Tank heat loss coefficient

[mfqd]

Hi,

I have a question regarding the calculation of the heat transfer coefficient of a tank filled with hot water.
The calculation i know how is made, but my question is othe:
I have the hourly energy loss registered, so i know how much energy this tanks loses in an hour in a continuous function mode. So, dividing this by the total surface area, i have the heat loss for square feet: BTU/hr/m2.
The heat loss coefficient is represented by BTU/hr/m2/K.
So, how can i determine the coefficient havin the hourly heat loss value?

thanks

 

[IRstuff]

So, what's the temperature difference

TTFN

FAQ731-376

 

[jmiles]

Ok, grammer aside, my biggest pet peeve is mixing units the way you have, either use metric or imperial, dont try to use both its just a mess. so that would be Watts per hour per square meter, and you also said in words square feet, but then used square meter in your notation... slightly confusing.

Now to your question, you have an hourly heat loss, i am assuming this is something you calculated? what are the conditions of your tank, are you adding heat or fluid to the tank? if so you can likely assume that you will have a steady state heat loss and your pretty much done.

The coefficient is telling you how much heat loss there is based on a temperature differential, as you lose heat from the tank the bulk temperature will drop and you will lose less heat as time passes as the tank and the surroundings come into equilibrium.

Now personally when i do tank heat loss calculations i base it on (T1-T2)/Rtot

Where T1 is the bulk temperature of the tank, T2 is the ambient temperature and Rtot is the ttoal thermal resistance.

For a simple radial system you can calculate the total resistance by taking the Internal convection, the cylindrical wall conduction and the external convection and summing them as a thermal circuit.

This calculation is fairly straight forward, but if you want to calculate the internal and external convection coefficients, then things get a bit stickier, id recommend Process Heat Transfer by Donald Kern if you an find a copy it helped me with alot of my heat transfer issues.

When you are calculating the thermal resistance you can either use areas of 1 m2 and then take your heat loss as a W/m2 or you can input the actual dimensions and just get a heat loss in W.

 

[mfqd]

Hi,

Thanks for the answers.

1) O apologise for the imperial/metric units mess. I know that SI: W/m2K and IP: BTU/hr.ft2.R

2) This is a big industrial facility and these are tanks wich have for example, sprinklers, and other devices in it and is unthinkable to start some kind of complicated calculations. So, i know the energy spent to maintain the tanks with a temperature of 55ºC during th day, and i will use this to predict the energy losses.

3) I know that the ambient temperature also matters for the calculations, but i have the values for the energy consumptions for all months, so this doesn't matter also.

4) What i've concluded by "jmiles" answer, is that Q=U*A*DTML and U should be = Q/(A*DTML).

Correct?


Thanks!!!

 

[jmiles]

Ok So i think i understand what your looking for here, (im going to use metric units as i find BTU's confusing for dealing with heat transfer)

You have the hourly heat loss, and you have the surface area of the tanks, so we know that heat loss for one hour = X measure in Joules we know the surface area Y measured in m2, and we know the time frame is 1 hours

Now a Watt is a joule per second, so Divide your ehat loss in Joules by 3600 and you will have the heat loss in Watts

now your Heat Flux is W/m2

If you have the average temperature inside the tank, and the average temperature outside the tank you can get a heat transfer coefficient by dividing your Heat Flux by the delta T in °K

This will give you a heat transfer coefficient in (W/m2)*K

 

[mfqd]

YES.

Correct. That was my idea, and what i want.
So, tnak you "jmiles" for the reply!
It was very useful to me to have your agreement on this.

THANKS!