Wednesday, April 20, 2011

ENGR 224 - HW #4

This HW covers CB chapter 6 or LT chapter 6.

The homework consists of 8 problems for a total of 45 pts.

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

Cengel & Boles: Ch 6:
6.79 - Effect of Source and Sink Temperatures on HE Efficiency - 6 pts
6.85 - Thermal Efficiency of a Geothermal Power Plant - 3 pts
6.107 - Carnot HE Used to Drive a Carnot Refrigerator - 6 pts
6.110 - Actual and Maximum COP of an Air-Conditioner - 8 pts
6.134 - Thermal Efficiency of Heat Engines in Series - 4 pts

Special Problems
WB-1 - "Show That" Problem Using the K-P Statement of the 2nd Law - 6 pts
WB-2 - Reversible, Irreversible and Impossible Power Cycles - 6 pts
WB-3 - A Reversible HE Used to Drive a Reversible Heat Pump - 6 pts

18 comments:

Dr. B said...

Problem 6.85
I used Cp = 4.22 k/kg to solve this problem.

Tee said...

It's beautiful weather outside... And I am looking the HW4....haizz.

Anonymous said...

I wonder on problem 6-110, is the 250 kJ/min and 900 W considered as the Qdot in?

So, in this problem, am I looking for the Wcycle and Qout?

Dr. B said...

Tee,
Sorry, bud. I hope you didn't spend too many daylight hours on Thermo this weekend.

Dr. B said...

6.110 , Anon , 2:22pm
The sum of 250 kJ/min and 900 W is the amount of heat that must be removed from the room in order to keep the room temperature constant. So, this must be equal to Qc.

Jayson said...

For WB-2 part c can you just say the refrigerator is now a heat pump instead and use the COP_HP instead of the COP_R formula since there is a Q_H given rather than a Q_C?

Dr. B said...

WB-2 , Jayson , 10:54pm
No. Refrigerators have a QH too. They reject heat into your kitchen. Just use the 1st Law to get Wcycle if you need to.

I hope this helps.

Anonymous said...

Problem 6.85

If I apply conservation of mass to this problem and assume that it is liquid water both entering and leaving the powerplant I get totally different results for the efficiency when I use 1-Qout/Qin versus Wnetout/Qin. Does conservation of mass not apply here?

Dr. B said...

6.85 , Anon , 7:44pm
Conservation of mass always applies unless we have a nuclear reaction.
But I don't understand the question. 210 kg/s of 150C geothermal water enters the hot "reservoir", gives up heat to the cycle and leaves the hot "reservoir" at 90C. I put reservoir in quotes because this is not a true thermal reservoir because its temperature changes from 150C to 90C.

I hope this helps.

Jayson said...

For problem 6.79 Is it TWO graphs that is wanted? and they are very similar to the once on Learnthermo and your slide show today?

Jayson said...

actually I think 6 plots never mind.

Anonymous said...

in 6-85, i don't understand the problem maybe. the temperature change occurs in the hot reservoir? how does it produce work without a cold reservoir? If 90 degrees is the cold reservoir, i Can use (Q/dt)-(W/dt)=(m/dt)(Delta-H_hat) to find the Heat input from the feed stream, right?

Jayson said...

6.85-Anon-10:06
I think you have to use Integral of Cp*dT for Delta H. After integration, since Cp is a constant you can do Mdot*Cp*(Tin-Tout). Use the Cp that Mr. B used- 4.22, but I think he messed up the units and it should be 4.22 kJ/kg*K not k/kg. At least I can not see how the calculations work if you use a k/kg for the units, cause it would not be dimensionally homogeneous!

Hope this helps.

WARNING; I am not 100% sure.

Dr. B said...

6.79 , Jayson , 9:39pm
I think you need 4 plots. Wmax vs TH, Eff vs TH, Wmax vs Tc and Eff vs TC. I used Excel to make 2 plots with 2 different y-axes. I can show you how to do this.

Dr. B said...

8.85 , Anon , 10:06pm
In this problem, the hot "reservoir" isn't a true reservoir. It is really a heat exchanger in which the geothermal water gives up heat energy to the power cycle and in doing so drops in temperature from 150C to 90C. So, apply the 1st Law to the geothermal water to determine QH.

The heat engine produces work from the heat it gets from the geothermal water and then rejects heat to the cold reservoir which is the environment at 25C.

I hope this helps.

Dr. B said...

6.85 , Jayson , 11:36pm
Oops. I did mess up the units and Cp = 4.22 kJ/kg-K.

Thanks for helping Anon out here, Jayson.

Kin Wai Kelvin said...

6.79
I have no idea of how to do it!?

Dr. B said...

6.79 , Kelvin , 3:48pm
It is a Carnot HE and you know QH, TH and TC. Calculate efficiency from TC and TH and then Wmax = WCarnot from the definition of efficiency. Do this many times in Excel and construct plots with the 6 curves described in the problem statement.

I hope this helps.