The tank below has a frictionless, adiabatic (no heat conduction) piston that se
ID: 2994244 • Letter: T
Question
The tank below has a frictionless, adiabatic (no heat conduction) piston that separates an ideal gas on the left from water on the right. The initial pressure and volume of the ideal gas is 50 kPa and 8 liters. The initial pressure and temperature of the water is 50 kPa and 25C. Heat is added slowly to the water until the volume of the ideal gas is 1 liter. During the compression of the ideal gas heat transfer maintains the ideal gas at constant temperature (initial temperature). If the mass of the water is 5 kg, determine the necessary heat transfer. There are various pieces of knowledge you have to bring together to solve this problem. First you must picture what is happening physically. Dont forget the work is: P is not constant.
Explanation / Answer
If the ideal gas is maintained at constant temperature, its final pressure is p1 V1/V2 = 400 kPa.
The original volume of the water (which is all liquid at the outset) would be (5 kg)/(997.02 kg/m^3) = 5.01494 L.
My density value in this case comes from NIST steam tables and is specific to water at 50 kPa and 25C.
In its final state, the water must have a volume of 12.01494 L and a pressure of 400 kPa.
Its specific volume is then 0.01201494 m^3 / (5 kg) = 0.002402988 m^3/kg.
At 4 bars, this is a mixture of liquid (v = 0.001084 m^3/kg) and vapor (v = 0.463 m^3/kg).
The vapor fraction x is found from (1-x)*0.001084 + (x)*0.463 = 0.002403 0.461916 x = 0.001319 x = 0.0028555
From this we can calculate the internal energy of the water in its final state: seems to be about 610 kJ/kg.
In the initial state (all liquid) the internal energy was the same as the enthalpy, 105 kJ/kg.
The increase in internal energy has been 505 kJ/kg. Q = W + delta-U. Finding the work done on the ideal gas isn't too hard, it's the integral from 8L to 1L of p dV = integral from 8L to 1L of (p1 V1) dV/V [for the gas] = (p1 V1) ln(1/8) = (50 kPa)(0.008 m^3)(ln 0.125) = -0.832 kJ
So the heat transferred to the water must be (505 kJ/kg)(5 kg) + (0.832 kJ) = 2525 kJ
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