Estimating the heat transfer coefiicent

[CraigH]

Homework Statement

A 2kW immersion heater for heating water in a large tank is to be made from a straight
horizontal sheathed element of diameter 10mm.

A sheathed element is an electrical resistance wire embedded in magnesium oxide powder and enclosed in a stainless steel tube, (there will be one in your kettle bent to a compact shape).

The tank has a thermostat that controls the water temperature at 60'C. If the surface temperature of the sheathed element is not to exceed 100'C, calculate the length of tube required assuming that convective losses from the element can be determined by the non-dimensional equation:

[itex] Nu = 0.53(Gr * Pr)^{0.25}[/itex]

For water at 75'C take
μ = 0.3783 ×10-3kg m-1s-1
cp= 4.191 kJ kg-1C-1
ρ= 974.7 kg m-3
k = 0.6633 W m-1C-1
β= 0.619 ×10-3K-1
g=9.81 m s-2

Homework Equations

[itex]\frac{dQ}{dt}=A*h*ΔT[/itex] Newtons law of cooling (definition of heat transfer coefficient)

[itex]Re = \frac{ρVL}{μ}[/itex] Reynolds Number (dimensionless)

[itex]Pr = \frac{μCp}{K}[/itex] Prandtl Number (dimensionless)

[itex]Gr = \frac{βgρ^{2}ΔTL^{3}}{μ^2}[/itex] Grashof Number (dimensionless)

[itex]Nu = \frac{hL}{k}[/itex] Nusselt Number (dimensionless)

The Attempt at a Solution

I know I have done the first part correctly, I have got the answer sheet right next to me. I just cannot see how the Grashof Number was calculated.

First Part. Calculate the length of the element needed to dissipate 2KW. (as a function of h)

from Newton's law of cooling:

[itex]\frac{dQ}{dt}=A*h*ΔT[/itex]

[itex]\frac{dQ}{dt}=(pi*d*L)*h*ΔT[/itex]

[itex]L = \frac{2000}{0.01*pi*40*h}[/itex] This is just the above equation rearranged with the known constants put in.

This equation gives the length at which a 10mm diameter element would transfer 2KW to the water. (when there is a temperature difference of 40 degrees and a heat transfer coefficient of h)

Second Part. Now we just need to calculate/estimate a value for the heat transfer coefficient between the element and the water.

To do this you need to:
Calculate the nusselt number with the equation given in the question.
Use the nusselt number to calculate the heat transfer coefficient.

4. my problem

The only problem is that you need to know the length of the element to calculate the Grashof number, and then you need it again to calculate the heat transfer coefficient from the nusselt number. So you need the length to calculate the length.

The information given in the question is all I was given in my question, and the relevant equations are all the equations in my data book which I think are relevent. There are no more equations that tell me the Gr Number.

In the answer sheet my tutor has done the first part the the same way I did the first part. And then he has just wrote:
Pr=2.390
Gr=1.6125*10^6
Nu=23.48
h=1558

I just can not see how he has calculated the Gr Number. Please help!

Thank you.

 

[CraigH]

Okay I now see how he has done it, but I still do not know the reason why he did this.
He calculated the Gr Number with a length of 1
Gr = 1612459.804
Then he calculated the Pr Number
Pr = 2.390253731
Then he calculated the Nusselt number using the equation given in the question.
Nu=23.48

Then he used the definition of the Nusselt number

Nu=(h*L)/k

With a length of 1, to work out h

23.48*k*L =h

h = 15.574284

Then for some reason he has multiplied this by 100

h = 1558
I have no clue at all why these steps were done. Why did he use a length of 1? you can't work out the "nusselt number per meter" because the Grashof Number is proportional to the length cubed. Also, why was the value of h 100 times bigger than what I calculated it as? he could have used a length of 100 in the Nu formula but I don't see why.

 

[CraigH]

by the way:
h = 1558
Is the value of heat transfer coefficient that gives the correct answer when put into the rearranged Newtons law of cooling formula. So this answer is correct.

 

[Chestermiller]

In this equation for natural convection heat transfer from a cylinder, the parameter L is supposed to be taken as the diameter of the cylinder, not the length of the cylinder.

 

[CraigH]

Oh, this would make a lot of sense. So the L^3 in the Grashof Number and the L in the Nusselt Number equation are both diameter.

Are you sure about this? Half of my exam will involve using these numbers, so I really need to understand them. In my lectures I struggled to keep up with what the tutor was saying so I'm just going through question papers and past exams.

Thank you so much for answering. If this is true you have definitely just saved me from failing my exam. I appreciate the help a lot.

P.S sorry for asking for a confirmation. I don't doubt your knowledge of the subject, it is just that this is a very important issue for me. If I have been using the wrong number in an equation that will be used in most of the questions in the exam it could cause the difference between a fail and a half decent mark.

 

[Chestermiller]

Yes, I'm sure about it.

 

[CraigH]

Thank you. I have been using the diameter as L in other questions and it is giving me the correct answer.

I still do not understand why you use the diameter though.

The L in the Re, Gr, and Nu Number is known as the effective length. According to my data book the L in the Nusselt Number is nearly always the same as the L in the Re and Gr Numbers.

