESTION 1 1.1 State the function of each of the following devices: 1.1.1 The Orstat apparatus (2) 1.1.2 The bomb calorimeter (3) 1.2 State the function of governors. (2) 1.3 Name THREE types of governors. (3) [10] QUESTION 2 A mass of gas occupies 1,2 m3 at 72 °C and at a pressure of 92 kPa. The gas is compressed to a volume of 0,16 m 3, according to the law PV1,374 = C. Take Cv = 0,719 kJ/kg.K and R = 0,269 kJ/kg.K. Calculate: 2.1 The mass of the gas (3) 2.2 The final temperature (6) 2.3 The change in internal energy (3) 2.4 The work done (3) [15] QUESTION 3 Feedwater with a temperature of 40,3 °C enters a boiler which is producing steam with a dryness fraction of 94% at a pressure of 2 350 kPa. The steam flows through a superheater at a constant pressure while the temperature increases to 260 °C. The steam delivery is 8,5 kg/kg fuel. The coal has a calorific value of 33 MJ/kg. The specific heat capacity for superheated steam is 2,14 kJ/kg.K. Calculate: 3.1 The heat absorbed by the coal in kJ/kg of fuel by the evaporator (4) 3.2 The heat absorbed by the superheater per kg of fuel (4) 3.3 The equivalent evaporation from and at 100 °C (3) 3.4 The efficiency of the installation (4) Copyright reserved Downloaded from hlayiso.com Please turn over [15]

(8190035) -4- T1790(E)(A11)T QUESTION 4 4.1 A 100 m3 volume of air at an absolute pressure of 200 kPa is saturated with a steam vapour at 54°C in a container. Calculate: 4.1.1 The mass of the air in the container (5) 4.1.2 The mass of the steam vapour in the container (3) 4.2 The arms of a simple watt governor are 250 mm long. At an average speed of 70 r/min the arms form an angle of 30° with the vertical spindle. The bores each have a mass of 1,5 kg. Calculate the centrifugal force acting on the governor. (7) [15] QUESTION 5 The analysis of a fuel according to the mass, is as follows: Carbon (C) = 84% energy value = 34 MJ/kg Hydrogen (H2) = 6% energy value = 142 MJ/kg Sulphur (S) = 7% energy value = 9 MJ/kg Oxygen (O2) = 3% Take the latent heat of steam formed in the exhaust gas to be 2 433 kJ/kg. The air temperature = 38 °C The dry exhaust gas temperature = 210 °C Calculate: 5.1 The higher heat energy value (6) 5.2 The lower heat energy value (5) 5.3 The mass of air required (4) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1790(E)(A11)T QUESTION 6 A single-stage, single-stroke reciprocal air compressor has a bore diameter of 200 mm and a stroke length of 300 mm. The crankshaft turns at 450 r/min. The free volume is 5% of the stroke volume. Assume N is 1,28 for compression and expansion. Air is sucked in at a pressure of 97 kPa and a temperature of 21 °C. The delivery pressure is 600 kPa. Calculate: 6.1 The free air delivery in m3/min (8) 6.2 The volumetric efficiency (2) 6.3 The delivery temperature (2) 6.4 Power required to drive the compressor (3) [15] QUESTION 7 The nozzle of a simple impulse gas turbine is positioned at an angle of 25° with the plane of the moving blades. Gas leaves the nozzle at a velocity of 850 m/s and the average blade velocity is 300 m/s. The inlet and the outlet angles of the moving blades are equal. There is a 10% loss in the velocity of the gas while it flows over the blades. 7.1 Construct a velocity diagram in the ANSWER BOOK (landscape) and indicate the values of ALL velocity (m/s) on the diagram. Use the scale 1 mm : 5 m/s NOTE: No marks will be given if the values are not indicated on the diagram and if the diagram is not constructed to scale. (8) 7.2 Determine the following from the diagram in QUESTION 7.1 7.2.1 The inlet and outlet angles of the moving blades (2) 7.2.2 The mass of gas used in kg/s when the turbine generates a power of 150 kW (2) 7.2.3 The diagram efficiency (3) [15] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

(8190035) -1- T1790(E)(A11)T POWER MACHINES N5 FORMULA SHEET 1. Q  W  U n 1 n 1 T2  P2  n V  17.      1  2. U  mC T T1  P1   V2  3. Q  mC T  1  1 T2  P2   V      1  T1  P1  18. 4. Q  mC p T  V2  19. h  h f  h fg V 5. Q  P1V1 ln 2 V1 20. h  hg  C p T  T V  21. h  h f  h fg  hg 6. S  m C ln 2  R ln 2   T1 V1  n  1  hsup  1 941  22. Vsup    7. W  P1V n  P  V2 8. W  P1V1 ln Vm V1 23.  Vg P1V1  P2V2 9. W  M n 1 24.  M m P1V1  P2V2 25. U = H - PV 10. W  1 1 2 26. gZ1  U1  P1V1  C1  Q  2 11. R  C p  C 1 2 Cp gZ 2  U 2  P2V2  C2  W 12.   2 C  m s (h2  h1 ) 27.   P1V1 P2V2 m  CV 13.  T1 T2  m s (h2  h1) 14. PV = mRT 28. EE   m  2 257 15. P1V1  P2V2 101,325 16. P1V1n  P2V2n 29. p  ( Bm  M m ) 760 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1790(E)(A11)T 100  8  30. m C  8 H 2  S  O2  23  3   y y 31. Cx H y   x  O2  xCO2  H 2O  4 2  O  32. H .C.V .  (CVC .C )  CVH 2  H 2  2   (CVs .S )  8  33. L.C.V .  H .C.V .  h fg (9H 2 ) (mw  me ) C p T 34. H .C.V .  mf  n 1   n 1   n   P2  n   n   P2  n  35. W  P1Ve      1  mRT1      1  n  1   P1    n  1   P1        1  1 Ve Vc  P2  n  c  .100  1     1  1     (rp ) n Vs  P1  36. Vs     V 37.   .100 Vs  38. Fc  m(C fe  C fi )  39. P  mU [Cwi  (Cwe )] 2U [Cwi  (Cwe )] 40.  .100 Cai 2 41. U = DN  42. m V  AC 43. (m  M ) g  m 2h  44. Vs  D2 L 4 45. θ1  tc  twi Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1790(E)(A11)T 46. θ2  tc  two θ1  θ2 47. Log.temp.diff.  θ n 1 θ2 P  48. Piso  P1 V1 n  2   P1   n 1  n  P  n 49. Pact  P1 V1    1  2 n 1  P    1   Piso 50. Niso   100 Pact Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE NATIONAL CERTIFICATE POWER MACHINES N5 11 AUGUST 2017 3 This marking guideline consists of 18 page: CHIEF MARKER: D SAROTI SIGNATURE: ST ~~ INTERNAL MODERATOR: NN NOMANDLA SIGNATURE: =~ - MCM: M DU PREEZ SIGNATURE: _ \\Aiice> DATE: 12 AUGUST 2017 Downloaded from hlayiso.com Copyright reserved

QUESTION 1 1.1 1.1.1 The Orstat apparatus is used to analyse” the dry products of combustion by volume. ” OR The Orstat apparatus is used to analyse the dry products of (2) combustion. ” 1.1.2 The bomb calorimeter is used to determine” the heat value (calorific value) vof fuels under ideal conditions. “ (3) 1.2 Governors are used to maintain an enginev at a constant speedv (2) OR Its main aim is to control” the desired speed of an engine. ” (2) 1.3 Watt governor / centrifugal governor” Porter governor” Hartnell governor’ Proell governor v Dead mass governors” Spring controlled governors” Inertial governors” Pickering governor” Mark any 3 (3x 1) (3) [10] QUESTION 2 Given: P; = 92 kPa V2 = 0,16 m3 Cv=0,719 kJ/kg.K R= 0,269 kJ/kg.K T= Vie 12m? = (724273) =345K 2.1 Mass of gas BY, = mT, RY, _ 92x1,2 0,269 x 345 =11896kg = (3) Downloaded from hlayiso.com

2.2 Py," = print V. 1,374 V, 1,374 12 \ = 92 = (2) ¥ =1465,95 v PV, =mRT, ee 2 mR _ 1465,95 x 0,16 1,1896 x 0,269 = 732,98 K v (6) ALTERNATIVE METHOD 2.2 T> _ a T WW, n-1 T=; @) v v T, = 345V x ayern’v Ty = 732,98KV ¥ (6) QUESTION 2.3 NB IF THE CANDIDATE USE (Ti -T2) APPLY POSITIVE MARKING TECHNIQUE 2.3 Change in internal energy AU =mC,AT =mG(t-4) > =1,1896x 0,719 x (732,97 — 345) v =3318447 vv (3) Downloaded from hlayiso.com

2.4 BY, ~ PV, Workdone = — — V — (1465,95 x 0,16) — (92x 1,2) ) 1374-1 =-33196k) vy (3) ALTERNATIVE METHOD 2,4 WD = mR (T, —T,) WD = 1,189 x 0,269 (345 — 732,98) v WD = —331,77kJ ¥ (3) [15] QUESTION 3 Given: Ms = 8,5 Pressure = 2350kPa Cy = 33 MJ/kg {ss = 260 °C Me = 1 x = 0,94 Cpss =2,14 kJ/kg.K Cv = 33 Md/kg 34 At 2350kPa A 403°C h, =h, + ty hy = 169 ki/kg V =947+ 0,94x 1 853) = 268882 KW kg v Heatabsorbed m, Qua= hh) Mm, 85 = 268882 — 169) v = 214147 kikg v (4) Downloaded from hlayiso.com

320 hah +O, (t.-4) At 2 350 kPa = 2800 +2,14(260-220,7) v= 20,7°C = 2884,102 ki/kg m Quy Hh, 7 hy) my = “(28841 02-—2688,82) v = 1659,897 ki /kg v (4) 3.3 The equivalent evaporator (EE) at 100 °C _ ms (h, —h,) im jx2257 y __ 8,5(2884,102 — 169) ~ 1x 2257 = 10,23 v (3) EE v NB FOR Q3.4 ALLOCATE FULL MARKS IF EFFICIENCY IS GIVEN AS 7 = 0,699 3.4 The thermal efficiency of the steam plant meh) 199% my CV v 8,5(2884,102-169) 633000 699% v qn = (4) ALTERNATIVE METHOD 3.4 2257 n= EEx— v 2257 Vv 1 = 10,23 VES 000 1 = 69,967%Y (4) [15] Downloaded from hlayiso.com

QUESTION 4 NB R WAS NOT GIVEN SO IF THE CANDIDATE GETS STUCK AT R, ALLOCATE FULL 5 MARKS FOR QUESTION 4.1.1. ALLOCATE FULL MARKS IF THEY ASSUME R WHICH IS IN REASONABLE RANGE OF 0,287Kj/kg Given: Pa = 185 kPa R = 0,287 kJ/kg.K L=250mm_ Speed = 75 r/min Va = 100 m3 Ta= 327°C m=1,5kg Angle = 30° 4.1 4.1.1 For the mass of air Air P=P.-P, At 15 kPa P,=P-P, , T=54 °C 200 -15 185 kPa sy MRT , M, = PVa Vv RT, _ 185 x 100 ~ 0,287 x 327 197 125 kg of air v (5) i PV aa ll 4.1.2 Themass of steamvapour _ total volumeof container At 15kPa 7 “ volumeof | kgof steam v Ve = 10,02 m/kg _ 100 10,02 =9,98 kgof steam vy (3) Downloaded from hlayiso.com

4.2 Cos 30° = a/250 v 250Cos 30° = a = 216,5mm v Now r Sin30° = 7/250 v r=125mm Vv Take moments about 0 (Fxa)=Wexr) v Fx0,2165 = (1,5 x 9,81) x 0,125 v F = 8,49 N at aconstant speed of 70 r/min v ALTERNATIVE METHOD 4,2 w 2nN - 60 W = 2xNX70/, 60 W = 7,333 rad/secY m 2 Tan$= ve mg g mw?r 2p Tan@ = =v mg g 2 Tang =~ g Tan@xg ro w2 Tan 30x 9,81 =——— _ = 105 mmv 7,332 g rr h=% = 183 mmv Foxh=mgxr Fo = mg x Downloaded from hlayiso.com (7)

0,105 Fo =15x9,81x00¥ Fo = 8,443 NV [15] QUESTION 5 5.1 O, HCV .=(C¥eCHCV 5, (1, - Ahr, S) v = (34 x 0,84) + 12{ 0.06 | +(9x0,07) Y = 28,6 + 142(0,0563) + 0,63 v =37,23 Milkg ¥ (6) 5.2 LCV.=H.CV.—hy OH;) ¥ 237,23 - (2,433%9% 0,06)” =37,23-1314 v =35,916 Miikg = y (5) 5.3 Dry product mass Air required = ~ icf +8H, + S- o, | v - (ass x 3} + (8 x 0,06) + (0,07 - 000] v Air required =12 kg v (4) [15] Downloaded from hlayiso.com

QUESTION 6 CONSIDER THE ALTERNATIVE METHOD ATTACHED USING STP. GIVE FULL CREDIT WHEN LEARNER USES NTP OR STP AS IT WAS NOT SPECIFIED Given: P1 = 97 kPa T1 = 294°C n = 1,28 D=200mm P2 = 600 kPa R = 0,287 kJ/kg.K L = 300 mm 6.1 Free air delivery 1 Fi=Ve yy09 214%) _-{)" SV Vs P 1 a1 4 220508 | coo v Vs 97 = 84,24% v Vs = GDL x 0,8424 = 7X02'*03 ogang y =0,007939 m> sy PY, PUY.) Tf, y ARV, ‘ PT; _ 97(0,007939)288 ~ 1013x294 = 0,007447 m* v Thus V,/min = 0,007447 x 450 v = 3,35 m/min Y v 6.2 For volumetric efficiency V, =—£x100 Na V. s _ 0,007939 0,00942 = 84,29% v 100 v Downloaded from hlayiso.com (8) (2)

6.3 nel T,_(P\r 7, |B ant P 0,28 97 = 437,98 K v 6.4 — i ; Pan to (2 -1 v v 0,28 — 7f 9007939 x 450 V/ 1,28 (x 8 60 0,28 )) 97 =12,93 kW v CONSIDER THE ALTERNATIVE METHOD ATTACHED Downloaded from hlayiso.com (2) (3) [15]

QyESTUN — & = sing & STP Gy\ Fe — te WN = Ke a VA Ng= 0,05 Cv, -\,) = 0,55 (0, 44) 2 6 coo tn Y P=tol,32S T=243Kk. | = ATX 0, 442%. HK 243 iol,e2s X aay = 0w0tm ” = ©eFxK USO = S168 we lic VA © be ec iw = S152 ¥ “9S Mos ©, SG = Sob @ | 6, act ' \ae-4 V 5 Oseqstn Vg ae Qe . Vu =Ne ke 4 ae wsUy] 7 jot var( “1% - ~ oa ner L sae ik vse << P38 Yam Yess ae = oy solasg,3 / 2Q-} (eys- Sis4qvw = 60142 7 J | | Torat [15 | ENE N t= _ OT aT — &F oT Downloaded from hlayiso.com

* MARKING GUIDELINE -9- T1790(E)(A11)T POWER MACHINES N5 4 W NB) TOLERANCE = t 215%. CONSIDER. THE RANGE OF QUESTION 7 ; VALUES PRWIDED AND 32055 mk, ALLOCATE FULL MARKS 7 < For THE CopeECT Cho (64.11 RANGE. (B12, Strrr/s — 328, Sona} | w <3 ia 33 Af .° a 5 z on E == Bn “9g Pt, ) o as 2 & 3 6 Ch(74 85mm) A =, 354,25 m/s (8) Downloxddedh fap lay 999m Please turn over Copyright reserved

7.2 7.2.1 Inlet angle (Cri) = 37° and outlet angle (Cre) = 37°v” 7.2.2 P=mUC,, P UC wh _ 150 000 v ~ 300 x (170,15 x 5) = 0,558 kg/ls v Range of P = (0,544kg/s_to 0,572 kg/s ) 723 0 UC 100 _ 2x 300(179,15 x5) 7 8507 =74,4% v x 100 Range of efficiency 7 = 72,54% to 76,26 % Downloaded from hlayiso.com (2) (2) (3) [15] TOTAL: 100

T1450(E)(A13)T NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 13 April 2017 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam tables (BOE 173) Superheated steam tables (appendix to BOE 173) Calculators and drawing instruments may be used. This question paper consists of 5 pages and a formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1450(E)(A13)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1450(E)(A13)T QUESTION 1 A kilogram of gas with a temperature of 21 °C is adiabatically compressed from 109 kPa to 1 MPa. The specific heat capacity at constant volume and constant pressure is 0,7 kJ/kg.K and 1 kJ/kg.K respectively. Calculate the following: 1.1 The volume before compression (6) 1.2 The volume after compression (5) 1.3 The final absolute temperature (3) 1.4 The work done per kilogram of gas (3) 1.5 The change in internal energy (3) [20] QUESTION 2 2.1 A steam boiler plant supplies 5 200 kg of wet steam at a pressure of 780 kPa every hour. The steam is 2% wet and the feed water temperature is 40,3 °C. The boiler uses 660 kg of fuel to supply 5 200 kg of steam and the fuel has a heat value of 30 MJ/kg. Calculate the following: 2.1.1 The boiler efficiency (7) 2.1.2 The equivalent evaporation from and at 100 °C (3) 2.2 An economiser is installed and the feed water temperature in QUESTION 2.1 is increased by 28,8 °C. The efficiency of the boiler increases by 7,25% and all other conditions remain the same. Calculate the mass of fuel saved every hour. (10) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1450(E)(A13)T QUESTION 3 A gas consists of the following volumetric analysis: CO = 33% H2 = 22% O2 = 12% CH4 = 15% N2 = 18% The atomic masses are as follows: C = 12 O = 16 H = 1 N = 14 Calculate the percentage analysis according to mass. [20] QUESTION 4 4.1 Name TWO types of axial flow turbines. (2) 4.2 State the purpose of an Orsat apparatus. (2) 4.3 State the functions of the following components of a steam boiler: 4.3.1 Air heater 4.3.2 Economiser (2 × 2) (4) 4.4 Explain the meaning of the following terms as used with governors: 4.4.1 Isochronism 4.4.2 Hunting (2 × 5) (10) 4.5 Name TWO types of air pumps. (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1450(E)(A13)T QUESTION 5 Steam leaves the nozzles of a single-stage impulse turbine at a velocity of 805 m/s. The nozzles are inclined at an angle of 20° to the plane of rotation and the blade exit angle is 18°. The turbine was tested and it developed 290 kW of power whilst consuming 1 kilogram of steam every second at a speed of 9 550 r/min. The wheel has a mean radius of 400 mm. Calculate the following: 5.1 The velocity of the blade (3) 5.2 The total whirl velocity (3) 5.3 Use a scale of 1 cm = 50 m/s and construct a velocity diagram in the ANSWER BOOK. Enter all the values (m/s) onto the diagram. NB: No marks will be awarded if the values (m/s) are not entered onto the diagram and if the diagram is not constructed to the given scale. HINT: Construct the diagram in landscape format in the ANSWER BOOK. (8) 5.4 Determine the following from the diagram: 5.4.1 The coefficient of friction 5.4.2 The diagram efficiency (2 × 3) (6) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

(8190035) -1- T1450(E)(A13)T FORMULA SHEET n 1 n 1 T2  P2  n V  1. Q  W  U 17.     1  T1  P1   V2   1  1 T2  P2   V  2. U  mCv T 18.      1  T1  P1   V2  3. Q  mCv T 19. h  h f  xh fg 4. Q  mC p T 20. h  hg  C p T V2 5. Q  P1V1 ln 21. h  h f  h fg  hg V1  T V  n  1  hsup  1941  6. S  m C v ln 2  R ln 2  22. Vsup     T1 V1  n  P  Vm 7. W  P1V 23. x Vg V2 M 8. W  P1V1 ln 24. x V1 M m P V  P2V2 9. W 1 1 25. U  H  PV n 1 P1V1  P2V2 1 2 10. W 26. gZ1  U 1  P1V1  C1  Q   1 2 1 2 gZ 2  U 2  P2V2  C 2  W 2 ms h2  h1  . 11. R  C p  Cv 27.  . m f CV ms h2  h1  . Cp 12.   28. EE  . Cv m f 2257 P1V1 P2V2 p  Bm  M m  101,325 13.  29. T1 T2 760 100  8  14. PV mRT 30. m C  8H 2  S  O2  23  3   y y 15. P1V1  P2V2 31. Cx H y   x  O2  xCO2  H 2O  4 2 P1V1  P2V2 n n 16. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1450(E)(A13)T  O  32. H .C.V .  (CVC .C )  CVH 2  H 2  2   (CVS .S )  8  33. L.C.V .  H .C.V .  h fg (9H 2 ) (m w  we )C p T 34. H .C.V .  mf  n 1   n 1   n   P2  n          1  n P n 35. W  P1Ve      1  mRT1  2  n  1   1 P  n  1   P1       1  Vc  P2  n  1 Ve 36.  c  .100  1     1  1     (rp ) n Vs Vs  P1     Va 37. a  .100 Vs . 38. Fc  m(C fe  C fi ) P  mU C wi   C we  . 39. 2U C wi   C we  40.  2 .100 C ai 1 41. KE  mv 2 2 42. U  DN . 43. mV  AC 44. (m  M ) g  m 2 h  45. Vs  D2L 4 46. 1  t c  t wi 47.  2  t c  t wo Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1450(E)(A13)T 1   2 48. log . temp. diff .   ln 1 2 P  49. Piso  P1V1 ln  2   P1   n 1    P1V1    1  n P n 50. Pact  2 n 1  P1     P 51.  iso  iso .100 Pact 2N 52.  60 g ( M  m) 53. h m  2 g 54. h 2 55. mgr  m 2 rh Copyright reserved Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA MARKING GUIDELINE ~ ee NATIONAL CERTIFICATE POWER MACHINES N5 13 APRIL 2017 & This marking guideline consists of 7 pageq—_| o CHIEF MARKER: D SAROTI SIGNATURE: < , =I INTERNAL MODERATOR: M KABINGA SIGNATURE: 4p MCM: M DU PREEZ SIGNATURE: DATE: 13 APRIL 2017 Mark the alternative responses and the ranges of values provided AP (Guerin 78) A (FL ont) Hifemape CMa vepe MS ) fe (Malai ) Ae | z | poet L=kaw * Downloaded from hlayiso.com Copyright reserved Please turn over

