How to Calculate Work Transfer in Steady Flow Conditions

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niles_uk
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Hi, first post so here goes:
I'm having a little trouble with a question on my tutorial, and I am not sure what to do. Here it is:
Steam under steady flow conditions has the following states:

Initial Conditions:
Pressure - 47 Bar
Spec Volume - 0.09957 m^3 / kg
Internal energy - 3420 kJ/kg
Velocity - 21 m/s

Final Conditions:
Pressure - 10 Bar
Spec Volume - 0.1636 m^3 /kg
Internal Energy - 2440 kJ/kg
Velocity - 38 m/s

Heat is added to the surroundings at the rate of 0.75 kJ/s, if the rate of steam flow is 0.36kg/s calculate the rate of work transfer

I have been given the formula:

(C1^2 /2) + u1 +p1v1 + q =(C2^2 /2) +u2 +p2v2 + w


thanks for any help in advance,
chris
 
on Phys.org
im guessing the
u is internal energy,
v is velocity
p is pressure (in N/m)
c is specific volume?

right/wrong?