Chapter4, Lesson C - Equilibration of a Tank and a Piston-and-Cylinder Device

4C-2 :	Equilibration of a Tank and a Piston-and-Cylinder Device								5 pts

A piston-and-cylinder device, B, is connected to tank A by a tube and a valve, as shown below. The volume of tank A is 1 m3. Initially, tank A contains saturated water vapor at 100 kPa. Tank B contains 1 m3 of water at 400°C, 300 kPa. The valve is opened and the water in both A and B is allowed to come to equilibrium.

a.)	Find the initial mass contained in each tank: mA1 and mB1.
b.)	If the final, equilibrium temperature is T2 = 200°C, determine the heat transfer and work for the process.

Read :	The key aspect of this problem is whether ANY water reamins in the cylinder, B, at equilibrium. If there is water left in B at the final state, it will exist at T2 and P2 = PB1 because the piston would still be "floating". The other key is that this is a closed system, so the mass of water in the entire system remains constant. We can use the Steam Tables and the given initial volumes to answer part (a). In part (b) the work is done at constant pressure as the piston descends. So it is not difficult to compute. Finally, solve the 1st Law to determine Q. This is possible because we know the initial and final states and the work !






Given :	PA1 =	100	kPa		Find :	mA1 =	???	kg
	VA1 =	1	m3			mB1 =	???	kg
	VB1 =	1	m3			Q =	???	kJ
	xA1 =	1	kg vap/kg			W =	???	kJ
	TB1 =	400	°C
	PB1 =	300	kPa
	T2 =	200	°C

Assumptions:		- The initial and final states are equilibrium states
		- The process is a quasi-equilibrium process.

Part a.)	We can determine mA1 because we know the volume of the tank and we can look up the specific volume of the saturated vapor that it contains.


		Eqn 1			NIST WebBook:		VA1 =	1.6939	m3/kg
							mA1 =	0.5904	kg

	We can use the same approach to determine mB1, but first we must determine its state.
	At 300 kPa, Tsat = 133.52°C. Since TB1 > Tsat, tank B initially contains superheated vapor.

			NIST WebBook:		VB1 =	1.0315	m3/kg

					mB1 =	0.9695	kg

Part b.)	In part b, there are two possibilities. At equilibrium, either B contains some water or it is completely empty.
	Case 1 - B is not empty: PB2 = PB1 because the piston is still floating.
	Case 2 - B is empty: VB2 = 0 and ALL of the water is in tank A.

	Let's test Case 1 first. Since T2 > Tsat at 300 kPa, the water would still be superheated vapor and the specific volume would be:
							V =	0.7164	m3/kg

	Therefore the total volume occupied by ths superheated vapor would be:

				Eqn 2			V2 =	1.117	m3

	Since this volume, which the total mass of water in the system occupies at PB1, is greater than the volume of tank A, we can conclude that all of the water could not fit into tank A at PB1. If P2 were less than PB1, the water would occupy even more volume and again would not fit into tank A. Some water must remain in the cylinder and, therefore, PB2 = PB1.





	Therefore:					VB2 = V2 - VA1 =		0.117	m3

	Calculate the PV or boundary work from:								Eqn 3

	But, since this process is isobaric:								Eqn 4

							W =	-265	kJ

	Finally, we need to apply the 1st Law to determine Q. Use all of the water in both containers as the system.

									Eqn 5

	Or:								Eqn 6

	Use the NIST WebBook and the ASHRAE Convention to determine all of the specific internal energies.

		mtot =	1.5598	kg		UA1 =	2505.6	kJ/kg
		U2 =	2651.0	kJ/kg		UB1 =	2966.0	kJ/kg

		ΔU	-219.5	kJ

Answer:		Q =	-484.3	kJ

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