. MARKING GUIDELINE QUESTION 1 1.1 R = 1.2 pyr = 1.3 AS Copyright reserved -2- POWER MACHINES N5 c,-¢, Vv 1-0,7 v O3k/kgK J MRT, MRT, Bk 1x 0,3x 294 109 0,80917m> oY = v 0,80917 te wv 1000 0.16614 m’ J P, 2 T2= 711 (ViNV2)"1 P T2 = 294 (0.809/0.172)°42° ' im T2 = 571.236 K ayy ! Range of T2 aS 553.81 K to 573.46K 29 (ee J 109 553,82 K J Downloaded from hlayiso.com T1450(E)(A13)T (6) Range of Y 1.4 - 1.43 for V2 = 0.166 m? to 0.176 m? (5) Please turn over

MARKING GUIDELINE -3- T1450(E)(A13)T POWER MACHINES NS 1.4 T, -T, WD = Mx Coa y-i 294 — 553,82 1,4-1 = —194,865 kJ/kg = Ix of 1.5 AU = MxC,(T)-T,) v = 1x 0,7(553,82-294) ¥ = = 181, 8744S /kg v Range of WD -194.865 kJ/kg to -195.382 kJ/kg Range of AU -194,865 kJ/kg to -195.382 kJ/kg Mark any of the two answers 181.874 kJ/kg or -194.865 kJ/kg to -195.382 kJ/kg QUESTION 2 21 244 hk = 169k kg hb =f + xhfe v 717 + (0,98 2050) v = 226k /kg 1 Ketel M,xCv = 67,15% 212 ap _ M(m-h) M, x 2257 M,(H, =h,) 100 5200(2726- 169) 100 660 x 30% 10° 5200(2726 — 169) 660 x 2257 v v Mark also 67% 1 v Vv v = 8,926 kg steam/kg fuel JV Downloaded from hlayiso.com Copyright reserved (3) (3) [20] (7) (3) Please turn over

MARKING GUIDELINE -4- POWER MACHINES N5 T1450(E)(A13)T 2.2 ho = —ODBOKI/kg@69l°c OV ¥ Yov Neat 67,15 + 7,25 = 744% J M, (fy ~h) 100 f v M = M,(h, ~h) 100 Nxete * CV 1 _ __ 5200(2726-289) 100 74,4x30x10° 1 = — 567,76kg/ hour v -. Fuel saved = 660 - 567,76 4 = 92,24 kg / hour / QUESTION 3 (a) % (b) % SYMBOL | according | Molecular mass axb according to mass to volume 33 x 28 5) 924 100 co 38%Y |}12+16=28% | 924 | 096% 44.08% 22x2=44/ 44 100, , He 22%Y\2x1=2 VV 2096171” 12 x 32 =| 384 100 ° O2 12%/|2*16=32 ¥ 384. | 996% 1 1832% 15 x 16 =| 240 100 4 CHs 15% Y | 124+(4x1)=16% | 240 | d96% THES! 18 x 28 S| so4 100 ° Ne 18% |2%14=28 Y | 504 | 396% 7 2405% TOTAL 100 % 2096 100 % Copyright reserved Downloaded from hlayiso.com (10) [20] [20] Please turn over

MARKING GUIDELINE QUESTION 4 5. T1450(E)(A13)T POWER MACHINES NS 44 ¢ Impulse turbine e Reaction turbine ¥ Jv 4.2 An Orsat apparatus is used for the chemical analysis of the products of fuel combustion. / 4.3 4.3.1 4.3.2 44 4.4.1 4.4.2 45 e Edward v By heating the air used in the combustion chamber the efficiency of the installation is increased and fuel consumption is reduced. / WA Or It increases the temperature of the air supplied to the combustion chamber / The economiser supplies “heated water to the boiler thereby increasing the efficiency of the boiler. A cold-water supply subjects the boiler to internal stresses due to contraction. / Vv Or It increases the temperature of feed water to the boiler / (2 * 2) A governor is isochronous when the equilibrium speed is constant for every radius of ball rotajon. The sleeve remains in its lowest position until the equilibrium speed ,is reached. The slightest increase in speed causes the governor balls to move out to their maximum radius, lifting the sleeve to its highest position. v The condition where,the speed of an engine, which is controlled by a governor, fluctuates continually above and below the, average speed. This is caused by a governor that is very sensitive sairpump V e Hamilton air pump/Rotating air pump ~ QUESTION 5 5.1 u _ Copyright reserved aDN v “60, 2x0.4x2x9550 ¥ 60 400m/s v Downloaded from hlayiso.com (2) (2) (4) (5) (6) (2) [20] (3) Please turn over

MARKING GUIDELINE 5.2 P Cc = w 5.3 Copyright reserved 6- T1450(E)(A13)T POWER MACHINES N5 MxUxC, P MxU 290 x 10° 1x 400 725m/s v (3) CFIFE7IM/S65,5cm) Range Cri= 438.75 mis 4 to461.26mis oY L aa € U a + Nv uw N cal ae = Od € ~ o u a > Rd oO » £ S So 2 __ 10 399.75 rs w/e , to 128.13 mis B rang 104° tp 106° Range Cfe = S 117 mis = to 123 mis (8) Downisaded from Rlayiso.com Please turn over

MARKING GUIDELINE -7- T1450(E)(A13)T POWER MACHINES NS 5 5.4.4 K _ Gre y Cri _ 390 v ~ ‘450 K range 0.845 - 0,867 to 0.889 Y (3) Downloaded from hlayiso.com Copyright reserved Please turn over

MARKING GUIDELINE -8- T1450(E)(A13)T POWER MACHINES N5 5.4.2 2xUxCw_ 100 v7 7 = eK Ci 1 2x 400x725 100 JV = xX 805? 1 = 89,5% V (3) [20] TOTAL: 100 Downloaded from hlayiso.com Copyright reserved

T1150(E)(N29)T NOVEMBER EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 29 November 2016 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Calculators and drawing instruments may be used. This question paper consists of 5 pages and 1 formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1150(E)(N29)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1150(E)(N29)T QUESTION 1 The density of a certain gas is 1,35 kg/m 3 at a temperature of 0 °C and 101,3 kPa. The specific heat capacity at constant pressure of this gas is 1,05 kJ/kg.K. 6,2 kg of this gas is compressed adiabatically from 0 °C and 101,3 kPa to 5,5 MPa. Calculate the following: 1.1 The gas constant (3) 1.2 The adiabatic index (5) 1.3 The initial volume (3) 1.4 The final volume (3) 1.5 The final thermodynamic temperature (3) 1.6 The work done during the compression (3) [20] QUESTION 2 2.1 The barometer and vacuum gauge reading of a surface condenser is 760 mm Hg and 685 mm Hg respectively. A turbine exhausts steam into this surface condenser. The gas constant for air is 0,287 kJ/kg.K and the temperature of the air extracted is 36,2 °C. Calculate the following: 2.1.1 The volume of the air extracted in m3/kg (9) 2.1.2 The volume of steam lost through the air ejector with each kilogram of air (5) 2.2 Name the FOUR methods of firing boilers. (4) 2.3 Name TWO types of steam boilers. (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1150(E)(N29)T QUESTION 3 A certain gas has the following volumetric analysis: H2 = 40% CH4 = 28% CO = 22% N2 = 10% 47% excess air is used to burn 100 m 3 of this gas. Air consists of 21% oxygen according to volume. Calculate the following: 3.1 The theoretical volume of oxygen required (13) 3.2 The actual volume of air used (2) 3.3 The volume of each dry product of combustion (5) [20] QUESTION 4 A single-stage, single-stroke air compressor has a stroke length of 320 mm and a bore diameter of 210 mm. The free volume is 6% of the stroke volume and the crankshaft turns at 450 r/min. The index for compression and expansion is 1,28 and the air is taken in at 98 kPa and 22 °C. The air is delivered at 610 kPa and the atmospheric conditions are 101,3 kPa and 15 °C. Calculate the following: 4.1 The volumetric efficiency at intake conditions 4.2 The volumetric efficiency with reference to free air delivery 4.3 The delivery temperature in kelvin 4.4 The effective swept volume per minute 4.5 The cycle power (5 x 4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1150(E)(N29)T QUESTION 5 An impulse turbine develops 110 kW at a speed of 8 000 r/min. The mean radius of the blade is 473 mm. The inlet and outlet blade angles are 32° each to the plane of the wheel. There is an 8% loss in velocity due to friction and the gas is discharged in an axial direction. 5.1 Calculate the blade velocity. (3) 5.2 Use a scale of 1 cm = 50 m/s and construct a velocity diagram in the ANSWER BOOK. Enter all the values (m/s) onto the diagram. NOTE: No marks will be awarded if the values (m/s) are not entered onto the diagram and if the diagram is not constructed to the given scale. HINT: Use the answer page in the landscape format to construct the diagram. (8) 5.3 Determine the following from the diagram: 5.3.1 The relative inlet velocity (1) 5.3.2 The relative exit velocity (1) 5.3.3 The inlet velocity (1) 5.3.4 The exit velocity (1) 5.3.5 The exit velocity of flow (1) 5.3.6 The inlet velocity of flow (1) 5.3.7 The mass flow of the gas per second (3) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

(8190035) - 1- T1150(E)(N29)T POWER MACHINES N5 FORMULA SHEET n 1 n 1 T2  P2  n V  1. Q  W  U 17.     1  T1  P1   V2   1  1 T2  P2   V  2. U  mCv T 18.      1  T1  P1   V2  3. Q  mCv T 19. h  h f  xh fg 4. Q  mC p T 20. h  hg  C p T V2 5. Q  P1V1 ln 21. h  h f  h fg  hg V1  T V  n  1  hsup  1941  6. S  m C v ln 2  R ln 2  22. Vsup     T1 V1  n  P  Vm 7. W  P1V 23. x Vg V2 M 8. W  P1V1 ln 24. x V1 M m P V  P2V2 9. W 1 1 25. U  H  PV n 1 P1V1  P2V2 1 2 10. W 26. gZ1  U 1  P1V1  C1  Q   1 2 1 2 gZ 2  U 2  P2V2  C 2  W 2 ms h2  h1  . 11. R  C p  Cv 27.  . m f CV ms h2  h1  . Cp 12.   28. EE  . Cv m f 2257 P1V1 P2V2 p  Bm  M m  101,325 13.  29. T1 T2 760 100  8  14. PV mRT 30. m C  8H 2  S  O2  23  3   y y 15. P1V1  P2V2 31. Cx H y   x  O2  xCO2  H 2O  4 2 P1V1  P2V2 n n 16. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) --2- 1- T1150(E)(N29)T  O  32. H .C.V .  (CVC .C )  CVH 2  H 2  2   (CVS .S )  8  33. L.C.V .  H .C.V .  h fg (9H 2 ) (m w  we )C p T 34. H .C.V .  mf  n 1   n 1   n   P2  n          1  n P n 35. W  P1Ve      1  mRT1  2  n  1   1 P  n  1   P1       1  Vc  P2  n  1 Ve 36.  c  .100  1     1  1     (rp ) n Vs Vs  P1     Va 37. a  .100 Vs . 38. Fc  m(C fe  C fi ) P  mU C wi   C we  . 39. 2U C wi   C we  40.  2 .100 C ai 1 41. KE  mv 2 2 42. U  DN . 43. mV  AC 44. (m  M ) g  m 2 h  45. Vs  D2L 4 46. 1  t c  t wi 47.  2  t c  t wo Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1150(E)(N29)T 1   2 48. log . temp. diff .   ln 1 2 P  48. Piso  P1V1 ln  2   P1   n 1   P2  n P1V1    1  n 49. Pact  n 1  P1     Piso 50.  iso  .100 Pact 2N 51.  60 g ( M  m) 52. h m  2 g 53. h 2 54. mgr  m 2 rh Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE NATIONAL CERTIFICATE NOVEMBER EXAMINATION POWER MACHINES N5 29 NOVEMBER 2016 This marking guideline consists of 8 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1150(E)(N29)T POWER MACHINES N5 QUESTION 1 1.1 PV = MRT PV R = MT 101,3 × 1 = 1,35 × 273 = 0,275 kJ /kg.K (3) 1.2 R = Cp − Cv Cv = Cp − R = 1,05 − 0,275 = 0,775 kJ /kg.K Cp γ = Cv 1,05 = 0,775 = 1,35 (5) 1.3 P1V1 = MRT1 MRT1 V1 = P1 6,2 × 0,275 × 273 = 101,3 = 4,5949 m 3 (3) 1.4 P1V1γ = P2V2γ 1  P γ V2 = V1  1   P2  1  101,3  1, 35 = 4,5949   5500  = 0,2384 m 3 (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1150(E)(N29)T POWER MACHINES N5 1.5 P2V2 = MRT2 P2V2 T2 = MR 5500 × 0,2384 = 6,2 × 0,275 = 769 K (3) 1.6 P1V1 − P2V2 WD = γ −1 = (101,3 × 4,5949) − (5500 × 0,2384) 1,35 − 1 = − 2416,39 kJ (3) [20] QUESTION 2 2.1 2.1.1 PT = (BM − VM ) × 101,3 760 = (760 − 685) × 101,3 760 = 10 kPa PS = ( 6 kPa @ 36,2 o C from steam tables ) Pa = PT − Ps = 10 − 6 = 4 kPa PaVa = M a Ra Ta M a Ra Ta Va = Pa 1 × 0,287 × 309,2 = 4 = 22,1851 m 3 /kg (9) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1150(E)(N29)T POWER MACHINES N5 2.1.2 Vs = ( 23,74 m 3 /kg @ 36,2 o C from steam tables ) M a × Va = M s × Vs M a × Va Ms = Vs 1 × 22,1851 = 23,74 = 0,9345 kg/kg of Air (5) 2.2 • Oil • Gas • Solid coal • Pulverised coal (4 x 1) (4) 2.3 • Fire-tube boiler • Water-tube boiler (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1150(E)(N29)T POWER MACHINES N5 QUESTION 3 3.1 Hydrogen to Water 2 H 2 + O2 = 2H 2O 2 m H 2 + 1m O2 3 3 = 2 m 3 H 2O 1 3 1 m3 H 2 + m O2 = 1 m 3 H 2O 2 CH 4  y y C x H y +  x + O2 = xCO2 + H 2O  4 2  4 4 CH 4 + 1 + O2 = 1CO2 + H 2O  4 2 CH 4 + 2O2 = CO2 + 2 H 2 O 1 m3 + 2 m3 = 1 m3 + 2 m3 Carbon Monoxide to Carbon Dioxide 2CO + O2 = 2CO2 2 m 3 CO + 1 m 3 O2 = 2 m 3 CO2 1 3 1 m 3 CO + m O2 = 1 m 3 CO2 2 Symbol % Vol Formula O2 CO 2 H2O N2 (m3) Reqd H2 40 40 1m3H 2 +0,5m3O 2 = 1m3H 2 O 40x0,5=20 0 40 0 CH 4 28 28 1m3CH 4 +2m3O 2 =1m3CO 2 +2m3H 2 O 28x2=56 28 28x2=56 0 CO 22 22 1m3C+0,5m3O 2 = 1m3CO 2 22x0,5=11 22 0 0 N2 10 10 None 0 0 0 10 TOTAL 100% 100m3 87m3 50m3 96m3 10m3 ∴ Theoretical volume O2 Re quired = 87 m 3 (13) 3.2 147 Actual volume of air required = 87 × 21 = 609 m 3 (2) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1150(E)(N29)T POWER MACHINES N5 3.3 CO2 = 50 m 3  79  N2 =  609 ×  + 10  100  = 491,11 m 3  21  Excess O2 =  609 ×  − 87  100  = 40,89 m 3 (5) [20] QUESTION 4 4.1 1  P n η vol = 1 + α − α  2  × 100  P1  1  610  1, 28 = 1 + 0,06 − 0,06  × 100  98  = 1,06 − 0,25035 = 80,965 % 4.2 Pa × η air P1 × η vol = Ta T1 P1 × η vol × Ta ηa = T1 × Pa 98 × 80,965 × 288 = 295 × 101,3 = 76,469 % n −1 4.3  P2  n T2 =   T1  P1  n −1 P  n T2 = T1  2   P1  1, 28 −1  610  1, 28 = 295   98  = 440 K Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1150(E)(N29)T POWER MACHINES N5 4.4 πD 2 Ve = η vol × × L× N 4 π × 0,212 = 0,80965 × × 0,32 × 450 4 = 4,0382 m /min 3  P  n  n −1 4.5 n Pcycle = × P1Ve  2  − 1 n −1  P1     4,0382  610  1, 28  1, 28 −1 1,28 = × 98 ×   − 1 1,28 − 1 60  98    = 14,829 kW (5 x 4) [20] QUESTION 5 5.1 π ×D× N U = 60 π × 0,473 × 2 × 8000 = 60 = 396 m / s (3) 5.2 On next page (8) 5.3 5.3.1 Cri = 505 m/s (1) 5.3.2 Cre = 465 m/s (1) 5.3.3 Ci = 865 m/s (1) 5.3.4 Ce = 247,5 m/s (1) 5.3.5 Cfe = 247,5 m/s (1) 5.3.6 Cfi = 267,5 m/s (1) 5.3.7 P = m × U × Cw P m = U × Cw 110 × 10 3 = 396 × 825 = 0,3367 kg / s (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- T1150(E)(N29)T POWER MACHINES N5 5.2 [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

T1150(E)(A10)T AUGUST EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 10 August 2016 (X-Paper) 9:00–12:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Drawing instruments must be used. Calculators may be used. This question paper consists of 5 pages and 1 formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.comPlease turn over

(8190035) -2- T1150(E)(A10)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1150(E)(A10)T QUESTION 1 The average blade velocity of a single-stage impulse turbine is 300 m/s. The relative inlet velocity of the gas is 600 m/s and the blade inlet angle is 20 o. Friction causes a 5% loss in velocity as the steam moves over the blades. The turbine does not develop any axial thrust and uses 90 kg gas every minute. 1.1 Use a scale of 1 cm = 50 m/s and construct a velocity diagram in the ANSWER BOOK. Enter ALL the values (m/s) onto the diagram. NOTE: No marks will be awarded if the values (m/s) are not entered onto the diagram and if the diagram is not constructed to the given scale. HINT: Use the ANSWER PAGE in the LANDSCAPE FORMAT to construct the diagram. (11) 1.2 Determine the following from the diagram: 1.2.1 The exit gas angle leaving the turbine (1) 1.2.2 The inlet gas angle (1) 1.2.3 The exit blade angle (1) 1.2.4 The power developed (3) 1.2.5 The diagram efficiency (3) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1150(E)(A10)T QUESTION 2 2.1 A combined separating and throttling calorimeter was used to determine the dryness fraction of the steam in the main steam pipe. The observations were as follows: Pressure in the steam main = 980 kPa Water collected in the separator = 0,5 kg/min Steam condensed after throttling = 10 kg/min Steam pressure after throttling = 240 kPa Temperature of steam after throttling = 135 oC Specific enthalpy of superheated steam = 2,05 kJ/kg.oC Calculate the dryness fraction of the steam in the mains. (9) 2.2 A boiler produces 15 500 kg of steam at a pressure of 2 200 kPa every hour. The fuel consumption is 1 560 kg/h whilst the feed water has a temperature of 39 oC. The equivalent evaporation of the boiler, from and at 100 oC is 10,6 kg steam/kg fuel and the calorific value of the fuel is 29 800 kJ/kg. Calculate the following: 2.2.1 The efficiency of the boiler (5) 2.2.2 The dryness fraction of the steam (6) [20] QUESTION 3 A single-acting, single-cylinder air compressor has a clearance volume which is 6% of the stroke volume. The air is taken in at a pressure and temperature of 96,5 kPa and 22 oC respectively, and delivered at a pressure of 772 kPa. The compression and expansion index is 1,35. The atmospheric pressure of the air is 101,4 kPa and the atmospheric temperature is 16 oC. The mass of air consumed is 45 kg/min and the gas constant for air is 0,286 kJ/kg.K. The compressor has a mechanical efficiency of 75%. Calculate the following: 3.1 The volumetric efficiency referred to atmospheric conditions (15) 3.2 The power of the motor required to drive the compressor (5) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1150(E)(A10)T QUESTION 4 4.1 Indicate whether the following statement is TRUE or FALSE. Choose the answer and write only 'true' or 'false' next to the question number in the ANSWER BOOK. 'The adiabatic process is a process which is carried out in such a way that heat is transferred into the system, but not out of the system'. (2) 4.2 Air with a mass of 0,3 kg has a volume of 0,175 m 3 and a pressure of 150 kPa. The air is then compressed to a pressure of 1 350 kPa, according to the law PV1,26 = C. The specific heat capacity of air at constant volume and constant pressure is 718,5 J/kg.K and 1,0045 kJ/kg.K respectively. Calculate the following: 4.2.1 The gas constant 4.2.2 The initial temperature of the air 4.2.3 The final temperature of the air 4.2.4 The change of internal energy of the air 4.2.5 The work done 4.26 The heat received or rejected by the air (6 x 3) (18) [20] QUESTION 5 The rotating balls of a porter governor each have a mass of 2,3 kg. The arms which are 330 mm long, forms an angle of 27o with the vertical and rotates at a mean speed of 125 r/min. The sleeve has a movement of ±30 mm from the centre position. Calculate the following: 5.1 The sleeve mass (9) 5.2 The sensitivity of the governor (11) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

