Question

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.

Answer 1

The pressure of water is 7.3851 kPa

Explanation:

Given data,

V = 150×$10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$  = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$, the containers volume is V, waters mass is m.

$v_{1}$ = 150×$10^{-3}$ $m^{3}$/1

$v_{1}$ = 0.15 $m^{3}$/ 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 $m^{3}$/ Kg and 0.13 $m^{3}$/ Kg.

$T_{1}$= 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$= 0.15 $m^{3}$/ Kg

In saturated water labels for $T_{2}$  = 40°C.

$v_{f}$ = 0.001008 $m^{3}$/ Kg

$v_{g}$ = 19.515 $m^{3}$/ Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$.

In saturated water labels for $T_{2}$  = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa