Answer:
[tex]\dot Q_{out} = 13369.104\,kW[/tex]
Explanation:
The turbine is modelled after the First Law of Thermodynamics:
[tex]-\dot Q_{out} - \dot W_{out} + \dot H_{in} - \dot H_{out} + \dot K_{in} - \dot K_{out} + \dot U_{in} - \dot U_{out} = 0[/tex]
The rate of heat transfer between the turbine and its surroundings is:
[tex]\dot Q_{out} = \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out} - \dot W_{out} + \dot U_{in} - \dot U_{out}[/tex]
The specific enthalpies at inlet and outlet are, respectively:
[tex]h_{in} = 3076.41\,\frac{kJ}{kg}[/tex]
[tex]h_{out} = 2675.0\,\frac{kJ}{kg}[/tex]
The required output is:
[tex]\dot Q_{out} = \left(8\,\frac{kg}{s} \right)\cdot \left\{3076.41\,\frac{kJ}{kg}-2675.0\,\frac{kJ}{kg}+\frac{1}{2}\cdot \left[\left(65\,\frac{m}{s} \right)^{2}-\left(42\,\frac{m}{s} \right)^{2}\right] + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m) \right\} - 8000\,kW[/tex][tex]\dot Q_{out} = 13369.104\,kW[/tex]
