A piston-cylinder device contains 0.1 [kg] of air at a pressure of 100 [kPa] and a temperature of 400 [K] that undergoes an expansion process. The volume of the piston-cylinder device expands from 1 [m3 ] to 3 [m3 ] at a constant pressure of 2,000 [kPa]. Then, as the piston-cylinder device expands from 3 [m3 ] to 5 [m3 ], the pressure linearly decreases from 2,000 to 1,000 [kPa]. (a) Determine the heat input. (b) Determine the work for the overall process

a) Q_in = 30 KJ

b) W_net, out = 7,000 KJ

Explanation:

Given:

- mass of air m = 0.1 kg

State 1: State 2: State 3:

- P_1 = 2000 KPa P_2 = 2000 KPa P_3 = 1,000 KPa

- T_1 = 400 K V_2 = 3 m^3 V_3 = 5 m^3

- V_1 = 1 m^3

Find:

(a) Determine the heat input.

(b) Determine the work for the overall process

Solution:

- Process 1: Constant pressure, Expansion Heat addition.

Using Ideal Gas Law for Air.

P_1*V_1 / T_1 = P_2*V_2 / T_2

Where, P_1 = P_2,

V_1 / T_1 = V_2 / T_2

T_2 = (V_2 / V_1) * T_1

T_2 = (3 / 1) * 400 = 700 K

The amount of Heat input Q_in:

Q_in = m*c_p * (T_2 - T_1)

Q_in = 0.1*1 * (700 - 400) = 30 KJ

The amount of work done by the system in the process:

W_out = P_1 * (V_2 - V_1)

W_out = 2000 * (3 - 1) = 4,000 KJ

- Process 2: Linear Expansion.

W_out = Area under the P-V graph

W_out = Area of Trapezium

W_out = 0.5 * (P_ 1 + P_2) * (V_3 - V_2)

W_out = 0.5 * (3000) * (2)

W_out = 3,000 KJ

- The total work for overall process is:

W_net, out = 4,000 + 3,000 = 7,000 KJ