One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Determine the initial temperature and the final pressure of the water.

The pressure of water is 7.3851 kPa

Explanation:

Given data,

V = 150×

m = 1 Kg

= 2 MPa

 = 40°C

The waters specific volume is calculated:

= V/m

Here, the waters specific volume at initial condition is , the containers volume is V, waters mass is m.

= 150×/1

= 0.15 / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 / Kg and 0.13 / Kg.

= 350+(400-350)

= 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

= = 0.15 / Kg

In saturated water labels for  = 40°C.

= 0.001008 / Kg

= 19.515 / Kg

The final state is two phase region < < .

In saturated water labels for  = 40°C.

= = 7.3851 kPa

= 7.3851 kPa

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