Thermodynamics, enthalpy change for steam

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Homework Statement


Steam at 200psia and 600 *F [state 1] Enters a turbine through a 3-inch-diameter pipe with a velocity of 10(ft)/(s). The exhaust from the turbine is carried through a 10-inch-diameter pipe and is at 5psia and 200 *F [state 2]. What is the power output of the turbine?

H1 = 1,322.6(Btu)/lbm
H2 = 1148.6(Btu)/lbm

V1 = 3.058ft^3 / lbm
V2 = 78.14ft^3 / lbm


Homework Equations



dH = (u^2 / 2*gc) + Q + W

The Attempt at a Solution



dH = -174Btu/lbm

174Btu/lbm = du/2gc + Q + W

mass flowrate = (A1*u1) / V1 == 0.1605 lbm/s
final velocity u2 = (m * V2) / A2 = 23.0 ft/s
du = 13ft/s

What do I do now? How do I calculate Q and W?
there are tables at the back of the book with properties of steam for various temperatures and pressures...some of the example problems I looked at make use of that, but I'm not quite sure how that helps here.
 

Answers and Replies

  • #2
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In the equation you are using (correct choice), W is the shaft work (i.e., the work done by the turbine per pound of steam =-work done on the turbine, per pound of steam). This is what you are solving for. Probably, a reasonable assumption is that the turbine operates adiabatically, so that Q = 0. You have already made a pretty good start at solving this problem.

Chet
 
  • #3
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But then what does the temperature change have to do with it? How can Q equal 0 when there's a temperature change? I don't have to use the temperatures in my calculations at all because Q=0?
 
  • #4
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But then what does the temperature change have to do with it? How can Q equal 0 when there's a temperature change? I don't have to use the temperatures in my calculations at all because Q=0?

Each pound of steam is expanding as it passes through the turbine and does work on the rotor. So its temperature is decreasing. You use the temperatures to get the inlet and outlet enthalpies, and thus the change in enthalpy. Q can, of course, be zero even if there is a temperature change (as you have already seen earlier in your thermo course).
 
  • #5
SteamKing
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The Q value is, I believe, heat added or lost to the surroundings. The turbine casing and steam piping will be insulated to minimize any heat transfer from the equipment to its surroundings.

The assumption of isentropic operation of the turbine may not necessarily be true. Since you have entrance and exit conditions for the steam, a check of the steam tables should provide entropy values at these points.
(Note: I checked with printed tables online and found that the turbine is not operating isentropically).
 
  • #6
SteamKing
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You know the change in enthalpy per pound of steam, and you know the flow rate of the steam in pounds per second. You can find the work equivalent of the change in enthalpy and then use the flow conditions to find out the rate at which work is being done by the turbine.
 
  • #7
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The Q value is, I believe, heat added or lost to the surroundings. The turbine casing and steam piping will be insulated to minimize any heat transfer from the equipment to its surroundings.

The assumption of isentropic operation of the turbine may not necessarily be true. Since you have entrance and exit conditions for the steam, a check of the steam tables should provide entropy values at these points.
(Note: I checked with printed tables online and found that the turbine is not operating isentropically).

Hi SteamKing. I hope you didn't think that I was implying isentropic operation. All I said was that the operation was probably close to adiabatic, so that Q would be essentially zero.

The OP should check the change in kinetic energy. I don't think that this will make a significant contribution, but it should at least be considered.

Chet
 
  • #8
SteamKing
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Hi SteamKing. I hope you didn't think that I was implying isentropic operation. All I said was that the operation was probably close to adiabatic, so that Q would be essentially zero.

The OP should check the change in kinetic energy. I don't think that this will make a significant contribution, but it should at least be considered.

Chet

No, the entropy values at entrance and exit are pretty close, but not identical. However, the difference in entropy should only affect the thermodynamic efficiency of the turbine.

Since this is a First Law type analysis, the Q will indicate the amount of heat exchanged across the boundary of the control volume around the turbine. In practical terms, Q is always minimized, if not eliminated entirely, by insulating the equipment, so that the heat in = work out + heat out.
 
  • #9
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Incidentally, that kinetic energy term in the original equation should really be the change in kinetic energy, and it should be on the same side of the equation as the ΔH. I still think that it going to turn out to be negligible compared to the enthalpy change.

Chet
 
  • #10
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So I solved it assuming adiabatic (Q=0) and assuming the kinetic energy term is negligible. But out of curiosity, how would I solve this problem if it WASN'T adiabatic? What if the turbine was transferring heat?
 
  • #11
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So I solved it assuming adiabatic (Q=0) and assuming the kinetic energy term is negligible. But out of curiosity, how would I solve this problem if it WASN'T adiabatic? What if the turbine was transferring heat?
Then you would need additional information such as the detailed geometry of the housing and the thickness and thermal characteristics of the insulation. You would also need to know the temperature outside the housing. You would be looking at conductive heat transfer through the insulation, and possible convective heat transfer from the outside surface of the insulation to the surrounding air.

I suggest that you calculate the kinetic energy term just to see how if compares with the enthalpy change.
 

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