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## Homework Statement

12kg of a fluid per minute goes through a reversible steady flow process. The properties of fluid at the inlet are p

_{1}= 1.4bar, ρ

_{1}= 25kg/m

^{3}, C

_{1}= 120m/s and u

_{1}= 920kJ/kg and at the exit are p

_{2}= 5.6bar, ρ

_{2}= 5 kg/m

^{3}, C

_{2}= 180m/s and u

_{2}

## Homework Equations

u

_{1}+ P

_{1}V

_{1}+ (C

_{1})

^{2}/2 + g.Z + Q = u

_{2}+ P

_{2}V

_{2}+ (C

_{2})

^{2}/2 + g.Z + W

u

_{1}+ P

_{1}V

_{1}and u

_{2}+ P

_{2}V

_{2}are both enthalpies

## The Attempt at a Solution

i)

h1-h2 = change in enthalpy

plugging in values gives (920kj/kg + 1.4*V) - (720kj/kg + 5.6*V)

920kj/kg + 1.4*V - 720kj/kg - 5.6*V

200kj/kg - 4.2*V

but how do I get what V is since it wasn't given and I believe it means specific volume. I probably got it wrong, educate me. And also for the second question how do I go about finding W if Q is not given. Thanks a lot [/B]