Answer:
work required = 205.59 kJ/kg
Explanation:
Given data:
Temperature of water vapor = 150°c
final pressure ( P2 ) = 1000 kPa
specific volume = constant
Determine work required in kJ/kg
we apply the equation below to resolve the problem
[tex]w_{rev} = v ( P1 - P2 )[/tex] ---- ( 1 )
next we have to find the value of the specific volume and saturation pressure of water vapor at 150°c using the saturated water-temperature table
v = specific volume = 0.39248 m^3/kg
P1 = saturation pressure = 476.16 kPa
substitute values into equation 1
[tex]w_{rev} =[/tex] 0.39248 ( 476.16 - 1000 )
= -205.59 kj/kg
hence work required = 205.59 kJ/kg
