Chapter5, Lesson C - Expansion of Steam Through a Throttling Valve

5C-4 :	Expansion of Steam Through a Throttling Valve								5 pts

Steam at 9000 kPa and 600°C passes through a throttling process so that the pressure is suddenly reduced to 400 kPa.

a.)	What is the expected temperature after the throttle ?
b.)	What area ratio is necessary for the kinetic energy change to be zero ?


Read :	We know the values of two intensive variables for the inlet steam, so we can determine the values of all of its other properties, including the specific enthalpy, from the steam tables. If changes in kinetic and potential energy are negligible and the throttling device is adiabatic, then the throttling device is isenthalpic. In this case, we then know the specific enthalpy of the outlet stream. The pressure of the outlet stream is given, so we now know the values of two intensive properties of the outlet stream and we can determine the values of any other property using the steam tables. Part (b) is an application of the 1st Law. The area must be greater at the outlet in order to keep the velocity the same because the steam expands as the pressure drops across the throttling device.








Diagram :







Given :	P1	9000	kPa		Find :	T1	???	°C
	T1	600	°C			A2 / A1	???
	P2	400	kPa

Assumptions :		1 -	The throttling device is adiabatic.
		2 -	Changes in potential energy are negligible.
		3 -	Changes in kinetic energy are negligible because the cross-sectional area for flow in the feed and effluent lines have been chosen to make the fluid velocity the same at the inlet and the outlet.



Equations / Data / Solve :

Part a.)	Begin by looking up the specific enthalpy of the feed in the steam tables.
	At a pressure of 9000 kPa, the saturation temperature is :						Tsat	303.4	°C

	Because T1 > Tsat, we conclude that the feed is superheated steam and we must consult the Superheated Steam Tables. Because 9000 kPa is not listed in the table, interpolation is required.


	At 600oC :		P (kPa)	H (kJ/kg)	V (m3/kg)
			8000	3642.0	0.04845
			9000	3633.7	0.04341		H1	3633.7	kJ/kg
			10000	3625.3	0.03837		V1	0.04341	m3/kg

	The 1st Law for a throttling device that is adiabatic and causes negligible changes in kinetic and potential energies is :

									Eqn 1

	Because the pressure drops in the throttling device and the feed is a superheated vapor, the effluent must also be a superheated vapor. So, to answer part (a), we must use the Superheated Steam Tables to determine the temperature of 400 kPa steam that has a specific enthalpy equal to H2.



	At 400 kPa :		H (kJ/kg)	T (oC)	V (m3/kg)
			3592.9	550	0.9475
			3633.7	568.6	0.9693		T2	568.6	°C
			3702.4	600	1.006		V2	0.96927	m3/kg

Part b.)	We need to use the definition of kinetic energy to determine how much the area of the outlet pipe must be greater than the area of the inlet pipe in order to keep the kinetic energy (and therefore the velocity) constant.



			Eqn 2						Eqn 3

	Because the mass flow rate at the inlet and outlet is the same, Eqn 3 simplifies to :								Eqn 4

	Next, we need to consider the relationship between velocity, specific volume and cross-sectional area.

									Eqn 5

	Now, substitute Eqn 5 into Eqn 4 to get :								Eqn 6

	Solve for the area ratio, A2 / A1 :								Eqn 7

							A2 / A1	22.328

Verify :	None of the assumptions made in this problem solution can be verified.

Answers :	T2	569	°C				A2 / A1	22.3

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