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3 years ago

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Water initially at 200 kPa and 300°C is contained in a piston–cylinder device 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 100 kPa
On the T−v diagram, sketch, with respect to the saturation lines, the process curves passing through the initial, intermediate, and final states of the water. Label the T, P and v values for end states on the process curves. Find the overall change in internal energy between the initial and final states per unit mass of water.

1 answer:

netineya [11]3 years ago

3 0

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

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