Heat transfer in a cylindrical tube

In summary, the conversation is about a person who is having trouble understanding a concept in their textbook. The concept involves a heat resistor passing through a cylindrical tube and the formula for it is R(heatresistor) = 1/2∏Lλ * ln (R1/R2). However, when the person follows this formula, they get a negative result because R1 is smaller than R2. They have searched through other sources but have not found any helpful information. They are hoping someone can explain to them the correct way to use R1 and R2 in the formula.
  • #1
5carola5
22
1

Homework Statement



I don't really understand something in my textbook. It says the heat Resistor (how to say that in English? I hope it's alright like this...) trough a cylindrical tube (from the inside to the outside or revers) would be: R(heatrisistor)=1/2∏Lλ * ln (R1/R2) And then my textbook gives the following picture (well oke a but nicer, but that's what I make of it)http://schattielover.deviantart.com/art/homework-314836756

But if I were to follow this I get a negative R every time (since R1 is smaller than R2 and so the ln gets negative right?), which I find very strange... So I searched trough other textbooks that I have and they had no information about it and searched trough the internet (I'm not very good at that ^^') and found nothing I can use.

So my question is: Can someone explain to me, nice and simple, what I should do with the R1 and the R2? (I probably am misunderstanding my textbook... I'm sure of it ^^')

Well thanks for reading all of this anyway, hope someone can help me with this. :)
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  • #2
I can't see your diagram but I think there is a mistake. R1 and R2 should be the other way around. See...

http://wiki.answers.com/Q/Formula_for_Heat_loss_in_a_pipe

Q= 2 pi k L(T1-T2)/ln(r2/r1)

where

k=conductivity of material of which pipe is made

T1= temperature of inside layer pipe
T2=temperature outside layer of pipe
L= length of pipe
r1= inner radius
r2=outer radius
ln=natural logrithem
 
  • #3
thank you so much! I can defiantly do something with that =D =D =D. I can be so stupid sometimes ^^'
 

Related to Heat transfer in a cylindrical tube

1. What is the main mode of heat transfer in a cylindrical tube?

The main mode of heat transfer in a cylindrical tube is convection. This is the transfer of heat through the movement of fluids or gases, such as air or water, within the tube. Convection is more efficient than conduction, which is the transfer of heat through direct contact between two materials.

2. How does the flow rate affect heat transfer in a cylindrical tube?

The flow rate of the fluid or gas within the cylindrical tube has a significant impact on heat transfer. A higher flow rate means that the fluid or gas is moving faster, resulting in more efficient heat transfer. This is because the faster-moving fluid or gas can carry more heat away from the surface of the tube.

3. What factors influence the rate of heat transfer in a cylindrical tube?

The rate of heat transfer in a cylindrical tube is influenced by several factors, including the temperature difference between the fluid and the tube, the thermal conductivity of the tube material, the fluid flow rate, and the length and diameter of the tube. These factors can be manipulated to increase or decrease the rate of heat transfer.

4. How does the shape of the cylindrical tube affect heat transfer?

The shape of the cylindrical tube can affect heat transfer in several ways. A wider diameter tube will have a larger surface area, allowing for more efficient heat transfer. Additionally, the shape of the tube can impact the flow of the fluid or gas, which can affect the rate of heat transfer. For example, a tube with a wider diameter may have a lower flow rate, resulting in slower heat transfer.

5. How can heat transfer in a cylindrical tube be optimized?

To optimize heat transfer in a cylindrical tube, several factors can be adjusted. These include increasing the flow rate of the fluid or gas, using materials with higher thermal conductivity for the tube, and increasing the surface area of the tube through changes in diameter or length. Additionally, insulating the tube can help to minimize heat loss and increase the efficiency of heat transfer.

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