Respuesta :
The pressure of water is 7.3851 kPa
Explanation:
Given data,
V = 150×[tex]10^{-3}[/tex] [tex]m^{3}[/tex]
m = 1 Kg
[tex]P_{1}[/tex] = 2 MPa
[tex]T_{2}[/tex] = 40°C
The waters specific volume is calculated:
[tex]v_{1}[/tex] = V/m
Here, the waters specific volume at initial condition is [tex]v_{1}[/tex], the containers volume is V, waters mass is m.
[tex]v_{1}[/tex] = 150×[tex]10^{-3}[/tex] [tex]m^{3}[/tex]/1
[tex]v_{1}[/tex] = 0.15 [tex]m^{3}[/tex]/ 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 [tex]m^{3}[/tex]/ Kg and 0.13 [tex]m^{3}[/tex]/ Kg.
[tex]T_{1}[/tex]= 350+(400-350) [tex]\frac{0.15-0.13}{0.1522-0.1386}[/tex]
[tex]T_{1}[/tex] = 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
[tex]v_{2}[/tex] = [tex]v_{1}[/tex]= 0.15 [tex]m^{3}[/tex]/ Kg
In saturated water labels for [tex]T_{2}[/tex] = 40°C.
[tex]v_{f}[/tex] = 0.001008 [tex]m^{3}[/tex]/ Kg
[tex]v_{g}[/tex] = 19.515 [tex]m^{3}[/tex]/ Kg
The final state is two phase region [tex]v_{f}[/tex] < [tex]v_{2}[/tex] < [tex]v_{g}[/tex].
In saturated water labels for [tex]T_{2}[/tex] = 40°C.
[tex]P_{2}[/tex] = [tex]P_{Sat}[/tex] = 7.3851 kPa
[tex]P_{2}[/tex] = 7.3851 kPa
The initial temperature and the final pressure of the water is "[tex]395.2^{\circ}\ C[/tex] and [tex]7.3851 \ kPa[/tex] "
Firstly, the particular load is calculated using the mass and volume of the water:
[tex]\to \alpha =\frac{V}{m}[/tex]
[tex]=\frac{150 \times 10^{-3}}{1}\ \frac{m^3}{kg}\\\\= 0.15 \frac{m^3}{kg}[/tex]
Using interpolation, the initial temperature is calculated from A-6 with the specified pressure and volume:
[tex]\to T_1 =T_1^{*} +\frac{T_2^*-T_1^*}{\alpha_2^*- \alpha_1^*} (\alpha- \alpha_1^*)[/tex]
[tex]= 350^{\circ} C + \frac{400-350}{0.15122 - 0.1386} \times (0.15 -0.1386)^{\circ} C\\\\ = 350^{\circ} C + \frac{50}{0.01262} \times (0.0114)^{\circ} C\\\\ = 350^{\circ} C + 50 \times 0.9033^{\circ} C\\\\= 350^{\circ} C +45.165 ^{\circ} C\\\\ = 395.165 \approx 395.2^{\circ}\ C\\\\[/tex]
The total pressure at the final temperature determined from A-5 is just the final pressure.
[tex]\to P_2 = 7.3851 \ kPa\\\\[/tex]
Find out more information about the initial temperature and pressure of the water here:
brainly.com/question/20913725
