Answer:
a) 158.4 HP.
b) 1235.6 °F.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:
[tex]m_{in}h_{in}=W_{out}+m_{out}h_{out}[/tex]
Whereas the mass flow is just the same, which means we have:
[tex]W_{out}=m_{out}(h_{out}-h_{in})[/tex]
And the enthalpy and entropy of the inlet stream is obtained from steam tables:
[tex]h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R[/tex]
Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:
[tex]s_{out}= 2.2096BTU/lbm-R[/tex]
Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.
Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:
[tex]h_{out}=1659.15BUT/lbm[/tex]
Then, the isentropic work (negative due to convention) is:
[tex]W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU[/tex]
And the real produced work is:
[tex]W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU[/tex]
Finally, in horsepower:
[tex]W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP[/tex]
Regards!
