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easier than my other question

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Smokey2007

Mechanical
Apr 2, 2008
16
asking a simplified question from the one yesterday. hopefully, if somebody can answer this, then i will be closer to answering the harder one, and this one does apply to my system.

if this is not easier, i guess there is no hope for me :(

lets assume
in my heat exchanger i will have 50 PSI of steam which equates to a approx internal boiler temp of 281 F.

the specific heat of the heat transfer fluid is .71 Btu/lb F at this temperature.

assuming a 1/8" stainless steel pipe thickness that is 24'' long with a .5" diameter which has a thermal conductivty of 18W/m-K will be transporting the heat transfer fluid,

what is the heat transferred?
 
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Smokey,

There's no way to know without some really detailed additional information. Here'w why: Heat transfer is a function of a number of factors all working simultaneously.
First, to come up with an overall heat transfer coefficient, you need to sum up several thermal resistances. These are (1) the tubside fluid film resistance. (2) the tubeside fouling factor, if any. (3) the metal resistance of the tube. (4) The shell-side fouling factor, if any, and (5) the shell side fluid film resistance.

The major resistances here are generally the two film resistances. However, in a conservative design, the fouling resistances can also be very significant, to the point where the "clean" overall coefficient is around twice the "fouled" coefficient.

These two film resistances are properties of the tube bundle geometry, as well as the fluid properties and their mass velocities through the tube bundle.

The next major component of the heat transferred per unit area is the effective temperature difference. In your case, the steam side is most likely isothermal, but the other side has a temperature difference that must be accounted for. That is, as heat is added to the fluid, the temperature increases, and the effective temperature difference is generally the average of the two terminal temperature differences. This is normally the Log Mean average, by the way, but an arithmetic average is usually OK for small temperature changes.

The third component of the amount of heat transferred is simple. It's the amount of surface available for heat transfer. By the way, in calculating the overall heat transfer coefficient, you need to base it all on the same heat transfer area, usually the outside surface of the tube.

In other words, Q = U X A X LMTD

Where
Q = Heat transferred per unit time

U = Overall Heat Transfer coefficient

A = Avaiable Area (Surface)

LMTD = Log Meat Temperature Difference

In your particular problem, the most important missing items are the flow of heat transfer fluid in pounds or Kg per hour, and the properties of the fluid at its temperatures. The important properties include viscosities, by the way.

Regards,

Speco (
 
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