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Overall Heat Transfer Coefficient


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#1 Dudesons123

Dudesons123

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Posted 18 August 2017 - 05:46 PM

Dear Scientist/Engineers,

I have a problem regarding a copper cooling coil and its overall heat transfer coefficient. A copper cooling coil with a 5 cm inner diameter is used to condense steam at a temperature 96.7C and pressure 0.9bar. Cooling water enters the coil at a temperature of 15C. The steam condenses at a rate of 0.1 kg s1.

 

At the bottom of the condenser is an opening of diameter 1 cm, where condensate exits and flows to a collection tank, which is open to the atmosphere at pressure 1 bar. The pipe connecting the two tanks has an inner diameter of 1cm and a total length of 20m.

 

The level of the liquid in the collection tank is located 2 m below the liquid level in the condenser. 

 

The heat transfer coefficient due to condensation on the outside of the cooling coil is 5000 Wm^-2 K^-1. If the mass flow rate of the cooling water is 5kg s^-1, what is the overall heat transfer coefficient of the cooling coil?

 

Liquid water:

  • heat capacity: 4.2 kJ kg^-1 K^-1
  • viscosity: 10^-3 Pa s
  • thermal conductivity: 0.6 W m^-1 K^-1
  • heat of vaporization: 2400 kJ kg^-1

 Use the Dittus-Boelter correlation:  

 

 Nu = 0.023Re^0.8Pr^0.4 

 

I have this following equation:       

ṁCpc(Tc2-Tc1) =UA*[((TH TC1)(TH TC2))/(ln((TH TC1)/(TH TC2)))]

 

So far I know that I have 2 unknowns which are the length of the cooling coil and water leaving the cooling coil. In addition to these missing parameters, I am also missing the diameter of the outer diameter of the cooling coil.

 

But I am not sure which other equation to use.

 

Should I assume that the wall has a negligible thickness (ignore the d0ln(d0/di)/2k from the equation below)?

1/U= 1/hi+diln(d0/di)/2k +(di/d0)/h0

Based on the inner surface area of the cooling coil.

 

Please help me, thanks.

 

Attached Files


Edited by Dudesons123, 19 August 2017 - 10:35 AM.


#2 breizh

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Posted 19 August 2017 - 06:58 AM

hi .

What about the condensation of steam : Mass flow rate * latent heat of the steam . ?

Why don't you look for engineering data about copper tube , Thickness and conductivity and the you may decide or not to simplify your equation? google should be your ultimate ressource .

 

Good luck .

Breizh



#3 Dudesons123

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Posted 19 August 2017 - 07:53 AM

Hi Breizh,

 

I am not given any data about the copper material in the question. All the information above are the only things given to me.

 

Thanks.



#4 breizh

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Posted 19 August 2017 - 08:22 PM

Hi ,

OK then but start to think as an engineer and find out what is necessary and not necessary to resolve a problem and explain the hypothesis you have been taking .

 

Good luck .

Breizh



#5 Pericles

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Posted 22 August 2017 - 06:37 AM

Hi Dudesons123

 

You also know these:

1) Q = steam flow rate * latent heat of steam 

    = 0.1 Kg/s * 2400 KJ/Kg

From this, the cooling water outlet (Tc2) can be calculated... (I calculated 25.8 C)

2) Like Breizh mentioned, look at engineering data for copper tubing. It looks like you might have a 2" copper tubing 

https://www.petersen...gory-s/1979.htm

3) Use your Dittus correlation to find/estimate h knowing that Nu = h*D/k; Pr and Re are known or can be calculated

4) Also UA = h*Pi*D*L =Q/LMTD, so you should be able to find U..

Good Luck!



#6 Dudesons123

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Posted 28 August 2017 - 05:45 AM

Hi Pericles,

 

Thanks for replying. I understand your point, however, I will still be missing the thermal conductivity of the copper material because I cannot use the internet to search up the engineering data for copper tubing. The information I have posted above were the only information to be used.  

 

Many thanks.

Dudesons123


Edited by Dudesons123, 28 August 2017 - 05:46 AM.





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