Answer:
a) [tex]m_{2} = 0.753\,kg[/tex], b) [tex]Q_{in} = 2122.963\,kJ[/tex]
Explanation:
A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:
State 1 - Gas-Vapor Mixture
[tex]P = 1500\,kPa[/tex]
[tex]T = 198.29^{\textdegree}C[/tex]
[tex]\nu = 0.02726\,\frac{m^{3}}{kg}[/tex]
[tex]u = 1192.94\,\frac{kJ}{kg}[/tex]
[tex]h_{g} = 2791.0\,\frac{kJ}{kg}[/tex]
[tex]x = 0.2[/tex]
State 2 - Gas-Vapor Mixture
[tex]P = 1500\,kPa[/tex]
[tex]T = 198.29^{\textdegree}C[/tex]
[tex]\nu = 0.06643\,\frac{m^{3}}{kg}[/tex]
[tex]u = 1718.12\,\frac{kJ}{kg}[/tex]
[tex]h_{g} = 2791.0\,\frac{kJ}{kg}[/tex]
[tex]x = 0.5[/tex]
The model for the rigid tank is created by using the First Law of Thermodynamics:
[tex]Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}[/tex]
Initial and final masses are:
[tex]m_{1} = \frac{V_{1}}{\nu_{1}}[/tex]
[tex]m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }[/tex]
[tex]m_{1} = 1.834\,kg[/tex]
[tex]m_{2} = \frac{V_{2}}{\nu_{2}}[/tex]
[tex]m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }[/tex]
[tex]m_{2} = 0.753\,kg[/tex]
a) The final mass within the tank is:
[tex]m_{2} = 0.753\,kg[/tex]
b) The total amount of heat transfer is:
[tex]Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}[/tex]
[tex]Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )[/tex]
[tex]Q_{in} = 2122.963\,kJ[/tex]
