Problem: 5 kg of steam contained within a piston-cylinder assembly undergoes an

Question

Problem:
5 kg of steam contained within a piston-cylinder assembly undergoes an expansion from state 1, where the specific internal energy is u1 = 2709.9 kJ/kg, to state 2, where u2 = 2659.6 kJ/kg. During the process, there is heat transfer to the steam with a magnitude of 80 kJ. Also, a paddle wheel transfers energy to the steam by work in the amount of 18.5 kJ. There is no significant change in the kinetic or potential energy of the steam. Determine the energy transfer by work from the steam to the piston during the process, in kJ.
Hint: Use thermodynamic first law.
Answer: between 300 and 400 kJ.

w = J Q

J= Proportionality constant (for BTUs, J= 7782 ft.lbf)

But I'm not really sure where to go from there..

Explanation / Answer

According to first law of thermodynamics,
Energy supplied = Energy stored + Workdone

this can be rearranged as,
Energy supplied = (- Energy utilised) + Workdone

Workdone = Energy utilised + Energy supplied

Here the internal energy of the steam is changed from
u1 = 2709.9 kJ/kg to u2 = 2659.6 kJ/kg
So in the process from 1 to 2 the internal energy of the steam got reduced or in other words, some part of the steam's internal energy is used to develop work

So contribution of internal energy to work output is
du = m (u1 - u2) = 5 (2709.9 - 2659.6) = 251.5 kJ
Total Energy utilised = 251.5 kJ

Now given that the heat input to steam is dQ = 80 kJ
Also the energy given to steam due to paddle is 18.5 kJ
setting the energy inputs due to heat and paddle together, we get
Total Energy supplied to steam as (18.5 + 80) = 98.5 kJ

Total Workdone = Energy utilised + Energy supplied

Total work that the stem can produce = 251.5 + 98.5 = 350 kJ

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