This HW covers CB chapter 7 or LT chapter 8.

The homework consists of 15 problems for a total of 71 pts.

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

Cengel & Boles: Ch 7:

7.127 - Power Requirement for an Air Compressor - 5 pts

7.131 - Analysis of an R-134a Compressor - 6 pts

7.146+ - Lost Work in a Heat Exchanger - 6 pts

Additional part c.) Determine the rate at which work is lost due to the irreversible nature of heat transfer in this process in kW. Assume the surroundings are at 20oC.

WB-1 - Back-Work Ratio of a Steam Power Cycle - 7 pts

Problem Statement :

Consider a steam power plant that operates between the pressure limits of 8 MPa and 20 kPa. Steam enters the pump as a saturated liquid and leaves the turbine as a saturated vapor. Determine the back work ratio (BWR is the ratio of the work delivered by the turbine to the work consumed by the pump). Assume the entire cycle to be reversible and the heat losses from the pump and the turbine to be negligible.

Hints :

Assume the cycle operates at steady-state and that the ump and turbine are reversible.

I suggest you make a table with 6 columns. The 1st column lists all the important properties you may need to evaluate for each steam in the cycle. Make a column for each of the 4 streams in the cycle. The last column should list the units for each variable. This table should help you keep track of what you know and what you still need to determine.

Make a nice process flow diagram. Make a nice TS Diagram. Include the 2-phase envelope and the two key isobars. Apply the 1st and 2nd Laws to the pump and turbine and you should be able to determine the BWR.

Two double-interpolations are required unless you use NIST or a plug-in for Excel or your calculator.

Ans.: BWR ~ 240

WB-2 - Polytropic Compression of N2 with Varying δ - 6 pts

Problem Statement : Nitrogen gas is compressed from 80 kPa and 27oC to 480 kPa by a 10 kW compressor. Determine the mass flow rate of nitrogen through the compressor assuming the compression process is …

a.) Isentropic, γ = 1.4

b.) Polytropic with δ = 1.3

c.) Isothermal

d.) Ideal, two-stage polytropic with δ = 1.3

Hints :

The compressor operates at steady-state.

Changes in kinetic and potential energies are negligible.

Flow work and shaft work are the only forms of work that cross the system boundary.

Nitrogen (N2) behaves as an ideal gas in all parts of this problem.

The heat capacities of N2 are constant and, therefore, γ is also constant.

The intercooler in part (d) cools the effluent from the first compressor back down to T1 before it enters the second compressor.

Parts (b), (c) and (d) should be plug-and-chug.

Ans.: a.) Mdot ~ 0.048 kg/s , c.) Mdot ~ 0.063 kg/s

WB-3 - Entropy Change, Heat Transfer and Irreversibilities - 7 pts

Problem Statement :

A closed system undergoes a process in which work is done on the system and heat transfer Q occurs only at temperature Tb. For each case listed below, determine whether the entropy change of the system is positive, negative, zero or indeterminate (you cannot tell for sure from the given information).

a.) Internally reversible process with Q > 0.

b.) Internally reversible process with Q = 0.

c.) Internally reversible process with Q < 0.
d.) Internal irreversibilities present with Q > 0.

e.) Internal irreversibilities present with Q = 0.

f.) Internal irreversibilities present with Q < 0.
Hints :
Consider the sign of each term in the defining equation for entropy generation.

WB-4 - Entropy Generation and Lost Work in a Nozzle - 6 pts

Problem Statement :

Oxygen, O2, enters a nozzle operating at steady-state at 3.8 MPa, 387oC and 10 m/s. At the nozzle exit, the conditions are 150 kPa, 37oC and 790 m/s.

a.) For a system that encloses the nozzle only, determine the heat transfer (kJ/kg) and the change in specific entropy (kJ/kg-K), both per kg of oxygen flowing through the nozzle. What additional information would be required to evaluate the rate of entropy production in this process ?

b.) Using an enlarged system boundary that includes the nozzle and a portion of its immediate surroundings, evaluate the rate of entropy generation (kJ/kg-K) and the rate of lost work (kJ/kg), both per kg of oxygen flowing through the nozzle. Assume that heat exchange at the enlarged system boundary takes place at the ambient temperature, 20oC.

Treat O2 as an ideal gas with variable heat capacities. Verify that the ideal gas assumption is valid.

Hints :

In part (a), use the 1st Law and the ideal gas property tables to determine Q.

