Heat transfer question 2

[SmartEngineer]

I have got a question by one of my friends as follows:
- In a heat exchanger, assuming the heating side on the shell side is constant, the question is that if the flow rate of the cold stream in the tube side is reduced, Is its outlet temperature is gone be high or low?
His answer to this question is that the outlet temperature of the cold stream in the tube side will get low because of the less surface area contact of the cold stream with the hot stream in the shell as the tubes will not be full of flow and then the contact area is less, hence the cold stream outlet temperature gets low.

My answer to this question is as follows:
The cold stream outlet temperature will be higher due to the following reason:
As the flow rate of the cold steam in the tube side is reduced, then the velocity will be reduced (coz area of tube is constant), by this the contact time is high leading to high temperature of the cold stream.

I know that there are other reasons of this question such as duty of the exchanger ..etc

What do you guys think?

Cheers

 

[djack77494]

SmartEngineer,
Unless you are condensing or boiling, your tubes should be full of liquid at all flowrates, so I don't understand your implication that the contact area will decrease.

When you say constant shellside heating, I assume you are saying that the shellside temperature does not change. That would be true, for example, if you were condensing steam at a given pressure on the shellside. Your source of heat would then be at a constant temperature.

Given the above assumptions, the tubeside outlet temperature will definitely go up as the flowrate is reduced. Look at the heat transfer equation:

Q = U * A * dT

A is totally unchanged by flowrate. U would not be affected significantly by modest flowrate changes. But Q would decrease, and therefore dT would decrease. The net result is that the tubeside outlet would be hotter.
Doug

 

[watwarrior]

The tube side outlet will get hotter if its flow is reduced (all other things remaining constant). This is just based on the overal time that the two fluids will be in contact for.
The simple explanation is that the longer the two streams stay in contact with each other, the closer they will be to an equilibrium point when they exit.

 

[Montemayor]

Smarty:

Doug is 100% correct in his response and he is right on top of this question because he uses the basic, fundamental Fourier rate equation to prove his point. I would hope more chemical engineers study his response as to logic, basic theory, and common sense. Far too many ChEs fail to understand the simplicity of the basics and the common sense applied and assumed in the final equation(s).

The main key to the question is that the basic assumption is that the tubes are full of liquid. In other words, there is no 2-phase or (worse) non-condensable gases present in the tubeside. How can a graduate engineer imagine that the liquid is literally “dripping” out of the tube side? This is where common sense should prevail. That’s the reason why all shellside liquid inlets should be on the bottom nozzle of a horizontal shell & tube unit and never at the top nozzle – to expel any trapped air or other non-condensable in the shellside. The basis of the heat transfer equation is rate of flow – not time – since this is a steady state operation, not batch. There is no equilibrium temperature involved. It is called the approach temperature and that is what Doug is inferring when he points to the LMTD decreasing.

Kudos to Doug.

 

[rmw]

Be careful here.

Doug's statement is based on his assumption that the "U" value doesn't change much, but that is not always true.

Basic heat transfer is a function of Time, Temperature and Turbulence. Temperature is a constant in this discussion. Time increases because of the slow down of the flow rate on the tube side, but, depending upon the dT, the loss of turbulence might have a profound effect on the heat transfer rate, especially of the flow rate reduction approached the laminar region.

One should look at the components of the "U" value to determne if flow rate is going to have a pronounced effect here.

rmw

 

[25362]

Holman's Heat Transfer brings an example of water heated by an oil to show when to apply the concept of NTU effectiveness. Effectiveness,

[ε] = actual heat transfer[÷]maximum possible heat transfer​

The question was, for a given heat exchanger, what would be the outlet water temperature if both fluids had the same inlet temperatures, but the water flow rate were reduced.

The answer in that particular example: water gets hotter. Although the flow rate of water was reduced by 41% the heat transferred was only 19% lower because the effectiveness of the exchanger rises at the lower flow rate.