(8190035) -1- T1150(E)(A10)T POWER MACHINES N5 FORMULA SHEET 1. Q  W  U n 1 n 1 T2  P2  n V  17.      1  2. U  mC T T1  P1   V2  3. Q  mC T  1  1 T2  P2   V  18.     1  4. Q  mC p T T1  P1   V2  V2 19. h  h f  h fg 5. Q  P1V1 ln V1 20. h  hg  C p T  T V  6. S  m C ln 2  R ln 2  21. h  h f  h fg  hg  T1 V1  W  P1V n  1  hsup  1 941  7. 22. Vsup    n  P  V2 8. W  P1V1 ln V1 Vm 23.  Vg P1V1  P2V2 9. W  n 1 M 24.  M m P1V1  P2V2 10. W  1 25. U = H - PV R  C p  C 1 2 11. 26. gZ1  U1  P1V1  C1  Q  2 1 Cp gZ 2  U 2  P2V2  C2 2  W 12.   2 C  P1V1 P2V2 m s (h2  h1 ) 13.  27.   T1 T2 m  CV 14. PV = mRT  m s (h2  h1) 28. EE   15. P1V1  P2V2 m  2 257 16. P1V1n  P2V2n 101,325 29. p  ( Bm  M m ) 760 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1150(E)(A10)T 100  8  30. m C  8 H 2  S  O2  23  3   y y 31. Cx H y   x  O2  xCO2  H 2O  4 2  O  32. H .C.V .  (CVC .C )  CVH 2  H 2  2   (CVs .S )  8  33. L.C.V .  H .C.V .  h fg (9H 2 ) (mw  me ) C p T 34. H .C.V .  mf  n 1   n 1   n   P2  n   n   P2  n  35. W  P1Ve      1  mRT1      1  n  1   P1    n  1   P1        1  1 Ve Vc  P2  n  36. c  .100  1     1  1     (rp ) n Vs Vs  P1      V 37.   .100 Vs  38. Fc  m(C fe  C fi )  39. P  mU [Cwi  (Cwe )] 2U [Cwi  (Cwe )] 40.  .100 Cai 2 41. U = DN  42. m V  AC 43. (m  M ) g  m 2h  44. Vs  D2 L 4 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1150(E)(A10)T 45. θ1  tc  twi 46. θ2  tc  two θ1  θ2 47. Log.temp.diff.  θ n 1 θ2 P  48. Piso  P1 V1 n  2   P1   n 1  n   P2 n 49. Pact  P1 V1    1  n 1  P    1   Piso 50. Niso   100 Pact Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE NATIONAL CERTIFICATE AUGUST EXAMINATION POWER MACHINES N3 10 AUGUST 2016 This marking guideline consists of 8 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1150(E)(A10)T POWER MACHINES N5 QUESTION 1 1.1 Cre = Cri × 0,95 = 600 × 0,95 = 570 m / s (11) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1150(E)(A10)T POWER MACHINES N5 1.2 1.2.1 42o (1) 1.2.2 13o (1) 1.2.3 21o (1) 1.2.4 P = M × U × Cw 90 = × 300 × 1095 60 = 492750 W = 492,75 kW (3) 1.2.5 2U × Cw 100 η = × Ci 2 1 2 × 300 × 1095 100 = × 885 2 1 = 83,884 % (3) [20] QUESTION 2 2.1 M x1 = M +m 10 = 10 + 0,5 = 0,952 Enthalpy before throttling (980 kPa) = Enthalpy after throttling (240 kPa) hf + x 2 hfg = hg + c p (t su − t s ) 759 + ( x 2 × 2017) = 2715 + 2,05(135 − 126,1) 2017 x 2 = 2715 + 18,245 − 759 x2 = 0,979 ∴Dryness Fraction = x1 × x 2 = 0,952 × 0,979 = 0,932 (9) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1150(E)(A10)T POWER MACHINES N5 2.2 2.2.1 Total Heat Transferred EE = 2257 ∴Total Heat = 2257 × EE = 2257 × 10,6 = 23924,2 kJ / kg Total Heat 100 η = × Calorific Value 1 23924,2 100 = × 29800 1 = 80,28 % (5) 2.2.2 Heat Gained = Heat Lost M s (h2 − h1 ) = Mf × Cv × η Mf × Cv × η h2 = + h1 Ms 1560 × 29800 × 0,8028 = + 163 15500 = 2570,778 kJ / kg hw = hf + xhfg hw − hf x = hfg 2570,778 − 931 = 1868 = 0,878 (6) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1150(E)(A10)T POWER MACHINES N5 QUESTION 3 3.1 V1 = Vc + Vs = 0, 06Vs + Vs = 1, 06Vs n n PV 1 1 = PV 2 2 1  P1  n V2 = V1    P2  1  96, 5  1,35 = 1, 06Vs    772  = 0, 227Vs ∴V2 − V3 = 0,227Vs − 0,06Vs = 0,167Vs n −1  P2  n T2 = T1    P1  1, 35 −1  772  1, 35 = 295   96,5  = 505,8 K PAV A P2 (V2 − V3 ) = TA T2 P2 (V2 − V3 ) × T A VA = PAT2 772 × 0,167Vs × 289 = 101,4 × 505,8 = 0,7265Vs V A 100 ∴η vol = × Vs 1 0,7265Vs 100 = × Vs 1 = 72,65 % (15) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1150(E)(A10)T POWER MACHINES N5 3.2 n Pcomp = × MR(T2 − T1 ) n −1 × × 0,286(505,8 − 295) 1,35 45 = 1,35 − 1 60 = 174,4 kW Pout 100 η mech = × Pin 1 Pout 100 ∴Pin = × η mech 1 174,4 100 = × 75 1 = 232,53 kW (5) [20] QUESTION 4 4.1 False , Q = 0J, therefore WD = ∆U (2) 4.2 4.2.1 R = Cp − Cv = 1,0045 − 0,7185 = 0,286 kJ / kg.K 4.2.2 P1V1 = MRT1 P1V1 T1 = MR 150 × 0,175 = 0,3 × 0,286 = 305,94 K n −1 4.2.3  P2  n T2 =   T1  P1  n −1 P  n T2 = T1  2   P1  1, 26 −1  1350  1, 26 = 305,94   150  = 481,44 K Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1150(E)(A10)T POWER MACHINES N5 4.2.4 ∆U = M × Cv(T2 − T1 ) = 0,3 × 0,7185(481,44 − 305,94 ) = 37,83 kJ 4.2.5 MR(T1 − T2 ) WD = n −1 0,3 × 0,286(305,94 − 481,44 ) = 1,26 − 1 = − 57,92 kJ 4.2.6 Q = ∆U + WD = 37,83 + (− 57,92 ) = − 20,09 kJ ∴ 20,09 kJ heat is rejected (6 x 3) (18) [20] QUESTION 5 5.1 h = L × cos 27° = 330 × cos 27° = 294 mm 2πN ω = 60 2 × π × 125 = 60 = 13,09 rad / sec hmω 2 ∴M = −m g 0,294 × 2,3 × 13,09 2 = − 2,3 9,81 = 9,51 kg (9) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- T1150(E)(A10)T POWER MACHINES N5 5.2 When sleeve rises 30 mm 0,03 h1 = 0,294 − 2 = 0,279 m When sleeve drops 30 mm 0,03 h2 = 0,294 + 2 = 0,309 m g ( M + m) ω1 = m × h1 9,81(9,51 + 2,3) = 2,3 × 0,279 = 13,437 rad / sec g ( M + m) ω2 = m × h2 9,81(9,51 + 2,3) = 2,3 × 0,309 = 12,768 rad / sec ω ∴ Sensitivity = ω1 − ω 2 13,09 = 13,437 − 12,768 = 19,57 : 1 (11) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

T1340(E)(A11)T APRIL EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 11 April 2016 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Drawing instruments, pens and a ruler Calculators may be used. This question paper consists of 5 pages, and 1 formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1340(E)(A11)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1340(E)(A11)T QUESTION 1 The mean blade diameter of a single-stage impulse turbine is 1 960 mm and it rotates at a speed of 3 898 r/min. The blade speed is 0,4 of the speed of the gas velocity leaving the nozzles. The nozzles are inclined at an angle of 22o to the plane of the wheel and the blading is symmetrical. The gas loses 12% velocity, due to friction, as it moves over the blades. The turbine uses 100 kg of gas every minute. 1.1 Calculate the peripheral speed of the blade (3) 1.2 Calculate the nozzle velocity of the gas (3) 1.3 Use a scale of 1 cm = 50 m/s and construct a velocity diagram in the ANSWER BOOK. Enter ALL the values (m/s) onto the diagram. NOTE: No marks will be awarded if the values (m/s) are not entered onto the diagram and if the diagram is not constructed to the given scale. HINT: Use the ANSWER PAGE in the LANDSCAPE FORMAT to construct the diagram. (8) 1.4 Determine the following from the diagram: 1.4.1 The relative velocity of the gas at the inlet of the moving blade (1) 1.4.2 The relative velocity of the gas at the outlet of the moving blade (1) 1.4.3 The velocity of the gas leaving the turbine (1) 1.4.4 The end thrust developed (3) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1340(E)(A11)T QUESTION 2 2.1 Define boiler efficiency. (4) 2.2 A boiler produces 18 000 kg steam every hour at a pressure of 2,5 MPa. The feed water to the boiler has a temperature of 37,7 °C. Fuel, which has a calorific value of 30 MJ/kg, is consumed at a rate of 1 775 kg every hour. The equivalent evaporation of the boiler from and at 100 °C is 11,2 kg steam/kg fuel. Calculate the following: 2.2.1 The efficiency of the boiler. (6) 2.2.2 The condition of the steam produced and state a reason based on the calculations. (6) 2.2.3 The dryness percentage of the steam. (4) [20] QUESTION 3 The density of a certain gas is 1,75 kg/m3 at a pressure of 115 kPa and 20 °C. 80 kJ of heat is used to heat 0,75 kg of this gas at constant pressure to a temperature of 126 °C. Calculate the following: 3.1 The characteristic gas constant (3) 3.2 The specific heat capacity of the gas at constant pressure (3) 3.3 The specific heat capacity of the gas at constant volume (3) 3.4 The change of internal energy (3) 3.5 The initial volume (2) 3.6 The final volume (3) 3.7 The work done during the heating process (3) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1340(E)(A11)T QUESTION 4 A single-acting, single-stage, reciprocating compressor runs at a speed of 650 r/min. The cylinder bore diameter is 220 mm and the piston stroke length is 310 mm. The clearance volume is 5,5 % of the swept volume and the compression index n = 1,35. Intake conditions are 98 kPa and 18 °C and the compression pressure is 650 kPa. The free air conditions are 101,325 kPa and 15 °C. Calculate the following: 4.1 The swept volume (3) 4.2 The effective swept volume (9) 4.3 The free air delivered in m3/min (5) 4.4 The volumetric efficiency (3) [20] QUESTION 5 A porter governor has a sleeve mass of 6 kg, whilst the rotating balls each have a mass of 1,8 kg. The arms of the governor are 330 mm long and they form an angle of 30° to the vertical. The coefficient of friction between the sleeve and its shaft is 0,33 and the minimum and maximum radii of rotation is 130 mm and 210 mm respectively. Calculate the following: 5.1 The minimum speed limit in r/min (12) 5.2 The maximum speed limit in r/min (8) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

(8190035) -1- T1340(E)(A11)T POWER MACHINES N5 FORMULA SHEET 1. Q = W + ∆U n −1 n −1 T2  P2  n V  17. =   =  1  2. ∆U = mCυ ∆T T1  P1   V2  3. Q = mCυ ∆T γ −1 γ −1 T2  P2  γ V  18. =  =  1  4. Q = mC p ∆T T1  P1   V2  V2 19. h = h f + χh fg 5. Q = P1V1 ln V1 20. h = hg + C p ∆T  T V  6. ∆S = m Cυ ln 2 + R ln 2  21. h = h f + h fg = hg  T1 V1  W = P1∆V n − 1  hsup − 1 941  7. 22. Vsup =   n  P  V2 8. W = P1V1 ln V1 Vm 23. χ= Vg P1V1 − P2V2 9. W = n −1 M 24. χ= M +m P1V1 − P2V 2 10. W= γ −1 25. U = H - PV R = C p − Cυ 1 2 11. 26. gZ1 + U1 + P1V1 + C1 + Q = 2 1 Cp gZ 2 + U 2 + P2V2 + C2 2 + W 12. γ = 2 Cυ • P1V1 P2V2 m s (h2 − h1 ) 13. = 27. η= • T1 T2 m ƒ CV 14. PV = mRT • m s (h2 − h1 ) 28. EE = • 15. P1V1 = P2V2 m ƒ 2 257 16. P1V1n = P2V2n 101,325 29. p = ( Bm ± M m ) 760 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -1- -2- T1340(E)(A11)T 100  8  30. m= C + 8 H 2 + S − O2  23  3   y y 31. C x H y +  x + O2 = xCO2 + H 2O  4 2  O  32. H .C.V . = (CVC .C ) + CVH 2  H 2 − 2  + (CVs .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (mw + me ) C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P2  n   n   P2  n  35. W = P1Ve     − 1 = mRT1     − 1  n − 1   P1    n − 1   P1        1  1 Ve Vc  P2  n  36. ηc = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1      Vα 37. ηα = .100 Vs • 38. Fc = m(C fe − C fi ) • 39. P = mU [Cwi − (−Cwe )] 2U [Cwi − (−Cwe )] 40. η= .100 Cai 2 41. U = πDN • 42. m V = AC 43. (m + M ) g = mω 2 h π 44. Vs = D2L Downloaded from hlayiso.com Please turn over 4 Copyright reserved

(8190035) -3- -1- T1340(E)(A11)T 45. θ1 = tc − twi 46. θ2 = tc − two θ1 − θ2 47. Log.temp.diff. = θ ln 1 θ2 P  48. Piso = P1 V1 ln  2   P1   n −1  n  P  n 49. Pact = P1 V1   − 1  2 n −1  P    1   Piso 50. Niso = • 100 Pact Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE NATIONAL CERTIFICATE APRIL EXAMINATION POWER MACHINES N5 11 APRIL 2016 This marking guideline consists of 8 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1340(E)(A11)T POWER MACHINES N5 QUESTION 1 1.1 πDN U = 60 π × 1,96 × 3898 = 60 = 400 m / s (3) 1.2 U = 0,4 × Ci U ∴Ci = 0,4 400 = 0,4 = 1000 m / s (3) 1.3 Diagram on next page. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1340(E)(A11)T POWER MACHINES N5 (8) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1340(E)(A11)T POWER MACHINES N5 1.4 1.4.1 Cri = 645 m/s (1) 1.4.2 Cre = 567,6 m/s (1) 1.4.3 Ce = 330 m/s (1) 1.4.4 Fc = m(cfi − cfe ) = 100 (375 − 325) 60 = 83,333 N (3) [20] QUESTION 2 2.1 The efficiency of any boiler is defined by the ratio of heat transferred to feed water in converting it to steam divided by the heat supplied by the complete combustion of the fuel. (4) 2.2 2.2.1 Total Heat Transferred EE = 2257 ∴Total Heat = 2257 × EE = 2257 × 11,2 = 25278,4 kJ / kg Total Heat 100 η = × Calorific Value 1 25278,4 100 = × 30000 1 = 84,26 % (6) 2.2.2 h1 = 158 kJ / kg Heat Gained = Heat Lost M s (h2 − h1 ) = Mf × Cv × η Mf × Cv × η h2 = + h1 Ms 1775 × 30000 × 0,8426 = + 158 18000 = 2650,692 kJ / kg At 2500 kPa the hg for dry steam is 2801 kJ / kg . ∴ The steam is wet , because hg is greater than h2 . (6) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1340(E)(A11)T POWER MACHINES N5 2.2.3 hwet = hf + xhfg hw − hf x = hfg 2650,692 − 962 = 1839 = 0,918 ∴ The steam is 91,8% dry. (4) [20] QUESTION 3 3.1 PV = MRT PV V R = = MT ρT 115 ×1 = 1, 75 × 293 = 0, 2243 kJ / kg.K (3) 3.2 Q = M × Cp (T2 − T1 ) Q Cp = M (T2 − T1 ) 80 = 0,75(399 − 293) = 1,0063 kJ / kg.K (3) 3.3 R = Cp − Cv Cv = Cp − R = 1,0063 − 0,2243 = 0,782 kJ / kg.K (3) 3.4 ∆U = M × Cv(T2 − T1 ) = 0,75 × 0,782(399 − 293) = 62,169 kJ (3) 3.5 M V1 = ρ 0, 75 = 1, 75 = 0, 42857 m3 Copyright reserved Downloaded from hlayiso.com Please turn over (2)

MARKING GUIDELINE -6- T1340(E)(A11)T POWER MACHINES N5 3.6 MRT2 V2 = P2 0,75 × 0,224 × 399 = 115 = 0,58289 m 3 or V1T2 V2 = T1 0, 42857 × 399 = 293 = 0,5836 m3 (3) 3.7 WD = P2 (V2 − V1 ) = 115(0,58289 − 0,42857 ) = 17,7468 kJ or WD = mR (T2 − T1 ) = 0, 75 × 0, 224 ( 399 − 293) = 17,81 kJ (3) [20] QUESTION 4 4.1 π D2 Vs (V1 − V3 ) = ×L 4 π × 0, 222 = × 0,31 4 = 0, 01178 m3 (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1340(E)(A11)T POWER MACHINES N5 4.2 V3 = 0,055(V1 − V3 ) = 0,055 × 0,01178 = 0,00065 m 3 V1 = Vs + Vc = 0,01178 + 0,00065 = 0,01243 m 3 1  P n V4 = V3  3   P4  1  650  1, 35 = 0,00065   98  = 0,00264 m 3 ∴Ve = V1 − V4 = 0,01243 − 0,00264 = 0,00979 m 3 (9) 4.3 Pf V f P1 (V1 − V4 ) = Tf T1 P1 (V1 − V4 ) T f Vf = Pf T1 98 ( 0, 01243 − 0, 00264 ) × 288 = 101,325 × 291 = 0, 00937 m3 ∴Free Air Delivered / min = 0, 00937 × N = 0, 00937 × 650 = 6, 0905 m3 / min (5) 4.4 V1 − V4 100 η vol = × V1 − V3 1 0,00979 100 = × 0,01178 1 = 83,1 % (3) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- T1340(E)(A11)T POWER MACHINES N5 QUESTION 5 5.1 Fµ = µ × Mg = 0, 33 × 6 × 9,81 = 19, 424 N h1 = L2 − r12 = 0, 332 − 0,132 = 0, 303 m g ( M + m) − F µ ω1 = m × h1 9,81( 6 + 1,8 ) − 19, 424 = 1,8 × 0, 303 = 10, 231 rad / s ω1 × 60 ∴ N1 = 2π 10, 231× 60 = 2×π = 97, 7 r / min (12) 5.2 h2 = L2 − r2 2 = 0,332 − 0, 212 = 0, 255 m g ( M + m) − F µ ω2 = m × h2 9,81( 6 + 1,8 ) − 19, 424 = 1,8 × 0, 255 = 11,153 rad / s ω2 × 60 ∴N2 = 2π 11,153 × 60 = 2 ×π = 106,5 r / min (8) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

