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 describes a thermodynamic process involving ammonia in a piston-cylinder assembly. Initially, the ammonia is at -20°C with a quality of 50%. It is then heated to a final state where the pressure is 6 bar and the temperature is 120°C. The pressure varies linearly with specific volume during the heating process. The question asks to determine the work done and the heat transfer during this process, both in kJ/kg.
Answer
Work done: $w = 26.12 \ kJ/kg$ Heat transfer: $q = 867.32 \ kJ/kg$
Answer for screen readers
Work done: $w = 26.12 \ kJ/kg$ Heat transfer: $q = 867.32 \ kJ/kg$
Steps to Solve
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Determine the initial specific volume ($v_1$) At the initial state, the temperature is -20°C ($T_1 = -20^\circ C$) and the quality is 50% ($x_1 = 0.5$). We can find the saturated liquid and vapor specific volumes ($v_f$ and $v_g$) at -20°C from thermodynamic tables for ammonia. $v_f = 0.001504 \ m^3/kg$ $v_g = 0.4071 \ m^3/kg$ $v_1 = v_f + x_1(v_g - v_f) = 0.001504 + 0.5(0.4071 - 0.001504) = 0.2043 \ m^3/kg$
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Determine the initial pressure ($P_1$) At $T_1 = -20^\circ C$, the saturation pressure $P_{sat}$ can be found from ammonia tables: $P_1 = P_{sat} = 2.911 \ bar$
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Determine the final specific volume ($v_2$) At the final state, the pressure is 6 bar ($P_2 = 6 \ bar$) and the temperature is 120°C ($T_2 = 120^\circ C$). Check if the final state is superheated vapor by comparing $T_2$ with the saturation temperature at 6 bar ($T_{sat} = 17.5^\circ C$). Since $T_2 > T_{sat}$, the final state is superheated vapor. From the superheated ammonia tables, find $v_2$ at $P_2 = 6 \ bar$ and $T_2 = 120^\circ C$: $v_2 = 0.2629 \ m^3/kg$
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Calculate the work done ($w$) Since the pressure varies linearly with specific volume, the work done per unit mass is given by: $w = \int_{v_1}^{v_2} P \ dv = \frac{1}{2}(P_1 + P_2)(v_2 - v_1)$ $w = \frac{1}{2}(2.911 + 6) \times 10^5 \ Pa \times (0.2629 - 0.2043) \ m^3/kg$ $w = \frac{1}{2}(8.911) \times 10^5 \times (0.0586) \ J/kg = 26115.5 \ J/kg = 26.12 \ kJ/kg$
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Determine the initial and final internal energies ($u_1$ and $u_2$) At the initial state, use the saturated liquid and vapor internal energies at -20°C from ammonia tables to find $u_1$: $u_f = 88.76 \ kJ/kg$ $u_g = 1303.6 \ kJ/kg$ $u_1 = u_f + x_1(u_g - u_f) = 88.76 + 0.5(1303.6 - 88.76) = 706.2 \ kJ/kg$ At the final state, from the superheated ammonia tables at $P_2 = 6 \ bar$ and $T_2 = 120^\circ C$: $u_2 = 1547.4 \ kJ/kg$
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Calculate the heat transfer ($q$) From the first law of thermodynamics for a closed system: $q = \Delta u + w = u_2 - u_1 + w$ $q = 1547.4 - 706.2 + 26.12 = 867.32 \ kJ/kg$
Work done: $w = 26.12 \ kJ/kg$ Heat transfer: $q = 867.32 \ kJ/kg$
More Information
The problem involves applying the first law of thermodynamics to a closed system undergoing a process where pressure varies linearly with specific volume. Thermodynamic tables for ammonia are essential to find the specific volume and internal energy at different states.
Tips
A common mistake is to confuse the units of pressure (bar) and specific volume ($m^3/kg$) when calculating work, forgetting to convert bar to Pascal. Also, failing to recognize the final state as superheated vapor and using incorrect property values from the saturated mixture tables can lead to errors.
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