The NTU method can also be used for the purpose of selecting the type of heat exchanger best suited for the heat-transfer objective.

An interesting discussion can be found in thread391-99241.

 

[UmeshMathur]

SmartEngineer:

In a liquid-liquid exchanger, there is no question that both sides must remain full of liquid, unless you got a stubborn gas pocket at startup that never cleared owing to bad design of exchanger internals.

While reduced flow rate will decrease the water-side film coefficient, the reduction in the (W*Cp)min to (W*Cp)max ratio in the effectiveness-NTU method, quoted by 25362, generally overwhelms the NTU factor [U*A/(W*Cp)min]. Therefore, for any exchanger geometry, you would see an increase in water side exit temperature, coupled with a reduction on the overall amount of heat transferred.

Incidentally, in my opinion, the effectiveness-NTU method is infinitely superior computationally to the traditional F-corrected-LMTD method. Also, there are readily available analytical equations for the exchanger effectiveness curves for most commercial geometries of interest, a claim that cannot be made for the F-factor correction charts required for the F-corrected-LMTD method.

A huge benefit of the effectiveness-NTU method is that (when U and Cp may be assumed constant), it can solve both design and rating problems without trial and error, unlike the F-corrected-LMTD method which ALWAYS requires trial and error. Besides, the effectiveness-NTU curves show clearly why, beyond some point, increasing surface area is useless in increasing the total heat transfer at fixed (W*Cp)min to (W*Cp)max ratio.

This makes me wonder why, in this day and age, ANYONE uses the old F-corrected-LMTD method at all.

 

[srfish]

UmeshMathur:

Where are these "readily available analytical equations for the exchanger effectiveness Curves" ? The Chemical Engineers' Handbook doesn't have them and I didn't see them in the Handbook of Heat Exchanger Design.

So I don't believe I will throw away the LMTD correction equations and charts just yet. Besides what do we put on TEMA specifications sheets on line 29 where it asks for MTD(Corrected)?

 

[UmeshMathur]

srfish:

Sorry for the delay in replying.

The definitive reference for the effectiveness-NTU method (developed in the early 1960s at Stanford University by Prof. Kays and his co-workers) is:

W. M. Kays and A. L. London, "Compact Heat Exchangers" (Krieger Publishing Co., Malabar, Florida, 3rd edition, 1984). Originally published by McGraw-Hill, 1984.

A detailed explanation of this method, including some newer parameters that have other benefits and are equivalent to the effectiveness factor E, is also available in:

G. F. Hewitt, G. L. Shires, and T. R. Bott: "Process Heat Transfer" (CRC Press, 1994)

Finally, most modern textbooks show both the equations and the effectiveness charts for the common heat exchanger configurations, e.g.:

J. P. Holman, "Heat Transfer" (McGraw-Hill, 1997, pp. 572-586)

Once you know the exchanger effectiveness E, you calculate the heat duty (Q) directly from the definition of E. The corrected MTD could then be back-calculated using the single formula:

(Corrected MTD) = Q/(U*A)

I strongly suggest a quick study of this method, which takes no more than about 30 minutes to comprehend. You will be amazed at the savings in time, especially if you are not doing things by computer (as you have analytical equations for most practical geometries).

The effectiveness graphs show you very quickly why adding surface beyond some reasonable limit is useless in increasing heat transfer.

Over a few hours, I developed solutions for standardized problems, without iteration, in a simple spreadsheet.

Sadly, in professional practice over the last 38 years, I have not found a SINGLE advantage of the traditional U*A*(Corrected MTD) method over the effectiveness-NTU method. In my opinion, the failure to discuss this well-established method in Perry's Handbook (even in the 7th edition) is a serious omission that should be rectified in the next edition.

I conclude with the following quote from Professor J.P. Holman (page 572. op cit): "The effectiveness method also offers many advantages for analysis of problems in which a comparison between various types of heat exchangers must be made for purposes of selecting the type best suited to accomplishing a particular heat-transfer objective".

Hope that helps.