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• # Calculation the temperature at certain length

Discussion in 'Calculations' started by Wmech, Apr 26, 2013.

1. ### WmechMember

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Hi, all

I recently join this forum and hopefully i can learn a lot from you guys who are well-experienced in the field.

A question about the temperature changing at certain length.

Here is the situation: A fan is placing behind a radiator and pushing 800 cfm of -10 ÂºC air towards the radiator (heat source) just like in a car system. The radiator generates approximately 60,000 BTU/h. What I am trying to find is the outlet temperature of warm air from the fan and the radiator. Also, what's the temperature at 3 feet away from the radiator?

I have some theories of how to attack this problem.

First way to counter this problem is to use the equation Q_dot = m_dot * cp * delta_T. Since I have the volume flow rate and the density of the air at -10 ÂºC, and also the specific heat of air, I can easily figure out the outlet temperature of the fan.

However, when the fan blows the heat out of the radiator, it must be taken into account some efficiency loss. In other word, there will be 60,000 BTU/h heat transfer rate if the efficiency is 100%. Therefore, the second way I would counter the problem is to use heat exchanger analysis (more specifically NTU method).

The problem is that the temperature and the fluid properties of inlet of heat source are unknown.

I am getting a sense that I have overlooked the problem and made it more complicated than it supposes to be. If there is any one who is more experienced with this, I am open all the suggestions and helps.

Thanks

2.
3. ### thermalMember

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Wmech, bonsoir

I sort of believe that you are on the right trace but lets get more profound.

The 60 000 BTU/h is a power. Is it generated by some kind of electrical coil or by a heat exchanger? In the first case you do not have to bother much. In the latter case you need to find out the temperature difference involved and create a number with the unit [BTU/(h*dT), which in my case, being European, would be [W/dT]. The best choice in your case would be dT = (warming liquid in – air in). The rest is just a matter of a proportional calculation.

Here is a site for air properties:
http://akemalhammar.fr/online/air_properties.html
The first approach would be to use an average value for T_air_in and T_air_out.

It is a fundamental principle that energy is conserved so there is no efficiency loss for the outgoing air. However it may happen that your heat exchanger has a bypass or that the exit is in free air which complicates the problem. In booth cases it is a matter of co-ejection, which is a very intricate phenomenon. In any case the temperature profile in the air will not be uniform, so defining it 3 feet above the outlet will depend on the size of your target.

I have personally tried to calculate this effect with the help of descriptions found in the literature several times but they have always failed considerably. Be aware!

4. ### WmechMember

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hi thermal

The heater is not generated by electrical coil. It's actually hot liquid (most likely hot water) running in a coil of pipes.

As you have suggested, the delta_T = Temperature of warming liquid in - Temperature of air in. However, I do not know the inlet temperature of the warming liquid which I understand it will make the calculation more complicated than it supposes to be.

After you mention to me about the temperature profiles of the system, it does make sense that it's hard to calculate the exit temperature at 3 feet away from the radiator. Maybe I should read up the temperature profile from my textbook.

The purpose of fan/radiator set-up is to heat up the air in an empty room, which excludes the mass of everything in the room (to make the calculation simpler).

Another way to find out the temperature at certain length is to use two equations: First equation --> Q_dot = m_dot * cp * delta_T and second equation --> Q_dot = delta_T / R where R is equal to Length / thermal conductivity of air * Area.

Use the first equation to find the exit temperature of air before the radiator. Then, use second equation to find the temperature at certain distance.

However, it is still awkward to me.

5. ### thermalMember

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Wmech, bonsoir

Watch out here, there are numerous definitions of dT. One is the difference between air_out and air_in which should be used in the energy equation Q_dot = m_dot * cp * dT.

Because there is a heat exchanger, a liquid that losses temperature and air that gains temperature, the driving force for that exchange is the temperature difference between these two fluids.

The equation for that heat transfer is Q_dot = Teta/R, where “Teta” is a temperature difference between the liquid and the air. For the “Teta” anything goes as long as the R, which has the unit [C/W] in european units and in american units [F*h/BTU] is defined for it.

Unless you know the inlet liquid temperature the problem can not be solved. You will have to make reasonably guesses!

6. ### WmechMember

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I agreed with you, thermal. Thanks for the clarification and suggestion.

7. ### vishwas12982Member

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Hello, isn't heat exchanger analysis used for fluids passing inside tubes?? Here by fan.. the fluid is passing on radiator.. so can't we use basic convection principle to calculate the outlet temperature? ? Newton's law of cooling

8. ### WmechMember

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As thermal explains to me, we can use Newton's cooling law to solve the temperature. I guess i am just making the problem more complicated than it supposes to me. I thought the 60,000 BTU/h heat from the radiator doesn't give you 60,000 heat output by the fan. That's why i thought this is a heat exchanger problem.

But yeah, i think we can use newton's cooling law to solve this problem.

9. ### thermalMember

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Wmech, bonsoir

It pleases me that you guys are using Q_dot and m_dot as variables. The “dot” is a time derivation so Q_dot=[Joules/s] =[Watt] and the m_dot=[kg/s]. During my studies, and this is a long time ago, my English, (I am Swedish), literature suggested all kinds of icons for these units but my professors all insisted on the “dot”.

It seems as if they are gaining acceptance. Furtherer more Q is heat energy [Joule], E is mechanical energy [Joule] and q and e are their surface densities, [Joule/m2]. I can spend hours on this intricate issue, which by no means is simple. But, it seems to me that the world in this case is developing sound policies towards comprehension.

10. ### WmechMember

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Hi thermal,

Actually it all thanks to my instructor. The way he taught us in thermodynamic was absolutely spot on. Though, i still need more practice in terms of application of applying proper theories. I should probably explain the differences between dot and non-dot in the first place. Thanks to you, i think people are more clear in terms of the purpose of dots.

11. ### vishwas12982Member

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Hi thermal,

I agree with you completely..