A piston-cylinder assembly contains ammonia, initially at a temperature of -20°C and a quality of 50%. The ammonia is slowly heated to a final state where the pressure is 6 bar and... A piston-cylinder assembly contains ammonia, initially at a temperature of -20°C and a quality of 50%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 120°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg. Read more

Steps to Solve

1. Determine the initial state properties

Given: $T_1 = -20^\circ C$ and $x_1 = 0.50$. We need to determine the specific volume $v_1$ and specific internal energy $u_1$ at the initial state. From saturated ammonia tables at $-20^\circ C$: $v_f = 0.001504 , m^3/kg$, $v_g = 0.2909 , m^3/kg$ $u_f = 88.77 , kJ/kg$, $u_g = 1287.8 , kJ/kg$

$v_1 = v_f + x_1 (v_g - v_f) = 0.001504 + 0.50(0.2909 - 0.001504) = 0.1462 , m^3/kg$ $u_1 = u_f + x_1 (u_g - u_f) = 88.77 + 0.50(1287.8 - 88.77) = 688.3 , kJ/kg$

2. Determine the final state properties

Given: $P_2 = 6 , bar = 600 , kPa$ and $T_2 = 120^\circ C$. From superheated ammonia tables at $600 , kPa$ and $120^\circ C$: $v_2 = 0.2658 , m^3/kg$ $u_2 = 1569.3 , kJ/kg$.

3. Calculate the work done

Since the pressure varies linearly with specific volume, the work done per unit mass is given by: $w = \int_{v_1}^{v_2} P , dv = \frac{1}{2}(P_1 + P_2)(v_2 - v_1)$

First, determine $P_1$. Since at $T_1 = -20^\circ C$ the ammonia is saturated, $P_1 = P_{sat} = 190.2 , kPa$ (from saturated ammonia tables).

$w = \frac{1}{2}(190.2 + 600)(0.2658 - 0.1462) = \frac{1}{2}(790.2)(0.1196) = 47.26 , kJ/kg$

4. Calculate the heat transfer

Apply the first law of thermodynamics for a closed system: $q - w = u_2 - u_1$ $q = u_2 - u_1 + w = 1569.3 - 688.3 + 47.26 = 928.3 , kJ/kg$