T1320(E)(N17)T NOVEMBER EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 17 November 2015 (X-Paper) 9:00–12:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Drawing instruments Calculators may be used. This question paper consists of 5 pages and 1 formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1320(E)(N17)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1320(E)(N17)T QUESTION 1 An impulse turbine develops 125 kW of power at a mean blade diameter of 805 mm and a speed of 9 490 r/min. The inlet and outlet angles of the blade is 30° and the coefficient of velocity is 0,8. The gas exits the turbine at an angle of 86o. 1.1 Use a scale of 1 cm = 50 m/s and construct a velocity diagram in the ANSWER BOOK. Enter all the values (m/s) onto the diagram. NB: No marks will be awarded if the values (m/s) are not entered onto the diagram and if the diagram is not constructed to the given scale. HINT: Use the answer page in the landscape format to construct the diagram. (13) 1.2 Determine the following from the diagram: 1.2.1 The mass flow of the gas (3) 1.2.2 The nozzle angle (1) 1.2.3 The diagram efficiency (3) [20] QUESTION 2 2.1 A gas is compressed adiabatically, to a pressure of 1 450 kPa, from a volume of 0,62 m3, a pressure of 169 kPa and a temperature of 19 °C. The density of the gas at STP is 1,38 kg/m3 and the specific heat capacity at constant volume is 0,716 kJ/kg.K. The temperature is 0 °C and the pressure is 101,3 kPa at standard temperature and pressure conditions. Calculate the following: 2.1.1 The value of the gas constant (4) 2.1.2 The value of the specific heat capacity at constant pressure (3) 2.1.3 The value of gamma (2) 2.1.4 The final absolute temperature (3) 2.1.5 The change in internal energy (4) 2.2 Define the TWO laws of thermodynamics. (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1320(E)(N17)T QUESTION 3 20 % excess air is used to completely burn a certain fuel, which consists of 85 % carbon and 15 % hydrogen, by mass. The atomic mass of the elements is as follows: Oxygen 16 Hydrogen 1 Carbon 12 Nitrogen 14 Calculate the volumetric composition of the products of combustion. [20] QUESTION 4 4.1 A steam turbine supplies a surface condenser with 6 800 kg of steam per hour, which is 25 % wet. The barometer reading is 750 mm Hg and the manometer reading is 667 mm Hg. The condensate and the air pump suction has a temperature of 37,7 °C. The cooling water enters the condenser at 22 °C and leaves at 42 °C. The air leakage is 0,8 kg for every 1 000 kg of outlet steam and the gas constant for air is 0,287 kJ/kg.K. Calculate the following: 4.1.1 The mass of cooling water required every hour 4.1.2 The capacity of the air pump in m3/min (2 × 7) (14) 4.2 The condenser in Question 4.1 is fitted with an air-cooling section which lowers the temperature from 37,7 °C to 21,1 °C. Calculate the following: 4.2.1 The new capacity of the air pump in m3/min (4) 4.2.2 The percentage saving in pump volume (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1320(E)(N17)T QUESTION 5 5.1 A watt governor has equal arms. The height of the rotating balls changes from 80 mm to 120 mm. Calculate the change in speed in r/min. (11) 5.2 The bore-to-stroke ratio of a single-stage air compressor is 1 : 1,7. The compressor takes in 1,2 m3 of air per minute at a temperature of 18 °C and a pressure of 105 kPa. The air is compressed according to the law PV1,35 = C and it is delivered at a pressure of 820 kPa. The compressor runs at 350 r/min and the gas constant for air is 0,287 kJ/kg.K. Calculate the following: 5.2.1 The power required to drive the compressor in kW (4) 5.2.2 The bore diameter in mm (3) 5.2.3 The stroke length in mm (2) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035 -1- T1320(E)(N17)T POWER MACHINES N5 FORMULA SHEET n −1 n −1 T2  P2  n V  1. Q = W + ∆U 17. =   =  1  T1  P1   V2  γ −1 γ −1 T2  P2  γ V  2. ∆U = mC v ∆T 18. =   =  1  T1  P1   V2  3. Q = mC v ∆T 19. h = h f + xh fg 4. Q = mC p ∆T 20. h = hg + C p ∆T V2 5. Q = P1V1 ln 21. h = h f + h fg = hg V1  T V  n − 1  hsup − 1941  6. ∆S = m C v ln 2 + R ln 2  22. Vsup =    T1 V1  n  P  Vm 7. W = P1 ∆V 23. x= Vg V2 M 8. W = P1V1 ln 24. x= V1 M +m P V − P2V2 9. W= 1 1 25. U = H − PV n −1 P1V1 − P2V2 1 2 10. W = 26. gZ 1 + U 1 + P1V1 + C1 + Q = γ −1 2 1 2 gZ 2 + U 2 + P2V2 + C 2 + W 2 m s (h2 − h1 ) . 11. R = C p − Cv 27. η= . m f CV m s (h2 − h1 ) . Cp 12. γ = 28. EE = Cv . m f 2257 P1V1 P2V2 p = (Bm ± M m ) 101,325 13. = 29. T1 T2 760 100  8  14. PV= mRT 30. m= C + 8 H 2 + S − O2  23  3   y y 15. P1V1 = P2V2 31. C x H y +  x + O2 = xCO2 + H 2O  4 2 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035 -2- T1320(E)(N17)T n n 16. P1V1 = P2V2  O  32. H .C.V . = (CVC .C ) + CV H 2  H 2 − 2  + (CVS .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (m w + we )C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P  n  n   P  n 35. W = P1Ve       − 1 = mRT1  2    − 1  2 n  −   1 1 P  −   P1  n 1      1  Ve Vc  P2  n  1 36. η c = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1     Va 37. ηa = .100 Vs . 38. Fc = m(C fe − C fi ) P = m U [C wi − (− C we )] . 39. 2U [C wi − (− C we )] 40. η= .100 C ai 2 1 41. KE = mv 2 2 42. U = πDN . 43. m V = AC 44. (m + M ) g = mω 2 h π 45. Vs = D2L 4 46. θ 1 = t c − t wi Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035 -3- T1320(E)(N17)T 47. θ 2 = t c − t wo θ1 − θ 2 48. log . temp. diff . = θ ln 1 θ2 P  48. Piso = P1V1 ln 2   P1   n −1  n   P2 n 49. Pact = P1V1   − 1  n −1  P1     Piso 50. η iso = .100 Pact 2πN 51. ω = 60 g ( M + m) 5.2 h= m ×ω 2 g 5.3 h= ω2 5.4 mgr = mω 2 rh Copyright reserved Downloaded from hlayiso.com

MARKING GUIDELINE NATIONAL CERTIFICATE NOVEMBER EXAMINATION POWER MACHINES N5 17 NOVEMBER 2015 This marking guideline consists of 9 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1320(E)(N17)T POWER MACHINES N5 QUESTION 1 1.1 π DN U = 60 π × 0,805 × 9490 = 60 = 400 m / s Cre Cri = K 480 = 0,8 = 600 m / s (13) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1320(E)(N17)T POWER MACHINES N5 1.2 1.2.1 P = M × U × Cw P M = U × Cw 125 × 10 3 = 400 × 935 = 0,334 kg / s (3) 1.2.2 180 (1) 1.2.3 2 × U × VW 100 Diag .η = × V12 1 2 × 400 × 935 100 = × 965 2 1 = 80,324 % (3) [20] QUESTION 2 2.1 Given: P 1 = 169 kPa P 2 = 1 450 kPa V 1 = 0,62 m3 T 1 = 292 K 3 C v = 0,716 kJ/kg.K Density = 1,38 kg/m T stp = 273 K P stp = 101,3 kPa 2.1.1 PV = MRT PV M 1 V R = but ρ = therefore = MT V ρ M 101,3 ×1 = 1,38 × 273 = 0, 269 kJ / kg .K (4) 2.1.2 R = C P − CV CP = R + CV = 0,269 + 0,716 = 0,985kJ / kg.K (3) 2.1.3 CP γ = CV 0,985 = 0,716 = 1,376 (2) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1320(E)(N17)T POWER MACHINES N5 γ −1 2.1.4 T2  P2  γ =   T1  P1  γ −1 P  γ T2 = T1  2   P1  1, 376 −1 1450  1, 376 = 292    169  = 525,37 K (3) 2.1.5 ∆U = M × CV (T2 − T1 ) (But P1V1 = MRT1 ) P1V1 = × CV (T2 − T1 ) RT1  169 × 0,62  =   × 0,716(525,37 − 292 )  0,269 × 292  = 222,896 kJ (4) 2.2 Joule's law states that mechanical energy and heat energy are mutually convertible. Clausius's statement, states that heat can flow from a hot substance to a cold substance unaided, but it is impossible for heat to flow from a cold substance to a hot substance without the aid of external work. (4) [20] QUESTION 3 Theoritical Mass of O2 Re qd . = (2,667 × C ) + (8 × H 2 ) = (2,667 × 0,85) + (8 × 0,15) = 2,26695 + 1,2 = 3,467 kg / kg Fuel 100 Theoritical Mass of Air Re qd . = O2 × 23 100 = 3,467 × 23 = 15 kg Air / kg Fuel Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1320(E)(N17)T POWER MACHINES N5 Pr oducts of Combustion by Mass / kg Fuel CO2 = 0,85 × 3, 667 = 3,11695 kg H 2O = 0,15 × 9 = 1,35 kg 20 Excess O2 = 3, 467 × = 0, 693 kg 100 77 120 N2 = 15 × × = 13,86 kg 100 100 TOTAL = 19, 02 kg % Analysis of Pr oducts by Mass 3,11695 100 CO2 = × = 16,39 % 19,02 1 1,35 100 H 2O = × = 7,1 % 19,02 1 0,693 100 O2 = × = 3,64 % 19,02 1 13,86 100 N2 = × = 72,87 % 19,02 1 TOTAL = 100 % Convert Products by Mass to Products by Volume Symbol % by Mass (a) Molecular a % by Volume Mass (b) b CO 2 16,39 16,39 0,373 100 = 0,373 × = 10,71 12 + (2x16) = 44 44 3,484 1 H2O 7,1 7,1 0,394 100 = 0,394 × = 11,31 (1x2) + 16 = 18 18 3,484 1 O2 3,64 3,64 0,114 100 = 0,114 × = 3,27 (16x2) = 32 32 3,484 1 N2 72,87 72,87 2,603 100 = 2,603 × = 74,71 (14x2) = 28 28 3,484 1 TOTAL 100 % 3,484 100 % [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1320(E)(N17)T POWER MACHINES N5 QUESTION 4 Given: M s = 6 800 kg/hr x = 0,75 t c = 37,7 oC B M = 750 mm Hg M M = 667 mm Hg Air Leakage = 0,8 kg/1000 kg t w1 = 22 oC t w2 = 42 oC R = 0,287 kJ/kg.K 4.1 4.1.1 PT = (BM − M M ) × 101,3 760 = (750 − 667 ) × 101,3 760 = 11 kPa hc = 158 kJ / kg (@ 37,7 o C ) Heat lost by Steam = Heat gained by Water M s (hf + xhfg − hc ) = M w × shc w (t w 2 − t w1 ) 6800[200 + (0,75 × 2388) − 158] = M w × 4,187(42 − 22 ) 12 464 400 = 83,74 M w Mw = 148 846,429 kg / hour (7) 4.1.2 PT = Ps + Pa Pa = PT − Ps = 11 − 6,5 = 4,5 kPa 0,8 × 6800 Air Leakage = 1000 × 60 = 0,09 kg / min Air Pump Capacity PaVa = MRTa MRTa Va = Pa 0,09 × 0,287 × 310,7 = 4,5 = 1,7834 m 3 / min (7) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1320(E)(N17)T POWER MACHINES N5 4.2 4.2.1 New Partial Pressure of Air Pa = PT − Ps = 11 − 2,5 = 8,5 kPa New Air Pump Capacity MRTa Va = Pa 0,09 × 0,287 × 294,1 = 8,5 = 0,8937 m 3 / min (4) 4.2.2 1,7834 − 0,8937 100 % Saving = × 1,7834 1 = 49,9 % (2) [20] QUESTION 5 5.1 At 80 mm height: g h1 = 2 ω1 g ω1 = h1 9,81 = 0,08 = 11,074 rad / s At 120 mm height: g ω2 = h2 9,81 = 0,12 = 9,042 rad / s Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- T1320(E)(N17)T POWER MACHINES N5 ω = 2πN ω1 × 60 N1 = 2π 11,074 × 60 = 2π = 105,749 r / min ω 2 × 60 N2 = 2π 9,042 × 60 = 2π = 86,345 r / min ∴∆ N = N1 − N 2 = 105,749 − 86,345 = 19,404 r / min (11) 5.2   n −1 5.2.1  n   P2  n WD / s ( Power ) = P1V1       − 1  n − 1   P1     1,2  1,35   820  1, 35  1, 35 −1 = 105 ×     − 1 60  1,35 − 1   105    = 5,7 kW (4) 5.2.2 V = Vs × N πD 2 1,2 = × L× N 4 πD 2 = × 1,7 D × 350 4 π × 1,7 D 3 × 350 = 4 1,2 × 4 D = 3 π × 1,7 × 350 = 0,136938 m = 136,94 mm (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -9- T1320(E)(N17)T POWER MACHINES N5 5.2.3 Stroke Length = 1,7 × D = 1,7 × 136,94 = 232,798 mm (2) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

T1300(E)(A14)T APRIL EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 14 April 2015 (Y-Paper) 13:00–16:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Drawing instruments, pens and a ruler Calculators may be used. This question paper consists of 5 pages, and a 3-page formula sheet. Downloaded from hlayiso.com

-2- (8190035) T1300(E)(A14)T DEPARTMENT OF EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 ________________________________________________________________________ INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write the answers to the questions, neatly and legibly, in your answer book. Downloaded from hlayiso.com

-3- (8190035) T1300(E)(A14)T QUESTION 1 1.1 A bomb calorimeter was used to test the higher heat value (HCV) of a fuel. The information on hand is as follows: Mass of fuel tested 0,7 g Mass of water in the calorimeter 2464 g Water equivalent of the calorimeter 327 g Temperature of the water before the test 18,5 oC Temperature of the water after the test 20,69 oC Specific heat capacity of the water 4,2 kJ/kg.K The lower heat value of the fuel 13,7% lower than the HCV Latent heat of the vapour 2453 kJ/kg Calculate the following: 1.1.1 The higher heat value of the fuel in MJ/kg (4) 1.1.2 The lower heat value of the fuel in MJ/kg (3) 1.1.3 The percentage of hydrogen content in the fuel (3) 1.2 A sample of coal has the following analysis, according to mass: Carbon: 86% (Heat value 35 MJ/kg) Hydrogen: 7% (Heat value 144 MJ/kg Oxygen: 7% The partial pressure of the steam in the exhaust gas is 5 kPa. Calculate the following: 1.2.1 The mass of air required to completely burn 12 kg of coal (4) 1.2.2 The higher heat value of the coal (3) 1.2.3 The lower heat value of the coal (3) [20] Downloaded from hlayiso.com

-4- (8190035) T1300(E)(A14)T QUESTION 2 The nozzles of an impulse turbine are inclined at an angle of 22o and the gas exits the nozzles at 540 m/s. The shaft rotates at 6976 r/min and the mean diameter of the wheel is 460 mm. There is no end pressure on the blades of the turbines and there is a 15% loss of velocity due to friction. The mass flow rate of the gas is 1 kg/s. 2.1 Use a scale of 1 cm = 30 m/s and construct a velocity diagram in the answer book. Enter all the values (m/s) onto the diagram. NB: No marks will be awarded if the values (m/s) are not entered onto the diagram and if the diagram is not constructed to the given scale. HINT: Use the answer page in the landscape format to construct the diagram. (14) 2.2 From the diagram, determine the following: 2.2.1 The power developed (3) 2.2.2 The kinetic energy that leaves the blades (3) [20] QUESTION 3 3.1 18,5 m3/min of air is supplied to a single-stage, double stroke air compressor at a pressure of 110 kPa and a temperature of 22 oC. The compressor runs at a speed of 400 r/min and delivers the air at a pressure of 820 kPa. Compression takes place according to the law PV1,35 = C and the clearance volume is 7% of the stroke volume. The gas constant for air is 0,287 kJ/kg.K. Calculate the following: 3.1.1 The stroke volume (10) 3.1.2 The temperature of the delivered air (3) 3.1.3 The power required to drive the compressor (3) 3.2 State Dalton’s law of partial pressure. (2) 3.3 Steam turbines are classified into two groups. Name the TWO groups. (2) [20] Downloaded from hlayiso.com

-5- (8190035) T1300(E)(A14)T QUESTION 4 A volume of gas is compressed polytropically to a pressure of 655 kPa. The gas is compressed according to the law PVn = C, from 0,114 m3 to ¼ of the original volume. The initial conditions of the gas are 110 kPa and 100 oC. The specific heat capacity at constant volume is 0,754 kJ/kg.K and at constant pressure is 1,05 kJ/kg.K. Calculate the following: 4.1 The mass of the gas (6) 4.2 The value of the index of compression (7) 4.3 The final, absolute, temperature (3) 4.4 The change of entropy (4) [20] QUESTION 5 5.1 3,5 kg of steam is expanded from 2 000 kPa and 300 oC to a pressure of 280 kPa. The specific heat capacity of superheated steam is 2,1 kJ/kg.K. The steam is 10% wet at the latter pressure. Calculate the total change of internal energy. (17) 5.2 State the THREE phases in which water can exist and give an example of each. (3) [20] TOTAL: 100 Downloaded from hlayiso.com

-1- -6- (8190035) T1300(E)(A14)T POWER MACHINES N5 FORMULA SHEET n −1 n −1 T2  P2  n V  1. Q = W + ∆U 17. =   =  1  T1  P1   V2  γ −1 γ −1 T2  P2  γ V  ∆U = mC v ∆T =  =  1  T1  P1  2. 18.  V2  3. Q = mC v ∆T 19. h = h f + xh fg 4. Q = mC p ∆T 20. h = hg + C p ∆T V2 5. Q = P1V1 ln 21. h = h f + h fg = hg V1  T V  n − 1  hsup − 1941  ∆S = m C v ln 2 + R ln 2  Vsup =   n  6. 22.  T1 V1  P  Vm 7. W = P1 ∆V 23. x= Vg V2 M 8. W = P1V1 ln 24. x= V1 M +m P V − P2V2 9. W= 1 1 25. U = H − PV n −1 P1V1 − P2V2 1 2 10. W = 26. gZ 1 + U 1 + P1V1 + C1 + Q = γ −1 2 1 2 gZ 2 + U 2 + P2V2 + C 2 + W 2 m s (h2 − h1 ) . 11. R = C p − Cv 27. η= . m f CV m s (h2 − h1 ) . Cp 12. γ = 28. EE = Cv . m f 2257 P1V1 P2V2 p = (Bm ± M m ) 101,325 13. = 29. T1 T2 760 100  8  14. PV= mRT 30. m= C + 8 H 2 + S − O2  23  3   y y 15. P1V1 = P2V2 31. C x H y +  x + O2 = xCO2 + H 2O  4 2 Downloaded from hlayiso.com

-2- -7- (8190035) T1300(E)(A14)T n n 16. P1V1 = P2V2  O  32. H .C.V . = (CVC .C ) + CV H 2  H 2 − 2  + (CVS .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (m w + we )C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P2  n  n   P2  n 35. W = P1Ve      − 1 = mRT1     − 1  n − 1   P1   n − 1   P1       1  Ve Vc  P2  n  1 η c = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs  P1  36. Vs    Va 37. ηa = .100 Vs . 38. Fc = m(C fe − C fi ) P = m U [C wi − (− C we )] . 39. 2U [C wi − (− C we )] 40. η= .100 C ai 2 1 41. KE = mv 2 2 42. U = πDN . 43. m V = AC 44. (m + M ) g = mω 2 h π 45. Vs = D2L 4 46. θ1 = t c − twi 47. θ 2 = t c − two -3- Downloaded from hlayiso.com

-8- (8190035) T1300(E)(A14)T θ1 − θ 2 48. log . temp. diff . = θ ln 1 θ2 P  49. Piso = P1V1 ln 2   P1   n −1  n  P  n 50. Pact = P1V1   − 1  2 n −1  P1     Piso 51. η iso = .100 Pact Downloaded from hlayiso.com

MARKING GUIDELINE NATIONAL CERTIFICATE (VOCATIONAL) APRIL EXAMINATION POWER MACHINES N5 14 APRIL 2015 This marking guideline consists of 9 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1300(E)(A14)T POWER MACHINES N5 QUESTION 1 1.1 1.1.1 (M w + We )C p × ∆t HCV = Mf = (2,464 + 0,327 )4,2 × (20,69 − 18,5) 7 × 10 − 4 = 36673,74 kJ / kg = 36,67 MJ / kg (4) 1.1.2 LCV = HCV × (100 % − 13,7 % ) = 36,67 × 86,3 % = 31,646 MJ / kg = 31,65 MJ / kg (3) 1.1.3  H  LCV = HCV −  hfg × 9 × 2   100  HCV − LCV 100 H2 = × hfg × 9 1 36,67 − 31,65 100 = × 2,453 × 9 1 100 = 0,22738 × 1 = 22,74 % (3) 1.2 1.2.1 100  8  Mass of air =  C + 8 H 2 + S − O2  23  3  100  8   =  × 0,86  + (8 × 0,07 ) − 0,07  23  3   = 12,1 kg / kg fuel ∴Total mass = 12,1 × 12 = 145,2 kg (4) 1.2.2 O2 HCV = (HV × C) + HV c H2 H2 + (HVs × S) 8 0,07 = (35 × 0,86) + 144 0,07 8 = 38,92 MJ (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1300(E)(A14)T POWER MACHINES N5 1.2.3 LCV = HCV (hfg × 9 × H ) 2 = 38,92 (2,424 × 9 × 0,07) = 37,393 MJ / kg (3) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1300(E)(A14)T POWER MACHINES N5 QUESTION 2 2.1 U = π × D× N Vr2 = Vr1 × K π × 0,46 × 6976 = = 390 × 0.85 60 = 168 m / s = 331,5 m / s (14) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1300(E)(A14)T POWER MACHINES N5 2.2 2.2.1 P = MUVw = 1 × 168 × 594 = 99792 W = 99,792 kW (3) 2.2.2 1 KE = MV 2 2 = 0,5 × 1 × 225 2 = 25312,5 W / kg = 25,313 kW / kg (3) [20] QUESTION 3 3.1 Given: V air = 18,5m3/min Double-Acting P 1 = 110 kPa T 1 = 295K N = 400 r/min P 2 = 820 kPa n = 1,35 V c = 0,07S v 3.1.1 V1 = V3 + S v = 0,07S v + S v V1 = 1,07S v 1 1 n P2 V4 = V3 P1 1 820 1, 35 = 0,07S v 110 = 0,31S v 2 V1 V4 = 18,5 m 3 / min 1,07 S v 0,31S v = 18,5 0,76S v = 18,5 S v / min = 24,3421 m 3 / min 24,3421 ∴ Sv = 400 × 2 = 0,030428 m 3 (10) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1300(E)(A14)T POWER MACHINES N5 n −1 3.1.2 T2  P2  n =   T1  P1  n −1 P  n T2 = T1  2   P1  1, 35 −1  820  1, 35 = 295   110  = 496,6 K (3) n  P2  n  n −1 3.1.3 P = P1V1   − 1 n − 1  P1     18,5  1,35   820  1, 35  1, 35 −1 = 110 × ×  ×   − 1 60  1,35 − 1   110    = 89,4 kW (3) 3.2 Dalton’s law of partial pressure states that the pressure of a mixture of gases equals the sum of the partial pressures of each individual gas at the same temperature, in the container. (2) 3.3 Axial – flow turbines Radial – flow turbines (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1300(E)(A14)T POWER MACHINES N5 QUESTION 4 Given: P2 = 655 kPa PV n = C V1 = 0,114 m 3 1 P1 = 110 kPa T1 = 373K V2 = of V1 4 Cv = 0,754 kJ / kg.K Cp = 1,05kJ / kg.K 4.1 R = Cp − Cv = 1,05 − 0,754 = 0,296 kJ / kg.K P1V1 = MRT1 P1V1 M = RT1 110 × 0,114 = 0,296 × 373 = 0,1136 kg (6) 4.2 1 V2 = of V1 4 = 0,25 × 0,114 = 0,0285 m 3 P1V1n = P2V2n n  V1  P2   =  V2  P1 V P2 n ln 1 = ln V2 P1 P  ln  2  ∴n =  P1  V  ln  1   V2   655  ln   =  110   0,114  ln    0,0285  1,78415 = 1,38629 Copyright reserved = Downloaded from hlayiso.com Please turn over (7) 1,287

