TITLE.PM5

GAS POWER CYCLES 691

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\M-therm\Th13-6.pm5

Hence,
Compressor work input, Wcompressor = cp (T 2 ′ – T 1 )
= 1.005 (529 – 288) = 242.2 kJ/kg
Turbine work output, Wturbine = cp (T 3 – T 4 ′)
= 1.11 (883 – 621.4) = 290.4 kJ/kg.
∴ Net work output, Wnet = Wturbine – Wcompressor
= 290.4 – 242.2 = 48.2 kJ/kg
Power in kilowatts = 48.2 × 16 = 771.2 kW. (Ans.)
Example 13.42. Calculate the thermal efficiency and work ratio of the plant is example
5.2, assuming that cp for the combustion process is 1.11 kJ/kg K.
Solution. Heat supplied = cp(T 3 – T 2 ′)
= 1.11 (883 – 529) = 392.9 kJ/kg
ηthermal = Net work output
Heat supplied

48.2

392.9

= = 0.1226 or 12.26%. (Ans.)

Now, Work ratio =

Net work output

Gross work output turbine

= 48.2

W

=

48 2

290 4

.

. = 0.166. (Ans.)
Example 13.43. In a constant pressure open cycle gas turbine air enters at 1 bar and 20°C
and leaves the compressor at 5 bar. Using the following data : Temperature of gases entering the
turbine = 680°C, pressure loss in the combustion chamber = 0.1 bar, ηcompressor = 85%, ηturbine
= 80%, ηcombustion = 85%, γ = 1.4 and cp = 1.024 kJ/kg K for air and gas, find :
(i)The quantity of air circulation if the plant develops 1065 kW.
(ii)Heat supplied per kg of air circulation.
(iii)The thermal efficiency of the cycle.
Mass of the fuel may be neglected.
Solution. Given : p 1 = 1 bar, p 2 = 5 bar, p 3 = 5 – 0.1 = 4.9 bar, p 4 = 1 bar,
T 1 = 20 + 273 = 293 K, T 3 = 680 + 273 = 953 K,

T(K)

293

1

2 ′

2 4 ′

4

3

s

1 bar

953

5 bar4.9 bar

Fig. 13.61