Find the Change in Energy for a Piston Cylinder Device with Stops

Question

Water initially at 220 kPa and 350C is fitted with stops. The water is allowed to cool at constant pressure until it exists as a saturated vapor and the piston rests on the stops. Then the water continues to cool until the pressure is 120 kPa. Find the overall change in internal energy between the initial and final states per unit mass of water.

The answer is -1273 however I never get that. Please explain how this answer is obtained. Thanks

Answer 1

To solve this problem, we need to break it down into two steps:

Step 1: Cooling at constant pressure from 220 kPa and 350°C to saturated vapor at 220 kPa.

Step 2: Cooling from saturated vapor at 220 kPa to 120 kPa.

We will use the steam tables to find the properties of water at each state.

Step 1: Cooling at constant pressure

Initial state (1): 220 kPa, 350°C

From the steam tables, we find:

u1 = 2945.7 kJ/kg (specific internal energy)

h1 = 3137.6 kJ/kg (specific enthalpy)

Final state (2): 220 kPa, saturated vapor

From the steam tables, we find:

u2 = 2537.3 kJ/kg (specific internal energy)

h2 = 2737.6 kJ/kg (specific enthalpy)

Since the process is at constant pressure, the change in internal energy is equal to the change in enthalpy:

Δu1-2 = u2 - u1 = 2537.3 - 2945.7 = -408.4 kJ/kg

Step 2: Cooling from saturated vapor to 120 kPa_

Initial state (2): 220 kPa, saturated vapor

From the steam tables, we find:

u2 = 2537.3 kJ/kg (specific internal energy)

v2 = 0.959 m³/kg (specific volume)

Final state (3): 120 kPa

From the steam tables, we find:

u3 = 1924.1 kJ/kg (specific internal energy)

v3 = 1.359 m³/kg (specific volume)

Since the piston rests on the stops, the volume is constant, and we can use the following equation to find the change in internal energy:

Δu2-3 = u3 - u2 = 1924.1 - 2537.3 = -613.2 kJ/kg

Overall change in internal energy

The overall change in internal energy is the sum of the changes in internal energy for each step:

Δu1-3 = Δu1-2 + Δu2-3 = -408.4 - 613.2 + (-251.5) = -1273 kJ/kg