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N6 POWER MACHINES SPOT ON R80 .00
N6 POWER MACHINES SPOT ON R80 .00, 269 pages. Read online or get the free PDF download.
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a
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
-T4460(E)(J29)T
ST EXAMINATION
REQUIREMENTS:
Calculators may be used.
This question paper sists:Of 6 pages and 1 formula sheet of § pages.
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(8190046)
-2- T1160(E)(J29)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
NOTE:
INSTRUCTIONS AND INFORM
If you answer more.
required number of.
be marked must be
je required number of questions, only the
jons will be marked. All work you do not want to
learly crossed out.
ION
only FIVE questiori
ALL the questions carefully.
er the answers cording to the numbering system used in this question
‘subsections of questions must
Use only BLUE or BLACK ink:
ALL sketches and diagrams must be done in pencil in the ANSWER BOOK.
Write neatly and legibly
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QUESTION 1
A two-stage single-acting air compressor is designed to receive 6 m*/min of free air at
100 kPa and 20 °C and deliver the air at 1 400 kPa. When the first-stage piston is at
the beginning of the compression stroke, the air in the cylinder is at 86 kPa and 35 °C .
Mechanical efficiency is 85 per cent. Compression for both stages follows the law
PV! 2 = C. Intercooling is complete and the compressor is designed for minimum work
done per cycle. Ignore free volume.
Assume R for air as 0, 287 kJ/kg.K and Cp as 1, 005 kJ/kg. K.
Calculate each of the following:
1.4
1.2
1.3
°C and at a tempo of
economiser, evaporator and
at a temperature of 37, 7 °C and
leaves it at 75, 9 °C: f the steam, leaving the boiler is 0,92. The
fuel consumption is 620 kg/h with a heat value of 38 Md/kg. The air is supplied at 25 kg
of air per kg of fuel. The atmospheric temperature is 22 °C, The inlet temperature of
the economiser is 480 °C. The heated steam is 2,04 kJ/kg.K and for the air
is 1,008 kJ/kg. K.
superheater. The su
2.1 Calculate each of the following by using steam tables:
2.1.1 Thermal efficiency of the boiler
2.1.2 Equivalent evaporation from and at 100 °C
2.1.3 Temperature of the exhaust gas at the outlet of the economiser
2.2 Draw up a heat balance in kJ/kg coal and as percentage to determine the
percentage heat lost to radiation.
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(8190046) -3- T1160(E)(J29)T
(5);
(5)
[20]
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(8190046) -4- T1160(E)(J29)T
QUESTION 3
The compressor and turbine of a closed-cycle gas-turbine plant receive air at 87 kPa
and deliver it at 620 kPa. The total volume entering the compressor is 1,57 m/s and
that leaving the compressor is 0,386 m/s. The temperature prior to compression is
15 °C and the highest temperature in the cycle is 800 °C. Assume isentropic
compression of the air and isentropic expansion takes place in the turbine.
Assume R for air = 0, 287 kJ/kg.K, y = 1, 4 and Cp = 1, 005 kJ/kg. K.
Calculate each of the following:
3.1 Air standard efficiency
3.2 Circulating mass of air
3.3 Work done in the com (4)
3.4 (4)
3.5 (4)
3.6 ompression is equal to
(20)
Calculate each of the following:
44 Critical pressure and the temperature of the gas at the throat (5)
4.2 Volume of CO2 flowing at the throat per second and its velocity in m/s (6)
43 Area of the nozzle in mm? and the diameter in mm at the throat (4)
4.4 Velocity and temperature at the exit if the Mach number is 1,65 ; Joh
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(8190046) -§-
QUESTION 5
T1160(E)(J29)T
A four-cylinder two-stroke cycle petrol engine operates on the constant volume cycle.
The following data is applicable:
Bore size
Stroke length per cylinder
Engine speed
Fuel consumption
Calorific value
Brake mean effective pressure
Compression ratio
Efficiency ratio
Gamma for air
Calculate each of the following
5.1
5.2 thermal efficiency
5.3
5.4
5.5
5.6
5.7 Mechanical efficie
5.8 Specific fuel consump!
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wou wa
It
oy ou
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80mm
90 mm
2 400 r/min
9,2 kgfh
44 000 kJ/kg
370 kN/m?
7:4
55 %
14
(3)
(2)
[20]
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(8190046) -6- T4160(E)(J29)T
QUESTION 6
A velocity-compounded two-stage impulse turbine has an average blade diameter of
1,5 m and it rotates at 31,83 r/s. The nozzle velocity is 652,5 m/s. The first row of
moving blades has a relative inlet velocity of 515 m/s. The gas leaves the fixed blades
at 350 m/s. The second row of moving blades has an inlet angie of 25 °. The gas
leaves the second row of moving blades at 75 m/s. The gas flow rate is 10 kg/s. There
is 10% loss of velocity across all blades due to friction.
6.1 Use a scale 1 mm = 5 m/s and construct velocity diagrams for the turbine in
the ANSWER BOOK. Indicate the lengths of ALL the lines as well as the
magnitude of the angles o lagrams.,
Determine each of the following the velocity diagrams:
6.2 6.2.1
QUESTION 7
8
38 kW to drive a refrigeration plant. The evaporator and
- 63 °C 27, 4 °C respectively. The plant converts
A compressor req S
condenser temperatures:
water at 11,3 °C to ice at -
The specific heat capacity of t
The latent heat of the ice is 334,
is 1,91 kJ/kg. K.
Calculate each of the following:
7.1 Refrigeration required in kJ to produce 1 600 kg of ice
7.2 Work done in kJ to produce 1600 kg of ice if the actual coefficient of
performance is 46 per cent of the ideal coefficient of performance
7.3 Mass of ice in kg produced by the plant in 24 hours
TOTAL:
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(10)
(9)
(5)
[20]
100
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Re
(8190046) -- T4160(E)(J29)T
FORMULA SHEET
Any applicable formula may also be used.
ENGLISH GENERAL AFRIKAANS
PV = mRT,
R=Cp-Cv
PV =k
PV" =k
PV’ =k
Q=AU + Av
: V,
As=m.Cp.In
P Vj
1
As=m.R.tn 2h
P,
Qz=m.Cp. AT
OQ=m.Cv. AT
Soy = Sg + Cp ln a
ny
Sie = Sq -Sy
S=Sp+xwy
fagy = Hg + CP lsu —1,)
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(8190046)
ENGLISH
Days = hy + xh
Vv,
ws = XV y
“2 T1160(E)(J29)T
GENERAL AFRIKAANS
2 iy, ~ 1941)
Vy = P, Py = hy + xh yy
r= Va Ke VB, =x,
V. &
work done.
lende fornnules vir
ef verrig (Av)
=m.Cp. AT’
xn P, an
B K, x4
n-t Pp
u-l
xn
= mR |r, not
pe thy Ops)
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(8190046) -3- T1160(E)(J29)T
ENGLISH GENERAL AFRIKAANS
Different formulae for Ferskillende formules
work done (Wd) vir arbeid verrig (Av)
= area of PV-diagram = area van PY-diagram
= work done firsi stage = avbeid verrig eerste
| work done second stadium + arbeid ver-
stage +... rig tweede stadium +
Wa roy = Wa, -Wal, Ay
ren — AV, — Avy
Want = Qnett A
nett = Quest
Different for
air standard.
cies (ASE)
Verskillende formiles
heat added — heat rejected warmle loegevoeg ~ warmte afgestaan
heat added warmte toegevoeg
Different volumetric Verskillende volumetriese
efficiencies, Oot rendememle, Qpot
_ Volune of air taken in _ Fotume lug ingeneem
Swept volume Stagvolune
_ Volume of free air __ Volume vrytug
Swept volume Slagvolume
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(8190046) -2- T1160(E)(J29)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
NOTE: lf you answer more.
required number of qj
ie required number of questions, only the
ons will be marked. All work you do not want to
5. ALL formulae used:must be writ in down.
6. Show ALL the intermed
7. Use only BLUE or BLACK ink.
8. ALL sketches and diagrams must be done in pencil in the ANSWER BOOK.
9. Write neatly and legibly
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pe nn
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(8190046)
ENGLISH
Different thermal
efficiencies, Oherm.
_ Wa
heat supplied
T1160(E)(J29)T
GENERAL AFRIKAANS
Verskillende termiese
rendemente, Oerm.
Ay
warnie loegevoeg
; opp , _ RD
' , = — Mei nM
rake therm mipjq CV reer ny 5 XW
IP ID
Mind. therm. = "hind term. WE
hind, therm. Inj) CV ind. term. WP
"eherm, =
__ it, (hs — hw)
: _ its as — iw)
Merm, ~ 7
mp x CV my, X WW
tT; BT
Ne =e
h-G bk
BP RD
Nmech, = TP. TD
Indicated aleurrendementverhouding
- Nind. ther) ind. term.
ASE LSR
Brake efficiency rat Remrendementverhouding
= "brake therm - Trem. term,
ASE LSR
Bp = 241% RD= 290
60 60
BP= Pyrake mean LANE RD= Lem gen, LANE
IP = Py. mean LANE ID = Pad. gem, LANE
m
ISEC = Lh ISBY = Uleie
ID
one Myth Moth
BSFC = ——— RSBV = bia.
BP RD
cop=—"! xyw ="
r-1 ,-1,
RE VE
COP =—— KYW =—
Wd : Ay
P=m.U.Avw D=m.U.Avw
Foy =m AV Peaks, =m. AV ¢
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(8190046)
ENGLISH
_ tits Fs ~ Aw)
5. T1160(E)(J28)T
GENERAL AFRIKAANS
2.U Av
2x10? (A, <n) + CP
= 210°C,
— Ms (hy ~ hy)
hv
mp x 2257 my X 2257
Nisa. = Wis, Miso, = Miso.
Wd poly, AV of
: _ Wa Ay
Nank, = “O Nrank, = O
tr
Ream, = 1- 7
h=u+t p¥
orn Cc?
sh Uy + HK +> + O- Ba +U + BK ++ =
Cc? c2
822 +U2 + Pry +4 Wed Zy + Uy + Phy + 7 tay
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
REQUIREMENTS:
This question paper consists of yages and 1 formula sheet of 6 pages.
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8190046) -2- T1350(E(M31)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
e@ required FIVE questions, only the first five
| work you do not want to be marked must be
3. stem used in this question
4. 8 of questions must
5.
6. Show ALL the intermediaté:steps.
7. Questions must be answeret
8. All sketches and diagrams must be done in pencil in the ANSWER BOOK.
9. Write neatly and legibly.
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8190046) -3- 71350(E)(M31)T
NOTE: Answer ANY FIVE questions in this question paper.
QUESTION 1
A two-stage, single-acting, reciprocating compressor takes in air at a tempo of
0,4 m/s. The compression index is 1,3. The delivery pressure is 800 kPa. The
rotational frequency of the compressor is 550 rimin. The initial pressure and
temperature for the low-pressure cylinder are 104 kPa and 19 °C respectively.
Conditions for maximum efficiency prevail and free volume is insignificant.
Assume R for air as 0, 287 kd/kg.K.
Calculate the following:
44 The intermediate (3)
1.2 The volume for th (6)
1.3 The diameter (D) of (4)
1.4 The er required to the mechanical efficiency
is 88 (7)
[20]
QUES]
In a boiler piant’e perheater, steam
is generated at a press : mperatul 50 °C at a rate of
5 400 kg/h. The dryn ) ate the super-heater is 0,9.
The temperatures of thi i ding the economiser are 44,8 °C
and 93,5 °C respectively. Th rate of 600 kg/h.
The heat value of the coal is 32M fic temperature is 24 °C.
Air is supplied at a rate of 15 kg of alr he temperature of the flue gas at
the outlet of the economiser is 210 ° heat capacity of the flue gases is
1,045 kJ/kg. K.
Calculate the following by using steam tables:
2.1 Efficiency of the plant (3)
2.2 Equivalent evaporation from and at 100 °C per kilogram of fuel (2)
2.3 Heat to the economiser in kJ/kg (2)
2.4 Heat to the evaporator in kJ/kg (3)
2.5 Heat to the superheater in kJ/kg (3)
2.6 Heat to the stack (chimney) in kJ/kg (2)
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8190046) -4- T1350(E)(M31)T
27 Draw up an energy balance in kJ/kg fuel, also as a percentage of the total
heat. (5)
[20]
AND/OR
QUESTION 3
A compression-ignition engine working on the dual cycle takes in two thirds of its total
heat supply at constant volume and one third at constant pressure expansion. The
compression is adiabatic. Heat is lost at a constant volume.
The following data is applic
Intake conditions =
Maximum pressure =
Compression ratio =
(5)
(2)
(5)
(2)
(2)
(4)
[20]
QUESTION 4
Air enters a nozzle at a pressure of 3 MPa, a temperature of 480 °C and a speed of
60 m/s. The air leaves the nozzle at a pressure of 600 kPa. The air flow rate through
the nozzle is 1,2 kg/s. The expansion from the inlet to the throat takes place with an
adiabatic efficiency of 90 percent while the adiabatic efficiency from the throat to the
outlet is 85 percent.
Assume gamma = 1,4; R = 0,287 kd/kg.K and Cp = 1,005 kJ/kg. K
Calculate the following:
(10)
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8190046) -5- T1350(E)(M31)T
4.2 The exit area in mm?
4.3 The Mach number
AND/OR
QUESTION 5
A ship is fitted with a turbo-charged two-stroke oil engine, with five single-acting
cylinders.
The following data is applicable to the engine:
Cylinder diameter
Stroke length per cylinder
Engine speed
Fuel consumption
86,5%
101,325 kPa
16°C
0,287 kJ/kg.
5.1
5.2
5.3
5.4
5.5 The volume of air induced per mit
5.6 The swept volume per minute |
5.7 The volumetric efficiency
AND/OR
QUESTION 6
A two-stage, velocity compound steam turbine has blading designed for an axial
discharge of the steam from the second row of moving blades. All the moving blades
are 30°. The blade circumferential speed, the nozzle angle and the fixed blade angles
are designed for a nozzle discharge velocity of 500 m/s.
The velocity coefficient for all blades is 0,9.
6.1 Use a lenghCoiereo arcleach ZaeouraL hd sryas speesoaa construct
(8)
(2)
[20]
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8190046) -6- 71350(E)(M31)T
velocity diagrams for the turbine in the ANSWER BOOK. Indicate the lengths
of ALL the lines as well as the magnitude of the angles on the diagrams and
calculate the scale.
6.2 Determine the following from the velocity diagrams:
6.2.1 The blade circumferential speed in m/s
6.2.2 The power developed per kg of steam per second in kW
6.2.3 The diagram efficiency
624 The axial thrust in KN
QUESTION 7
ween pressure boundaries
479 kJ/kg
462, 8 kJ/kg
110, 2 kJ/kg
51, 6 kJ/kg
Specific volume of dry satu 0,168 m*/kg
Temperature of the refrigerant i 31°C
Temperature of the refrigerant in evap! = -6°C
The methy! chloride enters the compress: 's a wet vapour, the condenser as a dry
saturated vapour and it leaves the condenser as a saturated liquid with no under-
cooling.
The actual coefficient of performance is 90,23 percent of the ideal coefficient of
performance.
The compressor bore size is 0,1285 m; it has a stroke length of 1,2 times the diameter
of the piston, a volumetric efficiency of 90 percent and it rotates at 420 r/min.
Calculate the following:
7.4 The ideal coefficient of performance and the actual coefficient of performance
7.2 The specific enthalpy of the refrigerant at the entrance to the compressor in
kJ/kg and the dryness factor
7.3 The swept Ome orhmcehpesr io ma pbrlarokesthe sweptyolume and
(10)
(3)
(5)
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(11046012) -1-
POWER MACHINES N6
FORMULA SHEET
Any other applicable formula may also be used.
ENGLISH GENERAL
PV, = mRI,
Cp-Cy
PV=c
PY" =c¢
PV'=c
O=AU + Wd
As=m. Rtn 2
P,
OQ=m.Cp. ar
Qz=m.Cv. AT
8,
ste
; r,
=S_+Cp In
p
Spy =S_— Sp
S=Sp 4x8 py
T1350(E)M31)T
AFRIKAANS
tl
oe
PV
PV" =k
PVi =k
Q=AU + Ay
Irs = Pg CD Clon 1.) °
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(11040012) 2 T1350(E)(M31)T
ENGLISH GENERAL AFRIKAANS
"la, -1941)
Figs = hy + We Vou = i Mas = hy + wpe
Vys = Vg r= 7 Ve Fis = XV,
¢
Verskillende fornniles vir
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(41040012) -3- T1350(E)(M31)T
ENGLISH GENERAL AFRIKAANS
Different formulae for Ferskillende fornules
work done (Wd) vir arbeid verrig (Ay)
= area of P¥-diagram = area von P¥-dicgram
= work done first stage = arbeid verrig eerste
1 work done second stadium + arbeid ver-
stage b... rig tweede stadium +
Waren = Wa, —-Wa, AV net = AY, — AVE
Wabyor = Qnett AV ner = Quott
Different formu
air stondar
cies (ASF,
Fsigllende fornules
vir lugstandaardrende-
vente (ESR)
heat added — heat rejected __
=1-
heat added 77 a xy ( winte loegevoeg
Different volumetric Verskillende volumetriese
efficiencies, Qt , rendemente, Ovo
__ Volume of atv taken in _ Folume lug ingeneem
Swept volume Slagvolume
_ Volune of free air _ Volume vrylug
Swepi volume Slagvolume
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(11040012) -4- T1350(E)(M31)T
ENGLISH GENERAL AFRIKAANS
Different thermal Verskillende termiese
efficiencies, Onerm. rendemente, Oerm.
= Wa 7 Ay
heat supplied warmie toegevoeg
, _ BP , _ RD
Tbvake therm. im pis fad Trem term. Mb) g X Www
IP iD
Tin. therm. > my, ®CV Mind. term. = Whig x WW
my (hs — bw) m, (hs — hw)
bah
hah
"Muech
Indicated. ency ratio
Brake efficiency rat
_ "erake therm.
ASE
_ Trem. term.
LSR
BP=2n IN
60
BP = Prrake mean LANE
IP = Pind. mean LANE
Ipc = Eth
IP
Mm
BSFC = £14
BP
cop -—11
Ty ~T,
cop = &
Wd
P=m.U.Avw
By =m AV;
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RD = Prey gan, LANE
ID = Pg. gou, LANE
ISBy = “bln
1D
RSBY = “bib
RD
7,
7 -T
VE
KYW =
Av
D=m.U,.AYw
KVWe=
gy =m MW
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a4 040012) -S- T1350(E)(M31)T
ENGLISH GENERAL AFRIKAANS
2.U . Aw
Ndia. = Re
. ia
: =1
+ P=PR 2 \7
+1
sop —tts (ty hry) py —ta ly I)
mp x 2257 my 2257
_ Wdiso, 3 = AViso
Miso. Wa yoty. “iso. AY pol,
Wd Av
"rank. ~ oO rank. ~ Oo
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(41040012)
C2
eZ, +U, + PR, + +=
+2
QZa +U, + PaVq + o +Wad
71350(E)(M31)T
feu
gZ,+U, + BM + 5 +05
2
Ly + Uy + RP, wy tay
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE |
NATIONAL CERTIFICATE
APRIL EXAMINATION
POWER MACHINES N6
31 March 2016
This marking guideline consists of 11 pages.
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MARKING GUIDELINE -2-
POWER MACHINES N6
QUESTION 1
14 P2= JPL P3
= 104-800
= 288,44 kPa
1.2
1.3
1.4
Vi = Ver/N = 0,4 x 60/550 = 0,04364 m?
m = Pi = 104x 0.04364 = 0.05416 kg
RT: 0,287x292
V2 = MRT = 0,05416 x 0,287 x292 = 0,01574 m*
P2288,44
Lo = Do
Vs2 = Vo = 1 D2 2. D2
4
0, 01574 = 1 (D2)5.
4
+ L2= D2 = 271,63 mm
pyr
P = 2 py 22) _]
n-1 B
43
Ts
= 2x13 104x0,4){ 28844)" _|
03 104
= 95,69 kW
“
act
Ul
Preg
a
95,69
0,88
= 108,74.4
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T1350(E)(M31)T
(6)
(7)
[20]
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MARKING GUIDELINE
QUESTION 2
2.1 At 44,8°C :
At 93,5°C :
At 1500 kPa
SW ptant
2.2
EEO =
2.3
Qzcon =
2.4 At 1 500 kPa:
h, =
Qaup =
2.5
Qsurme =
3.
POWER MACHINES N&
hit
he
4188 kJ/kg
302 kd/kg
& 350°C ch4 = 3148 kilkg
nou
anny -h) 100
m, xf, j
5400(3 148-188) 100
600x32000 1
83,25%
m, (ay _ A)
m, x 2257
5 400(3148 ~ 188)
600 x 2 257
11,803 kg. steantkg.. fuel
it, Ay _ h)
mM,
5 400(392 — 188)
600
1836.4/ / ke
hr = 845 kJ/kg and hig = 1 945 kJ/kg
hf + xhfg
845 +-(0,9 x1945)
2 595,5 Kilkee
m, (A, - h,)
my
§ 400(2 595,5 — 392)
600
19831,5.4/ fig
m, (hy = Ay)
My
5 400(3 148 — 2 595,5)
600
49725] ike
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T1350(E)(M31)T
(3)
(2)
(3)
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MARKING GUIDELINE
-4- T1350(E)(M34)T
POWER MACHINES N6
26 Oness = mCpar
= (15 +1) x1,045 (210 — 24)
= 3109,92 ki lkg (2)
2.7 Qin (kd 7kg) Qout (Kd /kg) %
Qruet = 32 000 | Qecon = 1 836 5,74
Qevap= 19 831,5 61,97
Qsurer =4972,5 15,54
QrLucas = 3 109,92 9,72
Unaccounted = 2 250,08 7,03
32 000 32 0CO 100 (5)
[20]
QUESTION 3
3.14 P
y = Cp/Cv = 1/0,712 = 1,4
3 8 4
-
2 ——
B lal NX s
1 _
= 806,26 K y
vey
f, = Pf +
V,
La
= toa29("°)
= 3 674,83.kPa (5)
3.2 . LP,
I, = 2
P,
_ 806,26x5 000
3674,83
= 1097..K (2)
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MARKING GUIDELINE
3.3
5.
POWER MACHINES N6
2
Q.5 ae
1
Qs-4 ~ 3
2 op 1 4
3 wnCvid,—-T,) = 3 mCp(T, - 75)
2x0,712 (1097 — 806,26) = i(7, — 1097)
T; 1511,01.K
B40 TN
r,
_ 1514,01x1
7 1097
= 13774.0°
3.6 yt
V;
O4
= isto 87)
13
615,62K
6 Pn |
8 ASE = Bah _x.,100%
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j-—
(F, -1,).+ 70 -B)
615,62 -.289
1- .
