TITLE.PM5

GAS POWER CYCLES 611

dharm

\M-therm\Th13-1.pm5

(v)Power of the engine, P :
Power of the engine working on this cycle is given by
P = (131.1 – 63.97) × (210/60) = 234.9 kW. (Ans.)
Example 13.4. A reversible engine converts one-sixth of the heat input into work. When
the temperature of the sink is reduced by 70°C, its efficiency is doubled. Find the temperature of
the source and the sink.
Solution. Let, T 1 = temperature of the source (K), and
T 2 = temperature of the sink (K)
First case :
TT
T

12

1

−

=

1

6

i.e., 6 T 1 – 6T 2 = T 1

or 5 T 1 = 6T 2 or T 1 = 1.2T 2 ...(i)

Second case :

TT

T

12

1

−−+[(70 273)] = 1

3

TT

T

12

1

−+ 343

=

1

3

3 T 1 – 3T 2 + 1029 = T 1

2 T 1 = 3T 2 – 1029

2 × (1.2T 2 ) = 3T 2 – 1029 ()Q TT 12 =1. 2

2.4T 2 = 3T 2 – 1029

or 0.6T 2 = 1029

∴ T 2 =^1029

06.

= 1715 K or 1442°C. (Ans.)

and T 1 = 1.2 × 1715 = 2058 K or 1785°C. (Ans.)
Example 13.5. An inventor claims that a new heat cycle will develop 0.4 kW for a heat
addition of 32.5 kJ/min. The temperature of heat source is 1990 K and that of sink is 850 K. Is
his claim possible?
Solution. Temperature of heat source, T 1 = 1990 K
Temperature of sink, T 2 = 850 K
Heat supplied, = 32.5 kJ/min
Power developed by the engine, P = 0.4 kW
The most efficient engine is one that works on Carnot cycle

ηcarnot =

TT

T

12

1

−

= 1990 850

1990

− = 0.573 or 57.3%

Also, thermal efficiency of the engine,

ηth = Work done

Heat supplied

=

0.4

(/)32.5 60 =

0.4 60

32.5

×

= 0.738 or 73.8%

which is not feasible as no engine can be more efficient than that working on Carnot cycle.

Hence claims of the inventor is not true. (Ans.)