In part (b), evaluate the entropy generation from its definition, using the Ideal Gas Property Tables and Gibbs 2nd Equation.

Lost work is just the product of Tsurr and Sgen.

Ans.: a.) Q ~ -30 kJ/kg , b.) Sgen ~ 0.2 kJ/kg-K and Wlost ~ 61 kJ/kg

WB-5 - Lost Work in an Air Compressor and HEX - 7 pts

Problem Statement :

Air flows through the compressor and heat exchanger in the system shown in the diagram. A separate liquid water stream (CP,W = 4.18 kJ/kg-K) also flows through the heat exchanger. The data given on the diagram are based on steady-state operation. Consider the air to be an ideal gas and neglect heat exchange with the surroundings as well as changes in kinetic and potential energies. Determine...

a.) The compressor power requirement in kW and the mass flow rate of the cooling water in kg/s.

b.) The rate of entropy generation in kW/K and the rate at which work is lost in kW for the compressor. Assume the temperature of the surroundings is 300 K.

c.) The rate of entropy generation in kW/K and the rate at which work is lost in kW for the heat exchanger. Assume the temperature of the surroundings is 300 K.

Hints :

Use the ideal gas EOS and the volumetric flow rate to determine the mass flow rate.

Use an energy balance to determine the work for the compressor in kW.

To determine the water flow rate, draw the control volume enclosing the heat exchanger. This control volume has 4 mass flows entering or leaving but no Q or W. An energy balance on this control volume yields the water flow rate.

Very important point- the air and the water DO NOT MIX in the heat exchanger !

Determine the change in enthalpy of the water using: ΔH = Cp ΔT and determine the entropy change of the water using ΔS = CP Ln[T2/T1].

Part (b) Don't forget about the cooling water when you calaculate the entropy generated.

Ans.: a.) WS ~ -50 kW, b.) Sgen,comp ~ 0.020 kW/K , Wlost,comp ~ 6 kW , b.) Sgen,HEX ~ 0.015 kW/K

## 31 comments:

On Problem 7-146 part b, I assumed that the heat exchanger is int. reversible, and adiabatic because there is no heat loss, thus it's isentropic. So, delta S=0. So, can I use the equation delta S=integral (Q/T) + Sgen?

7-127

To solve for T2 I'm using the equation (T2/T1) = (P2/P1)^((k-1)/k). I'm confused because I don't know what value to use for k from Table A-2b. Do I just pick and choose one at random?

7.146b , Anon , 7:44am

The SnT HEX is not internally reversible because the heat exchange inside the system occurs through a finite temperature difference. Therefore, yes you can use: dS/dT =integral (dQdot/T) + Sgen,dot to evaluate Sgen,dot.

7.127 , MUFfL3r , 10:34am

You could use that eqn to determine T2, but by doing so, you must assume that the heat capacities, and therefore gamma, are constant. There is nothing in the problem statement that implies this is true. So, I think you should do the best available method and use the variable, polynomial heat capacity equation. The good news is that with the ideal gas property table for air (pg 934 I think) you won't actually need to do any integrals. Just use the S-naught and H values you find in this table.

WB-1 should the TS diagram Include 3 isobars? I have the state where the steam enters the pump on the left side of the saturation curve because it is a saturated liquid, then because it is determined to be isentropic the process path goes straight up and I stopped at the top limit pressure of 8 MPa (800 kPa). Then going straight across the process goes outside the right side of the saturation curve so that when it goes straight down due to the isentropic turbine it ends at the right side of the saturation curve, because it is given that the steam comes out of the turbine as a saturated vapor. Is this a correct TS diagram? Is there a necessary 3rd Isobar? Or more than 3

?

Hey Thanks again for getting that program on my calculator working. It is so helpful and works great!

On WB-1 Is it a Rankine cycle? Cause that would make more sense than a Carnot cycle on a TS diagram.

For WB-4 part b it says to find the rate of work lost. the units you give are kj/kg but if it is a rate shouldnt it be kj/s not kj/kg?

WB-1 , Jayson , 5:34pm

No, just 2 isobars.

8 MPa = 8000 kPa.

Your description of the TS diagram sounds ok. I don't see what makes you think you need 3 isobars. The two isobars are P=20kPa and P=8000kPa. What other pressure is relevant ?

WB-1 , Jayson , 6:39pm

This cycle is actually a Rankine Cycle with superheat.

