10142005, 08:54 PM  #1 (permalink) 
Crazy

energy balance help..
I am really struggling with this question..
Liquid water at 60 bar and 250 degrees C passes through an adiabatic expansion valve, emerging at a pressure Pf, and a temperature, Tf. If Pf is low enough, some of the liquid evaporates. If Pf = 1 bar, determine Tf, and the fraction of the liquid feed that evaporates. Assume kinetic energy = 0 So, from the tables in the book, I think Tf would be 200 degrees celsius for 1 bar...? Then, the equation simplifies to become Q = delta H and delta H = mass * change in specific enthalpy. But I dont know the mass flow rate. The question suggests writing an energy balance, but without m, what can I do ? 
10152005, 12:30 AM  #2 (permalink) 
Psycho
Location: Virginia Beach, VA

They want to know the fraction of liquid that evaporates, not any absolute amount, you don't need a certain value for m. Pick your own values for m and run the math a couple of times, it should give you the same answers.
Since the liquid will evaporate until enough energy is carried off to cool the liquid to saturation temperature, Tf is going to be less than or equal to saturation temperature for Pf. I say "less than" only because To is significantly below Tsat for Po... pressure will have to drop to <40bar before evaporation starts.
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Simple Machines in Higher Dimensions 
10152005, 07:42 AM  #4 (permalink) 
pigglet pigglet
Location: Locash

danny:
you know Pf. If I'm understanding correctly, Adiabatic => Q = 0. Hin = Hout. You know Hin. You know Pf. Assume exit is saturated steam, I think. Ie. you know Tf from the steam tables. You therefore know the H value of sat. steam. Any extra energy goes to vaporization. All these are specific (molar / mass dependent), so you can work the balance without worrying about mass, or by assuming an arbitrary basis (I might suggest 1 mole or kg, depening on your values in the tables) as our binary friend above states. You've got the extra energy, and you know the heat of vaporization of water. I think you can calculate the fraction from that.
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10152005, 08:07 AM  #5 (permalink) 
Crazy

OK..
Liquid water at 60 bar and 250 degrees has H = 1085.8 Kj/Kg Pf = 1.0 bar, so Tf = 99.6 degrees, therefore H = 2675.4 Kj/Kg So, the excess that comes out is 2675.4  1085.8 = 1589.6 Kj/Kg so, assuming a mass flow of 1kg/hr, the extra energy would be 1589.6Kj/Kg which would go to vaporization. The answer is that 0.296 evaporated. How in the world did they arrive at that number ? 
10162005, 04:31 PM  #7 (permalink) 
pigglet pigglet
Location: Locash

Danny, I think there's a problem with the way you're interpretting your numbers. From what you posted, I can say You can't have this H_in = H_out *and* have different numbers. Thus, what I posted earlier is absolutely wrong in its assumptions: here is the correct technique.
Saturated steam assumes that you're at 100% vapor, right on the verge of forming the first droplet of liquid water. Look at the value of H_liquid for saturated conditions(there should some value similar to this for your exit P and T. One value for the enthalpy of the liquid, one for the value of the vapor. You know your "true" H value = it's specified by your inlet conditions. I'm guessing that the H_inlet value is between these two values of H_liq. and H_vap. at your exit condition. Interpolate between the H_liq and H_vap values, where x (amount vaporized) for the liquid is x=0, for vap its x=0.
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10162005, 06:02 PM  #8 (permalink) 
Psycho
Location: Virginia Beach, VA

Maybe this will help in your googling... You're calculating steam quality.
The key is that the amount of energy present in Xkg of 250C 60bar liquid will be equal to Xkg of 100C 1bar gas/liquid mixture. In: 250C 60bar Subcooled Liquid (hf) 1085.3 kJ/kg Out: 99.62C 1bar Saturated Liquid (hf) 417.46 kJ/kg 99.63C 1bar Saturated Vapor (hg) 2675.4 kJ/kg Solve the following system of equations: 1085.3X = 417.46Y + 2675.4Z X = Y + Z
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Simple Machines in Higher Dimensions 
10162005, 07:25 PM  #10 (permalink) 
Psycho
Location: Virginia Beach, VA

X = Y+Z
Y = XZ => 1085.3X = 417.46(XZ) + 2675.4Z 1085.3X = 417.46X  417.46Z + 2675.4Z 1085.3X  417.46X = 2675.4Z  417.46Z 667.84X = 2257.94Z Z = 667.84X/2257.94 Z = 0.295774X That 0.296 look familiar?
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Simple Machines in Higher Dimensions Last edited by 1010011010; 10162005 at 07:51 PM.. Reason: http://www.spiraxsarco.com/esc/default.asp?lang_id=USA&country_id=US 
10162005, 08:09 PM  #11 (permalink) 
Crazy

What I dont get it is, why did u use 2675.4 for the enthalpy ?
We have 2 streams coming out....evaporated water, and water. So I was using enthalpy values for evaporation and water. why did u use an enthalpy value for steam ? Last edited by danny_boy; 10172005 at 06:20 AM.. 
10172005, 08:12 AM  #12 (permalink) 
pigglet pigglet
Location: Locash

danny,
the 2675 kJ/kG is the number for 100% steam. the 417.5 kJ/kg number is for pure liquid. You could consider yourself to have two streams, the steam having the 2675 kJ/kg and the liquid having the 417 kJ/kg number independtly of one another, but to determine how much of the original amount is in each, you have to consider them together. Those H_liq and H_vap numbers basically tell you how much of the original energy is necessary to vaporize steam, at the exit conditions of your valve. Does that help?
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10172005, 11:56 AM  #13 (permalink) 
Psycho
Location: Virginia Beach, VA

Ignore slight changes in values, I'm using a different set of steam tables.
Enthalpy of Water (hf) 417.547 Enthalpy of Evaporation (hfg) 2257.63 Enthalpy of Steam (hg) 2675.18 hf + hfg = hg hfg (AKA Heat of Vaporization) is the energy you have to add to saturated liquid to change it into saturated gas. It's the enthalphy of the phase change, not of any particular state of matter. On some older references the subscript is actually "f→g" (that's supposed to be a little arrow)... somewhere along the way it just became "fg", which leads to confusion occassionally. In short... coming out the other end you have Water and Steam... not Water (noun) and Evaporation (verb).
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Simple Machines in Higher Dimensions 
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