This HW covers CB chapters 9-11 or LT chapters 9 & 10.

The homework consists of 7 problems for a total of 48 pts.

Please begin your question with the problem number you are asking about.

9.116 - Brayton Cycle with Regeneration - 8 pts

11.76 - Helium Gas Refrigeration Cycle - 6 pts

WB-1 : Brayton Cycle with Variable Heat Capacities - 6 pts

WB-2 : Effect of Turbine Feed T on Rankine Cycle Efficiency - 6 pts

Wb-3 : Special Rankine Cycle with Reheat and Regeneration - 8 pts

Wb-4 : Ammonia Cascade Refrigeration Cycle - 8 pts

WB-5 : Vapor-Compression Heat Pump - 6 pts

## 41 comments:

What are your office hours for the 3rd?

Camden 4:11 pm

My office hours are 8-9 am on Friday. I have to leave campus pretty promptly after class.

When you have a mixer or a splitter is it correct to assume that the two pipes going in have equal pressure for a mixer and that for a splitter both exit pipes have a equal pressure? Also do they have the same temperature and pressure and essentially all properties besides mass flow rate?

On WB-4 it says that 80 psia is the pressure of the mixer. Does that mean that P6 and P4 are both 80 psia or not?

On WB-4 it says that the sat vapor leaving the evaporator is at -20 F and later on it says that the sat vapor entering the second compressor is at 80 psia. When I looked up these values (a -20 F 80 psia sat vapor)on NIST it wasn't close to being a sat vapor( -20 f and 18.279 psia for sat vapor). Is it not a saturated vapor or am I misreading something or is NIST different from the book?

On WB-5 Im on part d and Im trying to calculate the isentropic efficiency of a value I already have. I have S hat 1 I just don't know if I keep the actual pressure to calculate the isentropic temperature or the other way around or if there is some other way

Camden, I kept the actual pressure of 9 bar and found the isentropic enthalpy from that and S hat 1. This gave me the right answer.

Camden 5:53pm

Yes, all streams entering and leaving a mixer or splitter must be at the same pressure.

In a splitter, all 3 streams have exactly the same properties. Only the mass flow rates are different (these are NOT intensive properties).

WB-4 , Camden , 6:38pm

Yes. P3 = P4 = P5 = P6 = 80 psia.

But remember that a flash drum is NOT a stream splitter. It is more than a tee in a pipe. Saturated liquid comes out the bottom and saturated vapor comes out the top.

WB-4 , Camden , 7:07pm

Stream 2 is at 20 degF. Stream 6 is saturated vapor at 80 psia. The confusion comes from the problem referring to the 2nd compression stage and I numbered that as compressor #1. I should fix my diagram. Make compressor #1 in the lower cycle and compressor #2 in the upper cycle. Then, the problem statement will make more sense.

I hope this helps.

WB-5 , Camden 7:38pm

You should use T3s to calculate H3s. Use T3 to calculate H3. Then, use H2, H3 and H3s to determine the isentropic efficiency of the compressor.

I hope this helps.

WB-5 , Anon 7:51pm

Yes, P3 = P3s = 9 bar.

WB-1 I am putting the flow diagram together and I am certain that there is a compressor and a turbine but is there two heat exchangers as well instead of condenser and boiler.

WB-1 , Jayson , 11:59am

Yes. There are also 2 HEXs in a Brayton cycle.

WB-1

I was able to get H2 and H4, but I'm kind of confused about what to do next. You say to find the net specific shaft work. Is that the same as:

Wnet = Wturb,out - Wcomp,in

On WB-5, it stated that compression is adiabatic to 9 bar and 60°C, so is it the temperature at stage 3?

For WB-2 I think I messed up my TS diagram. Should both the 3 and the 3S be inside the saturation curve? If not I am lost.

I think a better question might be is state 2 superheated?

I am confused on the equation that should be used for WB-2 Part B. For the graphs between 580 and 700 for quality and thermal efficiency.

WB-1

Ok can someone tell me where I'm messing up? I got:

Pr2 = 12.4368

H2s = 563.82

Pr4 = 9.41125

H4s = 519.32

And then to solve for the mass flow rate I applied the first law and got mdot = -W/(H3 - H2) where W = 15,000kW, H2 = 627.215 and H3 is in the IG air tables at 900K. But my answer is way off.

WB-1, Muffler,

You do not want to find the actual pressures. When you are finding Snaught using the ideal gas tables you simple plug in the ratio that is given in the problem statement. For instance the ratio is 8 to 1 so when using this equation: Snaught2-Snaught1-R/MW(Ln(P2/P1). You use the ratio as it fits. In other words instead of having P2/P1 in the natural log, you will either have an 8 or a 1/8.

Hope this was helpful.

What will your office hours be on monday?

In class you mentioned one of the answers listed was incorrect. Which problem was that?

WB-4

So is stream 2 at 20 degrees F or -20 degrees F??? You said one on the problem and the other here in the blog.

