Increase heat transfer 1

[JuanMC]

I have a heat exchanger, (shell & tubes) fan-cooled. It cools oil from 55 °C to 45°C. I would like to increase the net heat transfer with the minimum changes in the system. I have determined the additional heat stream that i need, but don't know how to relate it with the heat exchanger. This is what i've done:

1. I made tests and with Heat = mass flow x Cp x (Thout - Thin), i got the net heat transfer rate.
2. With this, applying LMTD (and F), I got the UA coefficient.

The problem is that the UA is the inverse of the thermal resistance of the system: it depends of the conductive and convection coefficients. The first one is related with the material of the tubes and the second, with the fluid's characteristics: viscosity, density, type of flow, but NOT temperature.
There is a way to decrease the air temperature, but first, i need to get how much would that help me.

I've tried to calculate it in the next way:

1. With the expermimental heat transfer, I got the UA (said before)
2. With UA, and different air temperatures, i calculate (again) the heat transfer rate (it is different of the inicial, due to the correction factor for cross flow exchangers)
3. Also, i get a new heat transfer rate. So, i calculate the "real" temperature for oil at the exit, loosing this heat.
4. Finally, i calculate the total heat lost by the oil with this "real" temperature.
5. I start again the process, until both values are the same.

I'll appreciate any help. I don't know if what i've done is ok.

Thanks!

 

[JuanMC]

I have the iteration process in a .xls sheet, if someone is interested.

Thanks again.

 

[dcasto]

There are three things you can do.
Get more area, get more U, get more air flowrate.

More area, add another bank.
More U, are the tubes clean inside and out (compare your UA to original design and hope the original design is correct.
Get more air flow, up the mtor size and re-pitch the blades. The GPSA handbook goes through a nice example of all these.

 

[insult2injury]

Decreasing the air temperature entering the exchanger will affect your LMTD not necessarily your UA if the properties of the air are relatively unchanged.

I2I

 

[JuanMC]

Yep, that is what i'm trying to determine by doing all that. I'm supposing that UA remains unchanged, but i need to determine the variation in heat transfer (by the variation of the air-inlet temperature), before selecting a way to modify the system.

dcasto, thanks. If i can't get enough heat exchange by reducing the air temperature, the next easiest modification is add another bank.

Let me re-edit my question:

To use the LMTD method, i must have all the temperatures (in/out for hot and cold fluids). I have them at the present operation rate... i can get the UA, but what i need to know is ¿how does affect the air temperature variation?

The only values i can know are the oil /air temperatures at the entrance... so, ¿how can i find the exit temperatures, if they are affected by the heat transfer rate?

I don't know if i'm going by the wrong way, using the LMTD method... should i go with NUT?

Thanks to everyone

 

[insult2injury]

From a programming perspective, I've favored the effectiveness-NTU or P-NTU methods, but they all are equivalent. No matter which method you choose, you should get pretty much the same answer.

I2I

 

[insult2injury]

Why don't you know the desired oil exit temperature?

I2I

 

[JuanMC]

I am not designing a new exchanger. I'm trying to increase the efficiency of the original exchangers, by cooling the air used by the fan. I could know this temperature (air at the entrance), but my desired oil exit temperature is the coolest possible with the minimum modifications.

 

[speco]

JuanMC,

In an air-cooled oil cooler the oil side generally controls the overall heat transfer coefficient. This assumes that the oil viscosity is pretty high and the oil is in laminar flow. Adding air flow does very little to increase cooling capacity.

So your approach of precooling the air is one possible solution. As you decrease the inlet air temperature, the outlet temperature of the oil also goes down. If the inlet temperature is not fixed, it will go down, too. However, assuming the inlet oil temperature is constant, you can estimate the increase in capacity be reiteration. Each time you lower the oil outlet temperature, you also increase the air temperature rise such that the two heat loads (air and oil) balance.

Then assuming that the U and A are constant (a pretty good assumption), you can solve for a balance using the new LMTD (Q = U X A X LMTD)

For more info on cooler performance problems try

http://www.stoneprocess.com

and click on "Why Doesn't My Cooler Work?"

Regards,

speco

 

[0707]

Is there any possibility to adjust the angle of the fans to overcome your air temperature flow needs? If so I think that is the way.

 

[25362]

Speco, are you sure that

in an air-cooled oil cooler the oil side generally controls the overall heat transfer coefficient ?

 

[0707]

25362

No I just think...But I am not sure

 

[JuanMC]

About this "in an air-cooled oil cooler the oil side generally controls the overall heat transfer coefficient"

I think that that is related to the material of the tubes and the fluid's characteristics: viscosity, density and type of flow (laminar or turb.) We need to consider the air density and its low specific heat (i'm not sure if "specific heat" is the english word for "Calor especifico"). So, even if i adjust the angle of the fans, an increase of the air flow only represents a small increase in mass/time, therefore a small increase in heat transfer.

Thanks for your help!.

I don't know if i should skip all this theory and just do the modifications, hoping that it works...

 

[speco]

25362,

I know it's not good to generalize too much in a forum like this. However, for most oils in this 55C to 45C temperature range, the viscosity is relatively high, making the flow laminar and the inside (tubeside) heat transfer coefficient very low. One would presume that this exchnager has finned tubes, which mulitply the effect of the low inside heat transfer coefficient. In this case, the oil side would be the controlling resistance.

However, if this were a hot oil cooler (say in the 200 to 300 C range), then the air side would be more equal to the oil side resistances.

Regards,

Speco