In a hypothetical air cycle, consisting of three processes, an adiabatic compression is followed by an isothermal expansion to the initial volume of compression. Finally a heat rejection process completes the cycle. (i) Draw the cycle on p-v and T-s diagrams and derive an expression for thermal

Question

In a hypothetical air cycle, consisting of three processes, an adiabatic compression is followed by an isothermal expansion to the initial volume of compression. Finally a heat rejection process completes the cycle.
(i) Draw the cycle on p-v and T-s diagrams and derive an expression for thermal efficiency of the cycle in terms of compression ratio r.
(ii) Also find the efficiency and m.e.p. of the cycle if the compression ratio is 14 and the induction conditions are 1 bar and 27°C. Take for air, [latex] c_{p} = 1.005 kJ/kg K and c_{v}[/latex] = 0.718 kJ/kg K.

(P.U.)

Accepted Answer

Given : r = 14 ; [latex]p_{1} = 1 bar ; T_{1} = 27 + 273 = 300 K ; c_{p} = 1.005 kJ/kg K ; c_{v}[/latex] = 0.718 kJ/kg K.

(i) The p-v and T-s diagrams of the cycle are shown in Fig. 45.

Consider 1 kg of air.
Consider adiabatic process 1–2 :

[latex] \frac{T_{2}}{T_{1}} = (\frac{v_{1}}{v_{2}})^{γ - 1} = (r)^{γ - 1} ∴ T_{2} = T_{1} (r)^{γ - 1} ; Q_{1–2} = 0[/latex]

Consider isothermal process 2–3 :

[latex] T_{2} = T_{3} [/latex]

[latex] Q_{2–3} = W = R T_{2} \ln(r)[/latex]

Consider heat-rejection (constant volume) process 3–1 :

[latex] Q_{3–1} = c_{v} (T_{3} - T_{1})[/latex]

Efficiency, η = [latex] \frac{Work done }{Heat added} = \frac{Q_{2–3} - Q_{3–1} }{Q_{2–3} } = 1 - \frac{Q_{3–1}}{Q_{2–3} } [/latex]

= [latex] 1 - \frac{c_{v} (T_{3} - T_{1})}{ RT_{2} \ln(r)} = 1 - \frac{c_{v} [T_{1} (r)^{γ - 1} - T_{1} ]}{R T_{1} (r)^{γ - 1} \ln(r)} [/latex]

= [latex] 1 - \frac{c_{v} }{R \ln(r)} [ 1 - \frac{1}{(r)^{γ - 1} }] [/latex].

(ii) [latex] Q_{added} = Q_{2–3} = R T_{2} \ln(r)[/latex]

where, R = [latex] c_{p} – c_{v}[/latex] = 1.005 – 0.718 = 0.287 kJ/kg K,

[latex] T_{2} = T_{1}(r)^{γ - 1} = 300 × (14)^{1.4–1}[/latex] = 300 × 2.874 = 862 K

(∵ γ [latex]= \frac{c_{p} }{c_{v}} = \frac{1.005}{0.718}[/latex] = 1.4)

∴ [latex] Q_{added} = 0.287 × 862 × \ln(14)[/latex] = 652.88 kJ/kg

[latex] Q_{rejected} = c_{v} (T_{3} - T_{1})[/latex] = 0.718(862 – 300) = 403.52 kJ/kg

η = 1 - [latex] \frac{Q_{rejected}}{ Q_{added} } = 1 - \frac{403.52}{652.88} [/latex] = 0.3819 or 38.19%

[ Alternatively : η = 1 - [latex] \frac{0.718}{0.287 × \ln(14) } \left\{ 1 - \frac{1}{(14)^{1.4–1}}\right\}[/latex]
= 1 – 0.9479 × 0.652 = 0.3819 or 38.19 % ]

[latex] W_{net} = Q_{added} - Q_{rejected}[/latex]

= 652.88 – 403.52 = 249.36 kJ/kg

Swept volume, [latex] v_{s} = v_{1} - v_{2} = v_{1} - \frac{v_{1} }{14} = \frac{13 }{14} v_{1} [/latex]

or, [latex] \frac{13 }{14} (\frac{RT_{1}}{p_{1}}) = \frac{13 }{14} × \frac{(0.287 × 1000)}{1 × 10^{5}} [/latex] × 300 = 0.7995 m³/kg

∴ m.e.p. = [latex] \frac{W_{net}}{ v_{s} } = \frac{249.36}{0.7995} [/latex] = 311.89 kN/m² or 3.1189 bar.

Question 8.43: In a hypothetical air cycle, consisting of three processes, ...

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