Thermodynamics III Exam Questions: VAAL University of Technology - Prof. Machete, Exams of Applied Thermodynamics

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VAAL UNIVERSITY OF TECHNOLOGY
FACULTY OF ENGINEERING
DEPARTEMENT: MECHANICAL
NATIONAL DIPLOMA: ENGINEERING: MECHANICAL
SUBJECT:
DATE:
DURATION:
THERMODYNAMICS III
NOV 2008
3 HOURS
EMTGA3-MAIN
EXAMINER: DW SPIRET
MODERATOR: PROF JH WICHERS
REQUIREMENTS:
Steam tables
INSTRUCTIONS:
Assume STP as 101.3 kPa and 15°C unless given
Answer all the questions as instructed
TOTAL:108
FULL MARKS 100
TfflS QUESTION PAPER CONSISTS OF:
(Properties of Water and Steam)
1 Front Page
2 Typed pages
3 Formulae sheets
DO NOT TURN THE PAGE BEFORE PERMISSION IS GRANTED
pf3
pf4
pf5

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VAAL UNIVERSITY OF TECHNOLOGY

FACULTY OF ENGINEERING

DEPARTEMENT: MECHANICAL

NATIONAL DIPLOMA: ENGINEERING: MECHANICAL

SUBJECT:

DATE:

DURATION:

THERMODYNAMICS III

NOV 2008

3 HOURS

EMTGA3-MAIN

EXAMINER: DW SPIRET

MODERATOR: PROF JH WICHERS

REQUIREMENTS:

Steam tables

INSTRUCTIONS:

Assume STP as 101.3 kPa and 15°C unless given

Answer all the questions as instructed

TOTAL: 108

FULL MARKS 100

TfflS QUESTION PAPER CONSISTS OF:

(Properties of Water and Steam)

1 Front Page

2 Typed pages

3 Formulae sheets

DO NOT TURN THE PAGE BEFORE PERMISSION IS GRANTED

VAAL UNIVERSITY OF TECHNOLOGY EMTGA3 2008 NOVEMBER (MAIN)

QUESTION 1

1 kg of air at 100 kPa and 600 K expands according to the law PV^16 = C. The final volume is three times larger than the initial volume.

Draw a P-V and T-S diagram and calculate the work, heat flow and entropy change. [13]

QUESTION 2

In a Joule cycle the pressure ratio is 10 and the upper and lower temperature limits are 773 K and 283 K respectively.

  1. Draw the P-V and T-S diagrams for this cycle. (2)
  2. Determine the unknown temperatures. (2)
  3. Determine the cycle work and the cycle heat. (6)
  4. Calculate the cycle efficiency and the work ratio for this cycle (2) [12]
QUESTION 3

A single stage compressor has a F.A.D of 5m^3 /min at 15°C and 101.3 kPa. The inlet conditions are 95 kPa and 20°C. The delivery pressure is 500 kPa. The compression index is 1.25 and the clearance volume is 2% of the swept volume. The compressor runs at 1450 rev/min and the stroke is twice the piston diameter.

Calculate: 1 The mass flow of air. (1) 2 The volumetric efficiency. (5) 3 The piston diameter. (5) 4 The power requirement of the compressor. (2) 5 The heat rejected from the cylinder. (2) 6 The isothermal efficiency. (2) 7 Draw a correctly labeled P-V diagram. (2) [19]

QUESTION 4

An eight cylinder 4-stroke engine has a bore of 229 mm and a stroke of 304 mm. The compression ratio is 14/1. The engine develops 500 kW at 750 rev/min. The mechanical efficiency is 90% and the volumetric efficiency is 78%. Fuel with a C.V. of 42 MJ/kg is supplied at an air/fuel ratio of 16/1. The ambient conditions are 100 kPa and 288 K. The IMEP for the pumping loop is 34.5 kPa. The exhaust gas temperature is 450°C and cp for exhaust gasses is 1.135 kJ/kg.K.

Calculate: 1 The brake thermal efficiency. (9) 2 The heat rejected from the engine. (4) 3 The IMEP for the power loop. (4) 4 The efficiency ratio if the cycle is based on the Otto cycle. (2) 5 Draw the P-V diagram for the Otto cycle. (1) [20]

QUESTION 5

A refrigeration plant is required to cool a cold room at -10°C with a cooling load of lOOMJ/h. The refrigerant to be used is Freon 12 (R 12). The refrigerant enters the compressor dry saturated. The temperature of the condensate is 5°C and it is 5°C undercooled. The heat rejected by the condenser is absorbed by water. The temperature rise of the water is 10°C. Assume isentropic compression

THERMODYNAMICS

General formulae

Q+W =AU

m [ Z,g + Vt C,^2 + h, ] +Q + W = m [ Zjg + 14 C 22 + h 2 ] h = u + pv m = CA/v

Vapour Model

U = Uf+X-Ufg S = Sf+Z.Sfg Sfg = Sg - Sf V = Vf + (^) XVfg

Ideal Gas Model

Y = Cv=R/(A.-l) R = cp - cv R = Ro/mo pV = mRT m = n.nio

= 8.3145 kJ.kmol'.K -1^ _v-_

Expansion law V=C PV=C PVn=C P=C

pi/r,=p2^r 2 P.V,=p 2 V 2 P,/p2=[V 2 /V,]n

v,/r,=V2^r 2

T,A' 2 =[V 2 A^l]n-^1 T,/r 2 =Pl/jp 2 *^1

Gas Process

Process Isochoric Isobaric

Isothermal

Isentropic Polytropic

Law V=C p=C pV=C pVY=C pVn=C

AU

w —

AH

AS

cv.ln (T 2 /T,) cp.ln (T 2 /T,) Q/T = R.ln(V 2 /Vi) 0 R.ln (V 2 A^,)+ c.ln (T 2 /T,)

-P(V 2 -V.)

-Glnfa/pj) -CrVj^-V^/Cy-l) (P2VrPiV,)/(n-l)

Q
AU
AH
-W

W[(n-y)/(y-l)l

Vapour Process

Process Isochoric Isobaric

Isothermal Isentropic

Polytropic

Other

Law V=C p=C T=C AS=Q= pVn=C pV=C None

AU AH

Tables Tables Tables Tables Tables

Tables

AS

Tables Tables Q/T 0 Tables

Tables

W
P(V,-V 2 )
AU-Q
AU

(P2V 2 - PlV,)/(n-l) -pV.ln(p,/p 2 ) AU-Q

Q
AU
AH
TAS
-W + AU
-W + AU

Steam plant

"Boiler = msxAh / mfxC.V. ^lpiant = msxAh^rt, / mfxC.V.

E.E. = ms/mf

E.E from & @ 100°C = E.E.xAh/

S.F.C = mf x3600 / (msxAh)

S.S.C. = ms x3600 / (msxAh)

= (tb-tc)/(feed heaters + 1)

Refrigeration

C.O.P= Q /W

C. O. P R E F - T , / ( T 2 - T , )

Reciprocating Compressors

W = nmR(T 2 -Ti)/(n-l)

rP = (P!/P 2 )1/z

Wis0 = mRTln(P 2 /Pi)

Tlvoi = v/vs = l+(V