MARKING GUIDELINE -8- T1300(E)(A14)T POWER MACHINES N5 4.3 P1V1 P2V2 = T1 T2 P2V2T1 T2 = P1V1 655 × 0,0285 × 373 = 110 × 0,114 = 555,26 K or n −1 P  n T2 = T1  2   P1  1, 287 −1  655  1, 287 = 373   110  = 555,26 K (3) 4.4  T   V  S1 − S 2 = m Cv ln 2  +  R ln 2   T1   V1   555,26   0,0285  = 0,1136 0,754 ln  +  0,296 ln   373   0,114  = 0,1136{0,29998 + (− 0,41034 )} = −0,01254 kJ / K (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -9- T1300(E)(A14)T POWER MACHINES N5 QUESTION 5 5.1 Given: M = 3,5 kg P1 = 2000 kPa t su = 300 o C x = 0,9 P2 = 280 kPa Cp = 2,1 kJ / kg.K U1 = h − P1V1 (@ 2000 kPa and 300 C ) o   0,231(h − 1941)  = {hg + Cp(t − t )} −  P × su s  1  sup    P1   = {2797 + 2,1(300 − 212,4)} − 2000 ×  0,231(3025 − 1941)    2000  = 2980,96 − 250,404 = 2730,556 kJ / kg U2 = h − P2V2 (@ 280 kPa and x = 0,9) = {hf + xhfg} − {P2 × xVg 2 } = {551 + (0,9 × 2170)} − {280(0,9 × 0,6460)} = 2504 − 162,792 = 2341,208 kJ / kg U 2 − U1 = 2341,208 − 2730,556 = −389,348 kJ / kg ∴ ∆U = 3,5(U 2 − U 1 ) = 3,5 × (− 389,348) = −1362,718 kJ (17) 5.2 Solid – eg Ice Liquid – eg Water Gas – eg Steam (3) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

T1310(E)(N26)N NOVEMBER EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 26 November 2014 (Y-Paper) 13:00–16:00 REQUIREMENTS: Steam Tables (BOE 173). Superheated Steam Tables (Appendix to BOE 173). Calculators may be used. This question paper consists of 5 pages and a formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1270(E)(N26)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1270(E)(N26)T QUESTION 1 1.1 State THREE advantages of a condenser. (6) 1.2 Every hour a surface condenser processes 6 500 kg exhaust steam, which has a dryness fraction of 0,83. Air leakage into the condenser is at a rate of 1,2 kg/1 000 kg of steam. The air pump suction pipe and the condensate have a temperature of 31 ̊C each. The barometer and vacuum gauge reading is 760 mm Hg and 662 mm Hg respectively. The temperature of the cooling water is increased by 21 ̊C after it passes through the condenser. The specific heat capacity of the water is 4 187 kJ/kg.K and R for air is 0,288 kJ.kg.K. Calculate the following: 1.2.1 The mass of cooling water required by the condenser every minute (8) 1.2.2 The capacity of the air pump in m3/min (6) [20] QUESTION 2 2.1 A certain gas has a density of 1,28 kg/m3 at 200 kPa and 267 ̊C. The law, PV1,29 = C, is used to expand 0,7 kg of this gas to 2,5 times its original volume, from 200 kPa and 267 ̊C. Calculate the following: 2.1.1 The characteristic gas constant (3) 2.1.2 The original and final volume of the gas (6) 2.1.3 The final pressure of the gas (3) 2.1.4 The final temperature of the gas (3) 2.1.5 The work done during the expansion (3) 2.2 State the function of a governor. (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1270(E)(N26)T QUESTION 3 Gas leaves a single-stage impulse turbine at an angle of 41°. Friction over the blading causes a 10% loss in velocity. The relative velocity of the gas at the inlet to the blades is 350 m/s at an angle of 30°. The blade experiences a peripheral velocity of 175 m/s, when 50 kg of gas flows through the turbine every second. 3.1 Use a scale of 1 cm = 25 m/s and construct a velocity diagram in the ANSWER BOOK. Enter ALL the values (m/s) onto the diagram. NB: NO marks will be awarded if the values (m/s) are NOT entered onto the diagram and if the diagram is NOT constructed to the given scale. (11) HINT: Use the answer page in the landscape format to construct the diagram. 3.2 Determine the following from the diagram: 3.2.1 The nozzle angle (1) 3.2.2 The nozzle velocity (1) 3.2.3 The exit angle of the moving blades (1) 3.2.4 The axial thrust (3) 3.2.5 The power developed (3) [20] QUESTION 4 4.1 18,04 kg of air is used for the complete combustion of a kilogram of fuel. The fuel has the following composition: Carbon = 87% Hydrogen = 2,5% Sulphur = 1% The balance of the fuel is non-combustible. (18) Calculate the percentages of the products of combustion. 4.2 Regulation C112 provides for the access and inspection openings of boilers. Name TWO reasons for these openings. (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1270(E)(N26)T QUESTION 5 5.1 A single-stage, double-acting, compressor must deliver 16 m3 of air every minute. The compressor receives the air at 101,3 kPa and 20 C ̊ and delivers it at a pressure of 900 kPa. The effective volume is 0,94 of the swept volume, and the speed of the compressor is 375 r/min. The index of compression is 1,3 and the mechanical efficiency is 79%. Calculate the following: 5.1.1 The volumetric efficiency (3) 5.1.2 The swept volume (4) 5.1. The temperature of the delivered air (3) 5.1.4 The power required to drive the compressor (6) 5.2 Regulation C97.7 refers. State the hydraulic test pressure for the following boilers: 5.2.1 Boilers which do NOT exceed the working gauge pressure of 500 kPa. 5.2.2 Boilers which exceed the working gauge pressure of 500 kPa. (2 x 2) (4) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -1- T1270(E)(N26)T FORMULA SHEET n −1 n −1 T2  P2  n V  1. Q = W + ∆U 17. =   =  1  T1  P1   V2  γ −1 γ −1 T2  P2  γ V  2. ∆U = mC v ∆T 18. =   =  1  T1  P1   V2  3. Q = mC v ∆T 19. h = h f + xh fg 4. Q = mC p ∆T 20. h = hg + C p ∆T V2 5. Q = P1V1 ln 21. h = h f + h fg = hg V1  T V  n − 1  hsup − 1941  ∆S = m C v ln 2 + R ln 2  Vsup =   n  6. 22.  T1 V1  P  Vm 7. W = P1 ∆V 23. x= Vg V2 M 8. W = P1V1 ln 24. x= V1 M +m P V − P2V2 9. W = 1 1 25. U = H − PV n −1 P1V1 − P2V2 1 2 10. W = 26. gZ 1 + U 1 + P1V1 + C1 + Q = γ −1 2 1 2 gZ 2 + U 2 + P2V2 + C 2 + W 2 m s (h2 − h1 ) . 11. R = C p − Cv 27. η= . m f CV m s (h2 − h1 ) . Cp 12. γ = 28. EE = Cv . m f 2257 P1V1 P2V2 p = (Bm ± M m ) 101,325 13. = 29. T1 T2 760 100  8  14. PV= mRT 30. m= C + 8 H 2 + S − O2  23  3   y y 15. P1V1 = P2V2 31. C x H y +  x + O2 = xCO2 + H 2O  4 2 n n 16. P1V1 = P2V2 Downloaded from hlayiso.com Copyright reserved Please turn over

(8190035) -2- T1270(E)(N26)T  O  32. H .C.V . = (CVC .C ) + CV H 2  H 2 − 2  + (CVS .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (m w + we )C p ∆T 34. H .C.V . = mf  n −1   n −1   n  2 P  n  n   2 nP  35. W = P1Ve      − 1 = mRT1     − 1  n − 1   P1   n − 1   P1       1  Ve Vc  P2  n  1 36. η c = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1     Va 37. ηa = .100 Vs . 38. Fc = m(C fe − C fi ) P = m U [C wi − (− C we )] . 39. 2U [C wi − (− C we )] 40. η= .100 C ai 2 1 41. KE = mv 2 2 42. U = πDN . 43. m V = AC 44. (m + M ) g = mω 2 h π 45. Vs = D2L 4 46. θ1 = t c − twi 47. θ 2 = t c − two Downloaded from hlayiso.com Copyright reserved Please turn over

(8011023) 3 T1270(E)(N26)N θ1 − θ 2 48. log . temp. diff . = θ ln 1 θ2 P  48. Piso = P1V1 ln 2   P1   n −1  n  P2  n 49. Pact = P1V1   − 1  n −1  P1     Piso 50. η iso = .100 Pact Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE NATIONAL CERTIFICATE NOVEMBER EXAMINATION POWER MACHINES N5 26 NOVEMBER 2014 This marking guideline consists of 9 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1270(E)(N26)T POWER MACHINES N5 QUESTION 1 1.1 - The treated water can be re-used in the boiler and this reduces cost. - A hot water supply is available which requires less energy for transformation to steam. -The pressure in the condenser is below atmospheric pressure which increases the thermal efficiency of the plant and the work done per kilogram of steam. (6) 1.2 Given: Ms = 6500kg / hour Air Leakage = 1,2kg / 1000kg steam tc = 31o C Bm = 760 mmHg Vm = 662 mmHg t 2 − t1 = 21o C x = 0,83 shc w = 4,187 kJ / kg.K R = 0,288 kJ / kg.K 1.2.1 Pt = (Bm − Vm ) × 101,325 760 = (760 − 662) × 101,325 760 = 13 kPa Heat lost by steam = Heat gained by water M s {(hf + xhfg ) − hc} = M w × shc w × ∆t 6500 {[214 + (0,83 × 2380)] − 130} = M w × 4,187 × 21 60 223101,6667 = 87,927 M w ∴M w = 2537,351 kg / min (8) 1.2.2 Pa = Pt − Ps = 13 − 4,5 = 8,5 kPa Ta = 31 + 273 = 304 K 1,2 6500 Ma = × 1000 60 = 0,13 kg / min Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1270(E)(N26)T POWER MACHINES N5 PaVa = M a RTa M a × R × Ta Va = Pa 0,13 × 0,288 × 304 = 8,5 (6) = 1,339 m 3 / min [20] QUESTION 2 2.1 Given: δ = 1,28 kg / m 3 m = 0,7 kg P1 = 200 kPa T1 = 540 K n = 1,29 V2 = 2,5V1 2.1.1 PV = MRT PV R = MT 200 × 1 = 1,28 × 540 = 0,289 kJ / kg.K (3) 2.1.2 P1V1 = MRT1 MRT1 V1 = P1 0,7 × 0,289 × 540 = 200 = 0,5462 m 3 V2 = 2,5V1 = 2,5 × 0,5462 = 1,3655 m 3 (6) 2.1.3 P1V1n = P2V2n n V  P2 = P1  1   V2  1, 29  0,5462  = 200   1,3655  = 61,332 kPa (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1270(E)(N26)T POWER MACHINES N5 n −1 2.1.4 P1V1 P2V2 T2  V1  = =   T1 T2 T1  V2  n −1 P2V2T1 V  T2 = T2 = T1  1  P1V1  V2  1, 29 −1 61,332 × 1,3655 × 540  0,5462  = or = 540  200 × 0,5462  1,3655  = 413,991 K = 413,991 K (3) 2.1.5 P1V1 − P2V2 W = n −1 = (200 × 0,5462) − (61,332 × 1,3655) 1,29 − 1 = 87,9 kJ or MR(T1 − T2 ) W = n −1 0,7 × 0,289(540 − 413,991) = 1,29 − 1 = 87,9 kJ (3) 2.2 The function of a governor is to control the desired speed of an engine. (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1270(E)(N26)T POWER MACHINES N5 QUESTION 3 3.1 Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1270(E)(N26)T POWER MACHINES N5 Vr2 K = Vr1 Vr2 = K × Vr1 = 0,9 × 350 = 315 m / s (11) 3.2 3.2.1 α = 20 o (1) 3.2.2 V1 = 510 m / s (1) 3.2.3 φ = 19 o (1) 3.2.4 Fc = m(Vf1 − Vf 2 ) = 50(177,5 − 105) = 3625 N = 3,625 kN (3) 3.2.5 P = MUC w = 50 × 175 × 600 = 5250000 W = 5250 kW = 5,25 MW (3) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1270(E)(N26)T POWER MACHINES N5 QUESTION 4 4.1 100  8  Theoritical mass of air reqd . =  C + 8 H 2 + S − O2  23  3  100  8   =  × 0,87  + (8 × 0,025) + 0,01 23  3   100 = × 2,53 23 = 11 kg air / kg fuel Mass of excess air = Actual air − Theoritical air = 18,04 − 11 = 7,04 kg 7,04 100 Percentage of excess air = × 11 1 = 64 % Mass of Pr oducts of Combustion : CO2 = 3,667 × 0,87 = 3,19 kg H 2O = 9 × 0,025 = 0,225 kg 64 Excess O2 = 2,53 × = 1,619 kg 100 SO2 = 2 × 0,01 = 0,02 kg 77 N2 = 18,04 × = 13,89 kg 100 18,944 kg Percentages of Pr oducts of Combustion : 3,19 100 CO2 = × = 16,839 % 18,944 1 0,225 100 H 2O = × = 1,188 % 18,944 1 1,619 100 Excess O2 = × = 8,546 % 18,944 1 0,02 100 SO2 = × = 0,106 % 18,944 1 13,89 100 N2 = × = 73,321 % 18,944 1 100 % Copyright reserved Downloaded from hlayiso.com Please turn over(18)

MARKING GUIDELINE -8- T1270(E)(N26)T POWER MACHINES N5 4.2 To provide suitable openings to clean and inspect all internal surfaces, longitudinal seams and circumferential seams. To provide at least one manhole to permit entry into the boiler. (2) [20] QUESTION 5 5.1 Given: Double − Acting M = 16 m 3 / min P1 = 101,3 kPa T1 = 293 K Ve = 0,94 Vs =16 m 3 P2 = 900 kPa N = 375 r / min n = 1,3 η mech = 79 % 5.1.1  P2  n 100 1 η vol = 1 + α − α   ×  P1  1 1  900  1, 3 100 = 1 + 0,06 − 0,06  ×  101,3  1 = 73,8 % (3) 5.1.2 Ve = Vη × Vs × N × 2 Ve Vs = Vη × N × 2 16 = 0,738 × 375 × 2 = 0,0289 m 3 (4) n −1 5.1.3 T2  P2  n =   T1  P1  n −1 P  n T2 = T1  2   P1  1, 3 −1  900  1, 3 = 293   101,3  = 485,05 K (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -9- T1270(E)(N26)T POWER MACHINES N5  P  n  n −1 5.1.4 n Pindicated = P1Ve   − 1  2 n −1  P1     16  900  1, 3  1, 3 −1 1,3 = × 101,3 ×   − 1 1,3 − 1 60  101,3     = 76,725 kW Pcomp = Pindicated × η 100 = 76,725 × 79 = 97,12 kW (6) 5.2 5.2.1 For gauge pressures which do not exceed 500 kPa, the hydraulic test pressure shall be double the authorised working gauge pressure. (2) 5.2.2 For gauge pressures which exceed 500 kPa, the hydraulic test pressure shall be 1,2 times the authorised working gauge pressure plus 400 kPa. (2) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

T1270(E)(A5)T AUGUST EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 5 August 2014 (Y-Paper) 13:00–16:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Drawing instruments are required. Calculators may be used. This question paper consists of 5 pages and a formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1270(E)(A5)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1270(E)(A5)T QUESTION 1 1.1 The blades of a single-stage, impulse turbine has a mean diameter of 960 mm. 100 kW of power develops at a speed of 4 755 r/min. The gas discharges in an axial direction and the coefficient of friction is 0,87. Both blade angles are 30o each to the plane of rotation. 1.1.1 Calculate the blade velocity in m/s (3) 1.1.2 Use a scale of 1 cm = 50 m/s and construct a velocity diagram in the ANSWER BOOK. Enter ALL the values (m/s) onto the diagram. HINT: Use the answer page in the landscape format to construct the diagram. NOTE: NO marks will be awarded if the values (m/s) are NOT entered onto the diagram and if the diagram is NOT constructed to the given scale. (8) 1.1.3 Determine the following from the diagram: (a) The nozzle angle (1) (b) The mass flow of gas in kg/s (3) (c) The diagram efficiency (3) 1.2 State the volume and enthalpy of 1 kg of steam which has a temperature of 500 °C and a pressure of 0,5 MPa. (2) [20] QUESTION 2 2.1 With the aid of steam tables, calculate the following: 2.1.1 The enthalpy found in 2 kg of steam which has a pressure of 480 kPa and is 7,3% wet. (3) 2.1.2 The enthalpy of a kilogram of steam which has a pressure of 1,5 MPa, a temperature of 396 °C and a specific heat capacity of 2,1 kJ/kg.K. (3) 2.1.3 The dryness fraction of wet steam which has an enthalpy of 2 650 kJ/kg at a pressure of 1,3 MPa. (3) 2.1.4 The heat required to change 2,3 kg of water from a temperature of 31 °C to dry saturated steam at a pressure of 1 000 kPa. (3) 2.2 Name THREE apparatus used to determine the dryness fraction of steam. (3) Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1270(E)(A5)T 2.3 A boiler plant supplies 5 600 kg of wet steam every hour at a pressure of 800 kPa. The steam is 9,5% wet and the supply water enters the boiler at 29 °C. The boiler consumes 11 kg of coal every minute, which has a heat value of 35 MJ/kg. Calculate the efficiency of the boiler. (5) [20] QUESTION 3 1 kg of fuel oil which has the formula C 12 H 4 completely burnt with the aid of 40% excess air. The atomic mass of carbon, oxygen and hydrogen is 12,16 and 1, respectively. Calculate the following: 3.1 The theoretical mass of air required to burn the fuel (8) 3.2 The actual mass of air used to burn the fuel (2) 3.3 The mass of each product of combustion (6) 3.4 The percentage analysis, by mass, of the combustion products (4) [20] QUESTION 4 A single cylinder, single-acting compressor takes in 56,4 m3 of air every hour. The air is delivered at a pressure of 900 kPa after it is received at a pressure of 103 kPa and 22 °C. The compressor does not have a clearance volume and the law of compression is PV1,35 = C. The stroke to bore ratio is 1,6 to 1 and the speed is 350 r/min. The electric motor experiences a power loss of 10% and the compressor experiences a power loss of 12%. Air has a specific heat capacity of 0,287 kJ/kg.K. Calculate the following: 4.1 The indicated power of the compressor (4) 4.2 The bore diameter in mm (8) 4.3 The stroke length in mm (2) 4.4 The power rating of the motor used to drive the compressor in kW (6) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1270(E)(A5)T QUESTION 5 5.1 Name TWO types of compressors. (2) 5.2 State Boyle's law. (4) 5.3 State Charles' law. (5) 5.4 A constant volume process was used to heat 3,3 kg of gas from a temperature of 21 °C and 0,87 m3 to a temperature of 137 °C. The specific heat capacity of the gas at constant volume is 0,718 kJ/kg.K and the gas constant is 0,289 kJ/kg.K. Calculate the following: 5.4.1 The quantity of heat transferred in kJ (3) 5.4.2 The final pressure of the gas after heating (6) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -1- - 6 - T1270(E)(A5)T FORMULA SHEET 1. Q = W + ∆U n −1 n −1 T2  P2  n V  17. =  =  1  2. ∆U = mCυ ∆T T1  P1   V2  γ −1 3. Q = mCυ ∆T γ −1 T2  P2  γ V  18. =  =  1  T1  P1   V2  4. Q = mC p ∆T 19. h = h f + χh fg V 5. Q = P1V1 ln 2 V1 20. h = hg + C p ∆T  T V  6. ∆S = m Cυ ln 2 + R ln 2  21. h = h f + h fg = hg  T1 V1  n − 1  hsup − 1 941  7. W = P1∆V 22. Vsup =   n  P  V2 8. W = P1V1 ln Vm V1 23. χ= Vg P1V1 − P2V2 9. W = n −1 M 24. χ = M +m P1V1 − P2V 2 10. W= 25. U = H - PV γ −1 1 2 11. R = C p − Cυ 26. gZ1 + U1 + P1V1 + C1 + Q = 2 1 gZ 2 + U 2 + P2V2 + C2 2 + W Cp 2 12. γ = Cυ • m s (h2 − h1 ) P1V1 P2V2 27. η= • 13. = m ƒ CV T1 T2 • 14. PV = mRT m s (h2 − h1 ) 28. EE = • 15. P1V1 = P2V2 m ƒ 2 257 101,325 16. P1V1n = P2V2n 29. p = ( Bm ± M m ) 760 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- - 7 - T1270(E)(A5)T 100  8  30. m= C + 8 H 2 + S − O2  23  3   y y 31. C x H y +  x + O2 = xCO2 + H 2O  4 2  O  32. H .C.V . = (CVC .C ) + CVH 2  H 2 − 2  + (CVs .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (mw + me ) C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P2  n   n   P2  n  35. W = P1Ve     − 1 = mRT1     − 1  n − 1   P1    n − 1   P1        1  1 Ve Vc  P2  n  36. ηc = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1      Vα 37. ηα = .100 Vs • 38. Fc = m(C fe − C fi ) • 39. P = mU [Cwi − (−Cwe )] 2U [Cwi − (−Cwe )] 40. η= .100 Cai 2 41. U = πDN • 42. m V = AC 43. (m + M ) g = mω 2 h π 44. Vs = D2L 4 Copyright reserved Downloaded from hlayiso.com Please turn over

