A closed piston-cylinder device contains 0.005m3 of saturated liquid water at a

Question

A closed piston-cylinder device contains 0.005m3 of saturated liquid water at a pressure of 150 kPa. Water is stirred and heated by a paddle wheel and a resistor placed in the device. After the resistor and paddle-wheel are turned on 45 minutes, one quarter mass of the water remains in the liquid phase. During this process, the pressure is kept constant. Given the paddle wheel work is 400kJ, there is a heat loss of 50kJ, and the current flows through the resistor is 8A, please determine the voltage of the resistor source.

Explanation / Answer

Initial state

V1 = 0.005 m^3

P1 = 150 kPa

Since it is saturated liquid state(From steam tables)

h1 = hf = 467.1 kJ/kg

u1 = uf = 466.9 kJ/kg

v1 = vf = 0.001053 m^3/kg

m = V/v1 = 0.005/0.001053 = 4.748 kg

Final State

Since 1/4th remains as liquid

Quality, x = 0.75

Since pressure is same

hf = 467.1 kJ/kg and uf = 466.9

hg = 2694 kJ/kg and ug = 2520

h2 = (1-x)*hf + x*hg = 2137.28 kJ/kg

u2 = (1-x)*uf + x*ug = 2006.73 kJ/kg

dU = m*(u2-u1) = 4.748*(2006.73-466.9) = 7311 kJ

From the first law

dQ1 heat added by the resistance circuit (V*i*t)

dQ2 is heat loss = (-50 kJ negative becuase it is heat loss)

dQ1 + dQ2 = dU + W

V*i*t -50*10^3 = 7311*10^3 + 400*10^3

V*8*45*60 = 7761*10^3

V = 359.31 V

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