(1097 ~ 806,26). + 1,4 (1 511,01 -1097)
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T1350(E)(M31)T
(5)
(2)
x..100%, = .62,47% (4)
[20]
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MARKING GUIDELINE
QUESTION 4
41
Te’ =
Noone =
90 =
Te
Pe =
Ce =
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-6-
POWER MACHINES N6
Ga
y+)
753 a
1441
627,5K
Te) 1
(753-Te) 100
(753 - 627,5) 1
640,05.
a
2 \r
ra
yti
14
od
3000/2
24
1 584,85 kPa
R Tc
P,
0,287x640,05
1584,85
0.11591.
2.000.Cp(7, — Te) +C,?
2 000:1,005(753 — 640,05) + 60? . = 480,2 mls
mVe
C
1,2¥0,11591
480,24
289,63 mim?
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T1350(E)(M31)T
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42
T2 =
Nowe =~
85 =
V, =
43
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-T- T1350(E)(M31)T
POWER MACHINES N6
re
P. Y
ra
P
Od
953, 20°)"
3000
415, 43K
(i -T;) 08
(Ge-T2) A
(640,05 -7,) 100
(640,05 — 475,43) 1
500,12.K
2000.Cp(1, -T,)+C)
2 000x1,005(753 — 500,12) + 60°. =715,46.m/s
RT,
£,
0,287x500,12
600
0,23922.m?
my,
C;
1,2x0,23922
715,4
401,26. min? (8)
1,489 (2)
[20]
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MARKING GUIDELINE
QUESTION 5
5.1
BPoo=
5.2
Nimech =
IP =
5.3
5.4
5.5
5.6
1 eRAKE THERMAL
V. =
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-8-
POWER MACHINES N6
(P, EACE
“60
ras s{“)s0,76) xT 12x5
60
4 687,07kW
BP 100%
IP
4687.07 | 100%
sy
5418, 58.4V
_ BP
inf | s.xCV
4687,07
} 050 «44 200
3 600
36,36%
x, 100%
x. 100%
2aNT
60 |
Qa
60
399,63 kNin
mRT,
p
1050
x 9 X0,287(16 + 273)
22.
101325
315,155.03 /min
D*xLxNeE
© (0,76)° xl,5x112x5
381,063.m? /min
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T1350(E)(M31)T
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(4)
(2)
(3)
MARKING GUIDELINE -9-
POWER MACHINES N6
5.7 V,
Tho. = 7, 20%
- 315,155 1 9%
381,063
= 82.70%
QUESTION 6
6.1
T1350(E)(M31)T
(2)
{20}
cwT = 48,1 cm
ai,= 17° pai = 30°
(— Bi,= 30°
Ri ¥ ¥
cfe = cae= 170m os : cri = 39cm | of = igem
i> Cars
ae 2° O75
lcs L m TT
owT = 6,4 cm |
|
6.2 6.2.1 U=3x 36,5 = 110 mis
6.2.2 P = m.u.CwT = 1 x 110 (660,65 + 233,6) =98,37 kW
6.2.3 _ 2UCw,
(Cai,)°
21 10(660,65 + 233,6).
= 00%
x, 100%
= 78,69%
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(3)
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MARKING GUIDELINE -10- T1350(E)(M31)T
POWER MACHINES N6
6.2.4 Faxial = m{(Cfir — Cfer) + (Cfi2 — Cfez)
= 1[(5,41 - 5) +(1,9- 1,7)
= 22,27 N (2)
[20]
QUESTION 7
7.4 Ideal C.0.P 9 = —*_ P(kPa
”T2=T1
267
04-267
= 267
a7
= 7,216
ity
Actual C.0.P = nx C.O.P 51,6 110.2 hi 4928479 (kak) h(kJ/kg)
= 0, 9023 x 7,216
=6,511 (3)
720 C.0.p =e
C.0.P = (he = hn) = hi ~he
6,541 = (479-1) = hi -110, 2
3118,769 + 110,2=hi +6511
32.28,965
hi =
TSAL
= 429,9 kJ/kg
ha = hs+ x1(ha hs)
_bi-he aap 546 age _
MI ecns « Genemmse ~ aia 7 292 (5)
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MARKING GUIDELINE “11- T1350(E)(M31)T
POWER MACHINES N6
7.30 yy=S@b = Ep*x1,20
_iganD® .i2unx(oazee
s a
= 0,002 m/s
Vs= 0 062 420 _ 0, 014 mis
oo
Ve = Vex "va = 0,014 x 0, 9 = 0, 0126 m/s (5)
14 Vie x1.Vgi = 0, 92 x 0,168 = 0,155 mv/kg
a Ye _9,0126 _ :
m = | Sass x 60 = 4, 88 kg/min (4)
75 Po = mr (ho- hi) = “2° (479 - 428, 9) = == (49, 1)
= 4kdis
Pm = Pc / Nec “is =5 kW (3)
[20]
TOTAL: 100
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
--71330(E)(N26)T
NOVEMBER EXAMINATION
© NATIO! SER \
POWER MACINES N6
(8190046)
26 November 2015:(X-Paper)
“9:00-12:00
REQUIREMENTS: Steam Ti (BOE 173) -
Calculators t
This question paper consists of 7 pages and 1 formula sheet of 6 pages.
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(8190046) -2- T1330(E)(N26)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACINES NG
TIME: 3 HOURS
MARKS: 100
NOTE: IF you answer more than th quired number of questions, only the
required number of questions willbe marked. All work you do not want to
be marked must be, clearly crossed out.
this question
8. All sketches and diagrams must be-done in pencil in the ANSWER BOOK.
9. Write neatly and legibly
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(8190046)
QUESTION 1
-3-
T1330(E)(N26)T
510 kg of gas per minute leaves the velocity compounded, two-stage, impulse turbine
with a relative velocity of 210 m/s.
The moving blades rotating at 3 220 r/min have an average diameter of 1 038 mm.
The inlet and outlet angles for all the moving blades are the same.
The velocity of flow at the turbine exit is 105 m/s.
The velocity coefficient for all the blades is 0,96.
1.41 Use a scale of 1 cm = 50 m/s and construct velocity diagrams for the turbine
in the ANSWER BOOK. Indicate the lengths of ALL the lines as well as the
magnitude of the angles-on'the-diagrams. (10)
1.2 Determine from the velocity diagrams:
1.2.1 The magnitude, of the moving blades angles (1)
122. the megnitude ofthe : rigle (1)
423° The magnitude of th (1)
The magnitude of the fixed blades inlet angles (1)
he nozzle velocity of the gas in m/s (1}
locity of the gas leaving the first stage in mis. a (1)
1.2.7 osity of the gas leaving the fixe blades in mis (4)
128 The axial thrust developed in the turbine in Nis (1)
1.2.9 The power deve sped by the turbine in KW 2
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(8190048) -4- T4330(E)(N26)T
QUESTION 2
A two-stage, double-acting, reciprocating compressor takes air in at a pressure of
100 kPa and a temperature of 21°C at a rate of 34 200 cm*/stroke.
The clearance volume for both stages is 4% of their stroke volumes.
The volumetric efficiency for the low-pressure cylinder is 95% and for the
high-pressure cylinder 96, 5%.
The pressure ratio for the low-pressure cylinder is 3, 8:1 and for the high-pressure
cylinder 3, 5: 1.
The intercooler reduces the air temperature by 96 °C.
The index (n) for compression and expansion for both stages is 1, 34.
The rotational frequency of the cormpr. is 320 r/min.
Take R for air as 0,287 kJ/kg. and Cp as 1 08 kJ/kg. K. “
Calculate the following:
2.1 r volume and the clearance volume for the
2.2 niet, the absolute
ery temperature
2.3 ffective swept
gh-pressure
24
QUESTION 3
Dry saturated steam at a pressure ’o
with a negligible velocity.
At the throat the pressure of the steam
sieam is 96% dry.
At the exit the pressure is 340 kPa, the actual dryness factor is 0,94 and the isentropic
dryness factor is 0,936.
ers a convergent-divergent nozzle
‘a, the diameter is 50,82 mm and the
Calculate by using steam tables only:
3. The specific enthalpy of the steam, the velocity in m/s, the specific volume
and the area in mm? at the throat of the nozzle as well as the mass flow rate
of the steam in kg/s
3.2 The actual specific enthalpy, the isentropic specific enthalpy at the nozzle
exit, the overall efficiency of the nozzle and the reheat factor
3.3 The velocity of the steam in m/s, the specific volume and the area in mm? at
the nozzle exit
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(6)
(2)
[20}
(9)
(7)
(4)
[20]
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(8190046) -5- T1330(E)(N26)T
QUESTION 4
A unit mass of air in a cylinder is compressed adiabatically from an initial pressure and
temperature of 99,3 kPa and 62 °C respectively to one eighteenth of its original
volume. Heat is then added at a constant pressure while the temperature increases by
1 224,2 °C.
The air is then expanded adiabatically to its original volume.
The cycle is completed by cooling the air at constant volume to the initial conditions.
Take gamma for air as 1,4 and Cp as 1,008 kJ/kg. kK.
Calculate the following:
41 The missing pressures in kPa and the absolute temperatures at the principal
points
4.2 The heat received, the heat rejected and the work done in kJ/kg of air
4.3 The thermal efficiency
QUESTION 5.
A steam- boiler plant supplies 13 300 kg of superheated steam per nour, at a pressure
of 2,5 MPa and a temperature of 260 °C to a turbine. :
erheated steam is 2,93 Kuikg, K.
The specific heat capacity of the:s
The feed water enters the econiom
to 124,7 °C.
Fuel with a calorific value of 31;
The evaporator consumes 66,4539
The moisture in the flue gases car
937% of the.Supplied heat away through the
chimney. e
The pressure at the chimney base is 100 kPa and the temperature is 100 °C.
The temperature in the boiler room is 20 °C.»
The heat loss unaccounted for is 8,545% of the supplied heat.
The specific heat capacity of water is 4,2 kJ/kg. K
The specific heat capacity of the dry flue gases is 1,045 kJ/kg.K.
Calculate by using steam tables only:
5.1 The specific enthalpy of the superheated steam, the thermal efficiency of the
plant and its equivalent evaporation from and at 100 °C
5.2 The dryness factor of the steam entering the superheater
5.3 The mass of the moisture in the flue gases per kg of fuel, the percentage of
the heat supplied carried away by the dry flue gases and the air-fuel ratio
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[20]
(8)
(4)
(8)
[20]
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(8190046) -6- T1330(E)(N26)T
QUESTION 6
A four-cylinder, four-stroke petrol engine operating on the Otto-cycle principle develops
a brake power of 72 kW at 3 000 r/min.
The ratio of the cylinder volume to the clearance volume is 8,5 :1.
The initial temperature of the cycle is 46 °C.
Heat is added at a rate of 1 029,82 kJ/kg of air.
The brake mean effective pressure is 800 kPa.
The index(n) for compression and expansion is 1,32.
Take gamma for air as 1,4 and Cv as 0,718 kJ/kg. K.
Calculate the following:
6.1 The swept volume, |
cylinder in cm’. (7)
6.2 (5)
6.3 The value of the characteris
airused © (5)
6.4 per kg of air (3)
[20}
QESTION 7
A vapour—compressio
operates between temperal
The compressor has a pis
volumetric efficiency of 89% and.
The wet vapour at the compressor.
The superheated vapour at the coi
1,15 kd/kg.K. e
The actual coefficient of performance of i
The following are extracts from the methy! chloride tables:
Saturation Specific Specific Specific entropy
temperature | volume (m%/kg) enthalpy(kJ/kg.K) (kJ/kg.K)
CC)
Vapour (vg) Liquid (hf)_|_ Vapour (hg) | Liquid (sf) | Vapour (sg)
4 0,058 54,7 463,8 0,217 8 1,738 4
36 0,189 418,3 480,7 0,438 2 1,610 4
Calculate the following:
71 The dryness factor of the refrigerant at the compressor inlet, the specific
enthalpy of the refrigerant before compression and after compression, the
temperature of the refrigerant after compression and the degree of the
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(8190046) -7- T1330(E)(N26) T
1.2 The specific entropy of the refrigerant after compression and after throttling. (5)
7.3 The swept volume of the compressor in m*/h, the volumetric flow of the
refrigerant in m@/h and the mass flow rate of the refrigerant in kg/h. (5)
HINTS: (i) Ta(sa-sb) = ha - hb
(ii) s after compression = s of saturated vapour at 36°C + Cp.In
[T after compression ~ T in condenser] [20]
TOTAL: 100
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(8190046) -1- T1330(E)(N26)T
FORMULA SHEET
Any applicable formula may also be used.
ENGLISH GENERAL AFRIKAANS
PY, = mRT,
R=Cp-Cy
PV—c PYV=k
PV" =¢ PV* =
PV'=c PV’ =k
Q=AU +Wd* = AU + Ay
Qz=m.Cp.AT
Q=m.Cv. AT
+ T,
Syn = Sg + Cp in —s
Ss
Spe =Se— Sy
S=S, +x yp
Poy = Fig + CP (lsu ~1,)
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(8190046) -2- T1330(E)(N26)T
-1
7 (hy, 1941)
Aus = hy + XN fy Poy = P Ing = hy + xIpy
Si
[7
Fans =F g atthe Pus = y
ia
Foa=P xh
Different formulae foi
work done (Wd) ,
a-l
Verskillende formules vir
aybeld yerrig (Ay)
” mT; (rye) -1
n-t
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(
I
(8190046) Be T1330(E)(N26)T
Different formulae for Verskillende formules
work done (Wd) vir arbeid verrig (Ay)
= area of PV-diagram = area van PV-diagrain
= work done first stage = arbeid verrig eerste
+ work done second stadium + arbeid ver-
stage +... rig tweede stadium +
WA yo = Wa, —Wa, AV = AV, — AVE
Wanete ~ Qnett AV nett = Qnett
Verskillende formules
ir lugstandaardrende-
Different formulae for.
air standard efficien:
cies (ASE)
-1 warmte toeg
_ heat added ~ heat rejected _
heat added
war
Verskillende volumetriese
rendemente, O,o1
Differem volumetric
efficiencies, Oot
__ Volume of air taken in __ Volume lug ingeneem
Swept volume Slagvolume
__ Vohune of free air __ Volume vrylig
Swept volume Slagvolume
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(8190046)
Different thermal
efficiencies, Oherm.
Wd
eat supplied
T1330(E)(N26)T
Verskillende termiese
rendemente, Oterm.
_ Av
warmite toegevoeg
BP _ RD
Nevake therm. = Mp i9 CF "rem term. = My) x
IP ID
Nind, therm. = ary Vind. ierm. =
my ts *CV My s XWW
_ an, (As — hw) hem = m, (hs — hw)
Therm. mp XC term. iiiy X WwW
7 wack 4 Be G _B-E
eR [2 Me =p ri
2~ Fi vend - Ly 27£
BP
Tech. = TP.
Indicated efficiency ratio
= Nind. (hevin:
ASE
Brake efficiency ratio
ndementverhouding
Remrendementverhouding
_ "brake therm _ "rem. term. Be
ASE ESRe*
Ti : L
BP=2n IN RD =2n NV
60 : 60
BP = Pyrate mean PANE “RD = P, LANE
IP = Pind. mean VANE
spc = EL
P
pspc = "Eth
BP
cop =
Ty - 7,
RE
cop - ke
Wal
Pa=m.U.AVw
Fy, =m. AV ¢
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rem gem.
ID = Fa. gem. LANE
i
ISBY = "bin
ID
RSBV = bl
RD
t
tT -T
KVW =
xvw ©
Ay
D=m_.U Aw
Fags, = Vp
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(8190046)
ye Ita Ue hy)
my x 2257
Wa,
Tiso, = Wa aa
poly.
Wd
rank. ~ Oy
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= AVigo,
Nisa. = A
¥ poli.
n Ay
rank. >
Q
L;
Nearn, = 1 7
h=ut pV
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T1330(E)(N26)T
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(8190048) 6
2
gh, + RK +S +Q=
c
82, + Uy + PyVo + Wd
T1330(E)(N26)T
2
Bh, +U + AY, +Lag-
2
BZ, + Uy + PF + >
+ Ay
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Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
AS
MARKING GUIDELINE
NATIONAL CERTIFICATE
NOVEMBER EXAMINATION
POWER MACHINES N6
26 NOVEMBER 2015
This marking guideline consists of 8 pages.
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MARKING GUIDELINE -2-
T4330(E)(N26)T
POWER MACHINES N6
QUESTION 1
i
Us a EM = WlyozeNar20)
oo Zo
i ts ba ‘5
éF ut ms BEih mm |
act Bey Bea ode, fe, 30°
hA2.0h 2 Q2?
P2- 3K, = 16,77
beleg, Pe, = 42,9°
HES. Cad. 111, Faafs
fag. Coe, = 346, Sint yp
PQd Ae, > 90,5 aa] s
bag,
Kays m| Ci Cfe,) } (Che, fe,
ot
18 [0 5 b4 Be) 4 lara aphe
Y
(Ms Mu Cw,
= 8 SKIS (4e48 4 34,5)
487 744 Rie MW, “
(io)
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MARKING GUIDELINE -3-
POWER MACHINES N6
QUESTION 2 P
2.1
2.2
2.3
2.4
Nvott = aol x 100%
Vst = £2 x 100%
nol
34 200 x10°9/0,95
T1330(E)(N26)T
0,036 m%/stroke
V4 = Voit Vor = Vor + 0, 04 Vs,
LP
= 1,04 Vsi
&
= 1,04 x 0,036
= 0,03744 mstroke
Vet = 0,04 Vs1 = 0,04(0,036) = 0,00144 mstroke
T2 = T2- reduction = 412,531 - 96 = 316,531 K
peed
T3=T2(E)"a = 316,531 (2) = 434,974 K
, .
= Pivel ye = 100x008 | 359.0 = 25,94 kg/min
RTL G, 287x294
Vel _ 9685571
nya 0965
Veo = =: 0,04 10 041,006 = 401,64 cm?
Ve2 = 0,04 vs2 = 0,04 x 10041,006 = 401,64 om?
Qin = mCp(T2- T2) = 2% x 4,008(96) = 41,836 kj/s
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MARKING GUIDELINE -4- T1330(E)(N26)T
POWER MACHINES N6
QUESTION 3
3.4 At900kPa: hi =hg=2773 kJ/kg
At 490 kPa: hfe = 637 kJ/kg, hige = 2 110 kd/kg , Voc = 0,3819 m3/kg
he = hfc + x hige = 637 0,96(2110) = 2 662,6 kJ/kg
Cc =,/2000(ha — he) = .f2000(2773 — 2662.6) = 469,894 mis
Vc = X Vg = 0,96 x 0,3819 = 0,3666 m3/kg
Ac =~ De? = © (50,82)? = 2028,426 mm?
ACK 2028.426 x10—6x469,894
= Sh = a = 2,6 kg/s
Ve 0,d666
m
(9)
3.2
At 340 kPa : hfs = 580 kJ/kg ,hfge = 2150 kd/kg ,Vga = 0,5385 m/kg
hz = hz + xq hfge = 580 + 0,94(2150) = 2601 ku/kg
he = hf, + x hfge = 580 + 0,936(2150) = 2 592.4 ki/kg
Moveran =( ht — ha)/( ht — he’) = (2773 -2601)/(2773 — 2592,4) = 95,238 %
% reheat = 100 — 95,238 = 4,762 % OR 0,04762 (7)
3.3 C2 =,/2000(h1 — h2) =./2000(2773— 2601) = 586,515 m/s
V2 = XaVge2
= 0,94(0,5385)
= 0,5062 m3/ kg
ame
Aa = C2
= 2,6<0,5062 x 10°
586,515
= 2243,967 mm? (4)
[20]
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MARKING GUIDELINE -5-
POWER MACHINES N6
QUESTION 4
41
4.2
43
Po = P3=Py (2) = 99, 3 ()!4 = 5 679,77 kPa
T= 11 (2) = 335 (#8)°4= 1 064,52 K
Ts = Te +1224,2 = 1064,52 + 1224,2 = 288,72 K
But V2 = V3
T2 Ts
(S)= we (2288,72)/(1064,52) = 2,15
23,
=. or 235
T1330(E)(N26)T
P4 = P3 ey" = 5679,77 = al = 289,981 kPa
Ta = (22) = (e228) = 978,284 K
Q2.3= mCp(T3— T2) = 1x1,008(2288,72 — 1064,52) = 1 233,994 kJ/kg
But y = Cp therefore Cv = Co
Cv ¥
Qs1 = mcw(Th — Ta) = 1x(=*)(335 -978,284) = - 463,164 KJ/kg
WD = Qreo - Qrej = 1233,994 — 463,164 = 770,83 kJ/kg
Mine = Oe) x100%
= WD/Qrec
= 770,83/1233,994 x 100 %
= 62,466%
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MARKING GUIDELINE -6- T1330(E)(N26)T
POWER MACHINES N6
QUESTION 5
5.1 At 2500 kPa:hf = 962 kJ/kg, hfg = 1839 kJ/kg ,jhg = 2801 kJ/kg & tss = 223,9°C
At 51,1°C : hiv = 214 kJ/kg & at 124,7 °C : hfe = 524 kJ/kg
Nes = hg + Cp (tes ts ) = 2801 + 2,93(260 - 223,9) = 2 906,773 kJ/kg
2906,773 -212°
nn= ms (ALY x 100%= 13300 (TE) 00m = 81,211%
maf xa
_ hse—kfwrl) _ 2506,773—214 =
EE = ms Cr) = 13300 ( 140082257 } 11,334 (8)
52 - (cetetrerare)
Nevap = Ms infest
xhfg = [Neves (72) + hfwe— hf] / hig
= [0,68453 — + 524 962]/1839= 1765,4416/1839
13300
= 0,96 (4)
5.3 At 100 kPa & 100 °C: hss = 2676 kJ/kg
tesa -C pwd:
Pmois = Mm [=—SEwae?y
pommolstx ce ] = O.O493Fx31500 = 0,6 kg/kg fuel
tese—Cpwtta) 2676—4.2(20}
Mm =
n dry flues = 100% - F)piant - Nmois - Plunaccounted
= 100% - 81,211% - 4,937% — 8,545%
= 5,307 %
= pee ues ier 4 m
-1
Cpglata) m
O,OS307xS 1500
=a 06-1
1,045 (100-203
= 19,596:1 (8)
[20]
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MARKING GUIDELINE -7-
POWER MACHINES N6
QUESTION 6
6.1 P
pp = Pate
et
Vs = LA= Se
PCE
6G 7208 6
= Sanoarnes X10
BOs 2600x5
= 900 cms 2
Vo=V3=Vo =
ta
= 8 = 120 cm
a
T1330(E)(N26)T
8.5-
Vi =Vs + Vc = 900 +120 = 1 020 cm?
