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Anonymous Poster

Heating Transfer

10/01/2010 2:58 AM

hello all, I have gone back to college after many years to study. While I have worked on many systems I do not have a qualification in enginering and that is what I am trying to get now. My area of interest is in heat transfer. We were asked a question in the course and I think I know the answer. I will give you the question and then I will give you my answer and the reason why I think I am correct.

Q: If you put a container of 1kg of water at 0degC into a container containing 10kg of water at 50degC what will the final temperature be (approximately) when the system finishes heat transfer. We have to make the following assumptions: We do not loose any heat to the outside or to through the container walls. (I don't know how to add a picture, so it is a bit difficult to explain in writing).

My answer is a that the water will be at around 45degC. My logic was that at 50degC the 10Kg of water has 2093KJ of heat energy and for every 4.186KJ that is transferred from the 10kg of water to the 1Kg of water will cause the 1Kg of water to increase by 1degC and cause the 10Kg of water to decrease by 0.1degC. Therefore at approximately 45degC both temperatures are the same and not heat transfer will occur.

We had the same question again but the temperature and mass of the heating water was now 100degC and 100kg. Using the same logic and assumptions I think the answer is approximately 99degC.

Based on the assuptions above is the answer, or more importantly, my logic correct? If my logic was right but I some how came up with the wrong answer then that is not a bit issue but if my logic was wrong and I came up with the correct answer then that would be a problem for me in the long run.

Please correct my usage of incorrect terms where appropriate. I am looking forward to your constructive feedback.

Thank you in advance for your patience and time.

Michael

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#1

Re: Heating transfer

10/01/2010 5:12 AM

Quick'n'dirty (kinda cheating) way of doing this (works because you've only got one specific heat to deal with):

(1 x 273 + 10 x (273+50) ) / 11 = 318.45..., take away the 273 gets back to 45.45...

(1 x 273 + 100 x 373) / 101 = 372.01, take away the 273 gets you 99.01

You're right that there won't be any heat transfer when they're at the same temperature, so it's equivalent to just mixing the 2 bodies of water.

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Anonymous Poster
#3
In reply to #1

Re: Heating transfer

10/01/2010 10:52 AM

JohnDG, thank you for your time and your reply. Regarding the comment on the specific heat our teacher mentioned this and we have been asked to look a the same question where the water is heating something other than water. We are just at the beginning of our studies so the examples that we are working on are very basic. One of the big issues for us at the moment is actually understanding the difference between temperature and heat, it is not something that we had to think about before.

I may not be any cleverer when I finish the course but at least I will realize how much I don't know .

Regards

Michael

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Anonymous Poster
#2

Re: Heating Transfer

10/01/2010 7:36 AM

Even easier

(0*1+10*50)/11 = 45.45

Since the 273.15 added on both the sides will eventually cancel out as you convert back into celcius.

Also note that though it is on the extremes- somehow it has not changed the state (liquid/solid/vapour) so the latent heat is not considered here.

Secondly the specific heat is taken as constant through the range which is not true in strict sense though the non-linearilty may be neglected here.

UD15

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Anonymous Poster
#4
In reply to #2

Re: Heating Transfer

10/01/2010 10:54 AM

UD15, thank you for the response. Hopefully I will actually understand your comments about being on the extremes and the non-linearity of the specific heat when I go further into the course.

Regards

Michael

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Anonymous Poster
#10
In reply to #4

Re: Heating Transfer

10/02/2010 5:59 AM

Look here where the specific heat is given at different temperature of water

In addition you may just like to have a look in wiki

This provides the same for semiconductors , Some theory

But for normal calculations these are not too important. The criticality comes when you work with wide range (eg super heated steam etc where the variaations do play a role.

UD15

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#5

Re: Heating Transfer

10/01/2010 11:21 AM

The best way to help you out is to recommend you to go to www.sprirax.com and download their information books on each related subject/s. This will help you further.

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Anonymous Poster
#6
In reply to #5

Re: Heating Transfer

10/01/2010 11:38 AM

Thank you ducon. I will look into that. I actually used the "Search all of CR4" and came up with some really great information. I found loads and loads of manuals and resources (I am reading "A Heat Transfer Textbook" at the moment and I have read the first 6 pages and it is really great, only 756 pages to go - don't understand it all/any yet but hopefully I will and I maybe able to help someone else out in the future as you have been doing here). The thread that I got the link for the book from was http://cr4.globalspec.com/thread/32885/Design-Guidelines-for-Heat-Exchangers.

Thanks to all of those that supplied this information.

Kind regards

Michael

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#7
In reply to #6

Re: Heating Transfer

10/01/2010 12:53 PM

Michael, I downloaded it, thanks. When you come on Boilers and steam, email me at duconems@yahoo.com and I'll try to help you as much as I can.

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#8
In reply to #6

Re: Heating Transfer

10/01/2010 3:43 PM

Michael - while you're around, why not register on CR4? Only takes a couple of minutes (and it's free ). You can get e-mail notifications when someone replies on your subscribed threads, exchange private messages (PMs) and all sorts of handy stuff.

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#9

Re: Heating Transfer

10/02/2010 3:12 AM

Use the following formula:

Tm= (M1xT1+M2T2)/(M1+M2)

Tm= MIXTURE TEMP ; M1= MASS OF WATER 1 ; T1= TEMP MASS OF WATER 1

M2= MASS OF WATER 2 ; T2 TEMP OF WATER 2

Good luck

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Anonymous Poster
#11

Re: Heating Transfer

10/02/2010 3:56 PM

Thank you all for your help. I will take JohnDGs advice and register. Looking forward to discussing further issues and hpefully I will be able to help someone at some stage,

Regards

Michael

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#12
In reply to #11

Re: Heating Transfer

10/02/2010 4:52 PM

That there's a good answer .

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#13
In reply to #11

Re: Heating Transfer

10/02/2010 5:13 PM

Hi Michael

I had a related question a while back and was in a similar situation. The learning curve was very steep and still is to be honest. It amazes me again and again how this forum can exchange information and get potential leads faster than whatsthename.

Your polite frankness and willingness to participate is very much appreciated here in CR4. Just showing interest in learning is contribution enough. Don't hold back, nobody is perfect.

Welcome to CR4, Ky.

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#14
In reply to #13

Re: Heating Transfer

10/02/2010 6:08 PM

There, another good answer

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#15
In reply to #14

Re: Heating Transfer

10/02/2010 6:26 PM

This is becoming a self esteem issue. My reply was really off topic and did not answer guests question.

The principles of heat exchange can get very complicated. Here is the prior link:

Wind Speed Cooling Factor?

Maybe Micheal can find something useful by following the links supplied by others. I'm nearly over it.

Thanks for the GA, I'm flattered, Ky.

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