-3- (8190035) -8- T1270(E)(A5)T 45. θ1 = tc − twi 46. θ2 = tc − two θ1 − θ2 47. Log.temp.diff. = θ ln 1 θ2 P  48. Piso = P1 V1 ln  2   P1   n −1  n  P  n 49. Pact = P1 V1   − 1  2 n −1  P    1   Piso 50. Niso = • 100 Pact Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE NATIONAL CERTIFICATE AUGUST EXAMINATION POWER MACHINES N5 5 Augusts 2014 This marking guideline consists of 8 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1270(E)(A5)T POWER MACHINES N5 QUESTION 1 1.1 1.1.1 πDN U = 60 π × 0,96 × 4755 U = 60 U = 239 m / s (3) 1.1.2 (8) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1270(E)(A5)T POWER MACHINES N5 1.1.3 A 17o (1) B P = MUCw P M = U × Cw 100 × 10 3 = 239 × 512.5 = 0,816 kg / s (3) 2 × U × Vw 100 C η = × V1 2 1 2 × 239 × 512,5 100 = × 535 2 1 = 85,6 % (3) 1.2 V su = 0,7108 m 3 / kg hsu = 3484 kJ / kg (2) [20] QUESTION 2 2.1 2.1.1 hwet = M (hf + xhfg ) = 2{634 + (0,927 × 2112 )} = 5183,648 kJ (3) 2.1.2 hsup = hg + C p (t su − t s ) = 2790 + 2,1(396 − 198,3) = 3205,17 kJ / kg (3) 2.1.3 hwet = hf + xhfg hwet − hf x = hfg 2650 − 815 = 1971 = 0,93 (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1270(E)(A5)T POWER MACHINES N5 2.1.4 hdry = M {(hf + hfg ) − hf w } hdry = M (hg − hf w ) = 2,3{(763 + 2014) − 130} or = 2,3(2777 − 130 = 6088,1 kJ = 6088,1 kJ (3) 2.2 Barrel Calorimeter Separating Calorimeter Throttling Calorimeter Combined Separating and Throttling Calorimeter (Any 3 x 1) (3) 2.3 h1 = 121 kJ / kg h2 = hf + xhfg = 721 + (0,905 × 2047 ) = 2573,535 kJ / kg M s (h2 − h1 ) 100 η = × mf × Cv 1 5600(2573,535 − 121) 100 = × 11 × 60 × 35 × 10 3 1 = 59,455% (5) [20] QUESTION 3 3.1 x = 12 y = 4  y y C x H y +  x + O2 = xCO2 + H 2O  4 2  4 34 C12 H 4 + 12 + O2 = 12CO2 + H 2O  4 2 C12 H 4 + 13O2 = 12CO2 + 2 H 2 O [(12 × 12) + (4 × 1)] + 13(2 × 16) = 12[12 + (2 × 16 )] + 2[(2 × 1) + 16] (144 + 4) + 416 = 528 + 36 (÷148) : 1 + 3,685 = 2,966 + 1,719 148 1 + 2,811 = 3,568 + 0,243 100 ∴ Theoritical mass of air reqd . = 2,811 × 23 = 12,22kg / kg of fuel Copyright reserved Downloaded from hlayiso.com Please turn over (8)

MARKING GUIDELINE -5- T1270(E)(A5)T POWER MACHINES N5 3.2 ∴ Actual mass of air reqd . = 12,22 × 1,4 = 17,11 kg / kg of fuel (2) 3.3 Pr oducts of Combustion CO 2 = 3,568 kg H 2O = 0,243 kg 40 O2 = 2,811 × 100 = 11244 kg 77 N2 = 17,11 × 100 = 13,175kg (6) 3.4 3,568 100 CO 2 = 3,568kg = × = 19,7 % 18,11 1 0,243 100 H 2O = 0,243 kg = × = 1,34 % 23,429 1 1,1244 100 O2 = 1,1244kg = × = 6,21 % 18,11 1 13,175 100 N2 = 13,175 kg = × = 73,71 % 23,429 1 18,11 kg (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1270(E)(A5)T POWER MACHINES N5 QUESTION 4 Given: M = 56,4 m3/hour P 1 = 103 kPa T 1 = 295K PV1,35 = C P 2 = 900 kPa N = 350 r/min Bore to Stroke Ratio = 1,6:1 η mech = 88% η motor = 90%  P  n  n −1 4.1 n   P = × P1V1  2  − 1 n −1  P1   56,4  900  1, 35  1, 35 −1 1,35 = × 103 ×   − 1  1,35 − 1 3600  103    = 4,694 kW (4) 4.2 Vol / min Vol / cycle = N 56,4 1 = × 60 350 = 0,002686 m 3 π × D2 Vol / cycle = ×L 4 π × D2 = × 1,6 D 4 Vol × 4 D = 3 π × 1,6 0,002686 × 4 = 3 π × 1,6 = 0,1288 m = 128,8 mm (8) 4.3 Stroke = 1,6 × D = 1,6 × 128,8 = 206,08 mm (2) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1270(E)(A5)T POWER MACHINES N5 4.4 Indicated Power η mech = Shaft Power Indicated Power ShaftPower = η mech 4,694 = 0,88 = 5,334 kW Shaft Power η motor = Input Power Shaft Power InputPower = η motor 5,334 = 0,9 = 5,93 kW ∴ Power rating of motor reqd . = 5,93 kW or 6 kW (6) [20] QUESTION 5 5.1 The reciprocating compressor The rotary compressor (2) 5.2 Boyle’s law states that at constant temperature, the pressure and volume of a certain mass of gas are inversely proportional. (4) 5.3 Charles’ law states that during the change of state of any gas in which the mass and pressure remain constant, the volume varies in proportion with the Absolute temperature. (5) 5.4 Given: m = 3,3 kg T 1 = 294 K V 1 = 0,87 m3 C v = 0,718 kJ/kg.K T 2 = 410 K R = 0,289 kJ/kg.K 5.4.1 Q = ∆U + W = M × C v (T2 − T1 ) + 0 = 3,3 × 0,718(410 − 294 ) = 274,85kJ (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- T1270(E)(A5)T POWER MACHINES N5 5.4.2 P1V1 = MRT1 MRT1 P1 = V1 3,3 × 0,289 × 294 = 0,87 = 322,285 kPa P1 P2 = T1 T2 P1T2 P2 = T1 322,285 × 410 = 294 (6) = 449,445 kPa [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

T1270(E)(A8)T APRIL EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 8 April 2014 (Y-Paper) 13:00–16:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Candidates need drawing instruments. Calculators may be used. This question paper consists of 5 pages and 1 formula sheet of 3 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1270(E)(A8)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1270(E)(A8)T QUESTION 1 2,682 kg of gas is expanded at a constant pressure of 680 kPa from 22 °C to 0,34 m3. The gas is then further expanded adiabatically to a pressure of 180 kPa. The specific heat capacity of this gas at a constant pressure and constant volume is 1,05 kJ/kg.K and 0,779 kJ/kg.K respectively. Calculate the following: 1.1 The volume at the commencement of the first process and at the end of the second process (10) 1.2 The absolute temperature at the end of each of the processes (6) 1.3 The total work done (4) [20] QUESTION 2 The clearance volume of a single-cylinder, single-acting, air compressor is 6,5% of the swept volume. Air is sucked in at 100 kPa and 23 °C and is delivered at 778 kPa. The compressor runs at 300 r/min and the stroke to diameter ratio is 1,6. The air is consumed at a rate of 46 kg/min whilst the compressor has a mechanical efficiency of 75%. The gas constant for air is 0,286 kJ/kg.K and the index n = 1,37. Calculate the following: 2.1 The volumetric efficiency (6) 2.2 The piston diameter and the stroke length (10) 2.3 The power required to drive the compressor (4) [20] QUESTION 3 3.1 50% excess air is used to burn 1 m3 of gas. The gas consists of the following: H 2 = 43% CH 4 = 23% CO = 22% N 2 = 12% Air contains 21% oxygen by volume. Calculate the following: 3.1.1 The volume of oxygen required (6) 3.1.2 The volume of air required to completely burn the gas (2) 3.1.3 The volume of each of the dry products of combustion (10) Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1270(E)(A8)T 3.2 Air pumps are used to remove air from the condenser to maintain a vacuum in the condenser. Name TWO types of air pumps. (2) [20] QUESTION 4 4.1 5 kg of gas is delivered to a single-stage impulse turbine every second. The nozzles are inclined at an angle of 19o and the velocity of the gas is 1 000 m/s. The gas loses 24% velocity due to friction over the blades. The outlet angle of the moving blade is 22° and the peripheral speed of the wheel is 400 m/s. 4.1.1 Use a scale of 1 cm = 50 m/s and construct a velocity diagram in the ANSWER BOOK. Enter all the values (m/s) onto the diagram. NB: No marks will be awarded if the values (m/s) are not entered onto the diagram and if the diagram is not constructed to the given scale. HINT: Use the answer page in the landscape format to construct the diagram. (8) 4.1.2 Determine the following from the diagram: A The inlet angle of the moving blades (1) B The power developed by the turbine (3) C The diagram efficiency (3) D The exit angle of the gas from the turbine (1) 4.2 Knowledge and skill is required to burn fuel efficiently. Give TWO reasons why fuel should be burnt efficiently. (2) 4.3 Name TWO types of dead weight governors. (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1270(E)(A8)T QUESTION 5 5.1 4,2% wet steam which is in a main line at a pressure of 2 MPa is passed through a separating and throttling calorimeter. The conditions in the throttling calorimeter are 140 °C and 200 kPa. The specific heat capacity for superheated steam is 2,09 kJ/kg.K. Calculate the dryness fraction as recorded in the separating calorimeter. (8) 5.2 A boiler, which has an efficiency of 73%, produces 5 500 kg of steam per hour at a temperature of 350 °C and 4 000 kPa. The water enters the boiler at a temperature of 39 °C. The calorific value of the coal used is 35 MJ/kg. Calculate the following: 5.2.1 The mass of coal consumed every minute (7) 5.2.2 The equivalent evaporation from and at 100 °C (3) 5.3 Name the TWO theoretical calorific values of a fuel. (2) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -6- -1- T1270(E)(A8)T POWER MACHINES N5 FORMULA SHEET 1. Q = W + ∆U n −1 n −1 T2  P2  n V  17. =   =  1  2. ∆U = mCυ ∆T T1  P1   V2  γ −1 3. Q = mCυ ∆T γ −1 T2  P2  γ V  18. =  =  1  T1  P1   V2  4. Q = mC p ∆T 19. h = h f + χh fg V 5. Q = P1V1 ln 2 V1 20. h = hg + C p ∆T  T V  21. h = h f + h fg = hg 6. ∆S = m Cυ ln 2 + R ln 2   T1 V1  n − 1  hsup − 1 941  22. Vsup =   7. W = P1∆V n  P  V2 Vm 8. W = P1V1 ln 23. χ= V1 Vg P1V1 − P2V2 M 9. W = 24. χ= n −1 M +m P1V1 − P2V 2 25. U = H - PV 10. W= γ −1 1 2 26. gZ1 + U1 + P1V1 + C1 + Q = 2 11. R = C p − Cυ 1 gZ 2 + U 2 + P2V2 + C2 2 + W 2 Cp 12. γ = • Cυ m s (h2 − h1 ) 27. η= • P1V1 P2V2 m ƒ CV 13. = T1 T2 • m s (h2 − h1 ) 28. EE = 14. PV = mRT • m ƒ 2 257 15. P1V1 = P2V2 101,325 29. p = ( Bm ± M m ) 760 16. P1V1n = P2V2n Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -7- -2- T1270(E)(A8)T 100  8  30. m= C + 8 H 2 + S − O2  23  3   y y 31. C x H y +  x + O2 = xCO2 + H 2O  4 2  O  32. H .C.V . = (CVC .C ) + CVH 2  H 2 − 2  + (CVs .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (mw + me ) C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P2  n   n   P  2 n   35. W = P1Ve     − 1 = mRT1     − 1  n − 1   P1    n − 1   P1        1  1 Ve Vc  P2  n  36. ηc = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1      Vα 37. ηα = .100 Vs • 38. Fc = m(C fe − C fi ) • 39. P = mU [Cwi − (−Cwe )] 2U [Cwi − (−Cwe )] 40. η= .100 Cai 2 41. U = πDN • 42. m V = AC 43. (m + M ) g = mω 2 h π 44. Vs = D2L 4 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -8--3- T1270(E)(A8)T 45. θ1 = tc − twi 46. θ2 = tc − two θ1 − θ2 47. Log.temp.diff. = θ ln 1 θ2 P  48. Piso = P1 V1 ln  2   P1   n −1  n   P2 n 49. Pact = P1 V1   − 1  n −1  P    1   Piso 50. Niso = • 100 Pact Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE NATIONAL CERTIFICATE APRIL EXAMINATION POWER MACHINES N5 8 APRIL 2014 This marking guideline consists of 9 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1270(E)(A8)T POWER MACHINES N5 QUESTION 1 Given: m = 2,682 kg Cp = 1,05 kJ/kg.K Cv = 0,779 kJ/kg.K Constant Adiabatic P1 = 680 kPa Pressure P2 = 680 kPa Process P3 = 180 kPa T1 = 295 K V2 = 0,34 m3 1.1 R  C p  Cv  1,05  0,779  0,271 kJ / kg.K P1V1  MRT1 MRT1 V1  P 2,682  0,271  295  680  0,31531 m 3 Cp   Cv 1,05  0,779  1,35 P2V2  P3V3 P2 V3  V2  P3 1  P2   V3  V 2   P3  1  680  1, 3 5  0,34   180   0,91 m 3 (10) The incorrect value of V2 was used instead of 0.34m3 Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1270(E)(A8)T POWER MACHINES N5 1.2 V1 V2 P2V2  MRT2  T1 T2 P2V2 V2  T1 T2  or T2  MR V1 680  0,34 0,34  295   2,682  0,271 0,31531  318,1 K  318,1 K P2V2 P3V3 P3V3  MRT3  T2 T3 P3V3 P3V3T2 T3  or T3  MR P2V2 180  0,91 180  0,91  318,1   2,682  0,271 680  0,34  225.365 K  225.365 K (6) 1.3 P2V2  P3V3 WD  P1 V2  V1    1  680  0,34  180  0,91  6800,34  0,31531     1,35  1   16,7892  192.571  209.36 kJ (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1270(E)(A8)T POWER MACHINES N5 QUESTION 2 Given: Vc = 6,5% P1 = 100 kPa T1 = 296 K P2 = 778 kPa N = 300 r/min Stroke:Dia = 1,6 M = 46 kg/min  = 75% R = 0,286 kJ/kg.K n = 1,37 2.1 V1  Vs  0,065Vs  1,065Vs V3  0,065Vs  P  n  100 1 1  P3  1, 37 V4  V3   OR  vol  1    2   1   P4   P1   1    778  1, 37  100 1 1  778  1, 37  0,065Vs    1  0,065   1   100   100   1  0,2906 Vs  77,44 % V1  V4 100  vol   V1  V3 1 1,065Vs  0,2906Vs 100   1,065Vs  0,065Vs 1  77,44 % (6) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1270(E)(A8)T POWER MACHINES N5 2.2 P1V1  MRT1 MRT1 V1  P1 46  0,286  296  100  38,942 m 3 / min Vol / min  V1  V4 38,942  1,065Vs  0,2906Vs 300 0,1298  0,7744Vs 0,1298 Vs  0,7744  0,16761 m 3  Vs   D 2  1,6 D 4 0,16761  4 D  3   1,6  0,511 m √  511 mm Stroke  1,6  511  817,6 mm (10) 2.3  n 1   n   P2  n P  P1Ve      1  n  1   P1      1,37   778  1, 3 7  1 , 3 7 1  100  0,1298     1  1,37  1   100   300 100  35,58   60 75  237,2 kW (4) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1270(E)(A8)T POWER MACHINES N5 QUESTION 3 3.1 Given: Excess Air = 50% N2 = 12% V = 1 m3 H2 = 43% CH4 = 23% Co = 22% 3 3.1.1 SYMBOL % Vol 3 Formula O2 Reqd in m CO2 in m Produced H2 43 0,43 2H2+O2=2H2O √ 0,43x0,5=0,215√ nil√ CH4 23 0,23 CH4+2O2=CO2+2H2O√ 0,23x2=0,46√ 0,23x1=0,23√ CO 22 0,22 2CO+O2=2CO2√ 0,22x0,5=0,11√ 0,22x1=0,22√ N2 12 0,12 nil nil nil 3 TOTAL 100 1m 0,785√ 0,45√ Volume of oxygen reqd. = 0,785 m3 (From Table Above) √ (6) All the ticks in red are half marks (12 half marks = 6 marks) 3.1.2 100 150 Vol of air reqd with 50% excess air  0,785   21 100  5,607 m 3 (2) 3.1.3 Volume of dry products of combustion: CO2  0,45 m 3 ( From Tables ) 21 Excess O2  5,607   O2 in Fuel 100  21    5,607    0,785  100   0,392 m 3 79 N2  5,607   N 2 in Fuel 100  4,43  0,12  4,55 m 3 (10) 3.2 The Edwards air pump The Rotary air pump (2) [20] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1270(E)(A8)T POWER MACHINES N5 QUESTION 4 4.1 4.1.1 19° (8) TOLERANCE: ± 10m/s for lengths; ± 1º for angles Half mark for each correctly drawn length & angle (8 half marks = 4) Downloaded from hlayiso.com Please turn over Half mark for each correctly labelled length & angle (8 half marks = 4) Copyright reserved

MARKING GUIDELINE -8- T1270(E)(A8)T POWER MACHINES N5 If a student labels in cm instead of m/s then they should only be credited for the diagram 4.1.2 A 29o (any answer between 28 o -31 o) (1) B P = M x Vw x U = 5 x 1 000 x 400 = 2 MW (any answer between 1.99MW – 2.01MW) (3) C 2  U  Vw 100    V12 1 2  400  1000 100   1000 2 1  80 % (3) (any answer between 79.6% – 80.4%) D 76o (any answer between 75 o -77 o) (1) 4.2  Fuel is expensive, and inefficient combustion means expensive wastage.  Inefficient combustion results in excessive pollution of the atmosphere with noxious gases. (2) 4.3  Watt Governor  Porter Governor  Proell Governor (Any 2 × 1) (2) [20] QUESTION 5 5.1 Enthalpy before Throttling (@ 2000 kPa)  Enthalpy after Throttling (@ 200 kPa) hf  x 2 hfg  hg  C p t su  t s  hg  C p t su  t s   hf x2  hfg 2707  2,09140  120,2   908  1889 x2  0,974 x  x1  x 2 x x1  x2 0,958  0,974 (8)  0,984 Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -9- T1270(E)(A8)T POWER MACHINES N5 5.2 Given:   73% Ms  5500 kg / hour t su  350 o C P  4000 kPa tw  39 o C Cv  35 MJ / kg 5.2.1 h1   163 kJ @ 39 o C  h2   3094 kJ / kg @ 350 o C & 4000 kPa  M s h2  h1  100    mf  Cv 1 M s h2  h1  100 mf   Cv   1 55003094  163 100   35  10 3  73 1  630,94 kg / hour 630,94 mf / min  60  10,52 kg / min (7) 5.2.2 M s h2  h1  EE  mf  2257 55003094  163  630,94  2257  11,32 (3) 5.3  A higher calorific value  A lower calorific value (2)  THE QUESTION IS AMBIGUOUS BECAUSE STUDENTS COULD [20] WRITE ACTUAL VALUES INSTEAD OF STATING.  CREDIT STUDENTS WITH 2 MARKS TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

