Answer: [tex]\dot m_{in} = 23.942 \frac{kg}{s}[/tex], [tex]\dot H_{out} = 39632.62 kW[/tex]
Explanation:
Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to [tex]\dot V = 1 \frac{m^{3}}{s}[/tex]. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:
Principle of Mass Conservation
[tex]\dot m_{in} - \dot m_{out} = 0[/tex]
First Law of Thermodynamics
[tex]- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0[/tex]
This 2 x 2 System can be reduced into one equation as follows:
[tex]-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0[/tex]
The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:
Inflow (Superheated Steam)
[tex]\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}[/tex]
The mass flow rate can be calculated by using this expression:
[tex]\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}[/tex]
[tex]\dot m_{in} = 23.942 \frac{kg}{s}[/tex]
Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:
[tex]h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}[/tex]
[tex]h_{out} = 1655.36 \frac{kJ}{kg}[/tex]
The enthalpy rate at outflow is:
[tex]\dot H_{out} = \dot m \cdot h_{out}[/tex]
[tex]\dot H_{out} = 39632.62 kW[/tex]
