Heat transfer area

[Covboy69]

Hi!
We've been 'messing' around with a new micro reactor type device.

We're now wanting to estimate the heat transfer coefficients for the device.

It basically a straight etched channel in a sheet of 316L SS. The heat transfer occurs throgh the base of the channel to a utility stream passing underneath.

The problem we are having is defining the heat transfer area.

For example, lets say the sheet of steel is 5cm wide and 2cm thick. The channel is 2cm wide and 1.8cm deep.
The utility stream covers the whol of the underside of the sheet (i.e. is 5cm wide.)

What is the Heat transfer area?

Is it simply the width of the base of the channel (i.e. 2cm x its length)
It it the area the utility fluid touches/cools (i.e. 5cm x its length)
or
Is it the area of the base and walls of the channel? (i.e. 2cm x its length + 2*(1.8cm * its length)) (The top of the channel is 'blocked' with an insulating material.)

Or is it something else??!!

All thoughts, comments, references, anecdotes Greatly received!!



Also, when calculating Q (for the Q =U *A *LMTD) is it correct/OK to us Q = the average of the energy gained by the utility and lost by the porcess stream??

 

[Twoballcane]

"It basically a straight etched channel in a sheet of 316L SS. The heat transfer occurs throgh the base of the channel to a utility stream passing underneath."


If this is like a fin being dipped into cool running fluid (sounds like a forced convection prloblem) the area I would take would be the part the fluid touches the end of the channel. But it looks like you many have two circuits happening. One is the forced convection with the fluid and natural convection with the part is exposed to air. However, most of the cooling would be with the forced convcetion.

"Also, when calculating Q (for the Q =U *A *LMTD) is it correct/OK to us Q = the average of the energy gained by the utility and lost by the porcess stream??"

Conservation of energy would dictate that Q out is Q in.




Tobalcane
"If you avoid failure, you also avoid success."

 

[magicme]

if i drew the sketch correctly...

the wall at the bottom of the channel is 0.2cm thick (2 - 1.8) .... I'd use that as the plate thickness for heat transfer calcs.

need to calculate the convection coefficients on both sides (channel and utility stream) and the t / k for the 0.2cm wall.

looks to me like most of the heat transfer will be thru this thin section, but I'd add a little to the width (make it maybe 1.10 * 2 = 2.2 cm to allow some additional transfer around the corners at the bottom of the channel.

and, as Twoballcane says Qin = Qout.

regards



------------------------------------
there's no place like gnome.

 

[vpl]

Yes Q_in = Q_out. However, if you've measured the inlet and outlet temperatures and the flow rates for both the utility and process side, you can calculate a Q for both sides with slightly different results. This is usually due to measurement error. If all your measurements have about the same degree of accuracy, using the "average" Q is probably best. If you know one measurement has more error in it than the others (often a flow rate), then calculate Q using the measurements with the least error.

Patricia Lougheed

Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.

 

[IRstuff]

How critical is the accuracy of your answer? That'll determine the degree of pain you'd need to achieve to get the answers.

There are 4 sides your reaction, neglecting the ends.

The top is exposed to air, so there's something on the order of 2-4 W/m^2-K of convection, plus radiated heat loss.

The two vertical sides has heat conducted and flowed to the bottom and external verticals. Assuming that the bottom is more efficient, you'll need to do a spreading resistance calculation, but you could potentially assume that it's equivalent to a single 1.5cm tall surface attached to the cooling channel through 2 cm of thickness.

Then, the bottom, which has been covered in others' postings.

TTFN

FAQ731-376