Past Exam Ques and Solution Part 1

7/23/2019 Past Exam Ques and Solution Part 1

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a

Past Exam Questions

Anutosh

School

of

MAE,



Tutorial 1 Question 3

http://www.absorblearning.com/med





•Figure 1 shows a cylinder containing

Ques #01 (MA2003- SEM I 2012/1

mixture with 1% of its volume in the liq

piston weights 100 kN and has a cross

system is exposed to the atmospheric

50 kN/m

3

(2)

(2ꞌ)

(1)

Figure 1: A Piston-cylinder arra

Q



Heat is added at a rate of 600 W ca

,piston touches the spring with a sprin

position, heat is removed. Due to hea

ꞌ,

amount of heat Q is applied to the sysmove up to position 3 and stop.

(a) Draw a P-v diagram of the syste

change from 1 to 3.(b) Calculate the initial temperature

mass of the fluid.

c Calculate the distance from os

at initial state to the spring) and the

(d) What is the final temperature at

2ꞌ

to 3.



(a) P-v diagram of the system a

the states changing from 1 to 3

3

’



P3



(a) Calculate the initial temperature, the

mass of the fluid.

Vliq + Vgas = V1 = 0.001 + 0.

Vliq = 0.001 m3, Vgas = 0.099

Force BalanceP1 Ap –Fp – Patm.Ap = 0 => P1 = Fp/Ap + P

vf @300 kPa = 0.001073 m3/kg

vg

@300 kPa = 0.60582 m3/kg

mliq = Vliq/ vf @300 kPa = 0.001/0.00107

mgas = Vgas/ vf @300 kPa = 0.099/0.6058



Total mass

= + = +q gas . .

Quality x = 0.163415/1.095 = 0.15

Initial TemperatureT1 = Tsat@300 kPa = 133.52 oC

(c) Calculate the distance from pos

at initial state to the spring) and theState 2

V2 = mt × vg@300 kPa = 1.095 × 0.605

L1 = V1/Ap = 0.1/0.5 => L1 = 0.2 m

L = V /A => L = 0.663373/0.5=

L1-2 = 1.327 – 0.2 = 1.127 m



Energy balance (states 1 and 2)

=> W = m h – hHere h2 = hg@300 kPa = 2724.9 kJ/kg and h

h = 561.43 + 0.15 × 2163.5 = 885.955 k

W

=> ∆

t = 1.095 (2724.9 – 885.955)=> ∆t = (2013.64 × 1000)/(600) = 3356 s

At Position 2ꞌ

h2ꞌ=561.43 + 0.9 × 2163.5 = 2508.58 kJ/kg

v2ꞌ= 0.001073 + 0.9 × (0.60582 – 0.001073



Wspring = 0.5 × (2.0-1.127)2 × 50

.

PF = FF / Ap = 50 × (2-1.127)/(0.5) = 87.3 kPa

P3 = 300 + 87.3 = 387.3 kPaP3

V3 = Ap ×( 2 + 0.2) = 0.5 × 2.2 = 1.1 m3

v = V /m = 1.1/1.095 = 1.00457 m3/kg

Wgas

= mt

P1

(v3

-v2

ꞌ) = 1.095 × 300 × (1.004

State 3 (P3 =387.3 kPa and v3 =1.00457 m

Usin su erheated Table of water we hav

h3 = 3640.9 kJ/kg

T3 = 571.61 oC



387.3 kPa = 0.387 MPa an

40.0387.088936.0 v

0.387 Mpa and 500 oC

.40.030.088936.018672.1

h 40.0387.05.3485 .

40.030.05.34856.3486



40.0387.000558.1 v

0.387 Mpa and 600 oC

.40.030.000558.134139.1

khh

39.370340.0387.03.3703

40.030.03.37033704

(At 0.387 Mpa, 1.00457 m3 /kg)

T 569280168.00492353.1

..

