Consider an air standard cycle in which the air enters the compressor at 1.0 bar and 20°C. The pressure of air leaving the compressor is 3.5 bar and the temperature at turbine inlet is 600°C. Determine per kg of air :
(i) Efficiency of the cycle, ( ii) Heat supplied to air,
(iii) Work available at the shaft, ( iv) Heat rejected in the cooler, and
(v) Temperature of air leaving the turbine.
For air γ = 1.4 and $c_{p}$ = 1.005 kJ/kg K.

Question

Consider an air standard cycle in which the air enters the compressor at 1.0 bar and 20°C. The pressure of air leaving the compressor is 3.5 bar and the temperature at turbine inlet is 600°C. Determine per kg of air :
(i) Efficiency of the cycle, ( ii) Heat supplied to air,
(iii) Work available at the shaft, ( iv) Heat rejected in the cooler, and
(v) Temperature of air leaving the turbine.
For air γ = 1.4 and [latex]c_{p} [/latex] = 1.005 kJ/kg K.

Accepted Answer

Refer Fig. 35.
Pressure of air entering the compressor, [latex]p_{1} [/latex] = 1.0 bar
Temperature at the inlet of compressor, [latex]T_{1} [/latex] = 20 + 273 = 293 K
Pressure of air leaving the compressor, [latex]p_{2} [/latex] = 3.5 bar
Temperature of air at turbine inlet, [latex]T_{3} [/latex] = 600 + 273 = 873 K

(i) Efficiency of the cycle, [latex]η_{cycle} [/latex]:

[latex]η_{cycle} = 1 - \frac{1}{ (r_{p})^{\frac{γ - 1}{γ}}} = 1 - \frac{1}{(3.5)^{\frac{1.4 - 1}{1.4}}} [/latex] = 0.30 or 30%. (∵ [latex]r_{p} = \frac{p_{2}}{p_{1}} = \frac{3.5}{1.0} = 3.5 [/latex])

(ii) Heat supplied to air :
For compression process 1–2, we have

[latex] \frac{T_{2}}{T_{1}} = (\frac{p_{2}}{p_{1}})^{\frac{γ - 1}{γ}} = (\frac{3.5}{1})^{\frac{1.4 - 1}{1.4}}[/latex] = 1.43

∴ [latex]T_{2} = T_{1} × 1.43 = 293 × 1.43 \simeq [/latex] 419 K

∴ Heat supplied to air, [latex]Q_{1} = c_{p} (T_{3} – T_{2}) = 1.005 (\dot{873} - 419) [/latex]= 456.27 kJ/kg.

(iii) Work available at the shaft, W :

We know that, [latex]η_{cycle} = \frac{Work output (W)}{Heat input (Q_{1})}[/latex]

or 0.30 = [latex] \frac{W}{456.27} [/latex] or W = 0.3 × 456.27 = 136.88 kJ/kg

(iv) Heat rejected in the cooler, [latex]Q_{2}[/latex] :
Work output (W) = Heat supplied ([latex]Q_{1}[/latex]) – heat rejected ([latex]Q_{2}[/latex])
∴ [latex]Q_{2} = Q_{1}[/latex] – W = 456.27 – 136.88 = 319.39 kJ/kg.

(v) Temperature of air leaving the turbine, [latex]T_{4} [/latex] :
For expansion (isentropic) process 3–4, we have

[latex] \frac{T_{3}}{T_{4}} = (r_{p})^{\frac{γ - 1}{γ}} = (3.5)^{\frac{1.4 - 1}{1.4}}[/latex] = 1.43

∴ [latex]T_{4} = \frac{T_{3}}{1.43} = \frac{873}{1.43} [/latex] = 610.5 K.

[Check : Heat rejected in the air cooler at constant pressure during the process 4–1 can also be calculated as : Heat rejected = m × [latex] c_{p} (T_{4} – T_{1})[/latex] = 1 × 1.005 × (610.5 – 293) = 319.1 kJ/kg]

Question 8.35: Consider an air standard cycle in which the air enters the c...

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