Something does not appear right with my following calculations. I am trying to determine how many BTUs will be transferred in one hour through a material and I am getting a huge number. I understand as "delta T" decreases so will the number of BTUs transferred, but in its initial state, this doesnt seem right. Can somebody please help me out.

given four 24" * 12" rectangle plates of 304L (stainless steel) that is 1/2" thick, how much heat will be transferred through it.

I used the following formula to calculate:

heat transfer coefficient
delta Q=h*A*delta T

k = thermal conductivity of the material = 18 W/(m*K)
t = thickness of material = .25" = .001588 meters

h = k/t = 11338 W/(m^2*K)

A = surface area = 4*height*length = 4" * 12" * 24" = 4 * .3048meters * .6096 meters = .74322 meters^2

delta T = 350F - 80F = 450K-300K=150K

delta Q = 11338W/(m^2*K)*.74322m^2*150K
delta Q = 1265751 W = 1265751 J/s

convert to BTUs (times by 3.41)
delta Q = 4316214 BTU/hr

did i do something wrong?? this number sounds way too high since the device is small and most heat exchangers I can find cannot supply this amount of BTUs.


2) how does flow rate affect this? i realize there is a point where it can flow so slow that the fluid no longer has BTUs to transfer, and it can flow too fast where the fluid does not have time to accept BTUs from heat exchanger/give off BTUs to solid.

so how would using 10 GPM(gallons per minute) effect the BTUs differently than 20 GPM?

2. when using the above equation Q=h*A*delta T(heat transfer coefficient) it gives me an answer in Watts which I can convert to BTUs. as delta T decreases so will Q, so

does this mean that if i calculate for Q and get 4316214 BTU/hr, after one hr of time, will 4316214 BTU/hr BTUs have been transferred? because as the BTUs are transferred, the delta T will decrease so therefore so will the Q.

or does this mean that 4316214 BTU/hr will be required to get the initial temp to equal the desired temp where delta T = 0??

At first your number looked pretty high to me too, but this number represents only the thermal conductivity of the metal itself. In other words, if you could apply heating and cooling coils directly to the surfaces of the metal and hold them 1 degree C apart, the heat flow would actually be 11338W/m^2.

However, heat must first get to the metal from the flowing fluid, and then must also exit from the metal into the surrounding air. (I'm assuming air as the outer fluid here. If it's something else, the calculation will be basically the same but with different numerical values.)

Therefore, there are 3 stages of heat flow, each with its own h: h1 is the convection coefficient from the hot fluid to the metal, h2 is the metal itself, and h3 is the convection coefficient from the metal to the cold fluid. The convection coefficients depend on lots of things like the flow rate, degree of turbulence, fluid composition, density, thermal conductivity, and specific heat. There are usually some large messy empirical formulas that are used to calculate these h values.

The total h is calculated by 1/h_total = 1/h1 + 1/h2 + 1/h3. Then you use the total h to get the actual heat transfer rate, which will be MUCH lower than just the rate you got for the metal itself, because the other h values are much smaller than the one for the metal. That's why you are seeing much lower heat transfer rates from actual heat exchangers; it's limited by the ability of the convection to transfer the heat, not the ability of the metal.

Hope that helps!

--Stuart Anderson

i greatly appreciate your help.

question though, i know my "k" for my fluid. how do i calculate the "h?"

h=k/t
h=.1347W/(m*K) / thickness of what??? is the "t" the diameter of the pipe or what?

once i calculate this "overall h," can i then get a heat exchanger that generates more BTUs than my new value for Q and be okay?

ps. i realize at this point i am neglecting heat loss through other parts in this process.

The thickness of the plate converts to 6.35mm instead of 1.588mm.

Your fluid probably limits the heat flow more than the plate does, I agree. There are few exceptions, for instance a sodium-potassium liquid conducts heat better than your 304L (but it would also dissolve it).

What equivalent thickness? Hey, that's the complicated part. If you hoped to use a measurable dimension here, forget it!

You will need to compute some characteristic numbers of your flow, see Prandtl, Reynolds and half a dozen more. From these numbers, you can deduce an equivalent thickness (which depends on the position along the flow) and a heat transfer.

Don't hope to understand it (or at least, to know the names and expressions and what to do with them...) in less than a day. Once you're easy with it, each case takes you a a handful minutes - though specialized people will do it in half a minute.

So... Find a good book and invest time in it. I know one by Sacadura, but is it available in English?

Just to give an intuitive idea of convection:

Your fluid conducts heat badly (but better than air does). It would be an acceptable insulator if it could stay still, like the cold air in a basement can stay still under the warm air of the basement (I mean, first floor in the US).

But your moving fluid has some turbulence which will efficiently replace the fluid in the vicinity by new fluid that was farther from the plate and hasn't got the temperature of the plate. So turbulence is the primary mechanism that enhances (a lot) heat exchanges between fluids and solids.

Bad luck for you, turbulence is nothing simple. No good theory. Just some empirical formulas. Even computer simulations should be taken with care here.

When designing heat exchangers, one possible direction is to maximize the contact surface and minimize the thickness of the fluids; you typically arrive to many small parallel pipes or plates, which may cost a lot to assemble and are susceptible to dirt. The other way is to take advantage of turbulence.

When you have become an expert in this activity, you may try to improve the cooler towers for power stations... In a 1400MW station, you need to evacuate 4000MW but lose as few °C as possible at this cold sink to improve the plant's efficiency - 30°C above ambient would be bad. Try to put figures on that undertaking: the amount of pipes is unaffordable, and alone the power requested for blowing cool air would be a huge loss for the plant. Interesting problem that needed original solutions.