Saturated water vapor at 150°C is compressed in a reversible steady-flow device to 1000 kPa while its specific volume remains constant. Determine the work required in kJ/kg.

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

----  ( 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

0.39248 ( 476.16 - 1000 )

       = -205.59 kj/kg

hence work required = 205.59 kJ/kg

efficiency of reversible power cycles operating between a hot re

explanation:

answer:

a

explanation: