Saturated water vapor at 160°C is compressed in a reversible steady-flow device to 1200 kPa while its specific volume remains constant. Determine the work required in kJ/kg.
The correct answer and explanation is :
To determine the work required in a reversible steady-flow device, we need to consider the conditions given in the problem:
- Initial state: Saturated water vapor at 160°C.
- Final state: The pressure is 1200 kPa, and the specific volume remains constant.
Step 1: Identify the Initial State Properties
From the steam tables for saturated water vapor at 160°C:
- The saturated vapor pressure at 160°C is approximately 474.35 kPa.
- The specific volume of saturated vapor ( v_{g} ) at 160°C is about 0.3928 m³/kg.
Step 2: Determine the Specific Volume
Since the specific volume remains constant throughout the process, we use the specific volume of the saturated vapor at 160°C, which is ( v = 0.3928 \, \text{m}^3/\text{kg} ).
Step 3: Use the Specific Volume to Find the Final Temperature
Since the volume is constant and the pressure changes, we can use the relationship between pressure and specific volume for an ideal gas, or we can check the steam tables at the final pressure (1200 kPa).
At 1200 kPa, the corresponding specific volume of steam in the superheated region is about 0.3928 m³/kg at a certain temperature. Using the steam tables, we find that the temperature corresponding to this specific volume and pressure (1200 kPa) is approximately 350°C.
Step 4: Apply the First Law of Thermodynamics for Steady-Flow Devices
In a steady-flow process, the work done is given by:
[
W = \int_{V_1}^{V_2} P \, dV
]
Since the specific volume ( v ) is constant, this simplifies to:
[
W = P \Delta V = P(v_2 – v_1)
]
But since ( v_1 = v_2 ), the change in volume ( \Delta V ) is zero, implying that no work is done. Therefore, in this case, the work required in the system is zero.
Conclusion:
The work required in this case is zero kJ/kg because the specific volume remains constant, and there is no volume change during the process. Thus, no work is done on or by the system in this steady-flow process.