tag:blogger.com,1999:blog-20953269.post4875776929549060861..comments2023-04-22T16:35:23.188-07:00Comments on Thermodynamics: ENGR 224 - HW #4Unknownnoreply@blogger.comBlogger18125tag:blogger.com,1999:blog-20953269.post-15398883716951436722011-05-04T16:22:44.921-07:002011-05-04T16:22:44.921-07:006.79 , Kelvin , 3:48pm
It is a Carnot HE and you k...6.79 , Kelvin , 3:48pm<br />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.<br /><br />I hope this helps.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-42423746912304529562011-05-04T15:48:26.846-07:002011-05-04T15:48:26.846-07:006.79
I have no idea of how to do it!?6.79<br />I have no idea of how to do it!?Kin Wai Kelvinhttps://www.blogger.com/profile/04222756481328268281noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-92195550494295115132011-05-04T08:09:03.418-07:002011-05-04T08:09:03.418-07:006.85 , Jayson , 11:36pm
Oops. I did mess up the u...6.85 , Jayson , 11:36pm<br />Oops. I did mess up the units and Cp = 4.22 kJ/kg-K.<br /><br />Thanks for helping Anon out here, Jayson.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-13618471099862313102011-05-04T08:07:05.750-07:002011-05-04T08:07:05.750-07:008.85 , Anon , 10:06pm
In this problem, the hot &qu...8.85 , Anon , 10:06pm<br />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.<br /><br />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.<br /><br />I hope this helps.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-12428216867017529882011-05-04T08:04:03.317-07:002011-05-04T08:04:03.317-07:006.79 , Jayson , 9:39pm
I think you need 4 plots. W...6.79 , Jayson , 9:39pm<br />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. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-47936158381359086002011-05-03T23:36:06.076-07:002011-05-03T23:36:06.076-07:006.85-Anon-10:06
I think you have to use Integral ...6.85-Anon-10:06 <br />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! <br /><br />Hope this helps.<br /><br />WARNING; I am not 100% sure.Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-2002532624741907272011-05-03T22:06:40.121-07:002011-05-03T22:06:40.121-07:00in 6-85, i don't understand the problem maybe....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?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-82983480134377124632011-05-03T21:44:39.679-07:002011-05-03T21:44:39.679-07:00actually I think 6 plots never mind.actually I think 6 plots never mind.Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-59779073290505911362011-05-03T21:39:51.157-07:002011-05-03T21:39:51.157-07:00For problem 6.79 Is it TWO graphs that is wanted? ...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?Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-35927063352608489952011-05-03T19:59:17.678-07:002011-05-03T19:59:17.678-07:006.85 , Anon , 7:44pm
Conservation of mass always a...6.85 , Anon , 7:44pm<br />Conservation of mass always applies unless we have a nuclear reaction.<br />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.<br /><br />I hope this helps.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-57147073966696546272011-05-03T19:44:01.984-07:002011-05-03T19:44:01.984-07:00Problem 6.85
If I apply conservation of mass to t...Problem 6.85<br /><br />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?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-45111385683968888382011-05-03T07:57:22.533-07:002011-05-03T07:57:22.533-07:00WB-2 , Jayson , 10:54pm
No. Refrigerators have a ...WB-2 , Jayson , 10:54pm<br />No. Refrigerators have a QH too. They reject heat into your kitchen. Just use the 1st Law to get Wcycle if you need to.<br /><br />I hope this helps.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-1108606811150468662011-05-02T22:54:45.981-07:002011-05-02T22:54:45.981-07:00For WB-2 part c can you just say the refrigerator ...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?Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-39062653009899542102011-05-02T10:26:43.956-07:002011-05-02T10:26:43.956-07:006.110 , Anon , 2:22pm
The sum of 250 kJ/min and 90...6.110 , Anon , 2:22pm<br />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.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-89835242467360138902011-05-02T10:24:32.974-07:002011-05-02T10:24:32.974-07:00Tee,
Sorry, bud. I hope you didn't spend too ...Tee,<br />Sorry, bud. I hope you didn't spend too many daylight hours on Thermo this weekend.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-62879061792111918562011-05-01T14:22:45.574-07:002011-05-01T14:22:45.574-07:00I wonder on problem 6-110, is the 250 kJ/min and 9...I wonder on problem 6-110, is the 250 kJ/min and 900 W considered as the Qdot in?<br /><br />So, in this problem, am I looking for the Wcycle and Qout?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-88420851837961490752011-04-30T17:14:42.306-07:002011-04-30T17:14:42.306-07:00It's beautiful weather outside... And I am loo...It's beautiful weather outside... And I am looking the HW4....haizz.Teenoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-81518908801809958052011-04-24T19:33:32.082-07:002011-04-24T19:33:32.082-07:00Problem 6.85
I used Cp = 4.22 k/kg to solve this p...Problem 6.85<br />I used Cp = 4.22 k/kg to solve this problem.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.com