Wednesday, May 10, 2006

HW #9, P#3 - NEW NEW NEW - Non-Ideal Binary Ammonia-Steam Rankine Power Cycle - 10 pts

A binary vapor power cycle consists of two Rankine Cyles with steam and ammonia as the working fluids. In the steam cycle, superheated vapor enters the turbine at 900 lbf/in2 and 1100°F. Saturated liquid leaves the condenser at 140°F. The heat rejected from the steam cycle is provided to the ammonia cycle. Saturated ammonia vapor enters the ammonia turbine at 120°F. Saturated liquid leaves the ammonia condenser at 75°F. Each turbine has an sentropic efficiency of 90% and the pumps operate isentropically. The net power output of the binary cycly is 7 x 107 Btu/h.

a.) Determine the quality of each turbine effluent.
b.) Determine the mass flow rate of each working fluid in lbm/h.
c.) Determine the thermal efficiency of the binary power cycle.

23 comments:

Anonymous said...
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Dr. B said...
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Dr. B said...

This is the new and hopefully better HW problem.

Dr. B said...

Graham 3:23 PM

I got more like Mdot,NH3=1.89Mdot,H2O
The difference may be due to slightly different H^ values or round-off, but I am not sure. You might want to check your calculations.

Yes, net Power output = Power produced-power consumed = WT1 + WT2 + WP1 + WP2. The solution uses
What = Vhat,satliq * deltaP
for the work for each pump.

Anonymous said...

I am looking at the diagram for a binary rankine power cycle from thermo CD and I'm having trouble matching it to the problem statement. In the diagram, there is no condenser in the high-temperature cycle (Did you mean heat exchanger?), and the low temperature cycle is connected to the boiler, although I get the impression from the problem statement that it shouldn't be.

Dr. B said...

Carina 5:45 PM
Yes, the heat exchanger serves the high-temperature cycle as a condenser.

In the Thermo-CD Notebook, I put stream 5 back into the boiler because this is often necessary.

However, in this HW problem, THIS IS NOT THE CASE ! Stream 5 goes straight into the Low temperature turbine. It does NOT go into the boiler.

The heat exchanger serves as the ONLY heat source for the low-temperature (NH3) cycle.

Good question. Sorry for the confusion.

Anonymous said...

Is problem 3A on the homework the replaced HW problem, or is it still the bad problem?

Anonymous said...

For the thermal efficiency, do we add the two efficiencies?

Anonymous said...

I am missing something in determing the state following the turbine. I understand that we use the isentropic efficiency to find the H^ for the real process. I also know that the entropy before and after the turbine are equal in the ideal case. But what state is it in here? Is it a saturated vapor? All I know about the last state is the entropy, which isn't enough to determine the enthalpy, etc.

Anonymous said...

nevermind I answered my question

Anonymous said...

Dr. B...

I got the quality values fine...for some reason...my equation for MdotA~MdotW is off by a power of 10...using these values for the SISO I get: MdotW(1065.69-108.69 Btu/lbm)=MdotA(178.44-126.91 Btu/lbm)...it leaves me with a coeffecient of 18.6 relating the two Mdot variables?...am I understanding that we use a system of two equations to get the actual values also?...did you find Qh of the cycle to also get the overall efficency? Thanks!

Dr. B said...

Graham 3:00 PM
Your method does not seem to work for me because you have 1 eqn in 2 unkns: Ma & Mw.
The keys to determining the Ma and Mw are Wcycle = 7x10^7 Btu/h AND the 1st Law as applied to the HEX. If you have all the H values, then these two equations have 2 unknowns: Ma and Mw.
Wnet= Wp1+Wp2+Wt1+Wt2 is the way to go. Keep in mind that Wp values are negative while Wt values are positive. So, you get Wnet in Btu/lbm.
I suggest you determine the Wpump values assuming incompressible fluids (Wpump = Vhat deltaP) or isentropic pumps using NIST.
I hope this helps, but I am not entirely sure what you did.

Dr. B said...

Greenpepper7 6:54 PM
Problem 3A is the new and "improved" problem.

Dr. B said...

Claire 8:28 PM
No. The efficiency of the ENTIRE cycle = Wnet / QH
Where: Wnet = Wp1+Wt1+Wp2+Wt2
and: QH = Q12 in my diagram on TCD Notebook page 236.

Dr. B said...

Jo 10:56 PM
You know the entropy and the pressure at each turbine outlet.
Draw a TS Diagram.
The boiler, condenser and heat exchanger can be assumed to be isobaric !

Dr. B said...

Kuchi 2:07 AM
The eqn you gave is the 1st Law for the HEX, right ? 2 unknowns.
The other eqn is:
Wnet = Wp1+Wt1+Wp2+Wt2 = 7x10^7 Btu/h
with the 1st law for each pump and turbine. Same 2 unknowns: Ma and Mw.
Solve simultaneously.
Mw ~ 125,000 lbm/h
Ma ~ 237,000 lbm/h

Anonymous said...

Dr. B,
You want us to use W = sum(pumps + turbs) but why cant we use Qh-Qc?

My set-up is as follows:
WhatH20 = (Qh-Qc)h20=497
WhatAmon = (Qh-Qc)Amon=486
Wnet = mH20(WhatH20) + mAmon(WhatAmon)

I solved and got the mAmon=1.89mH20

Anonymous said...

I think I understand what graham at 3:00 was trying to do.. I did it and got the right answer. Using the fact that mNH3=1.8923mH2O we know that mH2O(H2-HH3)+1.8923mH2O(H5-H6)=W (in Btu/s). My answer was 124k instead of 125k, but it seems like it's probably even more accurate than the other answer.

Anonymous said...

Brendan,
Note the work of 2 pumps need to be considered in calculating the Wnet.

Anonymous said...

Ian-
Good point, I added the two pump things to it and it worked perfectly

Anonymous said...

part b)
i have: mh20(hhat9-hhat8)+mnh3(hhat4-hhat3)=0
therefore
mnh3=-mh20(107.98-1066.009)/(178.8675-127.44)
or
mnh3=18.63mh20

where am I going wrong here?

I got the other equation correct, this one is wrong.

Dr. B said...

Brendan 5:02 PM
Yes, your method is correct and will yield the same answer.
Good work.

Dr. B said...

Kuester 8:02 PM
Yes, your work is correct.
I think the answers should be the same. Neither should be more accurate.
The 1.8923 came from applying the 1st Law to the HEX, right ?