T1450(E)(N15)T NOVEMBER EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 15 November 2013 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Calculators and drawing instruments may be used. This question paper consists of 5 pages, and a 3-page formula sheet. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -2- T1450(E)(N15)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -3- T1450(E)(N15)T QUESTION 1 1.1 Regulation C99 of the Factories Act, refers to the returns of boilers. State FIVE different occasions when a written notification and/or approval must be sought from the inspector for such return. (5) 1.2 State FIVE advantages of a mechanical stoker in coal-fired boilers. (5) [10] QUESTION 2 A certain gas at a temperature of 26 °C has a mass of 2,732 kg. This gas is compressed, isothermally, to 4 000 kPa from 140 kPa. The gas constant for this gas is 0,3 kJ/kg.K. Calculate the following: 2.1 The volume before compression (3) 2.2 The volume after compression (3) 2.3 The work done in kilojoules (3) 2.4 The heat gained or lost in kilojoules (3) 2.5 The change in entropy (S 2 – S 1 ) (3) [15] QUESTION 3 Water which has a mass of 8,5 kg and a temperature of 40,3 °C is used to form steam at a pressure of 980 kPa. The steam is then superheated to a temperature of 214 °C. The specific heat capacity of superheated steam is 2,1 kJ/kg.°C. Calculate the following, with the aid of steam tables: 3.1 The sensible heat/enthalpy (4) 3.2 The latent heat/enthalpy of evaporation (3) 3.3 The superheat (4) 3.4 The total heat required to superheat the steam (4) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -4- T1450(E)(N15)T QUESTION 4 4.1 A surface condenser has a vacuum-meter reading of 689,3 mm mercury and a barometer reading of 757 mm mercury. Wet steam with a dryness fraction of 0,93 is supplied to this condenser at a rate of 3 450 kg/h. The cooling water enters the condenser at 20,5 °C and leaves at 32,4 °C and the condensate has a temperature of 31 °C. The specific heat capacity of the cooling water is 4 187 J/kg. °C. Calculate the following: 4.1.1 The pressure in the condenser (3) 4.1.2 The mass of cooling water required by the condenser every minute (8) 4.2 The dryness fraction of wet steam is determined by using a separating calorimeter. For every 1,9 kg condensate which passes through the calorimeter, 0,36 kg of water is collected in the separator. Calculate the percentage of vapour in the steam. (4) [15] QUESTION 5 A sample of coal has the following analysis by mass: • Sulphur: 3% • Oxygen: 6% • Carbon: 84% • Hydrogen: 7% The atomic masses of the elements are as follows: • Hydrogen: 1 • Carbon: 12 • Oxygen: 16 • Sulphur: 32 Calculate the following for one kilogram of coal, from first principles: 5.1 The mass of each element (4) 5.2 The theoretical mass of oxygen required to burn the carbon (2) 5.3 The theoretical mass of oxygen required to burn the hydrogen (3) 5.4 The theoretical mass of oxygen required to burn the sulphur (2) 5.5 The total mass of oxygen required for complete combustion (2) Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -5- T1450(E)(N15)T 5.6 The theoretical mass of air required for complete combustion (2) [15] QUESTION 6 A single-acting, single-cylinder air compressor takes in 1,05 m3 of air, with a temperature of 16,5 °C, every minute. The air is received at a pressure of 102 kPa and delivered at a pressure of 720 kPa. The index of compression is 1,35 and the compressor does not have a clearance volume. The compressor operates at a speed of 325 r/min and it has a stroke-to-bore ratio of 1,45 : 1. The mechanical efficiency of the compressor is 90,5% and the gas constant for air is 0,287 kJ/kg.K. Calculate the following: 6.1 The cylinder diameter in mm (6) 6.2 The stroke length in mm (2) 6.3 The delivery temperature in ºC (3) 6.4 The power required to drive the compressor in kW (4) [15] QUESTION 7 An impulse turbine rotates at a speed of 50 r/s and it has a nominal-blade diameter of 668 mm. The outlet and inlet blade angles are 31° each and the turbine develops 120 kW of power. The gas is discharged in an axial direction and the loss in velocity due to friction is 9%. 7.1 Calculate the blade velocity of the turbine, in m/s. (3) 7.2 Construct a velocity diagram in the ANSWER BOOK (landscape) and enter ALL the values (m/s) onto the diagram. Use the scale 1 cm : 10 m/s NOTE: No marks will be given if the values are not indicated on the diagram and if the diagram is not constructed to scale. (9) 7.3 Determine the mass of the gas which flows through the turbine every minute. (3) [15] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -6-1- T1450(E)(N15)T POWER MACHINES N5 FORMULA SHEET 1. Q = W + ∆U n −1 n −1 T2  P2  n V  17. =  =  1  2. ∆U = mCυ ∆T T1  P1   V2  3. Q = mCυ ∆T γ −1 γ −1 T2  P2  γ V  18. =  =  1  4. Q = mC p ∆T T1  P1   V2  V2 19. h = h f + χh fg 5. Q = P1V1 ln V1 20. h = hg + C p ∆T  T V  6. ∆S = m Cυ ln 2 + R ln 2  21. h = h f + h fg = hg  T1 V1  W = P1∆V n − 1  hsup − 1 941  7. 22. Vsup =   n  P  V2 8. W = P1V1 ln V1 Vm 23. χ= Vg P1V1 − P2V2 9. W = n −1 M 24. χ= M +m P1V1 − P2V 2 10. W= γ −1 25. U = H - PV R = C p − Cυ 1 2 11. 26. gZ1 + U1 + P1V1 + C1 + Q = 2 1 Cp gZ 2 + U 2 + P2V2 + C2 2 + W 12. γ = 2 Cυ • P1V1 P2V2 m s (h2 − h1 ) 13. = 27. η= • T1 T2 m ƒ CV 14. PV = mRT • m s (h2 − h1 ) 28. EE = • 15. P1V1 = P2V2 m ƒ 2 257 16. P1V1n = P2V2n 101,325 29. p = ( Bm ± M m ) 760 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -7-2- - T1450(E)(N15)T 100  8  30. m= C + 8 H 2 + S − O2  23  3   y y 31. C x H y +  x + O2 = xCO2 + H 2O  4 2  O  32. H .C.V . = (CVC .C ) + CVH 2  H 2 − 2  + (CVs .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (mw + me ) C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P2  n   n   P  2 n   35. W = P1Ve     − 1 = mRT1     − 1  n − 1   P1    n − 1   P1        1  1 Ve Vc  P2  n  36. ηc = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1      Vα 37. ηα = .100 Vs • 38. Fc = m(C fe − C fi ) • 39. P = mU [Cwi − (−Cwe )] 2U [Cwi − (−Cwe )] 40. η= .100 Cai 2 41. U = πDN • 42. m V = AC 43. (m + M ) g = mω 2 h π 44. Vs = D2L 4 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) -8-3- - T1450(E)(N15)T 45. θ1 = tc − twi 46. θ2 = tc − two θ1 − θ2 47. Log.temp.diff. = θ ln 1 θ2 P  48. Piso = P1 V1 ln  2   P1   n −1  n   P2 n 49. Pact = P1 V1   − 1  n −1  P    1   Piso 50. Niso = • 100 Pact Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE NATIONAL CERTIFICATE NOVEMBER EXAMINATION POWER MACHINES N5 15 NOVEMBER 2013 This marking guideline consists of 9 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1450(E)(N15)T POWER MACHINES N5 QUESTION 1 1.1 • When the user of a boiler permanently ceases to use a boiler. • When the user transfers ownership of the boiler to any other person. • When a person acquires a boiler. • The user shall notify the inspector when a boiler is damaged. • A user who proposes to execute repairs to a boiler. • No repairs are to be executed on a boiler without the prior written approval of the inspector. • No modifications are to be effected to a boiler without prior written approval of the inspector. (Any FIVE) (5) 1.2 • The stoker grate is self cleaning. • Labour costs are reduced. • Poor quality coal can be used. • The even feeding of coal prevents smoke. • The correct stoker speed improves the efficiency of the boiler. (5) [10] QUESTION 2 Given: T1 (T2 ) = 299 K M = 2,732 kg P2 = 4000 kPa P1 = 140 kPa R = 0,3 kJ / kg . K 2.1 P1V1 = MRT1 MRT1 V1 = P1 2,732 × 0,3 × 299 = 140 = 1,75 m 3 (3) 2.2 P2V2 = MRT2 MRT2 V2 = P2 2,732 × 0,3 × 299 = 4000 = 0,06127 m 3 (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1450(E)(N15)T POWER MACHINES N5 2.3 V  P  WD = P1V1 ln 2  or WD = MRT1 ln 1   V1   P2   0,06127   140  = (140 × 1,75) ln  = (2,732 × 0,3 × 299 ) ln   1,75   4000  = − 821,26 kJ = − 821,542 kJ (3) 2.4 Q = WD + ∆U = −821 + 0 (T1 = T2 ) = −821 kJ (Re jected ) (3) 2.5   P   S 2 − S1 = M  R ln 1  + 0   P2     140   = 2,732 0,3 ln   + 0    4000   = 2,732 × (− 1,00572 ) = −2,74762 kJ / K (3) [15] QUESTION 3 Given: M = 8,5 kg t fw = 40,3 °C P = 980 kPa C p = 2,1 kJ / kg °C t su = 214 °C 3.1 hsensible = M (hf − h fw ) = 8,5(759 − 169 ) = 5015 kJ (4) 3.2 hlatent = M × hfg = 8,5 × 2017 = 17144,5 kJ (3) 3.3 hsu = M {C p (t su − t s )} = 8,5{2,1(2,4 − 179 )} = 624,75 kJ (4) 3.4 htotal = M {hg + C p (t su − t s ) − h fw } = 8,5{2017 + 2,1(214 − 179 ) − 169} = 16332,75 kJ (4) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1450(E)(N15)T POWER MACHINES N5 QUESTION 4 4.1 Given: Vm = 689,3 mm Hg Bm = 757 mm Hg x = 0,93 M s = 3450 kg / h t wi = 20,5 °C t wo = 32,4 °C t c = 31°C SHC w = 4187 J / kg .°C 4.1.1 101,3 P = ( Bm − Vm ) × 760 101,3 = (757 − 689,3) × 760 = 9 kPa (3) 4.1.2 hf = 183 kJ / kg (@ 9 kPa ) hfg = 2398 kJ / kg (@ 9kPa ) hc = 130 kJ / kg (@ 31°C ) Heat Gained (Water ) = Heat Lost ( Steam ) M w × SHC w × ∆t w = M s ( hf + xhfg − hc ) M w × 4,187 × (32,4 − 20,5) = 3450{183 + (0,93 × 2398) − 130} 7876833 Mw = 49,8253 = 158089,023 kg / hour ∴Mass of water reqd . per min ute = 158089,023 ÷ 60 = 2634,82 kg (8) 4.2 Given: M c = 1,9 kg M w = 0,36 kg Mc x = Mc + Mw 1,9 = 1,9 + 0,36 = 0,84 ∴ % of vapour in the steam = (1 − 0,84 ) × 100 = 16 % (4) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1450(E)(N15)T POWER MACHINES N5 QUESTION 5 Given: S = 3% H2 = 1 M coal O2 = 6 % C = 12 C = 84 % O2 = 16 H2 = 7% S = 32 5.1 3% S = 0,03 kg S 6 % O2 = 0,06 kg O2 84 % C = 0,84 kg C 7% H2 = 0,07 kg H 2 100 % = 1 kg Coal (4) 5.2 C + O2 = CO2 12 + (16 × 2) = [12 + (16 × 2)] 12 + 32 = 44 (÷12)1kg + 2,667 kg = 3,667 kg ∴Mass of Oxygen Re qd . to Burn Carbon = 0,84 × 2,667 = 2,24 kg O2 (2) 5.3 H 2 + O2 = H 2O 2( H 2 ) + O 2 = 2( H 2 O ) (2 × 1 × 2) + (16 × 2) = 2[(1 × 2 ) + 16] 4 + 32 = 36 ( ÷4) 1 kg + 8 kg = 9 kg ∋ Mass of Oxygen Re qd . to Burn Hydrogen = 0,07 × 8 = 0,56kg O2 (3) 5.4 S + O2 = SO2 32 + (16 × 2 ) = [32 + (16 × 2)] ( ÷32) 1 kg + 1 kg = 2 kg ∴Mass of Oxygen reqd . to Burn Sulphur = 0,03 × 1 = 0,03 kg O2 (2) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1450(E)(N15)T POWER MACHINES N5 5.5 Total mass of Oxygen reqd . = O2 c + O2 h2 + O2 s − O2 fuel = 2,24 + 0,56 + 0,03 − 0,06 = 2,77 kg O2 (2) 5.6 100 Theoritical mass of air reqd . = × 2,77 23 = 12,043 kg (2) [15] QUESTION 6 Given: V = 1,05 m 3 / min T1 = 289,5 K P1 = 102 kPa P2 = 720 kPa n = 1,35 N = 325 r / min Stroke to Bore Ratio = 1,45 : 1 η = 90,5 % R = 0,287 kJ / kg .K 6.1 Vol / min Vs = N 1,05 = 325 0,00323 m 3 / stroke Vs = Area × Stroke Length π 0,00323 = × D 2 × 1,45D 4 0,00323 × 4 D = 3 π × 1,45 = 0,14155 m = 141,55 mm (6) 6.2 Stroke Length = 1,45 D = 1,45 × 141,55 = 205,248 mm (2) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1450(E)(N15)T POWER MACHINES N5 n −1 6.3 T2  P2  n =   T1  P1  n −1 P  n T2 = T1  2   P1  1, 35 −1  720  1, 35 = 289,5  102  = 480,497 K = 480,497 − 273 = 207,5 °C (3)   n −1 6.4  n   P2  n Pout = P1Ve     − 1  n − 1   P1     1,05  1,35   720  1, 35  1, 35 −1 = 102 ×     − 1 60  1,35 − 1   102    = 4,542 kW Pout ∴Pin = η 4,542 = 0,905 = 5 kW (4) [15] QUESTION 7 7.1 U = πDn = π × 0,668 × 50 = 105 m / s (3) 7.2 V3 V2 = K 610 = 678 m / s See last page for velocity diagram. (9) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- T1450(E)(N15)T POWER MACHINES N5 7.3 P = M ×U × Cw P M = × 60 U × Cw 120 × 10 3 = × 60 105 × = kg / min (3) [15] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -9- T1450(E)(N15)T POWER MACHINES N5 QUESTION 7.2 VELOCITY DIAGRAM Copyright reserved Downloaded from hlayiso.com Please turn over

T1450(E)(J24)T AUGUST EXAMINATION NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 24 July 2013 (X-Paper) 09:00–12:00 REQUIREMENTS: Steam tables (BOE 173) Superheated steam tables (Appendix to BOE 173) Candidates will need drawing instruments. Calculators may be used. This question paper consists of 5 pages and 3 formula sheets. Copyright reserved dhet Downloaded from hlayiso.com Please turn over

(8190035) -2- T1450(E)(J24)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers according to the numbering system used in this question paper. 4. Write neatly and legibly. Copyright reserved dhet Downloaded from hlayiso.com Please turn over

(8190035) -3- T1450(E)(J24)T QUESTION 1 Draw a neat, labelled sketch of a combined separating and throttling calorimeter. [10] QUESTION 2 A certain gas which has a pressure of 175 kPa, a temperature of 20 ºC and a volume of 0,665 m3 is compressed adiabatically to a pressure of 1 396 kPa. The gas constant for this gas is 0,258 kJ/kg.K and the specific heat capacity at constant volume is 0,716 kJ/kg.K. Calculate the following: 2.1 The mass of the gas 2.2 The specific heat capacity at constant pressure 2.3 The index of compression 2.4 The absolute temperature of the gas after compression 2.5 The change in internal energy which takes place during compression. (5 x 3) [15] QUESTION 3 A steam plant which consists of an evaporator and a superheater, receives water at a temperature of 17,5 ºC. The specific heat capacity of the superheated steam is 2,1 kJ/kg.K. Calculate the following, with the aid of the steam tables: 3.1 The quantity of heat required to form 2,5 kg of dry saturated steam at a pressure of 1 200 kPa 3.2 The quantity of heat required to form 5,6 kg of wet steam which is 17% wet, at a pressure of 1 200 kPa 3.3 The quantity of heat required to form 8,8 kg of superheated steam at a pressure of 900 kPa and a temperature of 290 ºC (3 x 5) [15] Copyright reserved dhet Downloaded from hlayiso.com Please turn over

(8190035) -4- T1450(E)(J24)T QUESTION 4 Feed water enters a boiler plant at 37,7 ºC. The boiler plant supplies 6 500 kg of steam every hour at a pressure of 940 kPa. The plant consumes 750 kg of coal, with a heat value of 35 000 kJ/kg, every hour and the steam is 8% wet. Calculate the following: 4.1 The efficiency of the boiler (5) 4.2 The equivalent evaporation from and at 100 ºC (3) 4.3 The cost saving, in coal, per hour if the plant is fitted with an economiser which raises the temperature of the feed water by 32,9 ºC while the boiler efficiency is increased by 6,5%. ALL other conditions remain the same. The cost of the coal is R1 400/ton. (7) [15] QUESTION 5 7,5 kg of ethylene fuel which has a formula C 3 H 4 is completely burnt. Hydrogen has an atomic mass of 1, oxygen has an atomic mass of 16 and carbon has an atomic mass of 12, whilst air contains 23% oxygen and 77% nitrogen by mass. Calculate the following: 5.1 The mass of air required to completely burn 1 kg of the fuel (7) 5.2 The mass of air required to completely burn 5,8 kg of the fuel (2) 5.3 The products of combustion when 5,8 kg of fuel is burnt (6) [15] QUESTION 6 A reciprocating air compressor has a temperature of 30 ºC and pressure of 102 kPa at the end of the induction stroke. It has an effective swept volume of 3,84 m3/min and delivers the air to a receiver at a pressure of 587 kPa. The compressor does not have a clearance volume and the gas constant for air is 0,287 kJ/kg.K. The law of compression is PV1,25 = C. Calculate the following: 6.1 The mass of the air compressed every minute 6.2 The temperature at the end of compression 6.3 The actual power Copyright reserved dhet Downloaded from hlayiso.com Please turn over

(8190035) -5- T1450(E)(J24)T 6.4 The isothermal power 6.5 The isothermal efficiency (5 x 3) [15] QUESTION 7 The rotor of a single-stage impulse turbine rotates at 9 000 r/min and it has an effective diameter of 795,8 mm. 120 kg of steam, per minute, enters the turbine at a velocity of 1 050 m/s and at an angle of 21º. The steam loses 13% velocity due to friction and the outlet blade angle is 31º. 7.1 Construct a velocity diagram in the ANSWER BOOK (landscape) and enter ALL the values (m/s) onto the diagram. Use the scale 1 cm : 50 m/s. NB: NO marks will be given if the values are NOT indicated on the diagram and if the diagram is NOT constructed to scale. (12) 7.2 Determine the following from the velocity diagram: 7.2.1 The exit angle of the steam from the turbine (1) 7.2.2 The diagram efficiency (2) [15] TOTAL: 100 Copyright reserved dhet Downloaded from hlayiso.com Please turn over

(8190035) -6- -1- T1450(E)(J24)T POWER MACHINES N5 FORMULA SHEET 1. Q = W + ∆U n −1 n −1 T2  P2  n V  17. =  =  1  2. ∆U = mCυ ∆T T1  P1   V2  3. Q = mCυ ∆T γ −1 γ −1 T2  P2  γ V  18. =  =  1  4. Q = mC p ∆T T1  P1   V2  V2 19. h = h f + χh fg 5. Q = P1V1 ln V1 20. h = hg + C p ∆T  T V  6. ∆S = m Cυ ln 2 + R ln 2  21. h = h f + h fg = hg  T1 V1  W = P1∆V n − 1  hsup − 1 941  7. 22. Vsup =   n  P  V2 8. W = P1V1 ln V1 Vm 23. χ= Vg P1V1 − P2V2 9. W = n −1 M 24. χ= M +m P1V1 − P2V 2 10. W= γ −1 25. U = H - PV R = C p − Cυ 1 2 11. 26. gZ1 + U1 + P1V1 + C1 + Q = 2 1 Cp gZ 2 + U 2 + P2V2 + C2 2 + W 12. γ = 2 Cυ • P1V1 P2V2 m s (h2 − h1 ) 13. = 27. η= • T1 T2 m ƒ CV 14. PV = mRT • m s (h2 − h1 ) 28. EE = • 15. P1V1 = P2V2 m ƒ 2 257 16. P1V1n = P2V2n 101,325 29. p = ( Bm ± M m ) 760 Copyright reserved dhet Downloaded from hlayiso.com Please turn over

(8190035) -7- -2- T1450(E)(J24)T FORMULA SHEET (Continued) 100  8  30. m= C + 8 H 2 + S − O2  23  3   y y 31. C x H y +  x + O2 = xCO2 + H 2O  4 2  O  32. H .C.V . = (CVC .C ) + CVH 2  H 2 − 2  + (CVs .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) (mw + me ) C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P2  n   n   P2  n  35. W = P1Ve     − 1 = mRT1     − 1  n − 1   P1    n − 1   P1        1  1 Ve Vc  P2  n  36. ηc = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1      Vα 37. ηα = .100 Vs • 38. Fc = m(C fe − C fi ) • 39. P = mU [Cwi − (−Cwe )] 2U [Cwi − (−Cwe )] 40. η= .100 Cai 2 41. U = πDN • 42. m V = AC 43. (m + M ) g = mω 2 h Copyright reserved dhet Downloaded from hlayiso.com Please turn over

