Respuesta :
Our values,
[tex]V=380L = 380*10^{-3}m^3/s[/tex]
[tex]\dot{m} = 0.005kg/s[/tex]
[tex]P_1 = 3bar[/tex]
[tex]T_1=T_2=400\°c[/tex]
[tex]m_f=75\%[/tex]
We can know apply the properties of the superheated water,
[tex]P_1 = 3Bar[/tex]
[tex]T_1=T_2=400\°c[/tex]
So we have that
[tex]v_1=1.032m^3/kg[/tex]
And we can find the initial mass of the tank,
[tex]m_1 = \frac{V}{v}[/tex]
[tex]m_1 = \frac{380*10^{-3}}{1.032}[/tex]
[tex]m_1 = 0.368kg[/tex]
How the final mass is 75%, then
[tex]m_2 = 0.75*0.368 = 0.276 kg[/tex]
The time required is given by the change in the mass vs the rate of the mass, so
[tex]t= \frac{0.368-0.279}{0.005}[/tex]
[tex]t=17.8s[/tex]
Finally we have the specific volume in the tank, thatis
[tex]v_2 = \frac{V}{m_2} = \frac{380*10^{-3}}{0.276}[/tex]
[tex]v_2 = 1.3768m^3/kg[/tex]
From the properties of water in the table we can find the [tex]P_2[/tex], that is,
At [tex]v_2 = 1.378m^3/kg[/tex] and [tex]T_1 = T_2 =400\°c[/tex]
[tex]P_2 = 2.8bar[/tex]
Explanation:
The given data is as follows.
Tank volume (V) = 380 L = 0.38 [tex]m^{3}[/tex]
Initial pressure ([tex]P_{i}[/tex]) = 3 bar
Temperature (t) = [tex]400^{o}C[/tex]
Outlet mass flow rate ([tex]m_{o)}[/tex]) = 0.005 kg/s
Final mass ([tex]m_{f}[/tex]) = [tex]0.75m_{i}[/tex]
First, we will calculate the initial mass as follows.
[tex]m_{i} = \frac{V}{v_{i}}[/tex]
= [tex]\frac{0.38}{1.032}[/tex]
= 0.368 kg
and, [tex]m_{f} = m_{i} = 0.75 \times 0.368 = 0.276 kg/s[/tex]
also, [tex]\Delta m = 0.25m_{i} = 0.25 \times 0.368 = 0.092[/tex]
As, [tex]\Delta m = m_{o} \Delta t[/tex]
[tex]\Delta t = \frac{\Delta m}{m_{o}}[/tex]
= [tex]\frac{0.092}{0.005}[/tex]
= 18.4 s
[tex]v_{f} = \frac{V}{v_{f}}[/tex]
= [tex]\frac{0.38}{0.276}[/tex]
= 1.376 [tex]m^{3}/kg[/tex]
According to the table A-4, value of pressure at [tex]v_{f} = 1.376 m^{3}/kg[/tex] and T = [tex]400^{o}C[/tex] is 2.5 bar.
Therefore, value of [tex]-\Delta t[/tex] is 18.4 s, [tex]-v_{f}[/tex] is 1.376 [tex]m^{3}/kg[/tex] and [tex]-P_{f}[/tex] is 2.5 bar.
