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GAS POWER CYCLES 607

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(iii)Entropy change during the heat rejection process, (S 3 – S 4 ) :

Heat rejected = Heat added – Work done

= 243 – 130 = 113 kJ

673 K

313 K

12

4 3

Isotherms

Isentropics

T

S

Fig. 13.2

Heat rejected = T 3 (S 3 – S 4 ) = 113

∴ (S 3 – S 4 ) =

113 113

T 3 313

= = 0.361 kJ/K. (Ans.)

Example 13.2. 0.5 kg of air (ideal gas) executes a Carnot power cycle having a thermal
efficiency of 50 per cent. The heat transfer to the air during the isothermal expansion is 40 kJ. At
the beginning of the isothermal expansion the pressure is 7 bar and the volume is 0.12 m^3.
Determine :
(i) The maximum and minimum temperatures for the cycle in K ;
(ii)The volume at the end of isothermal expansion in m^3 ;
(iii)The heat transfer for each of the four processes in kJ.
For air cv = 0.721 kJ/kg K, and cp = 1.008 kJ/kg K. (U.P.S.C. 1993)
Solution. Refer Fig. 13.3. Given : m = 0.5 kg ; ηth = 50% ; Heat transferred during isothermal
expansion = 40 kJ ; p 1 = 7 bar, V 1 = 0.12 m^3 ; cv = 0.721 kJ/kg K ; cp = 1.008 kJ/kg K.
(i)The maximum and minimum temperatures, T 1 , T 2 :
p 1 V 1 = mRT 1
7 × 10^5 × 0.12 = 0.5 × 287 × T 1

∴ Maximum temperature, T 1 =

710 12

5 287

××^5

×

0.

0. = 585.4 K. (Ans.)

ηcycle = TT

T

12

1

− ⇒ 0.5 = 585 4

585 4

. 2
.

−T

∴ Minimum temperature, T 2 = 585.4 – 0.5 × 585.4 = 292.7 K. (Ans.)