Answer:
Temperature at the exit = [tex]267.3 C[/tex]
Explanation:
For the steady energy flow through a control volume, the power output is given as
[tex]W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})[/tex]
Inlet area of the turbine = [tex]60cm^{2}= 0.006m^{2}[/tex]
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume [tex]v_{1}[/tex]
as
[tex]v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg[/tex]
[tex]m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec[/tex]
for Ideal gasses, the enthalpy change can be calculated using the formula
[tex]h_{2}-h_{1}=C_{p}(T_{2}-T_{1})[/tex]
hence we have
[tex]W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})[/tex]
[tex]250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})[/tex]
Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000
evaluating the above equation, we have [tex]T_{2}=267.3C[/tex]
Hence, the temperature at the exit = [tex]267.3 C[/tex]