Respuesta :
Answer:
2285kw
Explanation:
since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.
Mathematically,
[tex]\\ E_{inflow} = E_{outflow}[/tex]
Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy [tex]\\ E_{inflow} = m_{1} h_{1}[/tex] .
[tex]\\[/tex]
[tex]E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}[/tex]
[tex] \\[/tex]
Where [tex] m_{1} h_{1}[/tex] are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa [tex]\\[/tex],
[tex]m_{2} h_{2}[/tex] are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively.[tex]\\[/tex],
and [tex]m_{3} h_{3}[/tex] are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively.[tex]\\[/tex],
We can now express write out the required equation by substituting the new expression for the energies [tex]\\[/tex]
[tex]m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}[/tex] [tex]\\[/tex]
from the above equation, the unknown are the enthalpy values and the mass flow rate. [tex]\\[/tex]
first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. [tex]\\[/tex]
It is more convenient to start from outlet 3 were we have a temperature [tex]100^{0}C[/tex] and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have [tex]\\[/tex]
[tex]h_{3}[/tex] = 2682.4 KJ/KG , at this point also from the table the entropy value ,[tex]s_{3}[/tex] value is 7.6953 KJ/Kg.K. [tex]\\[/tex]
Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. [tex]\\[/tex]
So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg.[tex]\\[/tex]
Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. [tex]\\[/tex]
Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e [tex]m_{1} = m_{2} + m_{3}[/tex] [tex]\\[/tex]
From the question the, the mass flow rate at the inlet [tex]m_{1}}[/tex] is 2Kg/s [tex]\\[/tex]
Since 5% flow is delivered into the feedwater heating, [tex]\\[/tex]
[tex]m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s[/tex] [tex]\\[/tex]
Also for the outlet 3 the remaining 95% will flow out. Hence
[tex]m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s[/tex] [tex]\\[/tex]
Now, from [tex]m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}[/tex] [tex]\\[/tex] we substitute values
[tex]W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}[/tex]
[tex]W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)[/tex]
[tex]\\[/tex]
[tex]W_{out} = 2285.19 kW[/tex].
Hence the power produced is 2285kW
