Chapter5, Lesson E - Discharging R-134a from a Tank Containing Water and Steam

5E-4 :	Discharging R-134a from a Tank Containing Water and Steam								6 pts

A 750 L rigid tank initially contains water at 250°C, 50% liquid and 50% vapor by volume. A valve at the bottom of the tank is opened and the liquid is slowly withdrawn at a constant flow rate. Heat transfer takes place such that the temperature within the tank is uniform and constant.

a.)	Calculate the mass of liquid, mass of vapor and the total mass of water in the tank at the inital condition.

b.)	Calculate the quality of the vapor-liquid mixture in the tank when half of the total mass has been removed from the tank.


c.)	Determine the amount of heat transfer required when half of the total mass has been removed from the tank. Note: this process can be considered a uniform-flow, unsteady process.


Read :	The key to this problem is that the process is an isothermal process. As a result, the properties of the liquid inside ths system and leaving the system are always the properties of saturated liquid at 250°C. As a result, this is a uniform state process. We can also assume it is a uniform flow process. If we further assume that changes in kinetic and potential energies are negligible and that no shaft work occurs, we can use the 1st Law to determine Q. Parts (a) and (b) require use of the steam tables or NIST Webbook and a working knowledge of the relationship between mass, volume and specific volume, but should not be difficult.







Given :	V	0.75	m3	T1 = T2 =		250	°C
	V1,vap	0.375	m3			523.15	K
	V1,liq	0.375	m3

Find :	a.)	m1,vap	???	kg	Diagram :
		m1,liq	???	kg
		m1	???	kg


	b.)	x2	???	kg vap/kg total

		when:

	c.)	Q12	???	kJ

Assumptions:
	1 -	Only satruated liquid water leaves the tank.
	2 -	The process is isothermal.
	3 -	Only flow work (no shaft work) crosses the system boundary.
	4 -	Changes in kinetic and potential energies are negligible.
	5 -	Uniform Flow: The properties and flow rate of the out let stream are constant over the cross-sectional area of the pipe and with respect to time.

	6 -	Uniform State: At all times, the properties of the outlet stream are the same as the contents of the system at that point in time.


Equations / Data / Solve :

Part a.)	Because the water liquid and vapor are in equilibrium with each other at all times throughout the process, they are always saturated. Therfore, we can determine the mass of liquid and vapor initially in the tank by looking up their specific volumes in the saturated steam tables or the NIST Webbook.



			Eqn 1						Eqn 2

	At 250°C :		Vsat vap	0.05013	m3/kg		m1,vap	7.481	kg
			Vsat liq	0.001251	m3/kg		m1,liq	299.8	kg

	We can determine the total mass of water in the system initially from:
							m1 = m1,vap + m1,liq

							m1	307.2	kg

Part b.)	We know that:			Eqn 3			m2	153.62	kg

	The key here is that we know both the volume and the total mass in the tank, so we can calculate the specific volume and use it to determine the quality from :


									Eqn 4

	The real key to this problem is that the process is isothermal. As a result, the properties of the saturated vapor in the tank remain constand and the properties of the saturated liquid inside the tank and flowing out of the tank also remain constant.



	We determine the overall specific volume at state 2 from:								Eqn 5


	Now, we can plug values into Eqns 5 & 4 :						V2	0.004882	m3/kg
							x2	0.07429	kg vap/kg tot

Part c.)	To determine Q, we need to apply the 1st Law for transient processes and open systems.
	Here, we assume that WS = 0, ΔEkin = ΔEpot = 0. The appropriate form of the 1st Law under these conditions is:

									Eqn 6

	Solving Eqn 6 for Q yields:								Eqn 7

	The specific enthalpy of the water leaving the system is the enthalpy of saturated liquid water at 250°C:
							Hout	1085.36	kJ/kg

	A mass balance allows us to determine mout :								Eqn 8

							mout	153.62	kg

	Next, we can determine U1 and U2 :								Eqn 9

									Eqn 10

				where:					Eqn 11

	Plugging values into Eqns 9 - 11 yields:						x1	0.02435	kg vap/kg tot

	At 250°C :		Usat vap	2602.4	kJ/kg		U1	1117.447	kJ/kg
			Usat liq	1080.39	kJ/kg		U2	1193.458	kJ/kg

	We are finally ready to put numbers into Eqn 7 to complete this problem.

		m2 U2		m1 U1		mout Hout
	Q =	183340	-	343325	+	166733	kJ

							Q	6747.6	kJ

Verify :	The assumptions made in this solution cannot be verified with the given information.

Answers:	a.)	m1,vap	7.48	kg	b.)	x2	0.0743	kg vap/kg total
		m1,liq	300	kg
		m1	307	kg	c.)	Q12	6750	kJ

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