Question

A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar and a quality of 0.2. The tank is heated at a constant rate while a pressure-regulating valve is used to maintain a constant tank pressure by allowing saturated vapor to escape. Kinetic and potential energy effects can be neglected. If the tank is heated until the quality reaches 0.5, determine the final mass (kg) within the tank and the total amount of heat transfer required (kJ).

Answer 1

Answer:

a) $m_{2} = 0.753\,kg$, b) $Q_{in} = 2122.963\,kJ$

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

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.02726\,\frac{m^{3}}{kg}$

$u = 1192.94\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.2$

State 2 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.06643\,\frac{m^{3}}{kg}$

$u = 1718.12\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.5$

The model for the rigid tank is created by using the First Law of Thermodynamics:

$Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}$

Initial and final masses are:

$m_{1} = \frac{V_{1}}{\nu_{1}}$

$m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }$

$m_{1} = 1.834\,kg$

$m_{2} = \frac{V_{2}}{\nu_{2}}$

$m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }$

$m_{2} = 0.753\,kg$

a) The final mass within the tank is:

$m_{2} = 0.753\,kg$

b) The total amount of heat transfer is:

$Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}$

$Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )$

$Q_{in} = 2122.963\,kJ$