I'm looking at the equations and their definitions and trying to see why you use the diameter instead of the length but I can't work it out.

[itex]Re = \frac{ρVL}{μ}[/itex] Reynolds Number (ratio of inertial force to viscous force)

[itex]Gr = \frac{βgρ^{2}ΔTL^{3}}{μ^2}[/itex] Grashof Number (ratio of work done by buoyancy force over Re^2 to inertial energy )

[itex]Nu = \frac{hL}{k}[/itex] Nusselt Number (ratio of actual heat flow to that just by conduction)

So, why would the diameter be used in these equations instead of length?

I have created a diagram to better see what is going on.
http://img163.imageshack.us/img163/6292/imersion2.png

Can you please explain why the;

inertial force [itex]ρVL[/itex]
work done by buoyancy force [itex]βgρ^{2}ΔTL^{3}[/itex]
actual heat flow [itex]hL[/itex]
(I know these aren't the actual equations for these things. I've just took these from the equations and definitions of Gr, Pr and Nu)

All use the diameter instead of the length of the cylinder for L.

Thank you

 

[Chestermiller]

The length parameter in heat transfer correlations is a "characteristic length" for the specific geometry under consideration. In your original problem statement, you said that the cylinder was horizontal, but in your figure, you showed a vertical cylinder. The heat transfer correlations for these two situations are different. For a horizontal cylinder, why would you think that the heat flux at the cylinder surface would depend on its length? If the cylinder were very short, that would be the case, but for a cylinder with a large L/D ratio, the local heat flux over the majority of the cylinder is not affected by the length, and only varies near the ends of the cylinder. And consider the case where the L/D ratio is infinite; in that case, there would not even be a length to use.

In summary, the heat transfer correlation and the characteristic length depend on the specific geometry. If you want to get a better understanding of all this (which is really very uncomplicated), get yourself a copy of Transport Phenomena by Bird, Stewart, and Lightfoot. First work on the early chapters where the viscous drag is evaluated analytically, and then expressed in terms of the Reynolds number. Then see the middle chapters, where the various aspects of heat transfer are considered, and expressed in terms of the Nussult Number.

Incidentally, in your figure, your mechanistic description of natural convection with a vertical heater is not correct. The hotter liquid near the element does not move directly outward, and the colder liquid away from the element does not move directly inward. The hotter liquid near the element rises until it approaches the lid, and then moves outward; it then turns downward when it approaches the outer surface of the container. The colder liquid away from the element moves downward until it approaches the bottom of the container, and then moves inward; it then turns upward when it approaches the element.

If you are going to do fluid mechanics and heat transfer in actual practice, you are going to need to flesh out your understanding. Transport Phenomena can help you with this.

 

[CraigH]

Thank you very much, this has explained a lot to me. I didn't notice that the cylinder was horizontal when I read the question. I'm going to look out for that in future. I might also have a look at that book in the future as this is quite an interesting topic, but for now I think I will just learn what I need to know to be able to answer the questions in the exam.

Thanks again.

 

[CraigH]

Sorry to bring this post up again, but my exam is now very soon and I still have a problem with determining which length to use as the characteristic length.

I'm pretty certain that the characteristic length is the length over which the fluid flows due to convection. (correct?)

Also, does the fluid always flow up vertically when the cause of the flow is natural convection? So the characteristic length will be the vertical length? (length of object in the z direction of a right handed coordinate system)

 

[Chestermiller]

Sorry to bring this post up again, but my exam is now very soon and I still have a problem with determining which length to use as the characteristic length.

I'm pretty certain that the characteristic length is the length over which the fluid flows due to convection. (correct?)

Also, does the fluid always flow up vertically when the cause of the flow is natural convection? So the characteristic length will be the vertical length? (length of object in the z direction of a right handed coordinate system)

Yes. You said the heating element is horizontal, so the characteristic length is the diameter. The flow is not always up vertically when the cause of the flow is natural convection. If cooling is involved, the flow is downward. Natural convection is caused by buoyancy effects associated with variations in density.

 

[CraigH]

Okay, this makes sense. However I'm struggling to apply this to other examples, in particular prismatic fins.

So say a rectangular prism is extruding horizontally from a heated base plate.

The convection of the air will be flowing vertically, and the only length of the prism in the direction of this flow is the thickness of the fin (d on this diagram). So will d be the characteristic length? And if the fins where extruding the base vertically then the characteristic length will be the length of the fin? (L on this diagram).
Because this suggests that fins will be much more efficient when placed vertically as the length of a fin is usually above 10 times the thickness of it. I don't know why but to me it seems strange that it will be more efficient just by re-orientating it.

And how about if the fins are extruding horizontally but there are many fins above and below it also extruding horizontally from the base plate? In this case the air will flow up from a lower fin but then have to flow across the length of the above fin, because the air cannot just flow straight through the fin. So will you use the thickness plus the lengths of the fins? (excluding the bottom fin)

 

[Chestermiller]

Typically, when you have multiple cooling fins, the natural convection effect is negligible, air is forced through the array of fins parallel to their surfaces. An example of this is a car radiator, where the fins are attached to the radiator tubes and act to increase the rate of cooling. The cooling fan blows air over the tube array and fins. I never heard of a practical situation like the one you describe.