Heat transfer 1

[doop4]

Thanks to all the valauble contributors in this website.

I was wondering if someone can help me with the following question.

I have water in a pipe at abt 23-24 degrees C and the temperature is not very stable since there is a pump that circulates this water into a closed loop pipe system. Hence I want to bring the temperature of the water down (usually by about a degree) in order to stabilise temperature in the system. For that I am hoping to add mains water into the system (which will be 10 degrees C or so)and wish to calculate how much of this water I would have to use to bring the water temperature down by about a degree or possibly less. The temperature increases by about 0.5 degrees every hour in the pipe systme but I am looking to have very precise constant temperatures throughout the whole day for experimental purposes. The water through a section of the pipe runs at about 5 kg per sec. So what I am having trouble calculating is that if for eg I wanted to bring the temperature of 5kg of water down from 24 degrees to 23.5 in say less than 10 minutes, how much of the "other water" (at 10 degrees C) will I need to do this?Is there an equation for this?

I would greatly appreciate a reply on this.

Thanks in advance

Dharshie

 

[katmar]

Dharshie,

The way you have posed this problem, the circulation rate is irrelevant. You need to know the total amount of water in your circulating system. Call this C kg. If the temp of this body of water increases at 0.5 deg C per hour then the energy absorbed is
C x 4.18 x 0.5 = 2.09 x C kJ/hour
(assuming specific heat of water is constant at 4.18 kJ/kg.deg C)

If you want to hold the system at 24 deg C by adding water at 10 deg C then the difference in energy between the 24 deg C water which is purged and the 10 deg C water which is added must be equal to the energy absorbed. Call this make-up rate (which equals the purge rate) M kg/hour. The heat absorption capacity of this added stream is
M x 4.18 x (24 - 10) = 58.52 x M kJ/hour

Since the heats must be equal, M = 0.036 x C

i.e. you must add 10 deg C water at a rate equal to 3.6% of your system volume every hour.

If there is some point in the circulating system where the water is hotter than 24 deg C, say at the point where the circulating water is returned to the holding tank, then it would be better to purge the hotter water and substitute this temperature in place of the 24 in the math above.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[doop4]

oh thank thank thank you so so much!!!!!!!

So can I ask you one last time to pls have a look my calculations below

flowrate of 9000kg/hr of water at 24.5 (so to bring it to 24, there is atemp change of 0.5 degrees)
hence energy absorbed per hour= 4.18x 0.5 x 9000=18810

Now I want to add ice to the water to cool it down to 24 degrees
hence energy given out by M kg/hr of ice would be
= 2.03 (sp heat cap for ice) x M x 24 (temp diff)= 48.72 M

hence M = 18810/48.72=386kg/hr that's abot 6 kg/min

this sounds a bit too large. Am i going wrong somewhere?

 

[katmar]

The cooling effect of the ice is its latent heat of melting at 0 deg C, plus the sensible heat of heating the water generated from 0 deg C to 24 deg C. The latent heat if ice is 333 kJ/kg. Therefore the heat absorbed by the ice is
M x (333 + (24 x 4.18)) = 433.3 x M kJ/hour

This must absorb the 18810 kJ/hour, so M = 18810 / 433.3 = 43.4 kg/h or 0.73 kg/minute.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[doop4]

thanks!