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[mac21]

I have been trying to solve the following problem given to me by my professor:

An organin liquid initially at 85 degree C is to be cooled to 20 degree C using a counter-current heat exchanger. Cooling water is available at 15 degree C which must not exceed 30 degree C on leaving the heat exchanger. The flow rate of the organic liquid is 15kg/s, which has a specific heat capacity of 2.8 J/g(degree C). Calculate the heat to be removed from the organic liquid ad the area required for the heat exchanger. Assume no fouling takes place.

initially I used q=mC(delta T) where m=flow rate C= specific heat capacity and T = temperature to solve for heat energy needed to be removed from the organic liquid. I found q = 2,730,000. I then solvd for the log mean temperature difference which i found to be 20.85 degree C.

With this information my original plan was to sub my value for q and log mean temperature difference into the equation q=UA(log mean temperature difference) where U is the over all heat transfer coefficient and A is the area of the heat transfe surface. The problem is that U and A are both unknown to me.

I am unsure if I took the right approach to solving this question or if I am simply missing some forumla.

Any help anyone can offer would be much appreciated.

[Art Montemayor]

Mac:

From what I can interpret of your description of how you got to where you are at, your application seems OK. Let me bolster your confidence to resolve this problem:

1. The amount of heat given up by the organic liquid is equal to the amount of heat taken up by the cooling water;
2. There is no other heat transfer or heat loss, except for the heat transferred between the two fluids;
3. The specific heat of the organic liquid (for the 65 ^oC range of heating it goes through) remains constant;
4. There is no phase change taking place in either fluid;
5. Therefore, the sensible heat transfer that takes place can be calculated by:

Q = W Cp (T2 – T1) = (15,000) (2.8) (65) = 2,730,000 J

Where,
W = 15 kg of organic liquid / sec
Cp = Specific heat of organic liquid = 2.8 J/g - ^oC
T2 = Inlet temperature of organic liquid = 20 ^oC
T1 = Outlet temperature of organic liquid = 85 ^oC

The only other relationship you have with respect to the required heat exchanger is Q = UA LMTD, which you are ready to apply. This is also OK and valid. You know the temperatures of the entering water and the exiting water, so you can solve for LMTD. All you can do now is solve for UA.

You are right; you don’t know the Overall Heat Transfer Coefficient (U) and so, if you are to calculate the required heat transfer area, you must assume a number for U.

Your logic and algorithm are valid in my opinion - although I would normally expect to assume the heat capacity of water remains constant over the smaller 15 ^oC range and knowing the water flow rate, I calculate Q. It is important to know the basis of the sensible heat transfer equation in order to have a feeling for how logical and accurate you can expect to be.