6.2
T= 7; (4)"" = 319 (28% = 632,71 K
Qe3 = mCv(T3— T2)
Ts= S347) = MBS + 632,71 = 2067 K
mcy 140,718
Ta= Ts(2)" = 2067 (2.)9# = 1 042,14 K
V4.
6.3 Cp = y.Cv = 1,4. x 0,718 = 1,0052 kJ/kg.K
R = Cp — Cv = 1,0052 - 0,718 = 0,287 kJ/kg.K
WD a 2RscacT2+T1) _ ixo2g7[2oe7—a04214-692,714515)
aml O32
= 637,81 kJ/kg
64
Q=WD + mCv(T2 ~- T1)
NRC TL FZ)
= BA. + mCu(T2- 71)
= feoser(si9—63271) 4x0,718(632,71 - 319)
0.32
= - 56,115 kJ/kg
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MARKING GUIDELINE -8- T1330(E)(N26)T
POWER MACHINES NG
QUESTION 7
P
54,7
118.3
463.8 4...
480,7
4 8 2
g § Se &
7A
Vegi O,0851
X == = 0,98
Ve nse
hy = he +x1(h7-he) = 54,7 +0,95(463,8 - 54,7) = 443,345 kJ/kg
= RE _ bh1i-bs
COP = wo hz-h1
_ hi-hs _ 443,345~118,3
he =hi + cop = 443,345 + “aS 519,808 kj /kg
he = hatCp(T2 — Ts)
2-h3
i. 519,9D8--280,7
T2 =T3+ cp
wis
= 309 +
= 343,007 K
Superheat = 343,007 ~ 309 = 34,007 °C
0.2 S2= Si =S3 + Cp.inS
=1,6104+1,15.In= 2 = 1,7305 kl/kg
Ss = Sy - “288 =1,7384 —
T1
463,8-118,3.
269
= 0,454 k] /ke
7.3 Vs =f D¢L = 3 (0, 098)? x 0, 105 x 340 x 60 = 16,157 m3/n
Vr = Nvo.Vs = 0,89 x 16,157 = 14,38 mh
mr = t= 2838 = 260,98 kg/h
Wei G,0554
TOTAL:
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100
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
ponsoo0eca
T1300(E)(J22)T
AUGUST EXAMINATION
REQUIREMENTS: Steam Tables (BOE 173)
‘Calculators may be used
“This question paper consists of 7 pages and 1 formula sheet of 5 pages.
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(8190046) -2- T1300(E)(J22)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACINES N6
TIME: 3 HOURS
MARKS: 100
NOTE: Hf you answer more than the required number of question: i
required number of questions will be marked. All work you'do not\want-to
be marked must be clearly crossed out.
INSTRUCTIONS AND INFORMATION
1. Answer any FIVE questions. ;
2. Read ALL the questions carefully...
3. Number the answers according
paper. ‘s
4, Questions may be
be kept together.
6. ALL formulae uged mus written down.
6. Show AL ‘i e interme jate steps.
7. { s-must be answered in blue or black ink.
8. “all Sketches and diagrams must be done in pencil in the ANSWER BOOK.
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(8190046) -3- T1300(E)(J22)T
Answer only FIVE of the seven questions.
QUESTION 1
A two-stage, single-acting, reciprocating compressor delivers 18 kg of air per minute.
The compression index and expansion index are both 1,34.
The delivery temperature is 235,227°C.
The swept volume for the low-pressure cylinder is equal to 25 times its. slearance
volume.
The rotational frequency of the compressor is 9,6 r/s.
The pressure ratio for the low-pressure cylinder is 4 : 1 and for the high Pressuie
cylinder 3, 8: 1. ae a
The clearance volume for the low-pressure cylinder is 1 200 cm?yi2e :
The initial pressure and temperature for the low-pressure cylinder are 95: KPa, and 24° c
respectively.
The volumetric efficiencies for the low-pressure and high-pressure cylinders are equal.
Take R for air as 0, 288 kJ/kg.K and Cp as 1, 008 kJ/kg
Calculate the following:
14 The absolute adiabatic temperature ) “the air after-compression in the low-
pressure cylinder, the absolute” temperatur after’ intercooling and the heat
absorbed by the intercooler i in
1.2 The power required to drive the comp!
1.3 The swept volumé in
volumetric efficienc:
“effective swept volume in m/s and the
r the. low: pressure cylinder. (3 x 2)
1.4 The effective swept vol n ‘m/s and the swept volume in m®/cycle for the
(2x2)
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(8190046) ~4- T1300({E)(J22)T
QUESTION 2
A steam-boiler plant, generating superheated steam at a pressure of 3 MPa and a
temperature of 350°C, was equipped with an air preheater to improve its efficiency.
Tests on the plant revealed the following information:
The mass of combustion moisture formed remained unchanged.
The enthalpy of the steam generated increased to 3 138 kJ/kg steam due to a slight
increase in temperature.
The rate of evaporation of steam improved from 77 400 kg/h to 78 690. kgf
The feed-water temperature remained unchanged at 132, 4°C.
The boiler room temperature remained unchanged at 20°C.
The mass of coal burnt per hour remained unchanged at.
value remained at 28,8 MJ/kg.
The specific heat capacity of the flue gases remained at.1,045
The specific heat capacity of the water remained at 4,2 kJAG. K
The air supplied per kg of coal burnt decreased from 34) kg. G}
The temperatureidf the flues at the chimney base" decrea mMi250°C to 150°C.
The pressure of the flues at the chimney base:femained un hanged at 150 kPa.
The heat lost in the combustion moi ure be! re. fitting ihe air preheater was
1 444,5kJ/kg fuel burnt.
600. kg ahd its calorific
Calculate by using steam tables only:,_
21 2.1.1 The mass. ofithe.combusti oisture formed in kg/kg fuel.
t inthe combustion moisture after the air preheater
was fitted in Jikg fugl.
2.2 2.2.1 The mass: cof t! ary flue gases formed in kg/kg fuel and the energy
ed carried awa by it in kJ/kg fuel before the air preheater was fitted.
88'of the dry flue gases formed in kg/kg fuel and the energy
atried away by it in kJ/kg fuel after the air preheater was fitted.
“, The thermal efficiency of the plant before the preheater was fitted.
The thermal efficiency of the plant after the preheater was fitted
and the increase in the efficiency.
2.4 Draw up a heat balance in kJ/kg fuel and as a percentage, after the’ air
preheater was fitted, to determine the percentage heat loss unaccounted for.
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(8190046) -5 T1300(E)(J22)T
QUESTION 3
The volumetric compression ratio of a single cylinder engine operating on the Otto
cycle principle is 9: 1.
The compression and expansion index are equal.
The pressure after combustion is 18,18 times the pressure after expansion.
The heat rejected during the exhaust stroke is 207,502 kJ/kg of air.
The temperature at the beginning of the compression stroke is 38°C.
Take R for air as 0,287 kJ/kg.K and Cv as 0,718 kJ/kg.K.
Calculate the following:
3.1 The index for compression and expansion.
3.2 All the missing absolute temperatures at the principal Bol
3.3 The heat received during combustion in kJ/kg
3.4 The heat rejected during compression in kik
3.5 The work done by the engine in kJ/k
3.6 The value of gamma and the air.siandard éfficiency. of the engine.
QUESTION 4
ssure, of 1 MPa ‘and a temperature of 230°C enters a
it leaves at a pressure of 200 kPa.
e inlet’ to the exit is 90% of the isentropic enthalpy
Superheated steam at a
convergent-divergent nozzle ai
The overall enthalpy drop fro!
drop due to friction.
At the throat the ar a is 2854 12 mm, the velocity of the steam is 500 m/s, the
pressure is 540 kPa, the steam’temperature is 164,4°C and the specific heat capacity
18:2;5 kJ/kg. K.
The isentropic enthalpy drop through the nozzle is 303 kJ/kg steam.
The indéx(n) for superhéated steam is 1,3.
Ignore the inlet velocity of the steam.
y U ing steam tables only:
44 The specific enthalpy of the steam at the throat and at the inlet.
4.2 The specific volume of the steam at the throat and the mass of steam flowing
in kg/s. (2x2)
4.3 The actual specific enthalpy drop through the nozzle, the actual specific
enthalpy, the velocity in m/s, the actual dryness factor, the specific volume
and the area in mm’ at the exit of the nozzle. (5 x 2)
44 The Mach number.
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(8190046) -6- T1300(E)(J22)T
QUESTION 5
The clearance volume of a single-cylinder engine operating on the dual cycle principle
is 300 cm.
The compression and expansion index are both 1,35.
The engine absorbed 1 148.8 kJ of heat per kg of air.
The volumetric expansion ratio is 6, 4: 1.
40% of the heat is absorbed during constant volume combustion and the remainder
during constant pressure combustion,
The temperature at the beginning of the compression stroke is 40°C.
The clearance volume is one fifteenth of the swept volume.
Take gamma for air as 1,4, Cp as 1,005 kJ/kg.K and Cv as 0,718 kJ/kg.K.
Calculate the following:
5.1 The swept volume and the missing volumes at the {
6.2 The missing absolute temperatures at the pri ips
6.3 The thermal! efficiency of the engine.
5.4 The work done by the engine in kJikg
QUESTION 6
A velocity compounded, _.two- tage pulse “turbine was designed for an axial
discharge. ,
The inlet and outlet angles oft inst tow of moving blades are both 25°.
The relative velocity at outlet from the first stage is 490 m/s.
The exit angle of the: fixed blade is Ar’.
The gas flow rate is 20, kg/s.
There is a% loss of velogi
‘oss all blades due to friction.
610 ks Use a iGth’ ‘of 40 mm for the average blade velocity and a length of 22 mm
. for the velocity of the gas leaving the turbine and construct velocity diagrams
“for the turbine in the ANSWER BOOK. Indicate the lengths of ALL the lines as
ell as: the magnitude of the angles on the diagrams and calculate the scale.
Detéimine the following from the velocity diagrams:
6.2 6.2.1 The nozzle angle.
6.2.2 The inlet angle for the second row of moving blades.
6.2.3 The outlet angle for the second row of moving blades.
6.2.4 The angle of the gas leaving the first stage.
6.2.5 The nozzle velocity in m/s.
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(8190046) -7- T1300(E)(J22)T
6.2.6 The velocity of the gas leaving the turbine in m/s. (1)
6.2.7 The average blade velocity in m/s. (1)
6.2.8 The power developed by the turbine in kW. (3)
[20]
QUESTION 7
A vapour-compression refrigerating plant, using methyl chloride as a_ refrigerant,
operates between temperature limits of -10°C and 34°C. f
The refrigerant is a wet vapour before compression with a specif
452,394 kJ/kg. 4
After compression the refrigerant has a temperature of 66°C.
The refrigerant is undercooled.by 5°C in the condenser,
The flow rate of the refrigerant is 6 kg/min.
The plant must produce ice at -4°C from water at 16°C.
The latent heat of fusion of ice is 336 kJ/kg.
The specific heat capacity of ice at -4°C is 1,9 kJ/kg. K
The flow rate of the cooling water is 19,2 kg/min. _
The specific heat capacity of water is 4,187 kJ/kg: K>
The refrigerant is 15,1% dry after throttling. .
At -10°C:
Specific enthalpy of saturated liquid : 45 Adi g.
Spevific enthalpy of dry saturated vapour
Specific entropy of saturated liqui id = 0,183
7 and the specific entropy at the compressor inlet and the
t capacity of the superheated vapour. (8)
The ‘specific enthalpy after compression, the specific enthalpy after
emperature of the cooling water in °C. (4x 2) (8)
7.3 The mass of ice in kg produced by the plant every 12 hours.
HINT: s after compression = s of saturated vapour at 34°C + Cp,In (T after
compression + T in condenser) (4)
[20]
TOTAL: 100
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(8190046)
POWER MACHINES N6
FORMULA SHEET
“1 71300(E)(J22)T
Any applicable formula may also be used,
ENGLISH GENERAL AFRIKAANS
PY, =mRi,
R=Cp-Cyv
-P
ve cy
PV=c
PV" =c
PV’ =c
Q=AU+Wa
As
As
As
Q
T,,
Soy = So +Cp In St
su g P L
_ ASS
OQ=
=am.Cyv. AT
my) Cy In Pcpan es
wR Mi
PB
=m.Cy in
j
=m. Cp tn 12
1
=m. Rin Hh
2
m.Cp. AT
Ss
fe = Sy ~ Sp
S=
SH
Sp txSx
+ CP lsu = 1,)
=h
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(8190046)
ENGLISH
Hyg = Np + Xryy
a
ws
=x,
2 T1300(E)(J22)T
GENERAL AFRIKAANS
aN joan
= A____ hyp = hp + xh
Pou P,, ns ‘f + XN
pa teile Bas = XV,
Different formulae for
work done (Wd)
‘illende formules vir
arbeid verrig (Av)
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(8190046) -3- T1300(E)(J22)T
ENGLISH GENERAL AFRIKAANS
Different formulae for Verskillende formules
work done (Wa) vir arbeid verrig (Av)
= area of PV-diagram = area van PV-diagram
= work done first stage = arbeid verrig eerste |
+ work done second stadium + arbeid ver-&
stage +... rig tweede stadium +
Wa new = We a, ~ Wa,
Waren = Quett
Verskillende formules
ir lugstandaardrende-
mente (LSR)
Different formulae for
air standard efficien-
cies (ASE)
te, -D+7? @ DI
_ heat added ~ heat reje x
heat added:
warmie toegevoeg ~ warmte afgestaan
warmte loegevoeg
Verskillende valumetriese
rendemente, Oot
Different volumetr
_ Volume lug ingeneem
Slagvolume
_ Vohune vrylug
Swept volume Slagvolume
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(8190046)
ENGLISH
Different thermal
efficiencies, Oherm.
Wa
7 heat supplied
BP
Tevake therm. ~ Mp jy % CV
IP
ind therm. = mp7,xCP
_ nm, (hs — hw)
Therm. = “mp x CV
GENERAL
T1300(E)(J22)T
AFRIKAANS
Verskillende termiese
rendemente, Oerm.
7 Ay
warmie toegevoeg
1 _ RD
rem tera, Mpis % WH
ind. term. ~ ° in tbls WW
BF
IT
Ne
BP
Tnech, > TP.
Indicated efficiency ratio
_ ind. therm.
ASE
Brake efficiency ratio
Ty -Ty
B-T,
i dikateurrendementverhouding
—_ Flind. term.
LSR
Remrendementverhouding
= "brake therm. _ Trem. term.
ASE ESR
Bp 27% Foor RD =27
7 60
IP = Prraviivean Ee ID = Ping, gen. LANE
isk Lh Ispy = Moth
ID
MELD. Myth
BSFC = a RSBV =—22"
BP RD
T,
CoP == KYW =
fh -T, hh
cop = R® xvw ==
Wd Ay
P=m.U,Avw D=m.U.A¥w
Fay om. AV, Foy, =n AV
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(8190046)
ENGLISH
‘a.
Drank, = ree
r+ BY ++ Q=
BL + Ug + Poy +2 oma
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-5- T1300(E)(J22)T
GENERAL AFRIKAANS
C, = 2x10 (Ay —h,) + C2
Cy = J2x 10? (y — fy) + CP
C, = 2x10 xCy G
EV = ms (Ay ~hy)
My ¥ 2257
Ay,
Niso. = 180.
AV poli,
Ay
Drank, = Oo
L;
Near, = 1 ~ ra
h=ut pV
Ce
eZ, +U, + BV; + Oe
Cc
B2y + Uy + Py top Ay
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE =|
NATIONAL CERTIFICATE
AUGUST EXAMINATION
POWER MACHINES N6
22 JULY 2015
This marking guideline consists of 9 pages.
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MARKING GUIDELINE -2- T1300(E)(J22)T
POWER MACHINES N6
Candidates must answer only FIVE questions.
QUESTION 1
Pp
14 aoe os
To = 1: (2) = = 297 (4) = 422,2K
= psy ES L oe -
Ta=T3(=) = = 508,227 Gas" = 362,2K y
Qin = mCp (T2-Ta) = 2x1,008 (422,2-362,2) = 18,144 kis 6)
120 Pe = ean (TUE ~ 1147 AD
= RESO (297 [(4)-= — 1] +362, 2[ (3,2)-" — i]}
= 92,358 kW. : (5)
13 Ver = 25Vs1 = 25x1200 =30000cm?
= 30 000 x 10 m?= 0,03 m®
Pi Ve, = mRTi
mRTi
Vet = PL = 18xb 288x297 = 0,27 m/s
66x55
twa = “x 100% = =x 100% = 93.75%
Vai G,0329,6
(3x2) (6)
14 Veg = MRTZ = Uexb2b6s86?4 = 9.0824 m%s (Note:P2=P1.k )
PZ 6004x553
(Note: ryort = Nyoi 2)
wo 2 = to x 100%
Ve2
Vs = Ryol2 x 100% = 208222302 = 9 00916 mcycle
93,75 256
(2x2) (4)
[20]
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MARKING GUIDELINE
QUESTION 2
21 2.1.1
2.1.2
2.2 2.2.1
2.2.2
2.3 2.3.1
2.3.2
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-3- 71300(E)(J22)T
POWER MACHINES N6
At 150kPa & 250°C :hss = 2973 kJ/kg
At 150kPa & 150°C: hss = 2773 kJ/kg
Qiost = Mm [hss — cpy(tatm )]
Mm = ——We = 1 444,5/(2 973 — 4,2 x 20) = 0,5 kg/kg fuel
hes—Cpwiltatm)
Quest = Mm [hss —cpa(tatn )] = 0,5[2773 - 4,2 x 20] =
1 344,5 kJ/kgfuel
mg= (Ma+1)-mm = (21,5 + 1)—0,5 = 22 kg/kg fuel
Qiost = MgXCPgxAt = 22 x 1,045 (250 — 20) = 5 287,7 kJ/kg fuel
Mg= (Mat 4)-Mm = (21 +1)-0,5 = 21,5 kd/kg fuel
Qiost = MgxCDgxAt = 21,5 x 1,045 (150 - 20) = 2 920,775 kJ/kg
fuel ,
At3000kPa and 350°C: hss = 3117 kJ/kg
At 1 32,4 °C : hf = 557 kJ/kg
Nth betore = Me[hss — hfwi]/(mmxCv) x 100%
= 77 400 [3 117 - 557] (8 600 x 28 800) = 80%
Minater = me[hss — hfwiJmeCv) x 100%
= 78 690 [3 138 ~ 557] (8 600 x 28 800) = 82%
Djincrease = 82-80=2%
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MARKING GUIDELINE . 4 T1300(E)(J22)T
POWER MAGHINES N6
24 Qin (kJ /kg} Qout (kJ /kg) %
28 800 Plant = 222 (3138-557) = 23 616,15 82
Qmoisture =1 3445 4,67
Qary = 2 920,775 10,14
Unaccounted =28 800 — 27 881,425 = 918,575 3,19
28 800 = 28 800 100
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[20]
QUESTION 3 P
3
3.1 vay"? _ (pa
@) =@)
2 “4
vayhe PB
ina) = (3) ;
Vv
_ Pa vayne _ _
ne=In (2)in(=)" = In 18,18/In 9 = 1,32 = ne (3)
3.2 To = Th ay
= 311 (9)0%
= 628,229 K
Qa1= m.Cv (Ti — Ta)
Ta = Ty - Qaafm.Cy = 311 - (-207,502/1 x 0,718) = 600k
Ts= Ta (4)""" = 600 (9) = 1 212K 6)
3.3 Qo3= m.Cv(T3—T2) = 1x 0,718(1212 — 628,229) = 419,148 kJ/kg (2)
3.4 Qi2 = mR(T1—Te) + m.Cv(T2- 71)
No - 1
= 41x0.287(311 — 628,229) + 1x0,718(628,229 —-311) = -56,744 kJ/kg
0,32 (3)
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MARKING GUIDELINE -5- T1300(E)(J22)T
POWER MACHINES N6
3.5 WD = mRiTe-Ta- To +Til = 1.x .0,287/1212 -G00 -628,229 + 311]
n-t 0,32
= 264,373 kdikg
3.6 Cp=R+Cv= 0,287 +0,718 = 1,005 kJ/kg. K
y= CplGv = 1,005/0,718 = 1,4
ASE = {1 —(2)" "yx 100% = [1 - (1/9)94] x 100% = 58,476 %
QUESTION 4
41 AtS40 kPa: hg = 2751 kdikg & ts = 154,8 °C
ho =hg + Cpliss—ts) = 2751 +2,5(164,4 — 14,8) = 2775 kulkg
Cc = fa000(RI — he)
i = he + Co/2 000 = 2775 + 50072 000 = 2775 + 125 = 2 900 kJ/kg
4.2 You 2 (MOBI) = 08 (RIBS) = 0.3564 mikg
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Ace ™™ | m= S882 = 2851,2x10%x500 = 4 kg/s
ce ¥o
0,3564 (2x 2)
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MARKING GUIDELINE 6 T4300(E)(J22)T
POWER MACHINES N6
4.3 hi — he= 303 ku/kg
ha — he = 0,9(hy — he) = 0,9(303) = 272,7 kd/kg
ha = -272,7 = 2 900 — 272,2 = 2 627,3 kJ/kg
=,(2000(272,7) = 738,512 m/s
At 200 kPa : hr = 505 kd/kg ,hfg = 2 202 kJ/kg & vg = 0,8854 m%/kg
ha = hr + xa(hfg)
Ma=ho-hr = 2627,3-505 = 0,9638
hfg 2202
V2 = Xa.Vg = 0,9638 x 0,8854 = 0,8533 m/kg
Ag = ~
= 0,060
738,612
= 4621,726 mm? (5x2) (10)
44 Mach number = C2/C.
= 738,512/500
= 1,477 (1)
[20]
QUESTION 5
5.1 Vso = 15 = VelV2 = VslV3 Pp
Vs = 15 Vc = 15x300 = 4500 cmS
Vs =V1=Vs + Vo = 4500+ 300 = 4800 cm?
