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A fluid with a pressure of 820 kPa, specific volume of
0.30 m/kg and a velocity of 150 m/s enters a device.
The work done by the fluid is 460 kJ/kg and the heat
loss due to radiation is 28 kl/kg. The exit condition of
the fluid are 130 kPa, 0.82 m/kg and 320 m/s.
Determine the change in internal energy.
W= 0-O PE -4Kr -AWA -AU
-4K -AWA -AU
USING THIS FORMULA
THERMODYNAMICS

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