8 kg of steam is contained in a piston-cylinder assembly with a paddle wheel, as

Question

8 kg of steam is contained in a piston-cylinder assembly with a paddle wheel, as shown in the figure. The paddle wheel draws a power of 0.5 kW, which is perfectly transmitted to the steam inside the piston-cylinder assembly. In a process that takes 42 seconds, the piston-cylinder undergoes an expansion from 50 bar to 1 bar. Moreover, in the process, 10.3 kJ of heat is absorbed (transferred from the surroundings to the steam). If the pressure-volume relationship for the steam is PVn = constant, what is the change in the internal energy of the steam? The initial volume is 1.8 liters. For this process, n=1.4. Give your answer in kJ and round to the nearest tenth. Assumptions:The system is closed and there is no change in the kinetic or potential energy of the system.

Explanation / Answer

Q =heat transfered = 0.5*42 Kj + 10.3 Kj
=31.3 KJ
p1 =50 bar , p2 =1 bar
v1 =1.8 liter , n= 1.4
p1v1^n =p2v2^n
V2^n = p1v1^n/p2 = 50*(1.8)^1.4/1 = 113.85

V2 = 29.43

work done =pv =(113.85/V^n) dv

              = 113.85[-(29.43)-0.4 +(1.8)-0.4]/(0.4)

              =151.41*100 kJ

           = 15141 kj

change in internal energy = 31.3 -15141 = -15109

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