You can make a path for a Carnot Cycle or a Rankine Cycle on a TS Diagram. The diagram has no natural preference.

WB-4b , Jayson , 12:53am

You can write the 1st Law as:

Qdot - Ws,dot = mdot * dHhat

or by dividing by mdot, as:

Qhat - Wshat = dHhat

Qdot is the rate of heat transfer per unit time.

Qhat is the rate of heat transfer per unit mass of the flowing fluid.

Both are often referred to as rates.

For WB-1 I fixed the 800 to 8000 kPa problem but now when I am doing the first law for the turbine and the pump, I think I am using the wrong equations. Are you supposed to use the one that has Wsdot/mdot = (gamma/gamma-1)((P2V2-P1V1). I really do not think I should that equation for the pump cause is there shaft work in a pump? I am lost as to which equations to use. I hope you can understand this question, my rhetoric is not what it could be.

Thanks

Please help me doing problem WB-4 part B. I've been stuck for hours trying to figure out what I should solve to get Sgen..

When I calculate mdot water for problem WB-5, is it okay if I calculate for the Qhead of air as the integral of Cp dT? Because the answer to the problem if I use constant Cp would be different than if I integrated Cp.

WB-1 , Jayson , 3:32pm

No. This is steam, not an ideal gas with constant heat capacities. Use the steam tables.

WB-4b , Anon , 4:22pm

Our latest form of the 2nd Law is:

deltaS = INT{ dQ/T } + Sgen

Solve for Sgen. If you completed part (a) then you already know both Q and deltaS. Tsys is given in the problem statement and con be considered to be constant, 20 degC.

Wlost = Tsurr * Sgen.

I hope this helps.

WB-5 , Anon , 5:47pm

When I determined the enthalpy change of the air in the heat exchanger, I just use the enthalpy values in the Ideal Gas Property Table on page 934. This table was generated by integrating the polynomial form of Cp. You can do what I did, or you can integrate the polynomial Cp equation. The results should be very similar.

I agree, do not assume constant Cp for air.

I hope this helps.

Dr. B - 6.04 PM

But which T should I use though for calculating dQ/T?

Do I use Tsurr?

Dr. B - 6.04 PM

Also, does that mean that I don't need to change the integral of dQ/t into the integral of (Cp/T)dT?

Gosh, all of these are starting to come into my head now.

WB2 part c/

I got m-dot negative (-11.36kg/s) .Since isothermal ==> T1=T2 ==>H2 = 310.52<H1... I think something's wrong over here.

Please, help me!

WB1

I started with drawing a TS diagram and making a table as "hint" part suggested, then I wrote down first law for 4 processes in the cycle.I am trying to understand what is the work delivered by turbine and what is work consumed by the pump?I think that is W(turbine)=W(3-4)+W(4-1)

W(pump)(should be "-")=W(1-2)+W(2-3).

Am I on the right track, Professor.

TThank you

I'm getting sgen=0.0392 kj/kg*k for part b of WB-5. I'm using deltaScompressor=Snought2-Snought1-(Ru/MW)ln(p2/p1).

For part B we're looking only at sgen from point one to point two right? I've checked my math and since everything comes from the table I really can't find my error.

Oops, I see my problem. I found deltaS-hat rather than S-dot. Nevermind...

WB-4b , Anon , 6:30pm

Use 20 degC because this is the temperature inside the expanded system at which the heat transfer occurs. Using this expanded system is equivalent to using Tsurroundings. As a result, the Sgen you get includes ALL of the irreversibilities and Sgen is Sgen,total.

I hope this helps.

WB-4b , Anon , 6:53pm

Yes. You do NOT have to change INT{dQ/T} into anything else because T is constant at 20C or 293K.

WB-5b , Justin , 8:35 & 8.37

Good work.

WB-2c , Tee , 7:23pm

Yep, mdot < 0 is a problem !

Ws = -(RT/MW)*Ln(P2/P1) for isothermal compression of an ideal gas.

I hope this helps.

WB-1 , Andrei , 7:55

If stream 1 is the boiler feed, then W23 is the work produced by the turbine and W41 is the Wpump. BWR = W23/-W41. Yes, Wpump <0 and Wturb >0. But there is no shaft work at the boiler or the condenser.

I hope this helps.

Thank you, Bill. Will be working on this one.

Bill,

I will come to your office at 8am to ask some questions.It is kind of hard to "show everything" through the BLOG what I mean in the process of solving a problem.

Andre

nice

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