Finding the Overall heat transfer coefficient

In summary, the conversation is about designing a fictitious heat exchanger using an existing heat exchanger as a model. The parameters of the existing heat exchanger are given, including its flow area, cooling power, and flow rate. The main objective is to determine the overall heat transfer coefficient as a function of water speed or Reynolds number, in order to determine the length of the fictitious heat exchanger. The conversation also discusses the need to measure certain parameters on the real heat exchanger in order to design the fictitious one.
  • #1
5carola5
22
1

Homework Statement



We are having a fictive, non existing heat exchanger with the following parameters:

Outer tube diameters: 20 & 12 mm
Wall thickness inner tube: 1mm, outer tube; 2 mm
flow area hot side: 0.0226m2, cold side; 0.0272 m2

The heatexchanger is going to cool some hot water, with cool water. The hot water is 55 degrees C to start with and we don't really care what it will become. The cool water is 15 degrees to start with and we don't want it to become more than 25 decrees Celsius. The heat exchanger is having max cooling power of 4kW (so Q=4kW) and the hot water is having a flowrate of 215 liter/hour

But we have to use an other heat exchanger with the following properties:

Outer tube diameters: 16 mm & 10 mm and the wall thickness of the inner tube is 12 mm

With that heat exchanger we are to determine the Overall heat transfer coefficient U (q=U*A*Tlm) as a function of the speed of the water (v) or Reynolds (Re=(vDρ)/μ. And then somehow we need to get the length of the non existing heat echanger from that...
So I have like no idea how to measure it... ^^' I think we can measure the flow rate of the water and the temperature at the beginning and ending. The heat exchanger is a counterflow heat exchanger. We were also given the following formula: q=m(with a dot on it)*Cp*(Tin -Tout)


Homework Equations



q=m(with a dot on it)*Cp*(Tin -Tout)

qin=qout

q=U*A*Tlm

Re=(vDρ)/μ

The Attempt at a Solution



I think it has something to do with the reynolds number, since if they are the same two flows are the some right? so you could use that one. μ and ρ would stay the same and D would chance if you use an other heat exchanger, but you can't really vary it, so the only thing in that one that you could vary to get the reynoldnumbers the same would be the speed (v)

q=U*A*Tlm

I'm having q and Tlm I want to know A and to get A I need U. (haha that has two meanings:P) but I really, really don't know how to get that one, all kind of thinks are going trough my head and the one making the most sense would be to first use the dementions of the non existing heat exchanger (because you know everything from that one) and then the only one you don't know would be U... but than you don't need to do the experiment to begin with and I just don't think that it would make sense... So I'm kinda out of ideas...

Someone any ideas how to get U? or some kind of tip or hind? Thanks so much in advance :)
 
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  • #2
There seems to be some information missing. The problem statement appears to imply that this is an annular flow heat exchanger, with one fluid flowing through an inner tube and a second fluid flowing through the annulus. Is this correct? Are the fluid flows co-current or counter current? Are you at liberty to select the flow arrangement? Which of the two fluids is flowing through the inner tube, and which through the outer? It looks like the person who posed this problem would like you to "play with" many different configiurations of this problem, and get the feel for what it is like to design a real heat exchanger. I don't understand the part about the second heat exchanger. How is that supposed to come in?
 
  • #3
Yes it is an annular flow heat exchanger. As a did say; it's a counterflow one. Yes we can select if we want counterflow or co-current, but for this assinment we need to select counterflow, we can chance the volumeflux (I really hope it's called like that in English) and the temperature. The hot fluit it going to the innertube.

There are two heat exchangers. One that is there in real live and one that is not. Now the teacher wants us to learn about scale models I think, so that's why we have to do measurements with one heat exchanger and then say something about the other...

Hoped that helped a bit and that now all infomration is there:) Thank you so much for trying to help :D:D
 
  • #4
OK. Now I get the idea of what you are trying to do. The fictitious heat exchanger does not exist yet, but you are trying to design it, and eventually, it might be built into a real heat exchanger. Your teacher is trying to give you some experience in how you can do experiments on a test heat exchanger to determine the information you need to design a new heat exchanger. He wants to give you a feel for how you would approach such an endeavor. You are going to have to do a bunch of calculations to figure out what experiments to do. This is a great exercise that he has assigned you.

Ask yourself, " what are the things I'm going to need to measure on the real heat exchanger to get the information I need for designing the fictitious heat exchanger?"

You are going to need to predict in advance the overall heat transfer coefficient for the fictitious heat exchanger, from knowledge of the cold fluid flow rate. To do this, you need to consider the three basic resistances that go into calculating the overall heat transfer coefficient. Please articulate for me what these three resistances are.

While you are thinking about that, think about this:
At the maximum cooling load, what is the minimum flow rate that the cooling water can have?

For this countercurrent arrangement, what is the lowest possible temperature that the hot fluid can exit at? What would the cooling load be if the hot fluid had this exit temperature? How does that compare with the maximum allowable cooling load?

These are all questions you should be asking yourself as part of the problem solving process. You need to be asking yourself these types of questions, and "playing with the numbers."
 
  • #5
Thanks so much for the answer. I will think about the thinks you said. Just for the record, we are not actually going to make the heat exchanger, we're just pretending. As for the overall heat transfer coefficient; for this part of the assignment we don't need to calculate that one jet (that will be the next part, if somehow I get this done:P). We just need to measure it. But all I could find about that was; Q=UATim which didn't help so much... Then when we are having 'U' we can get the length of the heat exchanger, but the big problem is that I have no clue what so ever on how to measure it. But well, today we are having that course again, so I guess I will just ask the teacher ^^

Thank you so much again for you help and I will really, really think about what you have said :)
 
  • #6
If you know the flow rates and the inlet and outlet temperatures, then you can calculate Q. You already know the heat transfer area, based, say, on the ID or the OD of the inner tube. From the inlet and outlet temperatures, you can calculate the LMTD. This is enough information to calculate U. You will have to run the experiment with a range of cold water flow rates, and a single hot water flow rate which gives you approximately the same Reynolds number in the annulus as that for the heat exchanger being designed.
 

What is the overall heat transfer coefficient?

The overall heat transfer coefficient, also known as U-value, is a measure of the rate at which heat is transferred through a material. It takes into account both conduction and convection heat transfer.

Why is it important to find the overall heat transfer coefficient?

Knowing the overall heat transfer coefficient is important in many engineering and scientific applications, such as designing efficient heat exchangers, insulation materials, and HVAC systems. It helps in optimizing the thermal performance of a system and reducing energy consumption.

How is the overall heat transfer coefficient calculated?

The overall heat transfer coefficient is calculated by dividing the total heat transfer rate by the temperature difference between the two sides of the material.

What factors affect the overall heat transfer coefficient?

The overall heat transfer coefficient is affected by several factors, including the thermal conductivity of the material, the thickness of the material, the surface area, and the type of fluid or gas on either side of the material. It also depends on the velocity and temperature difference of the fluid or gas.

Can the overall heat transfer coefficient be improved?

Yes, the overall heat transfer coefficient can be improved by using materials with higher thermal conductivity, increasing the surface area, and improving the flow and temperature difference of the fluid or gas. Additionally, using insulation materials can also help in reducing heat transfer and improving the overall heat transfer coefficient.

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