Vol exp ratio(Ver) = Vs/V4 = 6,4
Va = VslVer = 4800/6,4 = 750 cm®
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MARKING GUIDELINE -7- T1300(E)(J22)T
5.2
5.3
5.4
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POWER MACHINES N6
aynork
To= 11 (S)" = 313(16)°° = 826 K
Q23= m.Cv(T3—T2) = 0,4 Qr
T3 = O4Q7 + To = 04x11488 + 826 = 1466K
mCv 1x0,718
Q34= m.Cp(T4— Ts) = 0,6 Qr
Ts = O6Qr + T3 = O6x1148.8 +1466 = 2151,851K
mCp 1x1 ,005
Ts=Ta (EE)"°"? = 2151, 851 (16,4) = 4 123,697 K
(4x 2)
y = CpiCv = 1,005/0,718 = 1,4
Ts-Tt
> 2)
i (va-T2)4y(T3-T3) x 100%
Nth =
1123 697-313
= 1- — x 100%
(1466 -926)+ 1,4(2151,851—-146 6)
= 49,337%
WD = mR{(T4— Ta) +{ Ta=Ts—Te + Ti}
n-1
= 1x(1,005 -0,718){(2151 ,851 — 1466) + [2151,851 — 1123,697-826+313]}
0,35
= 619,266 kJ/kg
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MARKING GUIDELINE -8- T1300(E)(J22)T
POWER MACHINES N6
QUESTION 6
8 SCALE?
- € eter foal)
By (pape
fps i
jae?
fred Gug & Bl, db
6-303 Ga, 2,505 8
estes wifes
oo b
6 ab Cae eg mls
& ae? Ue 1 bOwnts
Pe 6-28 Pamleny tGu\y
oo) . em op h oe
is = 2001466 180.7) M40
oe
@
[173
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MARKING GUIDELINE -9- T1300(E)(J22)T
POWER MACHINES N6
QUESTION 7
P
a
302 T°
263-|-
0,183-]
1.616
7.1 hy = hy + x1(hs— hz)
x1 = (h1—h7)M( he— hz) = (452,394 — 45,4)(460,7 — 45,4) = 0,98
81 = S2= $7 + X1(Se— §7) = 0,183 + 0,98(1,762 -0,183) = 1,73 kJ/kg. K
$1 = S2= 83+ Cpe In(Tx Ts)
Cpo = gi-e3 sh. F9-1,616
ince) ine
= 1, 15 kd/kg
7.2 ho = hs + Cp(T2— Ts) = 480,1 + 1,15 (339 — 307) = 480,1 +36,8 = 516,9 kJ/kg
hs = he = hy + xe(ha—h7) = 45,4 + 0,151(460,7 — 45,4) = 108," 1 kJ/kg
Qecon = mr (h2— hs) = 6 (516,9 — 108,11) = 2 452,74 kJ/min
Qecon = mw.Cpw.Atw
Aty = Qecon/ mw.Cpw = 2 452,74/(19,2 x 4,187) = 30,51°C (4x 2)
7.3 Qevap = RE = mr (hy — he) = mi (hfgi + CpwAtw + Cp.Ati]
= __metbichs) , S(482.084 108A x60? 3 G99 347 kg/12hr
m= Eo
i hfgitcpwatw + Cpidti 9R6 44,187 (16)+-1,5(4)
TOTAL:
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conodo00g0
T1310(E)(A2)T =
APRIL EXAMINATION »
NATIONAL CERTIFIGATE...
POWER MACHINES Né
& (8190046), ;
_- PApril 2016 (¥
2015 (Y-Paper)
_ 13:00-16:00
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(8190046) -2- T1310(E)A2)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
NOTE: If you answer more than the required questions, only ‘the
questions will be marked. All work you do not want to be mai ed, must
be clearly crossed out. a
INSTRUCTIONS AND INFORMATION
1. Answer FIVE questions.
2. Read ALL the questions carefully.
3. Number the answers according to the ni abering system used in this question
paper. 2
4, Write neatly and legibly.
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(8190046) -3- T1340(E)A2)T
Answer FIVE of the seven questions.
QUESTION 1
A double acting, two-stage air compressor delivers 3 kg of air per minute at a pressure
of 1,5 MPa. The intake conditions are 98 kPa and 28 °C. The compression and
expansion index for both stages is 1,3. The volumetric efficiencies based on inlet
conditions are 92% for the low pressure cylinder and 90% for the high pressure
cylinder. The compressor rotates at 240 r/min. Intercooling at 360 kPa is complete
and the temperature after intercooling is 87°C. Mechanical efficiency is 85%..-:..,
Take R for gas as 0,288kJ/kg.K and Cp as 1,005 kJ/kg. K.
Calculate:
1.4 The power required to drive the compressor in kW (7)
1.2 The diameter of the low pressure cylinder jf the ‘oke i8.4,5 times the
diameter : i (7)
1.3 The heat transfer during compression in the fi (6)
[20]
QUESTION 2
A bailer plant comprises of an economiser aporator, superheater and air heater.
The feed water entering and_léaying. the-economiser has a temperature of 44 °C and
175 °C respectively. The.air for, combustion is heated from 15 °C to 150 °C. The
steam pressure is 300 kPa t the, entrance to the superheater the dryness fraction
is 0,98 ae
At the superheater outlet the temperature of the steam is 250 °C. The calorific value of
the fuel burned is 31 800kJ/kg: T ir fuel ratio is 20: 1. The steam is generated at
the rate of 9,5 .kg of fuel burne :
Take Cp for ‘air as 3 05 KOK and for flue gases as 1,045 kJ/kg.K
Calculate: ;
21 ‘hie heat transfer affected in each component per kg fuel burned (10)
2.2 The ‘plant efficiency (4)
2.3 The temperature of the flue gases leaving the plant (6)
[20]
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(8190046) -4. T1310(E)(A2)T
QUESTION 3
An internal combustion engine on the diesel cycle principle has a clearance volume
of 0,00041 m*, a piston diameter of 180 mm and the stroke length of 250 mm. Fuel is
cut off at 20 mm after TDC on the return stroke. The pressure and temperature after
polytropic expansion is 400 kPa and 1 150 K. The compression and expansion index
is 1,35 and 1,28 respectively.
Take R as 0,288 kJ/kg. K
Calculate:
3.1 The swept volume and cylinder volume in m?
3.2 The volume after constant pressure heat addition in.m®
3.3 The lowest and highest pressure in kPa _
3.4 The work done in kJ/cycle
3.5 The indicated power of the engine a
principle at 330 r/min with 4 cylinders.
QUESTION 4
our ‘oke é@ngine a Morse test was carried out. When
linders working, the brake power was 56 kW. The
h cylinder was cut. out in turn and the load reduced to
In a trial on a six cylind
running at full load anda
measured brake power-when
bring the engine back to its orig
Cylinder
Number ... 3 4 5 6
Brake," :
Power(kW) |. 44,2°° 44 43,9 443 441 43,7
The following are further recordings taken:
Indicated specific fuel consumption = 0,26 kg/kWh
Heat value of the fuel = 42 MJ/kg
Air fuel ratio by mass = 45,2: 1
Exhaust gas temperature = 400 °C
Specific heat capacity of exhaust gas = 1,1 kJ/kg. K
Cooling water temperature rise = 38 °C
Cooling water flow rate = 34 kg/min
Specific heat capacity of water = 4,187 kJ/kg.K
Normal air temperature = 18°C
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(8190046) -5- T1310(E)(A2)T
Calculate:
4.4 The mechanical efficiency
4.2 The mass of fuel used in kg/min
43 The mass of exhaust gas in kg/min
4.4 The heat lost to the exhaust gases in kJ/min
45 The heat lost to the cooling water in kJ/min
46 Draw up a heat balance in kJ/min and as percentage to determine pe centage
heat lost fo radiation. Assume that the heat lost to frictlonis aosorbed ‘bythe
cooling water. " :
QUESTION 5
rate of 5 kg/s to the exit
Steam expands through a convergent divergent nozzle a
i: 2,2 mm.
where the isentropic dryness factor is 0,94.and ‘the.diameter
At the entrance the superheated steamha
of 250 °C and the velocity is negligible.
s a:pressure of 4 500 kPa and a temperature
At the throat the steam has a.pressure of.820 kPa, a velocity of 500 m/s and a specific
heat capacity of 2,56 kJ/kg. with ex of 1,31.
ed ‘steam at the exit pressure is 0, 6684 mi/kg. The
The specific volume of-dry sat ¢
f the actual dryness factor.
isentropic dryness factor is 98,9
Calculate:
5.1 The:specific. enthalpy and temperature of the steam at the throat
5.2 : ‘The specific volume, the area in mm? and diameter in mm at the throat
ie actual dryness factor, the specific volume, the area in mm2, the velocity in
id the specific actual enthalpy at the exit
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QUESTION 6
A velocity compounded impulse turbine has two rows of moving blades with a row of
fixed blades between them. The nozzle delivers steam at 660 m/s and at an angle of
17° with the plane of rotation of the wheel. The first row of moving blades has an outlet
angle of 18° and the second row has an outlet angle of 36°. The row of fixed blades
has an outlet angle of 22°. The mean radius of the blade wheel is 155 mm and it
rotates at 4.000 r/min. The steam flow rate is 80 kg/min and its velocity is reduced by
10% over all the blades.
64 Use a scale of 1 mm = 5 m/s and construct velocity diagrams for the: turbine
and indicate the lengths of all the lines as well as the magnitude on the
diagrams. /
6.2 Determine the following from the velocity diagrams:
6.2.1 The axial thrust on the shaft in N
6.2.2 The total force applied on the blad
irection.of the wheel
in N :
6.2.3 The power developed by the turbine in:kW
6.2.4 The blading efficiency
6.2.5 The average blade.y
QUESTION 7
In an ammonia vapour éo yression refrigerator, the pressure in the evaporator is
276 kPa and the ammionia at: ‘entry tothe evaporator is 0,12 dry while at the exit
is 0,91 dry. During: sompression the work done per kg of ammonia is 170 kJ. The rate
of ammonia circulatis i min
Assume the compressor is single-acting runs at 120 r/min with a volumetric efficiency
of 80%.
Properties of ammo
Laient heat ( (hg) “1 340 kJ/kg
7.1 The actual coefficient of performance (COP)
7.2 The volume of vapour entering the compressor per minute
7.3 The diameter of the cylinder and stroke if the stroke and the diameter are the
same.
TOTAL:
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100
(8190046) -1- T1310(E)(A2)T
POWER MACHINES N6
FORMULA SHEET
Any applicable formula may also be used.
GENERAL
PY, =mRT,
R=Cp-Cyv
_P
Cy
PV=c
x
PV" =c
PVi=c
hh. ny Ryn
TA A
AU=m.Cyv. AT
Q=AU +Wad
Pp, Va.) 7
As =m) Cy. tn 2 + Cp in 4
| y 2 'p |
As =m.Cv In 22
Nigy = hy + Cp Osu = ty
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(8190046)
GENERAL
Pigs = By + XP yy
-2-
~ su ~ 1941)
Va = a P,
a hathe
Ke
Vio=—a? xl
Py =P xP;
oo. Peat
Pas ~* B
T1340(E)(A2)T
Different formulae for
work done (Wd)
=PxAV
=F In
7;
PY = Poke
n-1
_ PV = Poko
y-i
=m.Cp.AT
wn
= PV,
n Ve
Xi
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\
(8190046) -3- T1310(E}(A2)T
GENERAL
Differem formulae for
work done (Wd)
= area of PV-diagram
= work done first stage
+ work done second
stage +...
Wd yoy =Wa, Wel
Waren = Qnett
Different formulae for
air standard efficiencies (ASK)
peal
rye
Ne, D4 e? Dies
heat added ~ heat rejected _
heal added
Different volumetric
efficiencies,
vol
Volume of air taken itt
Swept volute.
_ Voliine.of “free air. . _
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(8190046) “4. T1340(E)(A2)T
GENERAL
Different thermal
efficiencies, Oherm.
Wd
~ heat supplied
BP
mys x CV
"brake therm, =
IP
mpjyx CV
"ind. therm, =
_ mt, (hs - hw)
"therm myx CV
BP
IP
Indicated efficiency ratio
mech. =
_ Vind. therm.
ASE
Brake efficiency ratio
_ Novake therm.
ASE
pp aon
60
BP = Pyate mean LANE
IP =P, -LANE
ind, inéan
COP = of
2-47
cop = 8B
Wd
P=m.U.AVw
Fy =m AV;
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(8190046) “Be 7134 O(E\(A2)T
GENERAL
_2.U Ave
Ndia. = ye
a
-1
Pah 2. ,
yt
y+i
Cy = f2 x10? (A, = he) + Ch
Cy = 2x 10? (Fy — hy) + CE
C, = J2x10 XC, (Hh -2)+ Cr
Cy = [2x10 xC, = B)+CF
_ nv,
A, == A
2 G
Nrank, = 0.
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(8190046) -6-
r
Nearu, =} r
1
h=u+t pV
ce
eZ, +U, + BY, + +O=
C2
BZo +Uy + PoVy +e Wad
T1310(E)(A2)T
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Department: —
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
| EE
MARKING GUIDELINE
NATIONAL CERTIFICATE
APRIL EXAMINATION
POWER MACHINES N6
2 APRIL 2015
This marking guideline consists of 10 pages.
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MARKING GUIDELINE +2- T1280(E)(A2)T
POWER MACHINES N6
QUESTION 1
P(kPa)
P3= 1 500k
P2= 360
Py=
Incomplete intercooling:
2. maemnRT Ss
ener yn — a] + EYE — 1]
Be
= 13xdxb 2882904 12) is 4 pls Ax8x0 288x360 en _ 4]
1.1 _emkT A.
G.axéO $B 6,3260 360
= 6, 5778 + 8, 76 = 15,338 kW
Preq = = 18,04 kW m
1.2 Vp = SERTH , 320-2882902 — 9005528 m/stroke
PL 2u 240298
Vsq= Ye = “SE = 0,006 m?
& 0,82
Vs, 72 Le na Sd
0,006 = 1, 47809 xa
d=0,172 m= 172 mm (7)
1.3 a! 28
Ta = 11 (2) = 30102): = 406,4 K
Qi2= ea +mCv (T2-T1) & Cv = Cp -R=1,005 -0,288 = 0,717 kJ/kg.K
=. 3x0,286( 301 —306,4) + 320,717 (406 4-301}
2x2 40x03 2x230
= -0, 6324 + 0,472 = -0,1604 kJ (6)
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MARKING GUIDELINE -3-
POWER MACHINES N6
QUESTION 2
2.1
2.2
2.3
From steam table: @ 3 000 kPa: hf = 1 008 kJ/kg
hfg = 1 794 kJ/kg
he = 184kJ /kg = he = 741 kJ/kg hs = 2858 kJ/kg
hp = hf +xhfg = 1 008 +0,98 x 1 794 = 2 766 kJ/kg
Qec = ms (he ~ he) = 9, 5 (741 -184) = 5 291, 5 kJ
Qo = ms (hp — he) = 9.5 (2 766 — 741) = 19 237 kJ
Qsh = ms (hs hy) = 9,5 (2 858 — 2 766) = 874 kd
Qan = ma.Cp. At = 20 x1,005 x (150 — 15) = 2713 kJ
mig (esha)
mf x2 Cv
n=
35 (2858-1863
da 31800
79,88%
Heat to flues = heat Supplied -Heat absorbed.
m x Cp x At =CV — Ms (hs — hc) + Ma xCpaxAta
24 x 1,045 (texn -15) =31 800 -9,5 (2 858 ~ 184) + 20 x 1,005 x 15
66585
t =—— +156
exh 31,945
= 320,24 °C
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MARKING GUIDELINE -4-
POWER MACHINES N6
QUESTION 3
3.1
3.2
3.3
3.4
3.5
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Yo = TEE mot8)?x025 5 9 90636 m?
4 4 ;
V1 = Vs +Vc = 0,00636 +0,00041 = 0,00677 m?
_ 20 - 20
V3 = V2 + =.Vs = 0,00041 +7 x 0,00636
= 0,00092 m®
P3 = Py cay" = 400 (2oue7?) 4,28
v3 G,.G0092
= §147,13 kPa
_ yayte (0.00041) 1,38
Py=P,(2)> = 5147,13 (2ctett) 435 = 116,82 kPa
P3&VB-PaVe P2Y2—-PIYL
ne 1 no-
WD/Cycle = P2 (V3 —V2) +
= 5147, 13 (0, 00092-0, 00041) + 5147,13 x — w G,00677
5147,13 x 0,00041 — 116,82 x 0,00677
0,35
= 2,625 + 7,2406 -3,76986
+ WD/Cycle = 6,096 kJ
IP = WD /Cycle x Cycle/second
= 6,096 x x4
= 134,11 kW
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MARKING GUIDELINE 5.
QUESTION 4
41
4.2
4.3
44
4.5
4.6
AIP,
AIP,
A IPs
AIP,
AIPs
AIPs
POWER MACHINES N6
= 56-44,2= 11,8 kW
= §6-44 =12kW
= 66-43,9=12,1 kW
= §6-44,3= 11,7 kW
= §6-44,1 =11,9kW
= §6~- 43,7 = 12,3 kW
AIP Total = 71,8 kW
m/min =
O16 X 71.8
oa x 100%
IP
= 2 x 100% = 77,99%
ILS
MSEC X EP
60
= 0,311 kg /min
60
Mexn/min = me_/min x (A/F +1)
0,311 x (15,2 + 1) = 5,038 kg/min
Qexh = MexnX Cpx AT
= §,038 « 1, 1 (400 ~ 18) = 2116,97 kJ/min
Qw = My .Cpw. AT = 34 x 4,187 x 38 = 5409,6 kJ/min
T1280(E)A2)T
Heat gained ((kJ/min) Heat supplied (kJ/min) Percentage
Q= mr x Cv BP = 56x60=3360 25,72
QW = 5 409.6 41,41
= 0,311 x 42000 Qen = 2 116,97 16,21
Unaccounted = 2 175,43 16,65
=13062
TOTAL = 13 062 100
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MARKING GUIDELINE -6-
POWER MACHINES N6
QUESTION 5
6.1 (superheated tables) :h, = 2 925 KJ/kg @
1 500 kPa and 250 °C
Cc =,/2000(h1 — he)
cc? = 500 2
2000 2600
© hy=he =
=125 kJ/kg
he = hy-125 = 2925-125 = 2 800 kJ/kg
hg = 2 768 kJ/kg @ 820 kPa
he = hg + Gp (tsup —ts )
. tsup -t, = a = corre = 12,5 °C
tsup = ts + 12,5: (t,=171,4°C @ 820 kPa)
= 171,4 +12, 5 =183,9 °C
5.2 oe trl fko-i941\ _ 6.51 (2800-1991
V8up? ( Pe ) LEL ( 210 )
= 0,2479 mi/kg
Ag = M2 = Sk8S3" = 0,002479 m*
ec 500
= 2479 mm?
Ac = © De? =2479 = = De?
. De = 56,181 mm
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MARKING GUIDELINE -T-
5.3
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POWER MACHINES N6
Xa" = 0, 9895. Xz = 0,94
X22 2 = 0,95
a= 0,5855,
Vo = Xe x Vg
= 0,95 x 0,6 684 = 0,635 m/kg
Ae =f D? = g (72, 2)°= 4094, 155 mm?
C2 = ME = = 7758,5 mis
C2 =,/2000(h1 — h 2)
= fen? 778 57
hy —hz = yoo 7 neoo 300,7kJ/kg
hz = hy — 300,7 = 2 925 — 300,7 = 2 624,3 kJ/kg
(Vg = 0, 6 684, P = 270 kPa)
(hf = 546kJ /kg, hfg = 2174 kJ/kg @ 270 kPa )
hz = hf + xhfg = 546 + 0,95 (2 174)
= 2614,3 kJ/kg
C2 =,/2000(h1 — h2)
= {(2000(2925 — 2611,3)
= 792m Is
iV 2 = 58 G635
A= C2 Fag
= 4 008,4 x 10° m?
= 4 008,4 mm?
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T1280(ENA2)T
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MARKING GUIDELINE -8- T1280(E)A2)T
POWER MACHINES N6
QUESTION 6
6.1
Lat$s ota,
=
& : £
% / g x
ee £ i
A Nd pea
ee 8 3
ey YY a Ne}
| 0
oe, A =
i. y i
i]
WD tap,
apa
a
WEES Bb
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MARKING GUIDELINE -9- T1280(E(A2)T
POWER MACHINES N6
6.2 U = 2 = 7 CERES = 64,9 mis
Cret = kCrin = 0,9 x 120 = 108 mm also Cre2=kCri2=0, 9 x 74 = 66,6 mm
Caiz = kCae; = 0, 9 x 95 = 85,5 mm
6.2.41 Fava m(A Cf - ACfe)
_ 8007 F)x5
60
=ON
6.2.2 Frang= m (ACw; - ACwe)
= B0(218- 121). 45
6G
=2240N
6.2.3 P = FrangxU
= 2 240 x 64,9
= 145,38 kW
6.2.4 _ 2UC ACHR)
~ re a
__ Bubd,9(1078 4 6059
~ 6602
= 50,1%
6.25 U=64,9mi/s
(5x2) (10)
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MARKING GUIDELINE ~10- T1280(E)A2)T
POWER MACHINES N6
QUESTION 7
71 Refrigerating effect = (qi — 9 2) x hig
= (0,91 — 0,12) x 1 340
= 1 0586 kJ/kg
COP = Refrigerating effect
Work done
— 1052.6
170
=6,23 (6)
7.2 Volume at suction =0,44x5,6x0,91
= 2,242 m/min
Volume of compressor = Volume at suction
Volumetric efficiency
= 2,242
0,8
= 2,803 m/min (7)
7.3 Volume/stroke = Volume of compressor
Revs per minute
= 2,803 = 0,02 336 m?
120
Volume /stroke = = : L=D
0,02336 = we
D9 = 0,029741
D = 3/0,029741
D=031m=310mm (7)
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TOTAL: 100
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higher education
& training
cocn0000n0
= Depariment:
= Higher Education and Training
ES
REPUBLIC OF SOUTH AFRICA
sists of 8 pages and a formula sheet of 6 pages.
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(8190046) -2- 11280(E)(J28)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
NOTE: If you answer more than the required number of questions,
required number of questions will be marked. All work you dg,not
be marked must be clearly crossed out.
INSTRUCTIONS AND INFORMATION
1. Answer FIVE of the seven questions.
2. Read ALL the questions carefully.
3. Number the answers according to th
paper. :
stem used in this question
4. Write neatly and legibly.
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(8190046) ~3- T1280(E)(J28)T
QUESTION 1
The intake conditions for the low pressure cylinder of a three-stage, double-acting,
reciprocating air compressor running at 340 r/min are 103 kPa and 29°C.
The pressure in the second intercooler is 1 632 kPa.
