Question

An air-standard cycle with constant specific heats at room temperature is executed in a closed system with 0.003 kg of air and consists of the following three processes:
1–2 v = Constant heat addition from 95 kPa and 17°C to 380 kPa
2–3 Isentropic expansion to 95 kPa
3–1 P = Constant heat rejection to initial state

The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4

a) Show cycle on P-v and T-s diagrams
b) Calculate net work per cycle in kJ
c) Determine thermal efficiency

Answer 1

Answer:

A) I attached the diagrams

B)W_net = 0.5434 KJ

C) η_th = 0.262

Explanation:

A) I've attached the P-v and T-s diagrams

B) The temperature at state 2 can be calculated from ideal gas equation at constant specific volume;

So; P2/P1 = T2/T1

Thus, T2 = P2•T1/P1

We are given that;

P2 = 380 KPa

P1 = 95 KPa

T1 = 17 °C = 17 + 273K = 290K

Thus,

T2 = (380 x 290)/95

T2 = 1160 K

While the temperature at state 3 will be gotten from;

T3 = T2 x (P3/P2)^((γ - 1)/γ)

Where γ = cp/cv = 1.005/0.718 = 1.4

Thus;

T3 = 1160 (95/380)^((1.4 - 1)/1.4)

T3 = 780.6 K

Now, net work done is given by the formula;

W_net = Q_in - Q_out

W_net = Q_1-2 - Q_3-1

W_net = m(u2 - u1) - m(h3 - h1)

W_net = m(u2 - u1 - h3 + h1)

From the first table i attached,

At T1 = 290K, u1 = 206.91 KJ/Kg and h1 = 290.16 KJ/Kg

At T2 = 1160K,u2 = 897.91 KJ/Kg

At T3 = 780K, h3 = 800.03 KJ/Kg

We are also given that m = 0.003 kg

Thus;

W_net = 0.003(897.91 - 206.91 - 800.03 + 290.16)

W_net = 0.5434 KJ

C) The thermal efficiency is given by the formula ;

η_th = W_net/Q_in

η_th = 0.5434/(m(u2 - u1))

η_th = 0.5434/(0.003(897.91 - 206.91))

η_th = 0.262