Respuesta :
Explanation:
The given data is as follows.
Pressure of steam at inlet of turbine, [tex]P_{1}[/tex] = 20 bar
Temperature at inlet of turbine, T = [tex]450^{o}C[/tex]
Pressure at outlet of turbine, [tex]P_{2}[/tex] = 10 bar
Mass flow rate of steam, m = 200 kg/min
Work produced by the turbine, [tex]W_{s}[/tex] = 1500 kW
Steam is heated at constant pressure to its initial temperature, i.e., temperature at outlet of heat exchanger, [tex]T_{3}[/tex] = [tex]450^{o}C[/tex].
(1) For an adiabatic turbine, the energy balance is as follows.
[tex]-W_{s} = m({H}_{2} - {H}_{1})[/tex]
where [tex]W_{s}[/tex] = work done by the turbine
m = mass flow rate of steam
[tex]H_{1}[/tex] and [tex]H_{2}[/tex] are the specific enthalpy of steam at inlet and outlet conditions of turbine.
Obtain the specific enthalpy of steam from Properties of Superheated Steam table
At 20 bar and [tex]450^{o}C[/tex], [tex]H_{1}[/tex] = 3358 kJ/kg
[tex]H_{2} = H_{1} - \frac{W_{s}}{m}[/tex]
[tex]H_{2} = 3358 kJ/kg - \frac{1500kJ/s \times 60 s/min}{200 kg/min}[/tex]
[tex]H_{2}[/tex] = 2908 kJ/kg
For P = 10 bar, [tex]H[/tex] =2875 kJ/kg for T= [tex]200^{0}C[/tex] and H = 2975 kJ/kg for [tex]T=250^{o}C[/tex]. Interpolate the values.
The temperature corresponding to P = 10 bar and [tex]H_{2}[/tex] = 2908 kJ/kg is T = [tex]216.5^{o}C[/tex]
Therefore, the outlet temperature is [tex]T_{2} = 216.5^{o}C[/tex].
(2) Energy balance on the heater is as follows.
Q = [tex]\Delta H[/tex] = [tex]m(H_{3} - H_{2})[/tex]
where, Q = heat input required by the steam
[tex]\Delta H[/tex] = specific enthalpy change
[tex]H_{3}[/tex] = specific enthalpy of steam at the outlet conditions of heat exchanger
At P = 10 bar and [tex]T_{3}[/tex] = [tex]450^{o}C[/tex], [tex]H_{3}[/tex] = 3371 kJ/kg.
Q = [tex]\frac{200 kg/min}{60 s/min} \times (3371 - 2908)kJ/kg[/tex]
[/tex]
Q = 1543.33 kJ/s
or, Q = 1543.33 kW
Therefore, the heat input required is Q = 1543.33 kW.