The clearance volume is 4% of the cylinder volume.
The swept volume for the low pressure cylinder is 0,0084 m*.
The index for compression and expansion is 1,32.
Intercooling is complete and the stage pressures are in geometri
1.1 Calculate:
1.1.1 The stage pressure ratio, the
the delivery pressure in kPa : (3)
1.1.2
(6)
1.1.3 (3)
1.44 (2)
1.1.5
(2)
1.1.6 (2)
ic efficiency of the low pressure cylinder (2)
[20]
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(8190046) -4- T1280{E)(J28)T
QUESTION 2
The following information was obtained from a test on a steam boiler plant:
The rate of evaporation = 7 350 kg/h
The pressure in the evaporator and superheater = 3 000kPa
The temperature of the superheated steam = 250°C
The temperature of the feed water 8h 7°C
The temperature of the water at the economiser exit OE '6°C
The mass of the coal used 750 kg/h
The temperature of the flue gases at entry to the econo! 353,33°C
The air supplied per kg fuel burnt 18 kg
The specific heat capacity of the flue gases 1,045 kJ/kg. K
The thermal efficiency of the plant = 80%
The dryness factor of the steam ai = 0,975
2.1
2.1.2
2.2
the plant to determine the heat unaccounted for.
(5)
[20}
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(8190046) -5- T1280(E)(J28)T
QUESTION 3
A four-cylinder, four-stroke petrol engine with a piston diameter of 75,15 mm and a
stroke length of 90,18 mm was tested while running at 4 800 r/min.
The brake mean effective pressure was 1 250 kPa.
The brake specific fuel consumption was 0,288 kg/kW.h.
The mechanical efficiency was 80%.
The calorific value of the fuel was 39 062,5 kJ/kg.
The clearance volume was 5% of the swept volume.
The value of gamma was 1,41.
3.1 Calculate:
3.1.1 The swept volume of each cyli \
and the indicate mean effectiv 2 ° (6)
3.1.2 The brake torque in Nm, ~ (2)
3.1.3 The mass of the fur
the indicated ther (6)
3.1.4 The volumetric ci
the indic: (6)
{20]
QUESTION 4
Air flows through
pressure of 23
ure is 475,454 kPa, the area is 1 534,6 mm?, the actual
ind the velocity of the air is 398,801 m/s.
Igriore the velocity of the air at the inlet and calculate:
4.1.1 The absolute temperature and the pressure in kPa at the inlet (5)
4.1.2 The absolute isentropic temperature, specific volume in m*/kg at
the throat and the percentage heat loss before the throat (6)
4.1.3 The absolute isentropic and actual temperatures, the velocity in
m/s and the area mm? at the exit. Assume the specific volume to
be 0,4154 m?/kg (8)
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(8190046) 6- 71280(E)(J28)T
[20]
QUESTION 5
A single-cylinder, single-acting engine working on the diesel cycle rotates at 300 r/min.
It has a piston with a diameter of 187 mm and a stroke length of 233 mm.
Fuel cut-off occurs at 10% of the swept volume after T.D.C.
At the start of compression the pressure is 106 kPa and the temperature is,2!
The index (n) for compression and expansion is 7,32.
The cylinder volume is 17 times the clearance volume.
R for air is 0,287 kJ/kg. K.
5.1 Calculate:
5.1.1 The swept volume, the clearanee
the volume after combusti
Jinder volume and
(5)
6.1.2 The missing absol
principal points (8)
5.1.3 The mass of ai
w (4)
[20]
5.1.4
QUESTION 6
A carbon dioxide r i ant Operates between limits of -4°C and 28°C.
The specific heat capacity of the superheated refrigerant is 2,5261 kJ/kg.K and that of
the saturated liquid refrigerant is 5,42 kJ/kg. K.
6.1 The following are extracts from carbon dioxide tables:
Saturation Specific enthalpy (kJ/kg) Specific entropy (kJ/kg/K
temperature
eC) Liquid Vapour (hg) Liquid (sf) Vapour (sg)
(hf)
4 348.6 83
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(8190046) T- 71280(E)(J28)T
28 [1786 [263.8 | 0, 638 0,922
Calculate:
6.1.1 The specific entropy in kJ/kg.K after compression and the dryness
factor of the refrigerant entering the compressor
6.4.2 The specific enthalpy of the refrigerant at the inlet to the
compressor and the specific enthalpy after compressing in kJ/kg
613 The specific enthalpy of the refrigerant before thro
the refrigerating effect in kJ/kg, the work done in
actual coefficient of performance
6.1.4 The power required fo drive the comp
mechanical efficiency is 83%
HINT: Entropy after compression
at
saturated vapour
Taf trae
28°C + Cp. In
QUESTION 7
A two-stage, velocity-compounde:
moving blades separated by
71 ‘he length of 35 mm for the average blade velocity and construct velocity
diagrams for the turbine in the answer book and calculate the scale. Indicate
the lengths of ALL the lines as well as the magnitude of the angles on the
diagrams.
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(8190046) -8- T1280(E)(J28)T
7.2 Determine from the velocity diagrams:
7.2.1 The average blade velocity in m/s
7.2.2 The nozzle velocity in m/s
7.2.3 The velocity of the steam leaving the first stage in rn/s
7.24 The velocity of the steam entering the second stage i
7.2.5 The velocity of the steam leaving the turbine in m/,
7.2.6 The nozzle angle
7.2.7 The energy developed by the turbine in k.
7.2.8 The blading efficiency
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100
(8190046) Ae T1280(E)(J28)T
FORMULA SHEET
Any applicable formula may also be used.
ENGLISH GENERAL AFRIKAANS
PV, = mY,
R=Cp-Cv
Cy
Y
Q=AU + Av
Q=m.Cp.Ar
Q=m.Cv. AT
AY
sit
= 8, +p In
5
Sy =Sg— Sy
S=S;,+ wy
A
SU
= hy + CP su -t,) PTO
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(8190046) ~10- T1280(E)(J28)T
-2-
ENGLISH GENERAL AFRIKAANS
PAN 1941)
Ins = hy + xh yy Vy = —_____—-
Poy
Vy HXV,
ws g
Different fornnilae for
work done (Wd)
a-l
mn
=—— MRE | (r,s) " 1
prada (ps)
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(8190046) -11- T1280(E)(J28)T
3.
ENGLISH GENERAL AFRIKAANS
Different formulae for Verskillende formules
work done (Wd) vir arbeid verrig (Av)
= area of PV-diagram = area van PV-diagram
= work done first stage = arbeid verrig eerste
+ work done second stadium + arbeid ver
Stage +... rig tweede stadium
Waren = Wa; ~ Wa,
Wayou = Qrete
Different formulae for
air standard efficien-
cies (ASE)
erskillende formules
ugstandacrdrende-
anente (LSR)
“Gp < D+? Ge DI
heat added — heat reject 1 _ warnite toegevoeg — warnite afgestaan
¥ ( B _ a9) warmte toegevaeg
Verskillende volumetriese
rendemente, Ovot
__ Volume lug ingeneem
Slagvolume
_ Volume vrylug
Swept volume ~ Slagvolune
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(8190046) 19.
ENGLISH GENERAL
Different thermal
efficiencies, Onerm.
Wd
. heat supplied
BP
mets x CV
"ovake therm. >
IP
T1280(E)(J28)T
AFRIKAANS
Verskillende termiese
rendemente, Germ
7 Ay
warmie loegevoeg
RD
my) XW
Trem term. =
Find. therm, = mypiax CV Tin. term, =
m, (hs — hw)
Therm, =
mp x CV
h- %-f,
rs ya
=F 1-1,
BP
Tmech. = 1p
Indicated efficiency ratio i kateurrendemeniverhouding
- Vind. therm. "ind. term.
ASE LSR
Brake efficiency ratio
Remrendementverhouding
= "brake therm. = "rem. term.
ASE LSR
BP = 29 EN , RD =n
60 60
BP = Porate mean RD = Prom gem. LANE
IP = Fya ID= Prd. gen. LANE
BSFC
qT,
cop =—!
27h
cop - Fé
Wad
P=m.U.AVw
Fa, =m AV,
ISBY = “bl
D
RSBV = Meih
RD
if
f-f,
KVW = VE
Ay
D=m.U. Aw
KYW =
ai
Fags. =m. AV ¢
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(8190046) -5- T1280(E)(J28)T
ENGLISH GENERAL AFRIKAANS
2.U .A¥w
Ndia. = we
Niso,
We
rank, = Oo Tank. = re
L;
=1-2
Nearn. va
h=u+ pV
.
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(8190046) -6-
C2
gZ, +U, + BV, + +O
C2
BZy +U, + PyVy +h + Wd
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T1280(E)(J28)T
C2
82, +0, + BY +S +=
C2
B2y +Uy + PP, +t Ay
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE
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NATIONAL CERTIFICATE
AUGUST EXAMINATION
POWER MACHINES N6
28 July 2014
This marking guideline consists of 9 pages.
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MARKING GUIDELINE -2- T1280(E)(J28)T
POWER MACHINES N6
QUESTION 14
t 3)
P (kPa) rs 4
1414 k= E = [ie = 3, 98 HP
1632
Po=Pixk = 103 x3, 98 = 409, 94 kPa
P4=P3xk= 1632 x 3, 98 = 6495, 36 kPa
103 .
11.2 Vi = Vet Ve= Vs +0, 04V1
0, 96 Vi = Vs
a Wie SE = 0, 00875 m?
Ve = 0, 04 Vi = 0, 04 x 0, 00875 = 0, 00035 m?
Va= Veta as =0, 00035 Sey zr = 0, 00099 m?
*.Ve = Vi~ Ve = 0, 00875 - 0, 00099 = 0,00776m? (6)
14.30 P= 22 Pe [(P2/P1) ™ - 1] x2x=
= 44,586W (3)
1.1.4 T2271 (PofPs) ™™ = 302 (3, 98) 2924.82 = 422, 12K, (2)
14.5 Qintereooter = MCp (Tz = Ti) == x 1,008 (422,12 ~ 302)
= 11,267 kJis Q)
14-6 Quateriaciet= $-Qintercooter = “M4 - 11,267 = 2,513 kdl,
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MARKING GUIDELINE -3- T1280(E)(J28)T
1.1.7
QUESTION 2
2.1.1
2.1.6
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POWER MACHINES N6
4.4.5 "yi = 22x 100%
BOOT 76
= oosy X 100%
= 92,4 %
At37,7°C =: hfwi = 158 kJ/kg
At 99.6°C > hfwe = 418 kJ/kg
Qeeon = 22 (hfwe-hfws) = 222 (418 — 158) = 9, 8 (260)
= 3548 kJ/kg
At 3000 kPa : hf = 1008kJ/kg & hfg = 1794 kd/kg
Qevap = 2 (hf — xhfg —hfw2) = 9, 8 [1008 + 0, 975 (1794) — 418]
= 22923, 67KS
At 3000 kPa & 250°C: hss = 2858 kJ/kg
Qe =22. (hss — xhfg ~ hf) = 9, 8 [2858 - 0,975(1794) - 1008]
= 988, 33 klilkg
me (hes—hfivi)
mf x Or
n=
=ms (hss _ htws) = 98(2958 -158) = 5,6 (27003
ae Be
mf x"
= 33075 kJékg
EE =ms(hss —hfw:)= 9,8 ( 2858 -158) = 9.8 (2700) = 11, 724
mf x 2257 2257 2257
Qecon = mCp (tin — tout )
tor = tin - ESP = 353,33- 2 _ = 353,93 - 128,33
(1842) 1,045
= 225°C
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(2)
(3)
(3)
(3)
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71280(E)(J28)T
MARKING GUIDELINE -4-
POWER MACHINES N6
2.2
Qin (kJ/kg) Qou (kJ/kg) %
33 075 Economiser 2548 | 7,704
Evaporator 22923,67 | 69,308
Superheater 988,33 | 2,988
Unaccounted 33075 — 26460 = | 20
6615
33.075 33.075 | 100
QUESTION 3
3.1.1 Vs=2 D?L == (0, 07515)? x 0, 09018
= 0, 0004 tn?
BP = PLACE = 1250 x 0,0004 x 4800 x 4
60 60 x2
= 80 kW,
Pim= Pom= 1250
n 0,8
= 1562, 5kRa
3.1.2 BP =.2n1N
60
T= 60 x BP = 60 x 80 x 10°
2nN 2x w x 4800
=159, 155 N.m
3.1.3 BSFC = mf/h
BP
mfls = BSFCx BP = 0,288 x 80
60 x 60 60 x 60
= 0, 0064 kg/g
ng = BP = 80. x 100%
mfisx CV 0,0064 x 39062,5
= 32%
UT _ 32 _
"r =a 3p 7 10%
3.1.4 Vetve _ vet0,0Sve _ L0bve =
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MARKING GUIDELINE
POWER MACHINES N6
AS E=[1= (1) x 100 % = [1 — (1/21) °4"] x 100%
= 71,3%
NT — 04
Ind. ratio = =x 100% = Trag X100%
= (6)
= 56, 1% [20]
QUESTION 4
41.4 Co = {2000 Cp (Ti — F2)
Ty = 82 4Te = MO * + 404, 11 = 78, 89 + 404, 11
2660Cp 200041,008
= 483K
:
Be 2 2 ay
pi ~ Cw
ra 24-4
Py = Po (e=)r-# = 475,454(F)e* = 900KPa (5)
4.1.2 ToT Cree) = 483(=) = 402,5K
Vow “= 1534.6 x 10° x398,801x60_ = 0,2448 mike
150
a= (1 oT) x 100%
_ 78,89. _
= (1-ZE5) x 100% = 2%, 6)
ot
you 2
4.1.3 T= TH) 7 = 4832) 4 = 330K
Tz = 1; -"(T1. 72’) = 483-0, 92(153)
= 342.24
Co= 4/2000 Cp (Ti —T2)
= +f 2000x1,008 (483 — 342,24)
= 582,703 m/s y
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MARKING GUIDELINE -6-
POWER MACHINES N6
mV2 _ 150-x0,415dx106
cz 60x53:-2,703
= 1949,492 myn?
Ag=
4.1.4 — Mach no = = 22288 = 1 336
Ge 398,301
QUESTION 5
5.4.4
P(kPa) N
Vs =% D2L =2 (0, 187)2x 0, 233 P2=Po 2 3
= 0, 0064 m&
Ve=Vi — Ve= 17Ve -Ve
Ve =16Vo= 0, 0064
106
Vo=V2= 0, 0064/16 = 0, 0004 m3
T1280(E)(J28)T
(8)
(1)
[20]
VieVa= 17Vo= 0, 0068 ms,
Va = 10%Vs + Vo
= 0, 1(0, 0064) +0, 0004 = 0, 00104 m?
5.1.2 P2=P3=P, S" =106(17)'*? = 4461, 68 kPa
Ta2Ts ("> = 205,5(17)99? =731 646 K,
VaO.T2 _ 0,001042731.646
Ta yr Feocos 1902.28. K.
Pa=Pa(a)" = 4461,68(0° =)? = 374,171 kPa
0.0068
000104.
Ta#Ts (2)*t= 1902, 28¢2*)** = 1043,001 K,
5.1.3 = BLE Vi 106 x0,0068
RTL ~ oy 287 x29 5.5
= 0,0085 kg/cycie.
300
= 0, 0085 x =~ =0, 0425 kg/s
=».
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MARKING GUIDELINE 7- T1280(E)(J28)T
POWER MACHINES N6
5.1.4 WD = mRI(Ts _ T2) + ae
a-l
= 0, 0425x0, 287[(1902, 28 - 731,646)
+ 1902,28-1043.091—731,646 +295,5.
0.32 |
= 0, 0425x0, 287[1170, 634 +1322, 009]
___...=.30,404 kW I 401
QUESTION 6
6.1.4 (5)
T
S2= $1 =S3 + Cp.In= 325,8
7 301
= 0,922 + 2, 5261.In=
a1
286 |
=1,122 kJ/kg ,
S1-S7 1422~0,912 08 | 4
x= S&-87 = qaestou2 = 0, $3 | \
6.4.2 hy=h7 +x1(hs-h7) | id
0,312 0,638 0,922
= 81,4 +0,93(318,6-81 4)
= 81, 44220, 6
= 302 kJfk
ha= hg+Cp(T2 — Ts) = 263,8 +2,5261(325,8-301)
=263, 8 + 62,447
= 326,447kJ/kg
6.1.3 he = ha — Cp(T4-Ts) = 178,6 — 5,42(301-296)
= 178, 6 -27,1
= 151, 5 kJ/kg
RE =h1-he=302 -151, 5=150,5kJ/kg
WD = he-h1=326,447-302=24,447kJ/kg
COP=S == = 6,156
WD 24.447
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MARKING GUIDELINE -8- T1280(E)(J28)T
POWER MACHINES N6
__ mir(h2-h1)
6.1.4 Pm =e
_ 18424,447)
60x0,83
= 8,836 kW
ae (2)
[20]
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MARKING GUIDELINE -9-
POWER MACHINES NG
QUESTION 7
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T4280(E)(J28)T
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MARKING GUIDELINE
7.2 7.2.1
7.2.2
7.2.3
1.24
7.2.5
7.2.6
7.2.7
7.2.8
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POWER MACHINES N6
U = 35x4 = 140 m/s
Vi = 148,5x4 = 594 mis
V2 = 78,7x4 = 314, 8 m/s
V3 = 74x4 = 296 mis
V4 = 17,5x4 = 70 m/s
ai=19,3° .
P = mu (Vwi +Vw2)
=1x140(204+7 1) x4
= 154 kJ/kg
n= 2UVwi/V:? x 100%
= 2x140x1100/594? x 100%
= 87,3%
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T1280(E)(J28)T
TOTAL:
(10)
[20]
100
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
oooooga000
T1280(E)(M31)T
APRIL EXAMINATIO
NATIONAL CERTI
onsists of 7 pages and 1 formula sheet of 5 pages.
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(8190046) -2- T1280(EKM31)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1. Answer any FIVE of the seven questions.
2. Read ALL the questions carefully.
3. Number the answers according i¢ he rity m used in this question
paper.
Write neatly and legibly.
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(8190046) -3- T1280(EKM31)T
QUESTION 1
The volumetric expansion ratio of an engine working on the dual cycle principle is
1:96.
The clearance volume is 6,25% of the cylinder volume and the swept volume is
4 500 cm’.
At the beginning of the adiabatic compression stroke the conditions indicate a pressure
of 86 kPa and a temperature of 18 °C.
The pressure after adiabatic expansion is 263,74 kPa.
Take gamma = 1,4, Cp = 1,008 kJ/kg.K and Cv = 0,72 kJ/kg. K.
Caiculate:
1.4 The cylinder volume, clearance volume and the
expansion in cubic centimetres (5)
1.2 All the missing absolute temperatures and the principal
points (9)
1.3 The heat received and the heat reje (4)
1.4 The thermal efficiency of the cycle (2)
[20]
QUESTION 2
Air enters a convergent-d i ‘at-A pressure of 2 800 kPa, a temperature of
450 °C and a velocity ofser
The diameter at the, 51mm.
The pressure at the exit i Pa.
aw is frictionless and the efficiency of the divergent part is 96%.
(3)
2.2 pressure in kPa, the absolute temperature, the velocity in m/s, the
ic volume in m/kg and the area in mm? at the throat of the nozzle (8)
2.3 The mass flow rate of the air in kg/s (2)
24 The absolute isentropic temperature, the absolute actual iemperature, the
velocity in m/s and the specific volume in m*/kg at the exit of the nozzle (7}
[20]
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(8190046) -4- 71280(E)(M31)T
QUESTION 3
A two-stage, single-acting, reciprocating air compressor delivers air to a receiver at a
pressure of 1 200,5 kPa.
The low-pressure cylinder has a piston with a diameter of 245 mm and a stroke length
of 350 mm.
The volumetric efficiency of the low-pressure cylinder is 90%.
The temperature at the beginning of the compression stroke for the low-pressure
cylinder is 30 °C.
The pressure in the intercooler is 343 kPa.
The compressor rotates at 355 r/min and the mechanical efficiency is 85'
The index for compression and expansion for both stages is 1,3.
Intercooling is complete for maximum efficiency.
Take R for air as 0,287 kJ/kg.K and Cp as 1,005 kJ/kg. K.
Calculate:
3.1 The swept volume and the effective s
cylinder in cubic metres (4)
3.2 The pressure before compressio: (2)
3.3 The work done in kW for the.
compressor in kW (4)
3.4. The mass of air delivered by (2)
3.5 (2)
36 (2)
3.7 (2)
3.8 i [ p olume of the high-pressure cylinder in cubic metres per
) 3
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(8190046) -5- T1280(E)(M31)T
QUESTION 4
A boiler plant burns 5 750 kg of coal with a calorific value of 33 MJ/kg per hour.
It generates superheated steam at a pressure of 3500 kPa and a temperature of
350 °C.
At the exit from the boiler the steam is 97% dry with a pressure of 3 500 kPa.
The feed-water temperature is 24,1 °C and at entry to the boiler the water has a
temperature of 83, 7 °C.
The temperatures at the entrance and the exit of the air preheater are 25..°C and
150 °C respectively.
The thermal efficiency of the plant is 80.12%.
The air fuel ratio is 20: 1.
The economiser receives 2 200 kJ/kg of heat.
The specific heat capacity of air is 1,005 kJ/kg.K.
41 Calculate by using steam tables only:
4.4.1
4.1.2
(4)
41.3 (2)
414
(4)
(7)
[20}
oke'eight-cylinder diesel engine was tested on a dynamometer
wing information was obtained:
=7,1 MW
= 45,5 MJ/kg
= 25:1
= 39, 01%
The mechanical efficiency = 80%
The fiow rate of the cooling water = 1 200 kg/min
The temperature rise of the cooling water = 65°C
The temperature of the exhaust gases = 423°C
The ambient temperature of the air = 18°C
The specific heat capacity of the exhaust gases = 1,05 kJ/kg. K
The specific heat capacity of the cooling water = 4,2 kJ/kg.K
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(8190046) -6- T1280(E)(M31)T
Calculate:
5.1 The brake torque in KN.m and the indicated power in MW (4)
5.2 The mass of fuel used in kg/s (2)
5.3 The brake thermal efficiency (2)
5.4 The indicated specific fuel consumption in kg/MW.h (2)
5.5 The mass of the exhaust gases in kg/min (2)
5.6 Draw up a table of the heat balance in MJ/min and percenta
the percentage heat loss. Assume that the heat to fri i ie
heat consumed by the cooling water. (8)
[20]
QUESTION 6
The average diameter of the moving blade: velocity-compounded,
impulse gas turbine, rotating at 2 107,5 r/
The inlet and outlet angles for the first ror re 25°.
