How to determine work by a fluid?

Question

A fluid enters a machine at 170 m/s, 800 kPa,and with a specific volume of 0.3 m^3/kg. The radiation heat loss is 25 kJ/kg. The internal energy decreases 485 kJ/kg. If the fluid leaves at 330m/s, 138 kPa, and 0.95 m^3/kg, determine the work done by the fluid.

Answer 1

For open systems the 1st law of thermodynamics takes the form, which is specific with respect mass passing through the machine:

Δ(h + g∙z + v²) = q + w

h enthalpy , g acceleration to earth gravity , z altitude, v speed, q heat transferred to fluid, work done by fluid.

Δ denotes the difference between outlet stream and inlet stream to the machine.

or in terms In terms of specific internal energy:

h = u + p∙V (V is here the specific volume)

=>

Δ(u + p∙V + g∙z + v²) = q + w

Therefore:

Δu + Δ(p∙V) + g∙Δz + Δv² = q + w

=>

w = Δu + Δ(p∙V) + g∙Δz + Δv² - q

height difference may be ignored. so:

w = Δu + Δ(p∙V) + Δv² - q

= -485kJ/kg + (138kPa∙0.95m³/kg - 800kPa∙0.3m³/kg) + ( (330m/s)² - (170m/s)² ) - (-25kJ/kg)

= - 485kJ/kg - 108.9kJ/kg + 80kJ/kg + 25kJ/kg

= -488.9kJ/kg

So the work done by the fluid is +488.9kJ/kg.