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12 April, 11:04

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

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  1. 12 April, 11:31
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    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
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