The relative inlet and outlet velocitie moving blades are 560 m/s
and 532 m/s respectively. ;
The outlet angle for the second row
turbine using a scale of 1 mm = 5 mis.
well as the magnitude of the angles on the
(10)
velocity in ms
ocity of the gas entering the fixed blades in m/s
he velocity of the gas leaving the fixed blades in m/s
The velocity of the gas leaving the turbine in m/s
The angie at the entrance to the second row of moving blades
6.2.6 The angle at the entrance to the fixed blades
6.2.7 The angle at the exit of the fixed blades
6.2.8 The relative velocity at the entrance of the second row of moving
blades in m/s
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(8190046) 7- T1280(EM31)T
6.2.9 The relative velocity at the exit of the second row of moving blades
in m/s
6.2.10 The diagram efficiency
(10x14) (10)
[20]
QUESTION 7
A methyl chloride refrigerating plant operating between pressure limits o
672 kPa is tested and the following information was obtained:
Temperature of the refrigerant in the condenser
Temperature of the refrigerant in the evaporator
= 110,2 kJ/kg
= 51,6 kJ/kg
= 0,168 m’/kg
t vapour. It enters the condenser as
a dry saturated vapour anc
undercooling.
The actual coefficie 90,23% of the ideal coefficient of performance.
The compressor has ‘diameter of 128,5 mm, a stroke length of 1,2 times
the diameter etric efficiency of 90% and it rotates at 420 r/min.
Calculate
(3)
nd the dryness factor (5)
The swept volume of the compressor in m’ per stroke, the swept volume and
the effective swept volume in m*/s (5)
74 The specific volume of the refrigerant at the entrance to the compressor in
m’/kg and the mass flow rate in kg per minute (4)
7.5 The compressor power in kJ/s and the power required to drive the
compressor in kW if the mechanical efficiency is 80% (3)
[20]
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(8190046) -1- T1280(E)(M31)T
POWER MACHINES N6
FORMULA SHEET
Any applicable formula may also be used.
ENGLISH GENERAL AFRIKAANS
PY, =mRT,
R=Cp-Cyv
_@
vey
PV=c
PY" =c
PV’ =c¢
O=AU +Wd O=AU + Av
“yen Pv cp itn Yo
R y,
m.Cy tn 22
1
As=m.Cp-n 2
¥,
As=m.R.tn Zh
Py
OQ=m.Cp.AT
Qz=m.Cv. AT
’ T,
Sou = Sg + IA
Ss
Spe = Sg —Sp
S=Sp+xSjy
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(8190046)
ENGLISH
Figg = hy + Xhyg
«2- T1280(E)(M31)T
GENERAL AFRIKAANS
a~l 1941)
A
Vou = P, Ays = hy + xh
rotiine Vag = Vy
Vi,=ia@ xh
P, = JP xPy
tm = 4 a
1
Different formulae for
work done (Wd)
illéide formutes vir
id verrig (Av)
n-l
xn a
= mRT, | Gps) ™ 1
n-l 1 f ‘os)
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(8190046)
ENGLISH
Different formulae for
-3- T1280(E)(M31)T
GENERAL AFRIKAANS
Verskillende forniules
work done (Wd) vir arbeid verrig (Av)
= area of PV-diagram = area van PV-diagram
= work done first stage = arbeid verrig eerste
+ work done second
stage +...
Wa nen = Wa, — Wel,
Wahren = Qnett
stadium + arbeid ver-
rig tweede stadium +
Different formulae for
air standard efficien-
cies (ASE)
heat added — heat rejecte
warmie loegevoeg ~ warmte afgestaan
heat added
warmte loegevoeg
Different volumetri
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Verskillende volumetriese
rendemente, Quot
_ Volume lug ingeneem
Slagvoluie
_ Volune vrylug
Slagvolume
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(8190046) 4. T1280(E)(M31)T
ENGLISH GENERAL AFRIKAANS
Different thermal Verskillende termiese
efficiencies, Onerm. rendemente, Qerm,
= Wa _ Av
~ heat supplied warmte toegevoeg
: _ BP _ RD
Novake therm. = mys XxCV Trem term. = Mb/s X ww
IP
Vind. therm. = mppy XCF Nind term. =
m, (hs — hw
iherm. ~ ims (hs — Ine) "erm.
my x CV
B-G T; -T,
Ne = 21 = 3 4
h-f B-t,
_BP
Mnech. IP
Indicated efficiency ratio
_ Mind. therm.
ASE
Brake efficiency ratio
Remrendemeniverhouding
- "ovale therm. = Trem. term.
ASE ESR
BP =2n IN RD=24 IN’
60 60
BP= Pyrake mean LAN RD = Prem gem. LANE
IP = Find. mean LANE, ID = Fd. gem, LANE
ISBV = Moin
ID
RSBY = Moth
RD
cop =." kewl
n-h Ti
COP = RE KVW = VE
Wa Ay
P=m.U .AVw D=m.U.AFw
Fy =a. A¥ Fats, =m. Vp
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(8190046) -5- T1280(E)(M31)T
ENGLISH GENERAL AFRIKAANS
2.U.APw
Ndia. = 7
an
=]
xy +1
Cy = 2x 10° (A —h,) + CP
Cy = 2109 (Fy ~ hy) + CP
EV = Ms As ~ hy)
My X 2257
Aviso
We = :
fo AY poh,
Av
Nrank, = oO
C c?
8% +U, + BR ++ O= 82 + Uy + A +>- +0 -
C3 Gy
#1 +02 +h + DS ivnloaded from hiayisé-coim *”
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE |
NATIONAL CERTIFICATE
APRIL EXAMINATION
POWER MACHINES N6
31 MARCH 2014
This marking guideline consists of 9 pages.
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T1280(E)(M31)T
-2-
MARKING GUIDELINE
POWER MACHINES N6
QUESTION 1
1.4 P(kPa)
Ve 2Vi ~Ve = Vi - 0,0825 V;
4.500 cm? = 0, 9375 Vs
VieV5 = = 4 800 om?
Vo=V2 =V3= 0, 0625 x 4800= 300 cm3
WS
W422 = 2% = 500 cms
36 9.6
V2=V3
vs= vy v(m)
(5)
12 py= py (42)"=86 (222) = 86 (16)!4= 4171252 kPa
Te= (C2) =291 (16) 94 = 882,147 K
Ts = Sets — 2OKTEX71 _ go2,403 K
f1 86
-1
TeTs (£2)" “= 892,423 (9,6) °4= 2 205,359 K
Ps=Py = Ps (S2)" = 263,74 (9,6) 1,4= 6 256,85 kPa
a Vb. TF — 9800 * 2205,359 _
Ty= SEE = MOUSSE = 1 393.215 K ©)
1.3 Qhree = mCv (Ts-T2) + mCp (Ta -Ts) }
= 1x 0,72 (1 323,215 - 882,147) + 4 x 1,008 (2 205, 359 - 1 323,215)
= 1 206,77 kdikg
Qe = mMCv (T1 =T5) = 1x 0,72 (291 - 892,423) = -433,025 kJ/kg (4)
14 rem = SES x 100% = AA E x 100% = 64,117% 0)
[20]
QUESTION 2
24 Cp=R+ Cv = 0,287 + 0,718 = 1,005 ki/kg.K
. §P _ 1,005 _
Y= a oe 714 (3)
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MARKING GUIDELINE 3. T1280(E)(M31)T
POWER MACHINES N6 i
2.2 _ 2a. 2.5
Po =P \Gareh # = 2800 (57)
= 1 479,189 kPa
= 602,5 K
Cy =,/2 #103 Cp(TI- Te) = 2000 x 1,005 (723 — 602,5)
= 492, 143 m/s
Vo =f = SEES = 0,1169 miikg
Ac = = de® = 7 (51)? = 2 042,821 mm?
a @)
23 m = ASSe = ROBIN HEE EE 18
¥o 01469
= 8,6 kg/s (2)
24 Ty.) = 723 CRE) = 4035 K
2806
4 = Te-T2
Te- T2
To =Tco—"®(Tc— Te’ ) = 602,5 - 0,96 (602,5 - 463,5) = 602,5 - 0,96 (139)
= 469,06 K
Co = (2000 Cp (71 — F2) = «(2000 x 1 ,005(723 — 469 06)
= 714,436 mis
Vo= RT2 . 9287 x469,.06 _ = 0, 2279 mkg
B2 590,674 (7)
[20]
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MARKING GUIDELINE -4- T1280(E)(M31)T
POWER MACHINES N6
QUESTION 3
3.4 P (kPa)
1 200,5 4 3
Vs= ; DAL =F (0,245)? x 0,35 HP
343 2
es
Vei = “vol x Vs = 0,9 x 0,0165 \\ 1
| |* Ve -vi-ve >|
= 0,01485 m® >
V5=Vr, vim?)
3.2 k= 12005 235
B2 3 ,
Ps 22 = 2 = 98 kPa
k 3,5
3.3 BBBIVel 7 ¢ PL. zt - MN
Pos [aD a
_ 2x8 n98 2001885 343 7 356
~ 0.3 es ys - 1) x
= 25,016 kW
Pa= 2 = SS = 29,431 kW
3.4 P, Ver = MRT1
_ Pre MN _ 59880,01485 355
"an *% g oa? a soa Keo
= 0,099 kg/s
3.5 Poy aaa\e
eT (2) n = 303 eS } 5
= 404,573 K
3.6 Qin. = mCp (T2—T1) = 0,099 x 1,005 (404,573 - 303)
= 10,106 kJ/s
3.7
Qu = (% — @ interc )x2 = (Bes — 10,106 x2
(12,508 - 10,108) x 2
4,804 kJ/s
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(4)
(2)
(2)
(2)
(2)
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MARKING GUIDELINE -5-
POWER MACHINES N6
T1280(E)(M31)T
3.8 Ve= mR
P2
_ 0,095 x60 x0,267 x 404,873
~ S55 “4d43
= 5,664 x10 * m3 (2)
[20]
QUESTION 4
41 At 24,1°C:hiws =101 kJ/kg
At 83,7°C: hfw2 = 351 kd /kg
4.1.1 .
Qecon = = (hfwe - hfwi}
_ Qeconx mf 2200x8750 _ 2200 25.750
mS * ewe —-hfwi ~ ssi-101SSCS*«S
= 50 600 kg/h (3)
4142 ne mip (hee—hbfwi)
mfx Cc¥
Res = BES + fis
= 08012 5750 489000 5 yay
s0600
= 30045 + 101 = 3 105,5 kJ/kg (4)
41.3 Qairpreneat = mCvAt
= 20 x 1,005 (150 - 25) = 20 x 1,005 (125)
= 2512,5 kJ/kg (2)
4.4.4 At3 500 kPa hf=1 050 ku/kg &hfg = 1752 kJ/kg
n — ma(hftxhig-hhvi} _ 56600 x [105040,97 (1F52)-1023
~ mix CY S750 x 33000
= 8,B (1050 +1699 44-101 = B.8x 2648 42 x 100%
33000 33600
= 70,625% (4)
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MARKING GUIDELINE -6- T1280(E)(M31)T
POWER MACHINES N6
4.2 Qin (kJ/kg) Qout (kJ/kg) %
33 000 Qecon = 2200 6,667
Qeyap = 8,8 [1 050 + 0,97 (1752) - 351] =21 106,27 | 63,958
Qeup = 8,8 (3 105,5 - 101) - 2 200-21 106,272 =3 133,328 | 9,495
Qairpreheater =2512,5 7,614
Unaccounted = 33 000 - 28 952,1 = 4 047,902 | 12,266
33 000 = 33 000 100
(7)
[20]
QUESTION §&
51 Bp = 2X
60
To Sor . Sara - 9995 kN
2nN inn iso ' .
®mech = 52 x 100%
Ip = 22. = 2422" - 9.875 MW
mec Ge , (4)
5.2 ps ath
icv
mffs = Ip. 887B x 1o®
Txey 08901 x 45500
= 0,5 kgfs (2)
5.3 Set = ar = See OR =x * mec
= 39, 01 x 0,8
= 31,21% =31,21% (2)
54 _omffh _ 0,560 60
SFO = = e765
= 202,817 kg/MW.h (2)
55 mg =mf (25+ 1) x60 = 0,5 (26) x 60
= 780 kg/min (2)
5.6 [Gin (MJ fin) Qou (MJ min) %
45,5x0,5x60 | BP =7,1x60 = 426 31,21
= 1365 Exhaust = 780 x 10°x 1,05 (423 -18) = 331,695 | 24,3
Coolant = 1 200 x 10% x 4,2 x 65 = 327,6 24
Unaccounted = 1365 - 1093,485 = 279,705 20,49
1 365 = 1365 100
(8)
[20]
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MARKING GUIDELINE -7-
POWER MACHINES N6
QUESTION 6
6.1
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T1280(E)(M31)T
(not to scale) (10)
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MARKING GUIDELINE -8- 71280(E)(M31)T
POWER MACHINES N6
6.2 U = TDN = mx 145% 2107.5 |
60 60 |
= 160 m/s
6.2.1 V1 = 141,6 x5 = 708 mis
6.2.2 V2= 78, 6x5 = 393 ms
6.2.3 V3=74,6x5= 373 mis
6.2.4 V4=21x5=105 m/s
6.2.5 63 = 30,6 °
6.2.6 pr=35°
6.2.7 a2 = 18°
6.2.8 Vr3 = 45, 3x 5 = 226.5 mis
6.2.9 Vre= 43 x 5 = 215 mis
6.2.10 An ene) x 100%
3 xt60 (9564388) 400%
?
2x 160% 1375
aes x 100%
708?
= 87,78% J (10)
[20]
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MARKING GUIDELINE -9-
POWER MACHINES N6
QUESTION 7
74 Ideal C.0.P = 7h.
_ 267
~ 30g-767
= 2
37
= 7,216
Actual C.0.P =x C.0.P
= 0,9023 x 7,216
= 6,511
T1280(E)(M31)T
81,6 410,2 Hr be2dire news) (3)
720 Cops =
C.0.P x (he— hn) = ht — ha
6,511 x (479 ~ he) = hy - 110, 2
3 118,769 + 110, 2 =hi + 6,591 hy
= 429,9 kJ/kg
hy = hs + x1(h4 — hs)
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MARKING GUIDELINE -10- T1280(E)(M31)T
POWER MACHINES N6
730 yet! 2 202x420
4
_ifxrD® _1,2x mx (41285 )%
~ 4 $
= 0,002 m/s
Vs = ee = 0, 014 m/s
Ve =VsX%wi =0,014x 0,9 = 0,0126 m¥s
7.4 Wa =x. - Var = 0,92 x 0,168 = 0,155 m/kg
m, = “8 = SSE x 60 = 4,88 kg/min
BS
05 Pe = mr (he- ha) = S88 (479 - 429,9) = 222 (49,1)
66
= 4kJ/s
— Pe 4
Pm = = yg TOkW
3 mech
TOTAL:
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(5)
(4)
(3)
[20]
100
|
i
|
|
i
;
|
i
i
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
TI460(E)(N21)T og
NOVEMBER EXAMINATION
‘This question paper consists of 7 pages and 1 formula sheet.
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(8190046) -2- T1460(E)(N21)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
NOTE: If you answer more than the required number of question
required number of questions will be marked. All wol
be marked, must be cleariy crossed out.
y
not. want'to
INSTRUCTIONS AND INFORMATION
4. Answer any FIVE questions.
2. Read AL the questions carefully.
3. Questions may be answered in.any. or ‘but subsections of questions must
be kept together.
4, Number the answers acedrding to. numbering system used in this question
paper.
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(8190046) -3- T1460(E)(N21)T
QUESTION 1
The following heat balance was drawn up after a 60 minute test on a single cylinder, 4-
stroke oil engine:
Q in fuel Q absorbed (kU/min)
(kJ/min)
4 800 Brake power 2 680
Friction power = 420
Cooling water = 1176
Exhaust gases = 1320
Unaccounted = 204
4 800 4800
The following information was also noted during the test:
Swept volume of engine
Total number of revolutions
Length of indicator diagram
Spring number
Cooling water used
Calorific value of oil used = 45 Muikg
Temperature rise of exhaust gases = 382°C
Length of pistol stroke = 1,21 x diameter
Specific heat capacity of cooling water = 4,2 kJikg.K
Specific heat capacity of exh 4,045 KJ/kg.K
Caiculate:
14
1.2
The diameter of the piston and the length of the stroke in mm.
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(10)
(3)
[20]
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(8190046) -4- T1460(E)(N21)T
QUESTION 2
A velocity compounded, 2-stage, impulse turbine is designed to discharge the gas
axially.
The inlet and outlet angles for the fixed blades are 36° and 17° respectively.
The relatively velocity at exit from the second row of moving blades is 1,125 times the
biade velocity.
The velocity of flow at the inlet to the first stage is 1,45 times the blade velocity...
The gas leaves the first stage with a relative velocity of 360 m/s. 4
The velocity coefficient for all the blades is 0,94.
The gas flows at a rate of 40 k/s.
2.1
the diagrams. Calculate the scale.
2.2 Determine from the velocity diagrams:
2.2.4 The nozzle angle
2.2.2 The inlet angle of the first wat. ioving blades
2.2.4
2.2.5
2.2.6
2.2.7
ne velocity of the gas at exit from the turbine in m/s
*.« The tangential force developed in the turbine in KN
The power developed by the turbine in KW
(10x2) (10)
[20]
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(8190046) -5- T1460(E)(N21)T
QUESTION 3
A gas enters a convergent-divergent nozzle at a pressure of 2 030 kPa, a temperature
of 627 °C and a velocity of zero and expands to a pressure of 598,9 kPa, at the exit.
The temperature loss in the convergent part is 150 °C.
The area at the throat is 1553,434 mm?
The Mach number for the nozzle is 1.3.
Take Cp for the gas as 1,005 kJ/kg.k.
Calculate:
3.1
the value of gamma, the pressure at the throat in kPa,
characteristic gas constant, the specific volume at the thr
3.2
QUESTION 4
The following readings were taken n during | a, test, on "constant pressure, open-cycle
gas turbine plant:
Air.temperature at compressor = 29°C
Ratio of pressures for the turbine” = 6:1
Air temperature at combustion chamber outlet = 677°C
Air temperature at plant ou : = 302,6°C
Mass of air circulating ~ = 20kg/s
= 845%
= 82,6%
= 86%
= 1,005 K//kg. K
= 1,4
‘The'adiabatic and actual absolute temperatures at the compressor outlet and
‘the work done on the compressor in kW.
42 The adiabatic and actual absolute temperatures at the turbine outlet and work
done by the turbine in KW.
43 The absolute temperature of the air at the combustion chamber inlet
44 The heat energy supplied in the combustion chamber in kJ/s.
45 The thermal efficiency of the piant.
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(7)
[20]
(2)
[20]
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(8190046) -6- T1460(E)(N21)T
QUESTION 5
A single acting, 3-stage, reciprocating compressor delivers 415 kg of air per minute to
an after cooler at a pressure of 6 720 kPa.
Intercooling is perfect and the stages form a geometric progression.
The compressor sucks air into the low-pressure cylinder at 405 kPa at a rate of
12,6 m1 min.
Take R for air as 0.288 kJ kg. K and Cp as 1,005 kJ kg. K.
The index for compression and expansion is 4,31.
Calculate:
6.1 The power to drive the compressor in kW. (3)
5.2 The ratio of pressure for the stage, the absolute tempe f
pressure cylinder, the absolute delivery temperature,"
intercooler in KW and the heat extracted by the water jacke
in kW. : (11)
5.3 The effective swept volume of the low pressure’eylinder, the effective swept
volume of the in yolume of the high
pressure cylinder in
240 rimin. (6)
[20]
QUESTION 6
In order to improve the e' ir-preneater was fitted.
During tests on th élantth following was noted:
{ steam increased from 2 705 kd/kg to 2 730 kJ due to a
rature.
The air supplie g of fuel decreased by 0,5 kg.
The feed wi erature remained unchanged at 78.7 °C
The rate of evaporation increased from 39 312 kg/h.
ofthe fuel burnt per hour remained unchanged at 42 000 kg.
ic value of the fuel burnt remained at 28.5 MJ/kg.
The temperature at the chimney base was reduced from 250 °C ta 160 °C.
The specific heat capacity for combustion moisture at 460 °C is 2,06 kJ/kg.K.
The heat lost to the moisture remained unchanged.
The boiler room temperature remained unchanged at 25°C.
The pressure of the gas at the chimney base remained unchanged at 450 kPa.
The mass of the combustion moisture remained unchanged.
The heat lost to the moisture in the flues and the dry flues before fitting the air
preheater was 4 577,4 kJ fuel and 4 702,5 kJ/kg fuel respectively.
The specific heat capacity of the dry flues remained at 1,045 kJ/kg. K.
The specific heat capacity of water remained at 4,2 kd/kg.K.
There was a 2% increase in efficiency after the preheater was fitted.
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(8190046) -T- 71460(E)(N21)T
Calculate by using steam tables only:
6.1 The mass of the air moisture in the flues per kg of fuel and the heat lost to the
moisture in the flues per kg of fuel after fitting the air-preheater. (6)
6.2 The mass of the air in the flues per kg of fuel before fitting the air-preheater
and the heat lost. (7)
6.3 The thermal efficiency of the plant before fitting the air-preheater and the
steam generated in kg/h after the air-preheater was fitted. (5)
6.4 The equivalent evaporation fram and at 100 °C of the im (2)
{20}
QUESTION 7
Carbon dioxide is used as a refrigerant in a vapour cof
The plant operates between limits of —12 °C and 24: °C
a dry saturated vapour.
The saturated liquid refrigerant leaves the
The entropy of the refrigerant after throttling is 0, 538 | oUKS. K
The cooling water temperature rise is
The specific heat capacity of the water | is:
The following are extracts fro}
SATURATION TEMPERA URE © SPECIFIC ENTHALPY | SPECIFIC ENTROPY
‘Liquid (4) | Vapour (hg) | Liquid (sf) _ | Vapour (sg)
56,62 322,66 0,222 1,241
155,58 282.58 0,560 0,986
hw The. dryness factor of the refrigerant at the compressor inlet, the specific
anthalpy at the inlet and the work done by the compressor in kJ/kg. (7)
7.2 The dryness factor of the refrigerant after throttling, the specific enthalpy after
throttling and the specific heat capacity of the saturated liquid refrigerant. (7)
7.3 The refrigeration effect in kJ/kg and the actual coefficient of performance. (3)
74 The mass of cooling water requires by the condenser in kg per refrigerant. (3)
[20]
TOTAL: 100
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(8190046) T1460(E)(N21)T
FORMULA SHEET
PY, =mRT,
ty. ny Bye
RW A
AU =m.Cyv AT
Q=AU+Wd Q=AU + Av
As =m cv in 24 cp.tin
A K
As =m_Cy.In
A
As=m.Cp.In 2
a
Asem. RAIn th
B,
Q=m.Cp. aT
Qzem.Cy. AT
© Tou
Soy = Sg + Ep In. oS
— (Figg 1941)
Pou Iiug = yp + Mpg
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MARKING GUIDELINE
NATIONAL CERTIFICATE
NOVEMBER EXAMINATION
POWER MACHINES N6
21 NOVEMBER 2013
This marking guideline consists of 9 pages.
