Thermodynamics Examples and Class test

Assignment – I & Class Test – I Examples

S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines

Assignment – I &

Class Test – I

Examples

Applied Thermodynamics & Heat Engines

S.Y. B. Tech.

ME0223 SEM - IV

Production Engineering



Example 1A turbine is supplied with steam at a gauge pressure of 1.4 MPa. After expansion in the turbine, the steam flows into a condenser which is maintained at a vacuum of 710 mm of Hg. The barometric pressure is 772 mm Hg. Express the inlet and exhaust steam pressures in Pascal (absolute).Take the density of mercury as 13.6 X 103kg/m3.

The Atmospheric Pressure,

P0 = ρ.g.z0 = (13.6 X 103).(9.81).(0.772)

kg/m3 m/sec2 mtr

= 1.03 X 105 PaInlet Steam Pressure,

Pi = [(1.4 X 106) + (1.03 X 105)] Pa

= 15.05 X 105 Pa

= 1.503 MPa…..Ans

Outlet Steam Pressure, (i.e. Condenser Pressure)

P0= (0.772 – 0.710).(9.81).(13.6 X 103)

mtr m/sec2 kg/m3

= 8.27 kPa…..Ans


S. Y. B. Tech. Prod Engg.

Example 2

Work done by Stirring Device upon the system;

W1 = 2 π T N = (2π).(1.275).(10,000) N-m rpm = 80 kJ

A piston and cylinder machine containing a fluid system has a stirring device in the cylinder. The piston is frictionless and is held down against the fluid due to atmospheric pressure of 101.325 kPa. The stirring device is turned 10000 revolutions with an average torque against the fluid of 1.275 N-m. Meanwhile the piston of 0.6 m diameter moves out 0.8 m. Find the net work transfer for the system.

W1

0.8 m

W2P = 101.325 kPa

System

This is Negative Work. Work done by the system on surrounding;

W2 = P.A.l = (101.325).(π/4).(0.6)2.(0.8) kN/m2 m2 mtr = 22.9 kJ

Net Work transfer is,W = W1 + W2 = (-80) + (22.9) = (-57.1) kJ…..Ans


S. Y. B. Tech. Prod Engg.

Example 3When a system is taken from state a to state b, in figure shown, along path acb, 84 kJ of heat flows into the system and the system does 32 kJ of work. (a)How much will the heat that flows into the system along path adb be, if the work done is 10.5 kJ?(b) When the system is returned from b to a along the curved path, the work done on the system is 21 kJ. Does the system absorb or liberate heat, and how much of the heat is absorbed or liberated?

Qacb = 84 kJ and Wacb= 32 kJ

a d

bc

Volume

Pre

ssur

eQacb = Ub – Ua + Wacb

Ub – Ua = 84 – 32 = 52 kJ

Qadb = Ub – Ua + Wadb

= 52 + 10.5 = 62.5 kJ….Ans (i)

Qb-a = Ua – Ub + Wb-a

= (-52) - 21 = (-73) kJ….Ans (ii)i.e. System liberates 73 kJ of Heat.



Example 4A cyclic heat engine operates between a source temperature of 800 oC and a sink temperature of 30 oC. What is the least rate of heat rejection per kW net output of the engine?

718.0

)273800(

)27330(1

1max

H

Lrev T

T

SourceTH = 1073 K

Wnet= 1 kW

QH

QL

Heat Engine

SinkTL = 303 K

Now,

kWQ

Q

W

SuppliedHeat

W

H

H

netnet

392.1718.0

1

718.0max

Hence, QL = QH – Wnet

= 1.392 – 1 = 0.392 kW….Ans



Example 5A domestic food freezer maintains a temperature of -15 oC. The ambient air temperature is 30 oC. If the heat leaks into the freezer at a continuous rate of 1.75 kJ/s what is the least power necessary to pump this heat out continuously?

sec/06.2258

303)75.1( kJ

T

TQQ

T

Q

T

Q

L

HLH

H

H

L

L

And, W = QH – QL

= 2.06 – 1.75 = 0.31 kW…Ans

Refrigerator cycle removes the Heat from the

Freezer at the same rate at which Heat leaks into it.

For Minimum Power Requirement;

AirTH = 303 K

Wnet

QH

QL

Heat Pump

Freezer TL = 258 K

QL = 1.75 kJ/sec



Example 6An ideal gas which obeys the equation PV=mRT is compressed in a piston – cylinder arrangement, such that the temperature remains constant. Derive an expression for the work done on the gas. Calculate the quantity of work when 2 kg of Helium is compressed from 1 atm, 20 °C to 1 MPa, holding the temperature constant.

2

1

1

2 lnln1

P

PTRm

V

VTRmWdV

V

TRmW

V

TRmP

V

V

Now, KkgJmolkgkg

KmolkgJ

MR ./5.2078

./4

../8314

R

.…AnsMJ

X

XK

Kkg

Jkg

P

PTRmW

789.2

100.1

100132.1ln.)293(.

.5.2078.)2(

ln

6

5

2

1

And,



Example 7A 1-gm quantity of Nitrogen undergoes the following sequence of quasi – static processes in a piston – cylinder arrangement.

(a)An adiabatic expansion in which the volume doubles.(b)A constant-pressure process in which the volume is reduced to its initial value.(c)A constant-volume compression back to the initial state.

Nitrogen is initially at 150 °C and 5 atm. Calculate the net work done on the gas in this sequence of process.

P1 = 5 atm = 5.066 X 105 N/m2

T1 = 150 °C = 423 K

m = 1 gm = 10-3 kg

For N2; KkgJmolkgkg

KmolkgJ

MR ./297

./28

../8314

R

Initial Volume : 345

3

1

11 1028.4

10066.5

)423).(297).(10(mX

XP

TRmV

Pre

ssu

re

Volume

1

23



Example 7….Contd

V2 = 2. V1 = 4.96 X 10-4 m3.

Adiabatic Process 1 – 2 :

25

4.15

2

112

/1092.1

2

1.)10066.5(

mNX

XV

VPPCVP

Adiabatic Work :

J

XXXX

VPVPW

)05.76(14.1

)]1048.2).(10066.5()1096.4).(1092.1[(

14545

112221



Example 7….Contd

Constant Pressure Process Work

J

XX

VVPdVPWV

V

62.47

]10)96.448.2[(.)1092.1(

)(..

45

23232

3

2

There is NO work done in Process 3-1, since V3 = V1 (i.e. Constant Volume process).

Total Work in the sequence of these processes is :

J

WWWW

)43.28(

)0()62.47()05.76(

212121

…Ans

Thermodynamics Examples and Class test

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