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# Thermodynamics Question

10 replies to this topic
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### #1 Blank03

Blank03

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Posted 15 June 2018 - 12:02 PM

Hi. so apparently I am reviewing for my exams hence it would be better to practice solving on your own.
so i started answering some questions and I got stuck with this problem. the question goes with answer and a "general solution" so I am on the part of deciphering it how does the  solution ended up to such value.problem is, we get different answers on getting delta H: i got -1513.69 btu/lbm, the solution goes with -99 btu/lbm. can someone please help me what have i done wrong?
Thank you very much.

### #2 Art Montemayor

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Posted 15 June 2018 - 03:18 PM

In order to determine how or where you went wrong in reaching your solution to the problem, you should submit your total, detailed calculations.  Think about it: how can our members help you find out what you did wrong if you don't submit your detailed calculations that they can review?  How else can we help you correct your errors????

I ask my self, year after year, why engineering students are either shy or reluctant to submit their presumed calculations for review when it is they who have asked for help in finding out where they can improve their performance?  It simply doesn't make common engineering sense.  I don't know whether it is a matter of not having any calculations made, laziness in preparing flow diagrams or legible calculations, or simply that students feel embarrassed about not being able to generate a logical, algorithmic, step-by-step method of correct calculations.  Our Forum members have always shown not only a willingness to help engineering students understand and dominate engineering solutions to engineering problems, but they have gone out their way to furnish the methodology, key learnings, and additional documentation showing methods of solution and additional sources for help.  Why should students logically place CONDITIONS on how they want help to resolve their problems.  That also doesn't make sense.   If help is requested and needed, one should be open-mined and totally receptive on what is being offered as helpful advice and comments all aimed towards reaching a correct resolution and constructive learning.

Our members are normally not going to work out the subject problem.  They will, however, take the time and make the effort to review the student's work and make whatever helpful comments and corrections are considered as leading to the final correct resolution.  Engineering studies are tough and demanding - if not the hardest - of university courses and it is vitally important to find the correct methodology of attacking problems as soon as possible.  There is nothing shameful or embarrassing about confronting what is identified as a difficult problem.  We have all been through that experience sooner or later.  That's why our members give their time and effort towards helping engineering student help themselves - but not by doing their work.

I hope these words are received as an effort to help you out as a student that is facing the same taxing and challenging situations that all our graduated Forum Members have faced themselves.  They have successfully undergone the same challenges presently confronted by you and they know, first-hand, how important it is to find the correct engineering method to solving these type of problems.  Helping them help you is the best thing you can do for your own benefit now and for your future as an engineer.

### #3 Blank03

Blank03

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Posted 16 June 2018 - 06:16 AM

Good day sir.
I have my solution there, posted. that is actually mine.

if you could see the written part there, that;s mine. the second picture.

the first picture goes with the solution online. if you could see the first picture, the solution is very "vague" and not as detailed as it is.

kind of like disappointing sir. THAT IS MY SOLUTION, on the second picture. i have uploaded it here. and i can see it sir here, right now.

Edited by Blank03, 16 June 2018 - 06:18 AM.

### #4 Blank03

Blank03

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Posted 16 June 2018 - 06:18 AM

should i re-attach the picture again even if the picture is already uploaded?

### #5 Blank03

Blank03

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Posted 16 June 2018 - 06:22 AM

here is the evidence of my solution and the VAGUE solution as reference.

i have uploaded it together with my query.

thank you sir,

### #6 Blank03

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Posted 16 June 2018 - 06:23 AM

i separated it, just in case, sir.

i can also send the picture on your email, as evidence.

### #7 Blank03

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Posted 16 June 2018 - 06:26 AM

i hope someone can tell me if they see the picture i attached here in this topic.

i don;t know, but I can see that whenever I send my reply here, or the original post, i can see the 2 pictures im talking about.

LET ME PLEASE KNOW IF YOU CAN SEE ALSO, thank you.

there is no use of just asking for a detailed correct solution if at the first place you don't actually know how it works.

I never done such thing.

### #8 Blank03

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Posted 16 June 2018 - 06:28 AM

my intentions are clean, I have actually written it with my own hands, my own solution.

I hope you can give me a chance by sending my solution and the problem as well on your email, so i can clean myself here.

### #9 Pilesar

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Posted 16 June 2018 - 08:49 AM

I do not see in your work where you give the answer you found to the question 'What is the temperature of the water?'

I will give you my 'back of the envelope' answer which is close by not exact to what the textbook answer is:

50 gpm * 500 lb/hr per gpm = 25000 lb/hr water flow.

-40000 Btu/min * 60 min/hr = -2,400,000 Btu/hr

Cp of water = 1 Btu/(lb F)

Temperature drop = -2400000/25000 * 1 = -96 F.

So final temperature is about 104 F.

I think you are reading the steam tables to include the heat of vaporization of the steam. This problem starts with water and ends with water. No steam. Find the heat capacity of the water using the saturated liquid values.

### #10 Saml

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Posted 16 June 2018 - 12:15 PM

my intentions are clean, I have actually written it with my own hands, my own solution.

I hope you can give me a chance by sending my solution and the problem as well on your email, so i can clean myself here.

Blank,

I know that the second solution is yours.

I tried to verify what you've done and I will tell you what are the difficulties I've found to trying to help you, until I almost gave up.

1) I am presbyopic, like many of the people with some experience (it is something that seems to come with it). So a low resolution pic of less than half a Megapixel, skewed, and low contrast is of no help. I had to save it and manipulate it a little to be able to read. No way I can read it on the 14 inches screen of my laptop. Solution: use some free apps to take pictures of your work with your cell phone (Office Lens or CamScanner are good ones. They straighten the image and provide  some image enhancement options.

2) Naming: in your class you many name Q, H and PE as you want. Not everyone names things the same. In fact the book solution has another naming for the same things. Like using a dot above the letter to mean something different from the plain letter. The way you wrote, force us to try to figure out what you meant. A couple of handwritten explanations of what the symbols mean and what you were doing would go a great way to facilitate us helping you. Remember: we are trying to help you, so the only think we ask is that you help us to help you.

3) Now, regarding your problem, you have 50 foot-pounds per pound for the elevation difference. That is 0.0642534 BTU/lbm. For some reason you multiplied by g and then assumed that the result is in BTU. 50 x 32.2 ft^2 lbm/s^2= 1610 is what you used as BTU. Don't worry. Unit conversion, specially if you are using customary units is a usual source of error see: https://en.wikipedia...Climate_Orbiter

There is no easy way out of this problem. The only solutions are: be careful, check and recheck, and have a feeling of dimensions. The later comes with practice and conscious thinking.  In your case, the energy to lift one pound 50 ft cannot be 1610 BTU. That would mean that something falling from 50 ft would heat by 1500-2000 F. Your experience will tell you that things don't heat that much when the fall 50 ft.

If this was useful, please post the solution here so we know the problem was solved. Thanks.

Edited by Saml, 16 June 2018 - 12:23 PM.

### #11 moatejr

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Posted 20 June 2018 - 04:30 PM

Where things begin to get wonky is the potential energy calculation.  g/gc*delta_z equates to a 50 lb_f-ft/lb_m, and there are 1.285^10-3 BTU/ ft-lb_f.  Hope this helps