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MARKING GUIDELINE -2- T1460(E)(N21)T
POWER MACHINES N6
QUESTION 1
vo
1.4 Bp = Ser = A 8 kW
60 60
rane v
[P= QsptOre 1680+420 _ 2100 _ 35 kW
gS a
Thaecn = = x 100% = x 100% = 80% «"
IP me ACE
60xIP BOXES XOX
¢* Taxexe Spioso = 800 kPa
P, = AaMey
La
PR exhg — T6CKRNY g Dy
Ag = = age = 870 mm
(10)
1.2 Qeo = My CP Aly
Aty = ee
WS nee Caw
_ BPX 60X60 28x 60 x 60 =G6Akg/hY
"ym = Tey O38 xc a0 7 = OF Kall
Q, 01
OR my = Le = a = 64kg/h
Qoxn = (tte + 1)my X CpgAty
a Bern ig 132060 ray
Ma = Tr cpghty Goapemaae 1 > 3l—4
= 30:1 (7)
“
1.3 vs = 20% = 4p? x Lap =p?
wf AE ne SPO 103 = 276,09 mm
L2inr L2inr
vw
L=1,21D = £21 x 276,09 = 334,069 mm (3)
[20]
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MARKING GUIDELINE -3-
POWER MACHINES N6
QUESTION 2
2.1
asain
Rrateat
Va bonis
Scales lon)
asa”
2 Ge w
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T1460(E)(N21)T
(10)
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MARKING GUIDELINE
2.2. 2.2.1
2.2.2
2.2.3
2.2.4
2.2.5
2.2.6
2.2.7
2.2.8
2.2.9
2.2.10
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-4-
POWER MACHINES N6
ai ga"
aa,
see?
e315 8%
G2 2S C7
G, 6 his
Oy, + ar, 3.
Uf + gak%s4 2 partie.
Vi = 269,8x42 ers,
Vg = 20,684 eats,
Chong = oe .
hang = 177 Meat Noigs
Feang = 77 Maat Mota}
4002 asajmia®
2 $0,028 kW
Pos Ux Peorg
160X #6,
XS ,08
2 gojaa kw
T1460(E)(N21)T
(10)
[20]
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MARKING GUIDELINE -5-
POWER MACHINES N6
QUESTION 3
v v vw
3.1 Cy = f 2000EpCT, — 7.) = 2000 1.005 x 150 = 549,09 m/s
T,-T, = AtY v ¥
2 T, = Ty — At = 900 ~ 150 = 750K
2
nen)
4s
R= ("= 2030(2.)** “= 1072412 kPa”
ye
R=Cp- Cv and yo
R= Cp -2= reves SPEED ya MOOD a 0,287 kj /kg.K
We a v= Seed 0,200 7 m3/kg
Ac= wey
m= ate = assaaevusagonse * = 4,25 kg/s”
vw vw
3.2 Cy = Mach No X C, = 1,3 X 549,09 = 713,817 m/s
Cy = (2000p T, = T)
h=t%- oe = 900 v ~ EI 646,5 Ke
cas’ os
TT 7507-6465» 103,5 .
= of we TSO GEE ee X 100% = 90%“
Naiv = pai 750-638" rng * 100% 90%
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T1460(E)(N21)T
(7)
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MARKING GUIDELINE -6- T1460(E)(N21)T
POWER MACHINES N6
QUESTION 4
1, = 575,6K r
Wi
NW
eee
Dre = 86% Combustion chamber
7 = 84,5% he = 82,6%
T= 302K
P,P, 6
m= Wkg/s Bess y= 1h Cp = 1005 kifkg.K
Fo ot
41 13 = 7, (2) © = 302°(6)"™ = 503,89 K~
mT,
Ty = 7, + EY = 202+ SE = 302 + 230,923 = 540,923 KY
The 0,845 °°
Cae ra ¥
P, = mCp(Ty, — T,) = 20 x 1,005(540,923 — 302) = 4.802,353 kW” (7)
y-a aa
42 som (Pye _ ft aa y
hah () = 950 () = 569,37 K
14-75 *
he Pad fo ov o¢
Ts = Tq — (Ty — Ts) = 950 — 0,826(950 - 569,37) = 950 — 314,4 = 635,6 Ky
f v vo vf
P, = mp(Ty — Ts) = 20 x 1,005(950 ~ 635,6) = 6 319,44 kW (6)
43 Nhe =e vo oe
&. Ty = Ty + Aye (Ty — Ta) = 540,923 + 0,86(635,6 — 575,6) = 540,923 + S16
= 592,523 K”
(3)
a ¥ vw v
44 Qeo = mCp(T, — Tz) = 20 X 1,005(950 — 592,523) = 7 185,288 ki /s 2)
v ¥ ,
45 nye = 8 x 100% = eee = X 100% @)
= 21,114%”
[20]
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MARKING GUIDELINE -T-
POWER MACHINES N6
QUESTION 5
5.1
aod
P= ZL Ave, ee )* _ ie
ont
— 3x1 31 12,6 | (6720
2481 Y 105 x BE 2e| (zane ~
O31 60 LOS
= D131
“O31
6720
105
228 105 x oni |(
= 108,534 kW
5.2 [Pet 3fe720-" y
= we SE ee gy
Ts 7 105 4
P,Ve, = mRkT,
AT = Seen = 306,25 Kv
mot aL oat
Th=TleTl at (2) Typ.) = 306,25" (42
= 425,186 KY
ov wv v
Qine = mEp(T ~ Ty) = Ex 1,005 (425,156 ~ 306,25)
= 29,875 kW ©
Quy = Orne = EES" — 29,875 = 6,303 RW
5.3 ov ¥
Ve, = 4% 220,21 x = 0,052 5 m3/stroke
s N 240
malty» _ 15%0,288%306,25
Ve, =
a 60x4x105~
= 13,125 x 10-43 stroke”
MRT, _ 15X0,2.88X306,25 ” 5 By .
eet Ys Se 5x stroke +
Ves 7 eananan ios?” 3,281 2 10-3 m3 /strok
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T1460(E)(N21)T
(3)
(6)
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MARKING GUIDELINE -8-
POWER MACHINES N6
QUESTION 6
6.1 At 150 kPa and 250°C: fig, = 2.973 Kd /ko
6.2
6.3
64
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Ab 160 Pai hg = 2696 k{/kg and by, = 113,3°C
BEFORE Qin = Mm[hise — CwAtylY
sity = 2 aia |
hn = edly = goa 7 O55 kg /kg fuel
¥ ¥
AFTER igs = hy + Cotes ~ te) = 2696 + 2,06(160 ~ 113,3) = 2 792,202 kI/kg
a a s y v
v v ¥
Qm = Mmlhss — CwAty] = 0,55[2792,202 ~ 4,2(25)] = 1 477,961 ky /keg
BEFORE Q,
Qg A702,5 ¥
Epytt, L0aS@S0~25)~ 20 ky
Mg = Mg + 1— Ma” ¥ yv v
A Mg = Mg ~ L4 My = 20-14 0,55 = 19,55 kg/kg fuel
v ¥
AFTER My = Mg ~ 0,5 + 1— my = 1955-05 +1—-0,55 = 19,5 kg/kg fuel
¥ ¥ v v
Qg = MglpgAty = 19,5 x 1,045(160 — 25) = 2750,963 kf/kg fuel
At 78,7°C: hyy, = 330 kJ/kg
_ mfitse~hpo,} __ 39322|2705-330]> = “
BEFORE =n, = ee 100% = gasppn 7 100%=78%
AFTER Tin = 78+ 2= 80%”
aes[heeRpwyd
Nn =
myxcr
UX Mm pXCV — OB+k4200x28500 ¥
Ms =a = Gorcaaa 39.900 kg/h
_ PsfPss—lywi} » _ 39900~12730-330)-"
AFTER BE wypX2257 2002257
= 10,102”
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T1460(E)(N21)T
(6)
(7)
(2)
[20]
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MARKING GUIDELINE -9-
POWER MACHINES N6
QUESTION 7
P (hay
-8L..]
‘ i ‘5. y/tg) i
52,62 155,58 282,58322,66 0.222063 0.9881,241
0,538
7.4
¥v ra
hy = he + 2,(hy — lig) = 56,62 + 0,75(322,66 ~ $6,62) = 256,15KI/Ieg
wv xv a
WD = hy — hy = 282,58 — 256,16 = 26,43 ly fleg
we S886 __ OS3BY0,222° _ 0,516 ,
02 *s ated (241-022 = rig 994
vox + ¥
hg = hg + X5(hy — hg) = 56,62 + 0,31(322,66 — 56,62) = 139,092 kI/kg
fis = hy — Cp(Ts — Ta)
we Bgchs __ 155,50~139002-" 16,488 _
p= hoy Spore ge MAKI KG
RE = hy..hs* 256,15 — 139, 092 = 117,092 = 117,058 kI/kg
COP = RE = 117,058 “= 4.429” = 4,429
WD = 26,437
TA
Qecon = hy — Ny = My X she x AtY
— 2acity _ 282,58-139,092~" _ 143.408
= thexae apewiege egg = OSB kg /kg
w 69,72
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T1460(E)(N21)T
stkifka.R)
(7)
(3)
(3)
[20]
TOTAL: 100
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Department:
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REPUBLIC OF SOUTH AFRICA
Calculators may be used.
ndid ; tes will require drawing instruments.
This question paper consists of 8 pages and 5 formula sheets.
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(8190046) -2- T1460(E)(J26)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES N6
TIME: 3 HOURS
MARKS: 100
af:questions, only the
ork'you do not want to
NOTE: If you answer more than the required num
required number of questions will be marked.
be marked, must be clearly crossed
INSTRUCTIONS AND INFORMATION
1. Answer any FIVE questions,
ns of questions must
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(8190046) -3- 11460(E)(J26)T
ANSWER ANY FIVE QUESTIONS.
QUESTION 1
Dry saturated steam with a temperature of 179 °C enters a convergent-divergent
nozzle with a negligible velocity.
At the throat the steam has a pressure of 500 kPa, it is 96,5% ary and the diameter is
50,891 mm.
The pressure at the exit is 350 kPa and the actual dryness factor-of tes steam is 0, 94.
The reheat factor of the nozzle is 4,5%
Calculate by using steam tables only:
1.4. The specific enthalpy of the steam, the velocity in m/s, ‘the. Specific volume and
the area in mm*at the throat of the n nozéle
steam in kg/s (9)
1.2. The actual specific enthalpy, the
dryness factor of the steam at-th (8)
1.3 -The velocity of the steam h (3)
. [20]
QUESTION 2
and for the second stage itis 3,8:1.
of the second stage is 87 °C.
Take R for air
Calculate:
2.1 The masé of air delivered in kg/s, the effective swept volume of the second
stage in cm%, the swept volume of the second stage in cm* and the diameter
of its piston in cm (10)
2.2 The clearance volume and the cylinder volume of the second stage in cm® (4)
2.3 The power required to drive the compressor in kW (4)
2.4 The absolute delivery temperature of the air (2)
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(8190046) A.
QUESTION 3
T1460(E)(J26)T
A six-cylinder, three-litre, four-stroke petrol engine was tested on a rope brake and the
following information was noted:
15 minutes
2 940 rfmin
Duration of test
Rotational frequency of engine
Mean diameter of brake
Net load on rope
Indicated power developed
Diameter of each piston
Brake thermal efficiency
Calorific value of fuel
Air-fuel ratio
boi Wn ou WoW WT
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(4)
(3)
(3)
(2)
(2)
[20]
(8190046) -5- Ti460{E)(J26)T
QUESTION 4
A velacity compounded impulse turbine consists of two rows of moving blades
separated by a row of fixed blades.
The average diameter of the moving blades is 975 mm and it rotates at 3 330 m/s.
The inlet and outlet angles of the first row of moving blades are both 26 °.
The exit angle of the fixed blades is 18°.
The relative velocity of the gas at the turbine inlet is 590 m/s.
The inlet and outlet angles of the second row of moving blades'are
There is a 5% loss of velocity across all blades due to fricti
The gas flow rate is 25 kg/s.
4.4 Use a scale of 1 mm = 5 m/s and construct elocity di : $-for the turbine in
the ANSWER BOOK.
Indicate the lengths of ALL the lines as well asthe magnitude the.angles
on the diagrams : (10)
4.2 Determine from the velocity diagra
~The power developed by the turbine in MW
S (10)
[20]
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(8190046) -6- T1460(E)(J26)T
QUESTION 5
A steam plant consisting of an economiser, evaporator and a superheater, generates
38 700 kg of superheated steam per hour at a pressure of 2 MPa from every 4 500 kg
of coal burned.
The specific heat capacity of the superheated steam is 1,08 kJ/kg.K.
The calorific value of the coal is 30 Mu/kg.
The air fuel ratio is 18: 1.
The thermal efficiency of the plant is 82,78% :
The temperature of the feed water entering the evaporator is.99,6 °C."
The temperature of the flue gases at the chimney base i is 20:
The boiler room temperature is 25 °C.
The temperature of the flue gases leaving the evapotétnk is 468
The temperature of the flue gases leaving the superheater is 3254
The specific heat capacity of the flue gases is 1,0: ko. Ke
Calculate by using steam tables only:
5.1 5.1.1 The heat absorbed by the ¢ omiser in kdlk coal *
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(5)
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(8190046) -7- T1460(E)(J26)T
QUESTION 6
A single cylinder engine operating on the diese/ cycle principle has a piston with a
diameter of 89 mm.
It has a cylinder volume of 600 cm*.
The temperature after combustion is 2,4 times the temperature after compression.
The volumetric compression ratio is 15 : 1.
The volumetric expansion ratio is 6,25 : 1.
The expansion index is 1,3.
The compression index is 1,32.
The heat received during combustion i is 1 007,412 kJ/kg gas
Take Cp for the gas as 1,005 kJ/kg.K and Cv as 0,718:kJ/kg:k
Calculate:
6.4 (3)
6.2 (3)
63
(7)
64 (3)
65 (4)
[20]
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(8190046) -8- T1460(E)(J26)T
QUESTION 7
Methy! chloride is used as a refrigerant in a vapour-compression refrigerating plant
operating between temperatures limits of -6 °C and 30 °C.
The refrigerant is a wet vapour with a specific enthalpy of 446,35 kJ/kg, at the
compressor inlet and it is superheated by 15 °C during isentropic compression.
The refrigerant is condensed in the condenser but nor undercooled.
The refrigerant flows at a rate of 1,8 kg/min.
The temperature rise of the cooling water is 15 °C.
The specific heat capacity of the cooling water is 4,2 kJ/kg.
The following are extracts from methyl chloride tables:
Saturated temperature (*C} | Specific enthalpy.A/kg' Specific. entropy (k/kg.K)
Liquid (A) eVapour:(h,) iquid Vapour (sz)
6 516 462.8. "1,746
30 108,6, 4/87 1,627
Calculate:
71 The dryness factor o' mpressor inlet, the specific
heat capacity of the
after compression
ct in kdikg @ work done by the compressor in kJ/kg
of performance.
Sarierconipresston ~ § of saruraredvapourar 30°% +.Cp.lel? arrarcompresston *
in condenser,
TOTAL:
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(10)
(4)
(3)
[20]
100
(8190046) -1- T1460(E)(J26)T
POWER MACHINES N6
FORMULA SHEET
Any applicable formula may also be used.
ENGLISH GENERAL
PY, =mRT,
R=Cp-Cv
PV=c
PV" =¢
PV" =c
O=AU + Wa Q=AU + Av
cy “In Poy Cp In ey
ra Ky
Av=m.Cy In
A
V;
As =m.Cp In
ey
As=m.Rtnh
A
Qz=m.Cp. AT
Q-=m.Cv.AT
f,
=S,+Cp. In
8 f;
AY
SH
Sy =Se-S+
S=Sp +See
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(8190046)
ENGLISH
My = hp + xhyy
Vous = XV,
ws g
-2-
GENERAL
nT q., -1941)
V,, =! —____
SH Py
ee
v.
T1460(E)(U26)T
AFRIKAANS
as = Ay + Xyy
Different formulae for
work done (Wd)
““Verskillende formules vir
arbeid-verrig (Av)
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(8190046) -3-
ENGLISH GENERAL
Differem formulae for
work done (Wd)
= area of P¥-diagram
= work done first stage
+ work done second
stage +..,
WA yoy = Wa, — Wa,
Ware = Qneit
T1460(E)(J26)T
AFRIKAANS
Verskillende formules
vir arbeid verrig (Av)
= area van PV-diagram
& arbeid ver-
tweede stadium + ...
7
~ Av,
Different formulae for
air standard efficien-
cies (ASE)
- mente (ESR)
Ve ‘skillende formules
vir lugstandaardrende-
_ heat added’ ~ heal rejecte
ag warmie loegevoeg ~ warmie afgestaan
warnile foegevoeg
_ Volume of air
Swept volume
_ Volume of free air
Swept volume
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Verskillende volumetriese
rendemente, Oroi
_ Volume lug ingeneem
Stagvolume
_ Volume vrylug
Slagvolume
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(8490046) -4- T1460(E)(J26)T
ENGLISH GENERAL AFRIKAANS
Different thermal Verskillende termiese
efficiencies, Oherm. rendemente, Oerm.
_ Wd 7 Av
heat supplied warnite toegevoeg
; BP
Ibrake therm. — Mp Is x CV
1 _ IP
ind. therm. fs xCV
\
1 _m, (hs — hw)
thernt. my x CV
i
_B-T .
No = Lf; Ne
_ BP
mech. TP
Indicated efficiency ratio dikateurrendementverhouding
Tind. term.
ISR
Renwrendementverhouding
_ Vind. therm.
ASE ;
Brake efficiency ratio
_ Trem, term.
LSR
_ Novake therm.
RD =2n IN
60
RD = Prom gem, LANE
= Fad. mean ID = Pra, gem, LANE
m
ISFC = 2th spy — MoLh
IP ID
aa fib Mein
BSFC =——— RSBY =
BP RD
cop =" xew =
Ty -7 tT -7;
RE VE
COP = KVW =—
Wad Ay
P=mU,AVw D=m.U. AVw
Fy =m. AV Fas, = 0. AV
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_ (8190046) 5 T1460(E)(J26)T
ENGLISH GENERAL AFRIKAANS
2.U.AVw
Re
a
-t
¥ yt)
Ndia, >
_ ils (As —hy)
My X2 257
Tank. = Trank. = Ty
2 C2
82, + U + BR ++ = BA + + BY +4 =
C3 2
Cy
BZ + Uy + Py + +Wa 824, + Uy + Py + = + Av
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REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE — |
NATIONAL CERTIFICATE
AUGUST EXAMINATION
POWER MACHINES N6
26 July 2013
This marking guideline consists of 8 pages.
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MARKING GUIDELINE -2- T1460(E)(J26)T
POWER MACHINES N6
QUESTION 1
1.4 At ATO: h, = BIT kf fg = hy
At 500 kPa: hy = 640 kf/kgi hy, = 2107 kj/kgit, = 151,8°C and V, = 0,3747 wf fkeg
vv v v
A, = hy + xhpg = O40 + 0,965(2107) = 2 673,255 kf /kg
v v v A
C, = ¥'2000(k, — h,) = §2000(2776 — 2673,255) = 453,31 m/s
v v v
Vi = x¥, = 0,965 (0,3747) = 0,3616 m'/kg
v v
A, = fd = : (50,894)? = 2 034,098 mun?
— mv,
Ae & v
A.C, 2034,098 x 4076 x 453,34"
m= = = = 2,55 kgfs
A 0.36164 (9)
1.2 At 350 kPa: hy = 584 ki/kgihy, = 2147k/kg and V, = 0,524 nm? /kg
v v v v
hy = hep + xq (hp) = 584 + 0,94(2147) = 2 602,18 kj /keg
voev
= The
Mor mab where "ox = 100% —% reheat = 100 —4,5 = 95,5%
hy—Ay . 2776 —2602,18
aL =hy- = 2776" — = 2776 — 182,01 = 2.595,99 YI fhg
os 0,955
2 = hy + Og)
hi —h, 2593,99— 584+" 2009,99
x; = = - = = 0,936
Bre 2147+ 2147 (8)
13 v v v v
Cy = f2000(h, — h,) = f2000(2776 — 2602,18) = 589,64 m/s
v v
V, = x¥, = 0.94 x 0.524 = 0.4926 mi /kg (3)
; [20]
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MARKING GUIDELINE -3- 71460(E)(J26)T
POWER MACHINES N6
QUESTION 2
2.4 PyVe, = maT y p 3
5 2
=0,25 kg/sy
6 4
MART 5, GOW" _ O,2SKO287 XBE0~" 5, 60
Vea = Py No 40405 ao” Vv
=15 375 cm /cycley
Ven
Mora Fan
Vs, = 2% = B88" _ 46 o00em! v
2 aye, 8 9605"
v
Vs="D*L=" Dex 12D =" DF
Co. .
_ favs _ 8 laxieooor -
Pp Py) 12" Qj Lane 25,701 cm
(10)
Vey = {= Fypin)¥5p — f1-0.9609) 160007" 270,562 om) ¥
(pay 1 (3,8) RE 1
v v v
V, = Vs,+¥e, = 16000 + 370,562 = 16 370,562 on? (4)
2.3 Po= tS PyVe, [(rps,) = ~ 1] + sea hal el rms) * - i}
= [2s 105” x S28 [arr cart +23 (ex 108) 8 le, gy’ if} 82 wy,
= (8.977 4 10, 301}4= (19, a78)4.