IN WB-3 I found the specific work of the cycle, The specific heat Q12 which is Qh, also i have Hhat for all the values necessary for that huge equation you gave us that has the Mdot5/mdot1 in it. Do you need to find Qhdot somehow? Cause I tried using specific values to get the thermal efficiency but I got a higher number. Also when finding mdot1 do you need to find mdot5 first, and if so could you lead me in the right direction? Anyone?

anonymous 4:37

its WB -4 part b

I got Ws23 = -72180 btu/h and Ws67 = -78180 btu/h but idk if thats right

WB-1

Ok so here's what I'm doing. First lookup H1 and Pr1 in the IG tables of air at 310K.

Solve for Pr2 by using the eq. Pr2 = (P2/P1)Pr1 where P2/P1 = 8. I got Pr2 = 12.4368

Then interpolate the answer you get for Pr2 to find the H2s value. I interpolated between 11.10 and 12.66 and got H2s = 563.82.

Then solve for H2 by using the isentropic efficiency eq for a compressor:

η = (H2s - H1) / (H2 - H1)

H2 = H1 + ((H2s - H1)/η)

H2 = 627.215

Do the same thing for T = 900K and solve for all those values just like above.

H3 = 932.93

Pr3 = 75.29

Pr4 = 9.41125

H4s = 519.32364

H4 = 577.23

Then use the first law mdot = -W/(H3 - H2) and use 15,000kW for W. Its not working! grrrr

Jayson 5:00

All i did to find mdot1 was Wsdot = mdot*Wshat. Since your given Wsdot and you found Wshat with the numerator of the huge previous equation you can solve for mdot. So in order to solve for m1 you only need to find Wcycle which requires the value of the ratio of mdot5/mdot1 not mdot5 itself. Also no you do not need to solve for Qhdot, only Qhhat because thats the denominator of the huge equation after the simplification of the problem, and is equal to Hhat2 - Hhat1. I dont know if this last part helped or not though maybe I misunderstood

WB-1 , Anon 4:42pm

The next thing to do is write the 1st Law on the expansion valve in the lower cycle to determine H1. Then, write the 1st Law on the Evap to determine Mdot1 (for the lower cycle).

I hope this helps.

WB-5 , Anon , 5:12pm

Yes, state 3 is 9 bar and 60 degC. State 4 is 9 bar and x=0.

WB-2 , Jayson , 6:42pm

3S must be directly below 2 and 3 must be on the same isobar as 3S, just at a higher value of S (to the right). It is not obvious ahead of time whether either 3 or 3S or both are inside the 2-phase envelope or not.

In this problem, it turns out that both 3S and 3 are inside the 2-phase envelope.

WB-2 , Jayson , 6:28pm

Yes state 2 is a superheated vapor.

WB-2b , Jayson , 6:55pm

You don't need a single eqn here. You need to repeat everything you did for part (a) (with T2=580) six more times with T2=600,620... Excel will make this much less painful.

WB-1 , MUFfl3r , 7:57pm

I assume your stream 2 comes out of the compressor and stream 4 comes out of the turbine. If so, your H2S and H4S values look OK. I did not use Pr.

The 15000kW is the net work for the cycle, including the turbine and the compressor. I think you only considered the W output of the turbine.

I hope this helps.

Jayson, 3:37 pm

My office hour will be 8-9am on Monday.

Anon , 4:37pm

Yes, My answers for WB-4, part (b) were wrong. The correct answers are: Wc1,act ~ 72,000 Btu/h (upper)

Wc2,act ~ 78,000 Btu/h (lower)

WB-4 , Anon , 4:58pm

Stream 2 is at -20 degF.

WB-3 , Jayson , 5:00pm

You cannot determine the thermal efficiency from Wdot and Qdot because the mass flow rate is not the same everywhere. Particularly, the mass flow rate is not the same through both turbines.

eff = Wdot,cycle / QdotH

Wdot,cycle = Wdot,pump + Wdot,HP + Wdot,LP

Wdot,cycle = m1*What,pump + m1*What,HP + m5*What,LP

QdotH = m1*QhatH

eff = [ What,pump + What,HP + (m5/m1)*What.LP ] / QhatH

So, all you need to know is the value of m5/m1.

You can evaluate m4/m1 by writing mass and energy balances on the feedwater heater. m5/m1 = 1 - m4/m1.

I hope this helps.

WB-1 , MUFfl3r , 5:52pm

It all looks good to me. I think you just need to consider the NET work for the cycle is Wturb + Wcomp, where Wcomp < 0.

Best of luck.

WB-3 , Camden , 5:00pm

I think Camden got this right.

In part (a) you don't need to know or determine any mass flow rates. You just need to determine the ratio m5/m1.

In part (b) you can determine all the mass flow rates because you are given Wdot,cycle.

on 76 I used TP^(y-1)/y=TP^(y-1)/y to solve for isentropic temperatures for T2s and T4s.Is this totally wrong because a constant heat capacity is not given? Also if this is the case what is the first step to solving this I'm kinda of lost. Im assuming it has something to do with the pressure ratio. Additionally is there any assumptions for this problem because I'm not really seeing any?

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