Answer:
-313.7 KW
Explanation:
Given
Fluid: Steam
Inlet 1:
Pressure [tex]P_{1}[/tex] = 2MPa
Temperature [tex]T_{1}[/tex]= 360°C
Velocity [tex]V_{1}[/tex] = 100 m/s
Exit 2:
Saturated vapor
Pressure [tex]P_{2}[/tex]= 0 .1 MPa
Velocity [tex]V_{2}[/tex] = 50 m/s
Mass flow rate m = 15 Kg/s
The elevation of the inlet [tex]Z_{1} -Z_{2}[/tex] = 3 m
Gravitational force g = 9.81 m/s^2
Power developed W = +7 MW
Required
a) The area at the inlet [tex]A_{1} [m^2][/tex]
b) The rate of heat transfer between the turbine and its surroundings [tex]Q[KW][/tex]
Assumption
Constant average values.
Steady flow.
Solution
a) specific volume for steam at [tex]P_{1}=2 Mpa[/tex] and [tex]T_{1}=360 C[/tex] from table
[tex]v_{1}=0.1411 m^3/kg\\[/tex]
Mass flow rate could be defined by.
[tex]m=\frac{AV}{v}[/tex]
Then area at inlet could be calculated as following.
[tex]A_{1} =\frac{mv_{1} }{V_{1} }=0.021 m^2[/tex]
b)
Energy equation could be defined by.
[tex]Q-W=m[(h_{2} -h_{1} )+\frac{V_{2}^2-V_{1}^2 }{2}+g( Z_{2}-Z_{1} )[/tex]
specific enthalpy for steam at [tex]P_{1}=2 Mpa[/tex] and [tex]T_{1}=360 C[/tex] from table
[tex]h_{1} =3159.3 kJ/kg[/tex]
specific enthalpy for steam at [tex]P_{2}=0.1 Mpa[/tex] and in saturated vapor state from table
[tex]h_{2} =32675.5 kJ/kg[/tex]
Then The rate of heat transfer between the turbine and its surroundings could be calculated as following.
[tex]Q=W+m[(h_{2} -h_{1} )+\frac{V_{2}^2-V_{1}^2 }{2}+g( Z_{2}-Z_{1} )[/tex]
=-313.7 KW
As the sign is negative it means that the heat transfer is from turbine to the surrounding.