(8190035) -8- -3- T1450(E)(J24)T FORMULA SHEET (Continued) π 44. Vs = D2L 4 45. θ1 = tc − twi 46. θ2 = tc − two θ1 − θ2 47. Log.temp.diff. = θ ln 1 θ2 P  48. Piso = P1 V1 ln  2   P1   n −1  n  P2  n 49. Pact = P1 V1   − 1    n −1  P    1   Piso 50. Niso = • 100 Pact Copyright reserved dhet Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE NATIONAL CERTIFICATE AUGUST EXAMINATION POWER MACHINES N5 24 JULIE 2013 This marking guideline consists of 8 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -2- T1450(E)(J24)T POWER MACHINES N5 QUESTION 1 [10] QUESTION 2 Given: P1 = 175 kPa T1 = 293 K V1 = 0,665 m 3 P2 = 1 396 kPa R = 0,258 kJ / kg.K C v = 0,716 kJ / kg.K 2.1 P1V1 = MRT1 P1V1 M = RT1 175 × 0,665 = 0,274 × 293 = 1,45 kg (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -3- T1450(E)(J24)T POWER MACHINES N5 2.2 R = C p − Cv Cp = R + Cv = 0,258 + 0,716 = 0,974 kJ / kg .K (3) 2.3 Cp γ = Cv 0,974 = 0,716 = 1,36 (3) 2.4 γ −1 γ T2  P2  = P  T1  1 γ −1 λ P  T2 = T1  2   P1  1, 36 −1 1396  1, 36 = 293  175  = 507,682 K (3) 2.5 ∆U = M C v [T2 − T1 ] = 1,45 × 0,716[507,682 − 293] = 222,883 kJ (3) [15] QUESTION 3 3.1 Given: t w = 17,5° C P = 1 200 kPa M = 2,5 kg hds = hg − hf w = 2 782 − 74 = 2 708 kJ / kg ∴hds ( for 2,5 kg steam ) = 2 708 × 2,5 = 6 770 kJ (5) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -4- T1450(E)(J24)T POWER MACHINES N5 3.2 Given: M = 5,6 kg x = 0,83 P = 1 200 kPa hwet = hf + xhfg − hf w = 798 + (0,83 × 1 984 ) − 74 = 2 370,72 kJ / kg ∴ hwet ( for 5,5 kg steam ) = 2 370,72 × 5,6 = 13 276 kJ (5) 3.3 Given: M = 8,8 kg t su = 290 °C P = 900 kPa C p = 2,1 kJ / kg.K hsu = hg + {C p (t su − t s )} − hf w = 2 773 + {2,1(290 − 175,4 )} − 74 = 2 939,66 kJ / kg ∴ hsu ( for 8,5 kg steam ) = 2 939,66 × 8,8 = 25 869 kJ (5) [15] QUESTION 4 Given: t fw = 37,7°C M s = 6 500 kg / hour P = 940 kPa M f = 750 kg / hour H v = 35 000 kJ / kg x = 0,92 4.1 h1 = 158 kJ / kg h2 = hf + xhfg = 751 + (0,92 × 2 023) = 2 612,16 kJ / kg M s (h2 − h1 ) 100 ∴η = × M f × Hv 1 6 500 (2 612,16 − 158) 100 = × 750 × 35 000 1 = 60,77 % (5) 4.2 M s (h2 − h1 ) EE = M f × 2257 6 500 (2 612,16 − 158) = 750 × 2 257 = 9,428 (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -5- T1450(E)(J24)T POWER MACHINES N5 4.3 h1 = 296 kJ / kg (@ 70,6 °C ) η = 60,77 + 6,5 = 67,27 % M s (h2 − h1 ) 100 Mf = × η × Hv 1 6 500 (2 612,16 − 296 ) 100 = × 67,27 × 35 000 1 = 639,43 kg / hour 1 400 ∴Cost saving = × (750 − 639,43) 1 000 = 1,4 × 110,57 = R154,80 / hour (7) [15] QUESTION 5 Given: M = 7,5 kg C3 H 4 H =1 C = 12 O = 16 5.1  y  y C x H y +  x + (O2 ) = x (CO2 ) +  [H 2O ]  4 2  4 4 C 3 H 4 +  3 + (O2 ) = 3(CO2 ) +  [H 2O ]  4 2 (12 × 3) + (1 × 4 ) + 3(16 × 2 ) = 3[12 + (16 × 2 )] + 2[(1 × 2 ) + 16] 40 + 96 = 132 + 36 (÷ 40) 1kg C3 H 4 + 2,4 kg O2 = 3,3 kg CO2 + 0,9 kg H 2O 100 ∴Mass of Air Re qd . to burn 1kg of fuel = × 2,4 23 = 10,435 kg Air (7) 5.2 Mass of air reqd . to burn 5,8 kg of fuel = 10,435 × 5,8 = 60,523 kg (2) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -6- T1450(E)(J24)T POWER MACHINES N5 5.3 CO2 = 3,3 × 5,8 = 19,14 kg H 2O = 0,9 × 5,8 = 5,22 kg 77 N2 = 60,523 × = 46,603 kg (6) 100 [15] QUESTION 6 Given: T1 = 303 K P1 = 102 kPa Vs = 3,84 m 3 / min PV 1, 25 = C P2 = 587 kPa R = 0,287 kJ / kg .K 6.1 P1V1 = MRT1 P1V1 M = RT1 102 × 3,84 = 0,287 × 303 = 4,5 kg / min (3) n −1 6.2 T2  P2  n =   T1  P1  n −1 P  n T2 = T1  2   P1  1, 25 −1  587  1, 25 = 303   102  = 430 K (3)  P  n  n −1 6.3 n Pact = P1V1   − 1  2 n −1  P   1   3,84  587  1, 25  1, 25 −1 1,25 = × 102 ×   − 1 1,25 − 1 60  102    = 13,679 kW (3) Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -7- T1450(E)(J24)T POWER MACHINES N5 6.4 P  Piso = P1V1 ln 2   P1  3,84  587  = 102 × × ln  60  102  = 11,424 kW (3) 6.5 Piso 100 η iso = × Pact 1 11,424 100 = × 13,679 1 = 83,51 % (3) [15] QUESTION 7 7.1 πDN U = 60 π × 0,7958 × 9 000 = 60 = m 375 / s V3 = V2 × 0,87 = 622 m / s See last page for velocity diagram. (12) 7.2 7.2.1 β = 64 ° (1) 7.2.2 2UV w 100 η = × (V1 ) 2 1 2 × 375 × 1 140 100 = × 1 050 2 1 = 77,55 % (2) [15] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com Please turn over

MARKING GUIDELINE -8- T1450(E)(J24)T POWER MACHINES N5 Copyright reserved Downloaded from hlayiso.com Please turn over

(8190035) T1490(E)(N15)T T1490(E)(N15)T NOVEMBER 2012 NATIONAL CERTIFICATE POWER MACHINES N5 (8190035) 15 November (X-Paper) 09:00 – 12:00 REQUIREMENTS: Steam Tables (BOE 173) Superheated Steam Tables (Appendix to BOE 173) Calculators may be used. Candidates will require drawing instruments, pens and a ruler. This question paper consists of 6 pages and a 3-page formula sheet. Downloaded from hlayiso.com Copyright reserved NOV 2012 Please turn over

(8190035) -- 2 -- T1490(E)(N15)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE POWER MACHINES N5 TIME: 3 HOURS MARKS: 100 ________________________________________________________________________ INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers correctly according to the numbering system used in this question paper. 4. Write neatly and legibly. Downloaded from hlayiso.com Copyright reserved NOV 2012 Please turn over

(8190035) -- 3 -- T1490(E)(N15)T QUESTION 1 1.1 What is meant by the specific heat capacity of an object? (2) 1.2 What is the purpose of governors? (2) 1.3 Name TWO types of steam turbines. (2) 1.4 What is the purpose of an Orsat apparatus? (2) 1.5 Name TWO types of steam boilers. (2) [10] QUESTION 2 0,6 kg of a gas is at a pressure of 1,5 MPa and a temperature of 300 °C. This gas is expanded reversibly and politropically to a pressure of 150 kPa. The following data is applicable to this gas: − Index of expansion = 1,24 − Gas constant = 0,287 kJ/kg.K − Specific heat capacity at constant pressure = 1,005 kJ/kg.K Calculate the following: 2.1 The temperature of the gas after expansion (3) 2.2 The work done (5) 2.3 The heat flow (4) 2.4 The change in entropy during the expansion for 1 kg of steam (3) [15] QUESTION 3 4 637,5 kJ of heat is dissipated from 1,75 kilograms of wet steam which is at a pressure of 2 000 kPa. The supply water which is used to form the steam enters the evaporator at 26,5 °C. The wet steam is then superheated to 350 °C. The specific heat capacity of the water is 4 187 J/kg.K. Calculate the following: 3.1 The dryness fraction of the wet steam (7) 3.2 The change in enthalpy from wet steam to superheated steam for 1 kg of steam (3) 3.3 The increase in the specific volume of the steam for 1 kg of steam (5) [15] Downloaded from hlayiso.com Copyright reserved NOV 2012 Please turn over

(8190035) -- 4 -- T1490(E)(N15)T QUESTION 4 10 800 kg of steam is generated every hour in a steam plant which consists of an evaporator, superheater and an economiser. The feed water enters the economiser at 29 °C and the evaporator at 72 °C. The steam which leaves the evaporator is 50% wet. The pressure and temperature of the superheated steam is 3 000 kPa and 250 °C. Coal, which has a heat value of 35 MJ/kg, is used at a rate of 15 kg/min. Calculate the following for 1 kg of fuel: 4.1 The quantity of heat transferred to the water in the economiser (4) 4.2 The quantity of heat transferred to the water in the evaporator (4) 4.3 The quantity of heat transferred to the water in the superheater (3) 4.4 The efficiency of the steam plant (2) 4.5 The equivalent evaporation from and at 100 °C (2) [15] QUESTION 5 5.1 A certain sample of fuel contains the following elements by mass: − Oxygen 3,7% − Hydrogen 9,2% − Carbon 84,6% − Sulphur 2,5% The calorific value of sulphur, carbon and hydrogen is 9,5 MJ/kg, 35 MJ/kg and 145 MJ/kg respectively. The partial pressure of the steam is 6,5 kPa in the exhaust gases. Calculate the following: 5.1.1 The higher calorific value of the fuel (4) 5.1.2 The lower calorific value of the fuel (4) Downloaded from hlayiso.com Copyright reserved NOV 2012 Please turn over

(8190035) -- 5 -- T1490(E)(N15)T 5.2 A fuel was tested, using a bomb calorimeter, and the following data was recorded: − Mass of fuel used 0,73 g − Water equivalent of the bomb 50 g − Mass of the water in the calorimeter 2,5 kg − Minimum temperature of the water 24,767 °C − Maximum temperature of the water 28,482 °C − The quantity of hydrogen in the fuel 15,4% − The partial pressure of the steam in the exhaust gases 5,5 kPa − The specific heat capacity of water 4,187 kJ/kg.°C Calculate the following: 5.2.1 The higher calorific value of the fuel in MJ (4) 5.2.2 The lower calorific value of the fuel in MJ (3) [15] QUESTION 6 0,04 m3/s of air is drawn into a single-acting, single-cylinder, reciprocating air compressor, at a pressure of 105 kPa. The air is then delivered to an air receiver at a pressure of 625 kPa, according to the law PV1,32 = C. The compressor has a mechanical efficiency of 82% and a motor transmission efficiency of 88%. The stroke to bore ratio is 1,4 to 1 and the compressor runs at 8 r/s. Calculate the following: 6.1 The diameter of the piston (6) 6.2 The indicated power (3) 6.3 The shaft power (3) 6.4 The power of the electric motor required to drive the compressor (3) [15] Downloaded from hlayiso.com Copyright reserved NOV 2012 Please turn over

(8190035) -- 6 -- T1490(E)(N15)T QUESTION 7 The nozzles of an impulse turbine supply 5 kg of gas at an angle of 19°, every second. The gas leaves the nozzle at a speed of 1 000 m/s. The mean blade speed is 475 m/s and the outlet angle of the moving blades is 23°. There is a 9,5% frictional loss in the relative part of the blades. 7.1 Use a scale of 1 mm = 5 m/s and construct a velocity diagram in the ANSWER BOOK (landscape) and insert ALL the values (m/s) on the diagram. (5) 7.2 Determine the following from the diagram: 7.2.1 The inlet angle of the moving blade (2) 7.2.2 The exit angle of the gas (2) 7.2.3 The driving force on the wheel in kN (2) 7.2.4 The axial thrust on the wheel in N (2) 7.2.5 The power developed by the turbine in MW (2) [15] TOTAL: 100 Downloaded from hlayiso.com Copyright reserved NOV 2012 Please turn over

(8190035) T1490(E)(N15)T POWER MACHINES N5 FORMULA SHEET n −1 n −1 T2  P2  n V  1. Q = W + ∆U 17. =   =  1  T1  P1   V2  γ −1 γ −1 T2  P2  γ V  2. ∆U = mC v ∆T 18. =   =  1  T1  P1   V2  3. Q = mC v ∆T 19. h = h f + xh fg 4. Q = mC p ∆T 20. h = hg + C p ∆T V2 5. Q = P1V1 ln 21. h = h f + h fg = hg V1  T V  n − 1  hsup − 1941  6. ∆S = m C v ln 2 + R ln 2  22. Vsup =    T1 V1  n  P  Vm 7. W = P1 ∆V 23. x= Vg V2 M 8. W = P1V1 ln 24. x= V1 M +m P V − P2V2 9. W= 1 1 25. U = H − PV n −1 P1V1 − P2V2 1 2 10. W = 26. gZ 1 + U 1 + P1V1 + C1 + Q = γ −1 2 1 2 gZ 2 + U 2 + P2V2 + C 2 + W 2 m s (h2 − h1 ) . 11. R = C p − Cv 27. η= . m f CV m s (h2 − h1 ) . Cp 12. γ = 28. EE = Cv . m f 2257 P1V1 P2V2 p = (Bm ± M m ) 101,325 13. = 29. T1 T2 760 100  8  14. PV= mRT 30. m= C + 8 H 2 + S − O2  23  3   y y 15. P1V1 = P2V2 31. C x H y +  x + O2 = xCO2 + H 2O  4 2 n n 16. P1V1 = P2V2 Copyright reserved NOV 2012 Downloaded from hlayiso.com Please turn over

(8190035) -8- T1490(E)(N15)T  O  32. H .C.V . = (CVC .C ) + CV H 2  H 2 − 2  + (CVS .S )  8  33. L.C.V . = H .C.V . − h fg (9 H 2 ) ( m w + we )C p ∆T 34. H .C.V . = mf  n −1   n −1   n   P2  n  n   P2  n 35. W = P1Ve        − 1 = mRT1     − 1    n − 1   P1   n − 1   P1       1  Ve  Vc  P2 n  1 36. η c = .100 = 1 −   − 1 = 1 + α − α (rp ) n Vs Vs  P1     Va 37. ηa = .100 Vs . 38. Fc = m(C fe − C fi ) P = m U [C wi − (− C we )] . 39. 2U [C wi − (− C we )] 40. η= .100 C ai 2 41. U = πDN . 42. m V = AC 43. (m + M ) g = mω 2 h π 44. Vs = D2L 4 Copyright reserved NOV 2012 Downloaded from hlayiso.com Please turn over

(8190035) -9- T1490(E)(N15)T 45. θ1 = t c − twi 46. θ 2 = t c − two θ1 − θ 2 47. Log .temp.diff . = θ ln 1 θ2 P  48. Piso = P1V1 ln 2   P1   n −1  n  P  n 49. Pact = P1V1   − 1  2 n −1  P1     Piso 50. η iso = .100 Pact Downloaded from hlayiso.com Copyright reserved NOV 2012 Please turn over

MARKING GUIDELINE NATIONAL CERTIFICATE POWER MACHINES N5 15 NOVEMBER 2012 This marking guideline consists of 10 pages. Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -2- T1490(E)(N15)T POWER MACHINES N5 QUESTION 1 1.1 The specific heat capacity of an object is the quantity of heat required to raise the temperature of 1 kg of the object by 1 °C. (2) 1.2 A governor is a device which controls the speed variation in an engine, (2) 1.3 Impulse Turbine Reaction Turbine (2) 1.4 An Orsat apparatus is used for the chemical analysis of the products of combustion, of a fuel. (2) 1.5 Water tube boiler Fire tube boiler (2) [10] QUESTION 2 Given: m = 0,6 kg P1 = 1,5 MPa T1 = 300o C + 273 = 573 K n = 1,24 P2 = 150 kPa R = 0,287 kJ / kg .K C p = 1,005 kJ / kg .K n −1 2.1 T1  P1  n =   T2  P2  T1 T2 = n −1  P1  n    P2  573 = 1, 24 −1  1500  1, 24    150  573 = 1,56152 = 367 K (3) Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -3- T1490(E)(N15)T POWER MACHINES N5 2.2 P1V1 = MRT1 MRT1 V1 = P1 0,6 × 0,287 × 573 = 1500 = 0,06578 m 3 P1V1n = P2V2n 1  P1  n V2 =   × V1  P2  1  1500  1, 24 =   × 0,06578  150  = 0,42126 m 3 P1V1 − P2V2 ∴WD = n −1 = (1500 × 0,06578) − (150 × 0,42126) 1,24 − 1 = 147,8375 kJ (5) 2.3 Cv = Cp − R = 1,005 − 0,287 = 0,718 kJ / kg .K ∆U = M × Cv × ∆T = 0,6 × 0,718 × (367 − 573) = −88,7448 kJ ∴Q = ∆U + WD = −88,7448 + 147,8375 = 59,0927 kJ (4)  T   P  2.4 ∆S =  Cp ln 2  −  R ln 2   T1   P1    367     150   = 1,005 × ln   − 0,287 ln    573     1500   = (− 0,44775) − (− 0,66084) = 0,21309 kJ / kg (3) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -4- T1490(E)(N15)T POWER MACHINES N5 QUESTION 3 Given: Ms = 1,75 kg P = 2000 kPa t w = 26,5 °C tsup = 350 °C shcw = 4187 J / kg .K 3.1 ∆hwet = 4637,5 ÷ 1,75 = 2650 kJ / kg hw = shcw × t w = 4,187 × 26,5 = 111 kJ / kg hwet = ∆hwet − hw = 2650 − 111 = 2539 kJ / kg But : hwet = hf + xhfg hwet − hf ∴x = hfg 2539 − 908 = 1889 = 0,863 (7) 3.2 hsup = 3138 kJ / kg (@ 2000 kPa and 350 °C ) ∴ ∆h = hsup − hwet = 3138 − 2539 = 599 kJ / kg (3) Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -5- T1490(E)(N15)T POWER MACHINES N5 3.3 Vwet = xV g = 0,863 × 0,09954 = 0,0859 m 3 / kg 0,231(hsup − 1941) Vsup = P 0.231(3138 − 1941) = 2000 = 0,13825 m 3 / kg ∴ Increase in specific volume = Vsup − Vwet = 0,13825 − 0,0859 = 0,05235 m 3 / kg (5) [15] QUESTION 4 Given: h1 h2 h3 h4 Econo Evaporator S/Heater tw1=29ºC tw2=72ºC x=0,86 P=3000kPa tsup=250ºC M s = 10 800 kg / hour M f = 15kg / min = 900kg / hour CV fuel = 35MJ / kg 4.1 h1 = 121 kJ / kg (@ 29°C ) h2 = 302 kJ / kg (@ 72°C ) Ms ∴ hecono = (h2 − h1 ) Mf × (302 − 121) 10800 = 900 = 2172 kJ / kg of fuel (4) Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -6- T1490(E)(N15)T POWER MACHINES N5 4.2 h3 = hf + xhfg (@ 3000 kPa ) = 1008 + (0,5 × 1794 ) = 1905 kJ / kg Ms ∴hevap = (h3 − h2 ) Mf × (1905 − 302 ) 10800 = 900 = 19236 kJ / kg of fuel (4) 4.3 h4 = 2858 kJ / kg (@ 3000kPa and 250°C ) Ms hsup = (h4 − h1 ) Mf × (2858 − 1905) 10800 = 900 = 11436 kJ / kg of fuel (3) 4.4 M s (h4 − h1 ) 100 η = × M f × CV f 1 10800(2858 − 121) 100 = × 900 × 35 × 10 3 1 = 93,84 % (2) 4.5 M s (h4 − h1 ) η = M f × 2257 10800(2858 − 121) = 900 × 2257 = 14,552 kg steam / kg fuel (2) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -7- T1490(E)(N15)T POWER MACHINES N5 QUESTION 5 5.1 Given: O2 = 3,7% CVs = 9,5 MJ / kg H2 = 9,2% CVc = 35 MJ / kg C = 84,6% CVH 2 = 145 MJ / kg S = 2,5% Ps = 6,5 kPa (CVs × S ) + (CVc × C ) + CVH 2  H 2 − O2   5.1.1 HCV =   8  = (9,5 × 0,025) + (35 × 0,846) + 145 0,092 − 0,037     8  = 0,2375 + 29,61 + 12,6694 = 42,5169 MJ / KG (4) 5.1.2 LCV = HCV − (hfg × H 2 × 9 ) = 42,5169 − (2,413 × 0,092 × 9 ) = 40,5189 MJ / kg (4) 5.2 5.2.1 (M w + We )C p × ∆t HCV = Mf = (2,5 + 0,05) × 4,187 × (28,482 − 24,767 ) 0.00073 = 54334,928 kJ / kg = 54,335 MJ / kg (4) 5.2.2 LCV = HCV − (hfg × 9 × H 2 ) = 54,335 − (2,42 × 9 × 0,154 ) = 50,981 MJ / kg (3) [15] Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -8- T1490(E)(N15)T POWER MACHINES N5 QUESTION 6 Given: V1 = 0,04 m 3 / sec P1 = 105 kPa PV 1, 32 = C P2 = 625 kPa η comp = 82 % η motor = 88 % n = 8r / s Stroke to Bore ratio = 1,4 to 1 6.1 V1 Vs = n 0,04 = 8 = 0,005 m 3 / stroke πD 2 But :Vs = ×L 4 πD 2 = × 1,4 D 4 Vs × 4 ∴D = 3 π × 1,4 0,005 × 4 = 3 π × 1,4 = 0,165673 M = 165,673 mm (6)  P  n  n −1 6.2 n Indicated Power = P1V1  2  − 1 n −1  P1      625  1, 32  1, 32 −1 1,32 = × 105 × 0,04   − 1 1,32 − 1  105   = 9,373 kW (3) 6.3 Indicated Power Shaft Power = η comp 9,373 = 0,82 = 11,43 kW (3) Copyright reserved Downloaded from hlayiso.com Please turn over

(MARKING GUIDELINE) -9- T1490(E)(N15)T POWER MACHINES N5 6.4 Shaft Power Motor Power = η motor 11,43 = 0,88 = 13 kW (3) [15] QUESTION 7 7.1 See diagram on next page (5) 7.2 7.2.1 35° (2) 7.2.2 90° (2) 7.2.3 Driving Force = Ms × Cw = 5 × 945 = 4725 N = 4,725 kN (2) 7.2.4 Axial Thrust = Ms (cfi − cfe ) = 5(325 − 200) = 625 N (2) 7.2.5 Power = Ms × u × Cw = 5 × 475 × 945 = 2244375W = 2244,375 kW = 2,244 MW (2) [15] TOTAL : 100 Copyright reserved Downloaded from hlayiso.com Please turn over

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