500600

h 643.348550015.563

Ener Balance state 2ꞌ to 3

643.348539.3703500600

Q = mt (h3 – h2ꞌ) + Wspring

Q = 1.095 × (3623.15 - 2508.58) + 19.05

Q = 1239.50 kJ



•Figure 1 shows a piston-cylinder-spring co

Ques #01 (MA2003- SEM II 2013/

. x - m o qu wa er – vapour m x u

the position shown, the piston is free to mo

and has a cross-sectional area of 50 cm2.

.

contact with a heat source of Q at 150 oC.causes the final pressure of the fluid to be

Spring

0

0.5 m

P

Q

Figure 1



(a) Draw the P-v diagram of the system and

from initial to final.

(b) Calculate the initial temperature and the

(c) Calculate the quality of water vapour wh

(d) Calculate the work done by the gas and

, .



(a) P-v diagram

P

P f

3

1 2

P i

v v

Wgas



(b) Calculate the initial temperaturethe fluid.

V 1 = 2.5 x 10-3

m3

2.0%20 x , A p = 50 cm2 = 0.005 m

2

Patm = 95 kPaInitial State

Force Balance

P1 A p –F p – Patm.A p = 0 => P1 = F p/A p + Patm. v f @125 kPa = 0.001048 m

3/kg

vg @125 kPa = 1.3750 m3/kg

3750.12.0001048.01 f g f vv xvv



Total mass

V 3

3

105.2

g

vm

1

.2758.0

T i = T sat @125 kPa = 105.97oC

82240.62.0444.361

fg f xhhh

(c) Calculate the quality of water vaptouches the spring.

a e

V 2 = V 1 + A p x 0.2 = 2.5 x 10-3

+ 0.005

10064.9105.3/ 33

22

mV v

001048.03861.02 f vv

001048.03750.12

f g vv



(d) Calculate the work done by the

Final State

P f = 500 kPa

375125500 P kPa

PP

s

A

sK

A

F P

.

25.05.7

.

K s P m

Final volume:

V 3 = V 2+ A p x 0.25 = 3.5 x 10-3

+ 0.00

33 ..33

At P f = 500 kPa, vg = 0.37483 m3/kg



Here kPagvv 500@3 so the final state is superheat

hgh3

3 .

kgmkPahgh

/524.0&500@3 3 =>5.0

0

6.30641.3168

6.30643

h

Energy balance (states 1 and 3):=> W = Wgas + Wspring

i f i PPvvmvvmPW 2313

2

1

2758.0524.012510064.9 gasW

3861.0524.010064.91 3

s rinW

W = 0.5156 kJ



(e) Calculate the amount of heat, Q

Energy Balance

Q = m (h3 – h1) + Wspring

89263.3067.10064.9 3 Q

= .



Ques #02 _b (MA2003- SEM II 20

ater vapour w t a ow rate o g s en10 MPa and 500 oC at a velocity of 80 m/s

and leaves at state 2 as saturated vapour a

.

fluid exits at state 3 [please see Figure 2 (bvelocity of 20 m/s. The pressure at state 3 i

•Write the overall mass and energy balance

to state 2 and from state 2 to state 3 under

conditions.

•Calculate the work output of the steam tur

•Calculate the heat transfer along the pipe



1.3375500&10@1

C MPag ohh kJ/kg

2706.3200@2 kPaghh kJ/kg

10m k /s

Under steady state conditions

321

Energy Balance (states 1 and 2)

21 E E

221

2

111

2hmW Qgz

V hmW Q out out inin

221

21

112

hmW QgzV

hmW Q out out inin

22

2

2

2

121

V V hhmW out



Energy Balance (states 1 and 2)

2

2222

W QgzhmW Q out out inin

2

22

2

W QgzhmW Q out out inin

2

233

32

hmgzhmQout

2

3

2

232

22

V V hhmQout



(ii)

kPagkPa f hh xhh 100@1100@100@3 .

2257.5.6.0417.51 h = 1772.0

Work output of the steam Turbine:

22

2

2

2

121

V V hhmW out

80

3.27061.337510

out W



(iii)

ea rans er a ong e p pe

2

3

2

2 V V

32

22

out

2

01.17723.270610out Q

9.9345out Q kW









Past Exam Ques and Solution Part 1

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