In a gas turbine power plant if maximum temperature in the cycle is 1200K and the minimum temperature in the cycle is 300K. The isentropic efficiency of compressor and turbine is 0.8 each, then what will be the value of optimum pressure ratio (approximately) for maximum work output in the cycle?
4.096

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Solution

The correct option is A 4.096
Optimum pressure ratio for maximum work cutput,
$(r)_{o p t} = {(η_{c} η_{T} \frac{T_{m a x}}{T_{m i n}})}_{c}^{\frac{γ}{c (γ - 1)}}$
Given $η_{c} = 0.8, η_{T} = 0.8, T_{m a x} = 1200 K, T_{m i n} = 300 K$
$(r_{p})_{o p t} = {(0.8 \times 0.8 \times \frac{1200}{300})}^{\frac{1.5}{2 \times (1.5 - 1)}} = (2.56)^{⎛ ⎝ \frac{1.5}{2 \times 0.5} ⎞ ⎠}$
$(r_{p})_{o p t} = 4.096$

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The correct option is A 4.096Optimum pressure ratio for maximum work cutput, (r)opt=(ηcηTTmaxTmin)γc(γ−1) Given ηc=0.8,ηT=0.8,Tmax=1200K,Tmin=300K (rp)opt=(0.8×0.8×1200300)1.52×(1.5−1)=(2.56)⎛⎝1.52×0.5⎞⎠ (rp)opt=4.096