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.
Understand the Problem
The question is asking to calculate the work and heat transfer for ammonia in a piston-cylinder assembly undergoing a thermodynamic process. The initial and final states are defined, and the process is described as a linear pressure-specific volume relationship. We need to use thermodynamics principles to find the work done and the heat transfer during this process.
Answer
$W = -17.42$ kJ/kg $Q = 954.45$ kJ/kg
Answer for screen readers
Work done: $W = -17.42$ kJ/kg Heat transfer: $Q = 954.45$ kJ/kg
Steps to Solve
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Determine the initial state properties
Given: $T_1 = -20^\circ C$ and $x_1 = 0.50$. We need to find $P_1$ and $v_1$ at this state using ammonia tables. From the saturated ammonia tables at $-20^\circ C$: $P_{sat} = 190.2$ kPa, $v_f = 0.001504 m^3/kg$, and $v_g = 0.62131 m^3/kg$. Therefore, $P_1 = 1.902$ bar. $v_1 = v_f + x_1(v_g - v_f) = 0.001504 + 0.50(0.62131 - 0.001504) = 0.3114 m^3/kg$. Also, we need $u_f = -54.54$ kJ/kg and $u_g = 1244.8$ kJ/kg. Then $u_1 = u_f + x_1(u_g - u_f) = -54.54 + 0.50(1244.8 - (-54.54)) = 595.63$ kJ/kg.
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Determine the final state properties
Given: $P_2 = 6$ bar and $T_2 = 120^\circ C$. We need to find $v_2$ and $u_2$ at this state. From the superheated ammonia tables at 6 bar and $120^\circ C$, we find $v_2 = 0.2673 m^3/kg$ and $u_2 = 1567.5$ kJ/kg.
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Calculate the work done
Since the pressure varies linearly with specific volume, the work done is given by: $W = \int_{v_1}^{v_2} P ,dv = \frac{1}{2}(P_1 + P_2)(v_2 - v_1)$. Converting $P_1$ to bars: $P_1 = 1.902$ bar. $W = \frac{1}{2}(1.902 + 6)(0.2673 - 0.3114) = \frac{1}{2}(7.902)(-0.0441) = -0.1742$ bar$\cdot m^3/kg$. Convert this to kJ/kg: $W = -0.1742 \times 100 = -17.42$ kJ/kg.
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Calculate the heat transfer
Apply the first law of thermodynamics for a closed system: $Q - W = \Delta U = U_2 - U_1$. Therefore, $Q = (U_2 - U_1) + W$. $Q = (1567.5 - 595.63) + (-17.42) = 971.87 - 17.42 = 954.45$ kJ/kg.
Work done: $W = -17.42$ kJ/kg Heat transfer: $Q = 954.45$ kJ/kg
More Information
The negative sign for work signifies that work is done on the system. The large positive value for heat transfer signifies a significant amount of energy input to the system as heat to achieve the change in state.
Tips
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Incorrectly reading property values from thermodynamic tables. Always double-check the units and the specific table you are using (saturated vs. superheated).
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Forgetting to convert units. In this case, ensuring that pressure is in the correct units (bar) and converting the work from bar$\cdot m^3/kg$ to kJ/kg is crucial.
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Getting the sign of the work wrong. Since the volume decreases, work is done on the system, hence negative.
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