The pressure of water is 7.3851 kPa
<u>Explanation:</u>
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