= 77,112 kw
(8)
OR Pe= i pmRr, lo p5;) i] ‘2 [(rps,) 1]
= 3 x 0,25 X 0,287 x 300 (4) ~ 1) +25 y 0,25 x 0,287 x 360 [eas - 1)
= = 35.907 444,203 = 77,11 kW (4)
2.4 Bok ee
rm =r, () n= 360°(3,8)° 38
= 508,882 Kv (2)
[20]
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MARKING GUIDELINE “4. T1460(E)(J26)T
POWER MACHINES N6
QUESTION 3
3.1 v vv Vv v
iBe}
_ QnT Hf _ Ini mgk)N _ 2a( 30,945,810 |x2940
60 60 . 60
— 2xe(2?d St5 72) 2980 7o7?
at ,
= S84 kv y
BP
RD B, LACE
. to v
wn SOBP GOKB EK"
b pace axt078 42540." where LAE = 31 =3 x 1077 3
= 1142857 kPa y (6)
oe x 100%) = 84%"
ico~
3.2 epee
Homeck ~ sp * 100%) =
p= ee = SE = 1 360,544 kPav
. Fmeck Coe
PLACE B LANE
pp = vee Mt
OR 60 6
po Sk, SN = 1 360,544 kPa
P yate ayig7ygoag (4)
3.3 ¥ v
ALAR xB x07 FH 3
= =e = yam
B nDls xloosa7}? ee 10 72,91 ram (3)
BP
34 y= v
mg jgx€V
BP 84-60 o
= — = — i3 ian’
My fe mpexeh ~ Ogeexaz0oar x15 Skgini5min (3)
3.5 ISFC = EA yom SRO 20
IP 15x 100~" 100
=02 kg/kW.hv (2)
3.6 Mage = (m, + my) X Wi min”
=(23" 4 on 1
=(23°41)K 5 = 24x45
=8kg/min’ (2)
[20]
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MARKING GUIDELINE -5- T1460(E)(J26)T
POWER MACHINES N6
QUESTION 4
“ra J ITH O97 HRB Bao oe"
. of
Al Ue “a5 = a / Aal GHarOy 7 38 VY
P19 ils 6
u
‘
4
Whos St, Cur |
[agen ee SS OO
uv a
4 Y= (49, BRS 2 746 SUS,
, a i
g Ve 2aa,an5 = 414,59 77/5 ,
ao
Le
we
6 Vave,7¥5 +392 sme,
y Fax m TM wee (even
cas [le41-4yse(aya- 23,05]
Va ee
€ 228 asds-aasayeliais 148)
& cas[ i876) cas la]
%) ’
ssi 2475 NN 6
a) pao’
ag
a
3 g Pe nm UVare ml (NesptNusn) °
225x170 (206,84 7%, JS
raed we a
2 ad xf FCs OS F 3S)
zasatto lu 432,8) x107*
2 6,088 MW IY (10)
‘ 1
SN
_ VEaead, trim (10) [20]
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MARKING GUIDELINE -6- T1460(E)(J26)T
POWER MACHINES N6
QUESTION 5
51 5.1.1 v v v wv v
Qacon = OMg F DCv(ty — tou) = (18 + 1)1,05(325 — 205) = 2,394 tifa
v v
5.1.2 Quay = (Mg + UCPltig — tone) = (18 + 1)1,05(465 ~ 325) = 2 793 kf /keg
v v v vv
5.1.3 Qaap = (CV-1e) — Pacon — Pru = (30 000 x 0,8278) — 2394 — 2793
=19 647k] /kg ¥
5.1.4 At2000 kPa: bh, = 908 ki/kg sh, = 1889KI/kg andzt, = 212,4°C
At 99,6°C: Any = 418 ky/kg
mM, x
Qevap = mally HRA g — Aye, |Y
- oe Zevag Xe __ iss? K4500 a “
xAyg = ae — hy + bey, = eae 908 +418
38700+"
= 1794,5445 = 1794,8345 = 0,954
tgg 1885"
(11)
Pe
52 Qo = TELA = Digg + CP (tes —B)]Y
cup NIT, __ 2793 x 4500 ww,
CPU tye — ty) = Se (1-2)ipg =n — (4 — 0,95) “1889
te = an + t, = — a 242,4"
= 426,027°C v (4)
5.3 , + a Pig Cb gl Fohim—Fatm)~" _ (18 41) 1osis0s 25")
MChinney = cv = a000D" x 100%
=11,97%V
= 100% — _ “
Yemaccounted va Went Menim
= 100 — 82,87 — 11,97”
= 5,250 (5)
[20]
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MARKING GUIDELINE -T- T1460(E)(J25)T
POWER MACHINES N6
QUESTION 6
6.1 Wy = Vy = 19%, = 15%
4
“8. — ratio
¥5
My 600-7 3
y=—= =96emn' ¥
a ratio 6,25"
1
Vv (3)
v v v
6.2 Y= Vile —1) = 405 — 1) = 40 x 14= 560 em)
Vs= = pTL
= Ms AO = + .
L= apt Txtae? 7 10 = 90 mm (3)
v v
6.3 Q2_3 =mCp(T, —T,) = 1X 1,005(2,4T, —T,) = 1005(1,47,) = 1 007,412
7 1 607,443 = 716 KY
* 1e4K1,005
vyoev v
T, = 2AT, = 2.4% 716 = 47184 KY
vv v
Reo 0,32
7,=7,(2)" =716(5) =301K
. v v ¥
_ % Agr 4 \O8
T, =Ty (2) =17184(4-) = 991,654«
v
Q,-, = mCp(T, — 7.) = 1X 0,718(301 — 994,654) = 495,89 ki/kg (7)
mR (T,,-Tp 3
6.4 Qa. = WD+ AU = a tm€v(Ts —T,) ¥
__ 1xG,287 (01
(Sei TE 4 x 0,718(716" — 401) = —372,203 + 297,97
= ~74,233 kf/kg Vv (3)
65 wo = mar|(t,- 7) +- Ev
Med ment
_ v -\ , 17184—991,656"" 746" 302
= 1X 0,287 [C7184 ~ 716") 4 ae |
= 0,287[1002,4 + 2422,4867 — 1296,875] = 0,287[2128,0117]
= 610,739 Kf /kgv (4)
[20]
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OHET “2 T1390(E\AQ)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
POWER MACHINES Né
TIME: 3 HOURS
MARKS: 100
NOTE: {f you answer more than the required number of questions, only the
required number of questions will be marked. All work you do not want to
be marked, must be Clearly crossed out.
INSTRUCTIONS AND INFORMATION
1. Answer any FIVE questions,
2, Read ALL the questions carefully,
3. Number the answers according to the numbering system used in this question
paper,
4 Questions may be answered in any order, but subsections Of questions must
be kept together.
5. ALL formulae used, must be written down.
8. Show ALL the intermediate steps,
7. ~ Questions must be answered in BLUE or BLACK ink.
8. ALL the sketches and diagrams must be done in pencil in the ANSWER
BOOK,
9. Write neatly and legibly.
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DHET 3. TA39O(EAS)T
QUESTION 4
‘, single-acting, three-stage, réciprocating compressor sucks air in af 100 kPa and
délivers.it-at a rate of 900 kg/h at a temperature of 105,002 °C,
The low pressure cylinder has.a diameter of 345 mm and its stroke length is 1,24 times
the dianieter of the piston.
The power required to drive the compressor is 87,05 kW,
The ratio.of pressure for all cylinclers is 3,4: 1.
The compression.and expansion index is 1,3.
The rotational frequency of the compressor is 324 r/min.
Take Cp for.air as 1,008 kJ/kg/K.
Calculate the following:
a
44 The effective swept volume of the low pressure eylinder in 2
eyele (4)
3
1.2 The swept volume in i and the volumeitic efficiency of the low pressure
eyele
cylinder (4)
1.3 The absolute temperature at the compressor inlet and the value of the
characteristic gas constant (8)
14 The clearance volume and the cylinder volume for the low-pressure cylinder
3
in@
cycle (4)
1.6 The heat extracted by the intercoolers in kW (3)
[20]
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veialih eamanon od me
DHET 4. TIS90(EYAQ)T
QUESTION 2
_ i %
ke ky
Qin be out ie
Economiser 5 8,256
i Evaporator = | 65,684
Superheater = 8,62
Moisture = 18.25
Dry flues = 6,35
Unaccounted =
2.4 Complete the heat balance above,
22 Calculate the following.by using steam tables only:
2.2.1 The thermal efficiency of the plant, the specific enthalpy of the feed
water entering the Sconomiser and its temperature and the specific
enthalpy of the -fead water entering the evaporator
2.2.2 The dryness factor of the steam entering the superheater
2.2.3 The normal temperature of the steam produced
QUESTION 3
A single-cylinder, four-stroke petrol Sngine working on the Otto cycle principle has a
volume compression ratio of 85:4.
The polytropic index for compression and expansion is 1,31.
The brake-power of the engine at 2 G00 timin is 11,9 kW.
The mechanical Officiency is 85%,
The length of the stroke is 1,25 times the diameter of the piston.
The indicated mean effective pressure is 4 000 kPa.
The initial pressure. of the cycle is 100 kPa,
The-temperature after expansion ig 2,8 times.the initial temperature,
Take gamma as 1,4. :
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7)
@)
(A)
(3)
[20]
per Be FAS9O(EARIT
Calculate the following:
The indicated power in kW (2)
3.2 The indicated work done in d/cycle (2)
3.3 The swept volume in m’, the diameter of the piston in mm and the length of
the stroke in mm (8)
3.4 Thevolumes in om®at the principle points (3)
3.5 The missing pressures at the principle points in kPa (5)
3.6 The air standard efficiency (2)
(20)
QUESTION 4
An engine working on the dual cycle principle has a volumetric compression ratio of
16:1.
The pressure and temperature at the beginning of the compression stroke are. 404 kPa
and 32°C respectively,
The compression index is 1,35.
The pressure-at the beginning of the expansion stroke is 1.4 times the pressure after
polytropic compression.
The pressure at the beginning of the exhaust stroke is 262,424 kPa.
The heat taken in-at constant pressure continues for ane thittieth of the stroke,
Take gamma as 4,4.
Calculate the following?
4A The missing absolute temperatures and pressures in. kPa at the principle
points ‘ (412)
42 The expansion index (4)
4.3 The ideal thermal efficiency of the cycle 4
(20)
QUESTION 5
Superheated steam enters & convergent-divergent nozzle at a pressure of 3 MPa, with
a temperature of 300 °C and no velocity and is expanded to a pressure of 860 kPa with
an actual dryness factor of 98%.
At the throat the pressure is 1,5 MPa, the specific heat capacity s 2,826 kulkg.K and
tha index for superheated steam is 1,3.
At the exit the isentropic dryness factor is 99,286% of the actual dryness factor.
The steam flows at a rate of 5,2 kgfs. .
The specific enthalpy.drop in the convergent part is 135,2 kJ/kg.
Do
Ronwrietht own loa ded fro m h layiso.com Dlanee firm over
THSS0(E ARF
© steam in ms, the Specific Sntialpy of the stéam, the
eMPerature of the steam the sp, Cifle vo, um)
, f the 5 eam, the area in mm?
and the diameter in-mm at the throat Of the Nozzle
(12)
5.2 The actual SPECific enthalpy ang the isentropic SP8Cifie 8nithalpy at the nozzte
®Xit, the actual enthalpy drop ang the Isentropic enthalpy drop through the
NOzzle and the nozzle efficiency (ay
[20]
QUESTION 6
Steam discharges axially frory 4 velocity compounded, iWo-stage, impulse turbine with
& Velocity of g Vs,
& inlet and &Xil angle of the first POW of. ma ING blades @8 Well as the inlet angle of
& Second Moving big eS are equal,
6 exit angle Of the S8cond row-of Moving blades ig 26°,
© @xial thrust ig develope 'n the secong Stage,
@ Velocity ®0efficient for all the bladas js 0,98,
he steam flow fate is 45 Kg/s.
6.4 Use g Scale of 1 mm 4 m/s and Construct Velocity diagrams forthe turbine-in
the ANSWE
Indicate the lengths OFALL the lines as wal as the Magnitude of the angles an
€ diagra, (10).
6.2 Determing the following from the velocity diagrams
6.24 The Nozzle angie
6.2.2 The.exit angle of the tixed blades
6.2.3 The inlet and exit angles of the first row of Moving bladas as well
88 the inlet angle of the Sécond row of Moving blades
6.2.4 The inlet angle of the fixed blades
6.2.5 The average blade velocity in ms
6.2.6 The nozzle Velocity jn ms
6.2.7 The Velocity of the Steam leaving the first Stage-in m/s
6.2.8 The blade Sfficienay
6.2.9. The power developed by the turbine jn MW (10)
[20]
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DHET
J TisIO(E} (AST
QUESTION 7
A yapour compression refrigeration plant uses carbon dioxide ae a refrigerant and
operates
between temperature limits of -8 °C and 34°C .
The compressor has a piston with a diameter of 120 mm, a stroke of 141,5 mm and it
rotates at 360 r/min.
‘The volumetric efficiency of the compressor is 90%,
‘The refrigerant enters the compressor as a dry saturated vapour and it enters the
condenser at a temperature of 64 °C.
The refrigerant leaves the condenser as a saturated fiquid with no undercasling.
At -8°C
a
The spacific volume of the dry saturated vapour = 0,072 =
The specific enthalpy of the dry saturated vapour = 184,1 kd/kg
The specific entropy of the dry saturated vapour = 0,700 8 kd/kg.K
At 34°C
The spacific enthalpy of the saturated liquid = 68,5 kdikg
The specific enthalpy of the dry saturated vapour = 201,41 kd/kg
The specific entropy of the dry saturated vapour = 0,684 2 kd/kg. K
Calculate the following:
7A
7.2
73
74
The specific heat capacity of the superheated vapour and the specific
enthalpy of the superheated vapour at the compressor outlet
a
The swept volume of the compressor in mY the volume flaw of the
s
4
refrigerant in ®” and the mass flow rate of the tefrigerani in kg/s
. Ss
The power required to drive ihe compressor in kW, the power required to
produce ice in kW and the actual coefficient of performance
The energy required to condense the refrigerantin kJ/min
HINT: s al 64 °C = 8y a. 34 °C + Gp in [7 af 64°C + T at 34 °C]
TOTAL:
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Rrnuricht ennentinct
(6)
6)
(6)
(4)
[20]
100
OHET
FORMULA SHEET
Any applicable formula may also be used.
ENGELS ALGEMEEN
2, = mT,
ReCp-cy
=
re Cy
PV ac
PY" =
PYT o6
n=l
Bo(wY fa
Rom) ~
aU = nt. Cy ar
Q=AU + We
asem[Ov.ta 2 scp.
1
£
22
4
As =m .Cy. in
« Vy
AS =m. Gp In 8
f
fi
Be
Qsm.Cp. ar
Q=m Cy ar
As=m. Rtn
So 8 + Cn Ti
Xp = 8, ~ Se
Sef, + xSq,
§
s
ou hy + Cbg, ~ty)
aed
”
4
K
_ aia
T1390(E)(Ag)7
AFRIKAANS
Py
uy
~
PV" =
PV? ok
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Os wright reserved
ALGEMLEN
Uy = 1940
Po
THSGO(ENAQ)T
AFRIKAANS
Bag =p + Xhryy
Vas = x¥,y
Different formulae for
work done (Wd)
= PRAY
KB
= BY Ins
nan
Phi Avs
~ antl
_ AR ~ Pa
y-1
=m,Cp. AT
= Be BY. & bL ry
a-1'° A
ant
Xt
=— RT \ Cog) "71
nel. i (ps)
Verskillende formules vir
arbetd verrig (Av)
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DHET
ENGELS
Different forinulae for
work cone (Wa)
“= area of PV-diagram
= work done first stage
+ work done second
Stage +...
Winey = Wed 1 Wd,
Wa,
edt
=O
nett
T1390(8)(AgyT
ALGEMEEN AFRIKAANS
Varskillende formules
vir arbeid verrig (Ay)
= area van PY-diagram
= arbeid verrig eerste
Stadiun + arheid ver.
rig tweede stadium +
Avggy = Ay, ~ Avy
Aree = Cheit
Different formulae Sor
alr standard efficien-
cies (ASE)
eat added ~ heat rejected
=]
Verskillende formules
vir lugstandaardyende-
mente (ESR)
SG, ~ Dry” TY
le
= farnite toegevoeg — warmte afgestaan
heat added
a
warntte toagevoeg
Different volumetric
efficiencies, Org,
Faluine of air taken in
ane ene ID
Swepe volume
Volume of free air
ae EE
Swept volume
Verskillende volumetriese
Fendemente, Oy
__ Votinne dug ingencem
Slagvolume
_. Volume veplug
© Slagvokune
i
=1~2/(2)r_,
ale
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“ooyrght reserved
DHET
TI390{EKAS)T
ENGELS ALGEMEEN AFRIKAANS
# it thermal Verskillende termiese
epficiencies, Onerm: rendemente, Oerm,
_ Wd - Ay
heat supplied warmte toegevoeg
n ___ BP _ gD
lorake therm. mye XP vein term. ir), x
_ IP x _ ID
Find. therm. = My /6 War int. tern, mh,
__ ia, (hs ~ hw) _ ims Chis — hw)
Pharm, * ve xCY Merm. iy » PP
hh TT B-
Ae = ao Th = Shot = po
-% T-y ty Fj
nen 22 “ng, =A.
tnecht. IP meg, ID
dudicated efficiency ratio Indikateurrendementverhouding
= Thind therm. = Find, term,
ASE LSR
Brake. efficiency ratio Remrendementverhouding
— Morake thenn, = Tremteran,
ASE LSR
RP ag LN. T=Fxr RD=22 1%.
60 60
BP = Pyrite mean LANE
IP = Find. mean LANE
_. Myrrh
Ke =—Lh
op
src = fle
BP
q,
T, -T,
cop - RE
Wd
P=m.U. Aw
F
COP =
Copyright reserved
‘ym. AP
"“T Downloaded fro
RD = Prom ger, LANE
ID=P, LANE
rid. gem.
ISBV we Tbh
iD
RSBY = “bth,
RD
KYW =
Ty ~T,
vw =e
Av
D=m.U,AKW
m hlayisd:éom
DHET THSSO(E\AgyT
ENGELS ALGEMEEN ALRIKAANS
_2.U AK
“dia, = ye
f.
2 \yrt
a
T, = 7 —~.
@ Gal
C2510 Gh ia
= 2x10" yaa
Cy = 2x10 xc, (G~B)+02
ay, mi,
Aye Ay = 2,
C we
nattinte 8 Ack
fy ~h, G~k,
A, ~My L-2
hy ~ hs Fim F,
get gedit
~ hy i~%
My Fy ~hy,) tts (Ay ~ A)
ER = is Rw) BV = 3 Ns Ty)
i yg X2IS7 Hy, x 2257
We Ay,
Tso, * em tS. Misq, RR
A noly. Vaoli.
thon = Wa _ AY
rank, oO Meantk. fe)
Ty
Tearn, = 1 — R
Ami + DF
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Copyright feserved
TAZOO(EMASIT ;
2 i
9 Cj -
By tU, + A +4 =
ce
‘ee PY, ty tee
ae ch A
GZy + Uy + PyVy 4S + Wed 82, 4+, + Bly bt dy
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4
i
{
MARKING GUIDELINE
wv
ates
=] =-
()
v ¥
“Inks nfs
hast
Pas) Pa
= dn PA/P 5 on (5/414) = anseresu(/202424(3an (16/4 flax 10/0 sy) = dv 5970,51/262.424)/(ln 1607
f= 432(
-6-
POWER MACHINES N6
T1390(E)(AS)T
15)
(4)
43 M=(h-s-TAya? 3 -Ta)+yiTia-T,3) ex 100%
FOZ 469 (— 305 (
. f “(126,86 (— 904
1
=|L~—
1110,762
= 06,114%
487,469
|x 100%
9.Q + 14€1690,20 (— 1126,66 ol *1°*
(4)
{20}
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MARKING GUIDELINE -7-
POWER MACHINES N6
QUESTION 5
v vw
BAC, = fioGoay— aS = of 200003523
Eee
= 520
ay
AL3 WO kPaand 300°C: hy, = Ay = 2905)
Ah hy ag v ¥
2
hy Ay ~ BA = 2955 ~135,2 = 2.95
AGL 600 kPa hk, = 2792 Rikg and &, =
he = hy, Cy Coty - v
TU380(E)AS)T
ato = (aye coe yy/ Cig.) + t8 = (2859.8 - 2792) /2,825-420 La = 24 + 2014
— 225,40
Ag = Dall ye/C,e = (5.2%0,1325)/520¢« [aor ts = 1325 ama [2c
dev EAT e+ cy) = Vila ides CR)
ay
= 41,07.
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.
B3
MARKING GUIDELINE 8 TISSO(E)AQ)T
POWER MACHINES N6
. fy
- Atze0 kPa: hye = = 2036—
aed ‘ RG
v w
hg = hy + whey ; ,9002036)
w wv
06 x O83 = 0573
vo
973(2036)
Rieke (
v
ve G4,90904 (8)
[20]
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MARKING GUIDELINE -9-
T1S90(E)(AS)T
POWER MACHINES N6
STION 6
ATerre
Vion? &S
6.28
2.8m | (10)
Pam U (Moor Moon)
a ASK 64 Gst,0¢ #3538)
is GSN bS [raggyxie
«= ye4a re (10)
[20]
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MARKING GUIDELINE - 10°
T1A90(EKAg)T
POWER MACHINES. N6
|
| ay F Fu
Cot ia
7
0.6842" 0,7008"
H8)/lnT,2/T)3 ) = (,7008" ( — 1.6842" (5/0 337(/3070 = 0,178 Efpkg. HC
v ¥ re vw v
hy hg + 07, — 7.) = 201,14 01780337 — 307) = 2011 40,17 8¢203
(6)
(6)
v v ve vo
73 Fos Mathys ~ Ag) = O420008,84 — 1840) - G12G2,34) = 2.6800 Ke
vw ¥ v v
Poy SOL, hy ag = L208) 68,8) = OLE 15.6) = 12,972 cur
COP
oR (6)
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|
|
MARKING GUIDELINE
wv
<44-
POWER MACHINES N6
vo
v
T1390(E)(A9)T
44 Qscon * Myfhg — Ay) = 0/12(206,44 — saa) BO = O,120137,94) x 80
Persentasiea
Percentage
TOTAL: 100
‘ MARKING GUIDELINE a TIS9OKENAS)T
POWER MACHINES N6
AUESTION 6
(40) [20]
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MARKING GUIDELINE - tO T1S90(E\AQ)T |
. POWER MACHINES.N6
MN PEF}
1
0,6842 0.7008"
7A = s34 Cp inly2/T.3 ¢
2 2:3)/GaT.2/7\2) = @,7008" C ~céeer" Min $37(/407() = GATS AN HOLA
~ Fy} 2011 + 1 78(837 — 207) = 2001+ a.1Teczed
(8) |
: 72
i
(8)
WS
oF (6)
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