TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 241

dharm

/M-therm/th5-1.pm5

(ii)The net work transfer, W = ηcarnot × Q 1 Q ηcarnot

W

Q

L =

N

M

O

Q

P
1
= 0.506 × 90 = 45.54 kJ. (Ans.)
(iii)Heat rejected to the sink, Q 2 = Q 1 – W [Q W = Q 1 – Q 2 ]
= 90 – 45.54 = 44.46 kJ. (Ans.)
Example 5.8. An inventor claims that his engine has the following specifications :
Temperature limits ...... 750°C and 25°C
Power developed ...... 75 kW
Fuel burned per hour ...... 3.9 kg
Heating value of the fuel ...... 74500 kJ/kg
State whether his claim is valid or not.
Solution. Temperature of source, T 1 = 750 + 273 = 1023 K
Temperature of sink, T 2 = 25 + 273 = 298 K
We know that the thermal efficiency of Carnot cycle is the maximum between the specified
temperature limits.

Now, ηcarnot = 1 –

T

T

2

1

= 1 –

298

1023

= 0.7086 or 70.86%

The actual thermal efficiency claimed,

ηthermal = Work done

Heat supplied

= ×××

××

75 1000 60 60

3 9 74500 1000.

= 0.9292 or 92.92%.

Since ηthermal > ηcarnot, therefore claim of the inventor is not valid (or possible). (Ans.)
Example 5.9. A cyclic heat engine operates between a source temperature of 1000°C and a
sink temperature of 40°C. Find the least rate of heat rejection per kW net output of the engine?
Solution. Temperature of source,
T 1 = 1000 + 273 = 1273 K
Temperature of sink,
T 2 = 40 + 273 = 313 K
Least rate of heat rejection per kW net out-
put :
For a reversible heat engine, the rate of heat
rejection will be minimum (Fig. 5.13)
ηmax = ηrev. = 1 –
T
T

2

1

= 1 –

313

1273

= 0.754

Now

W

Q

net

1

= ηmax = 0.754

∴ Q 1 = Wnet

0 754

1

..0 754

= = 1.326 kW

Now Q 2 = Q 1 – Wnet = 1.326 – 1 = 0.326 kW

Hence, the least rate of heat rejection = 0.326 kW. (Ans.)

Fig. 5.13