Question

Saturated water vapor at 140°C is compressed in a reversible, steady-flow device to 895 kPa while its specific volume remains constant. Determine the work required.

Answer 1

Answer:

The work required to compress the saturated water vapor to 895 kPa pressure is 130.9540 k J/Kg

Explanation:

Given data in question

temperature = 140°C

pressure  (P2) = 895 kPa

To find out

work required for compress saturated water

Solution  

We know the equation for reversible work for compress saturated water vapor

i.e.  

W =  $-\int_{1}^{2}vdP-\Delta ke - \Delta pe$

w is  reversible work, v is specific volume, P is water vapor pressure and

ke is kinetic energy and pe is potential energy

and in question we have given v is constant so ke and pe will be zero

so  

W =  $-\int_{1}^{2}vdP$

W =  -v( P2 - P1 )

we can given in question temperature = 140°C and use steam table "A-4 saturated water - temperature table"

at this water property P1 will be 361.53 kPa and v will be 0.50850 m³/kg

so put these value in above equation

W =  -0.50850( 104 - 361.53 )

W = 130.9540 kJ/Kg