To tackle this problem, we need to analyze the steam's behavior during the expansion process. We have steam at an initial pressure of 7 bar with a dryness fraction of 0.75, which means it is a mixture of saturated liquid and saturated vapor. The steam then expands isometrically (constant volume) to a pressure of 2 bar. Let's break down the calculations step by step.
1. Change in Internal Energy
The change in internal energy (\( \Delta U \)) can be calculated using the formula:
\( \Delta U = m(u_2 - u_1) \)
Where:
- \( m \) = mass of steam (1 kg in this case)
- \( u_1 \) = specific internal energy at the initial state (7 bar, 0.75 dryness fraction)
- \( u_2 \) = specific internal energy at the final state (2 bar)
First, we need to find the specific internal energy values from steam tables:
- At 7 bar, the saturation temperature is approximately 165.03 °C. The specific internal energy for saturated liquid (\( u_f \)) is about 700.2 kJ/kg and for saturated vapor (\( u_g \)) is around 2015.5 kJ/kg.
- Using the dryness fraction, we can calculate \( u_1 \):
\( u_1 = u_f + x(u_g - u_f) \)
\( u_1 = 700.2 + 0.75(2015.5 - 700.2) \)
\( u_1 = 700.2 + 0.75(1315.3) \)
\( u_1 = 700.2 + 986.475 = 1686.675 \text{ kJ/kg} \)
Next, we find \( u_2 \) at 2 bar. The saturation temperature is approximately 120.23 °C. The specific internal energy values are:
- \( u_f \) = 504.1 kJ/kg
- \( u_g \) = 2014.2 kJ/kg
Assuming the steam is still in a vapor state at 2 bar, we can take \( u_2 \) as \( u_g \):
\( u_2 = 2014.2 \text{ kJ/kg} \)
Now we can calculate the change in internal energy:
\( \Delta U = 1(2014.2 - 1686.675) \)
\( \Delta U = 327.525 \text{ kJ} \)
2. Change in Enthalpy
The change in enthalpy (\( \Delta H \)) can be calculated similarly:
\( \Delta H = m(h_2 - h_1) \)
Where:
- \( h_1 \) = specific enthalpy at the initial state
- \( h_2 \) = specific enthalpy at the final state
From the steam tables:
- At 7 bar: \( h_f = 660.5 \text{ kJ/kg} \) and \( h_g = 2015.3 \text{ kJ/kg} \)
Calculating \( h_1 \):
\( h_1 = h_f + x(h_g - h_f) \)
\( h_1 = 660.5 + 0.75(2015.3 - 660.5) \)
\( h_1 = 660.5 + 0.75(1354.8) \)
\( h_1 = 660.5 + 1016.1 = 1676.6 \text{ kJ/kg} \)
At 2 bar, the specific enthalpy values are:
- \( h_f = 504.5 \text{ kJ/kg} \)
- \( h_g = 2014.3 \text{ kJ/kg} \)
Assuming it remains in vapor form, we take \( h_2 = h_g \):
\( h_2 = 2014.3 \text{ kJ/kg} \)
Now we can calculate the change in enthalpy:
\( \Delta H = 1(2014.3 - 1676.6) \)
\( \Delta H = 337.7 \text{ kJ} \)
3. Work Done per Kilogram of Steam
For the work done during the isometric process, we can use the first law of thermodynamics:
\( Q - W = \Delta U \)
Where:
- \( Q \) = heat supplied (550 kJ/kg)
- \( W \) = work done
Rearranging gives us:
\( W = Q - \Delta U \)
\( W = 550 - 327.525 \)
\( W = 222.475 \text{ kJ} \)
Summary of Results
- Change in Internal Energy: \( 327.525 \text{ kJ} \)
- Change in Enthalpy: \( 337.7 \text{ kJ} \)
- Work Done per Kilogram of Steam: \( 222.475 \text{ kJ} \)
This breakdown illustrates how to approach the problem methodically, using steam tables and thermodynamic principles to derive the necessary values. If you have any further questions or need clarification on any of the steps, feel free to ask!
