# Thermal Physics problem: Heat transfer coefficient?

• rune
In summary, the conversation was about a thermal physics problem involving finding the temperature within a hollow cylinder with given temperatures at the inner and outer walls, and a given heat transfer coefficient. The equation to use was derived, and it was found that the resulting temperature does not depend on the heat transfer coefficient.
rune
I'm having a bit of trouble with a Thermal Physics excercise. I was wondering if anyone knew what the definition of the heat transfer coefficient was? I can't find anything about it in my textbook (Thermal Physics, 2nd ed., C.B.P. Finn).

More specifically, the assignment at hand says to find the temperature $$T(r)$$ in a hollow cyllinder with length $$L$$ (where $$r$$ is the distance from the axis of the cyllinder, so that the cyllinder is symmetric about this axis), inner radius $$r_1$$, outer radius $$r_2$$, inner wall is kept at temperature $$T_1$$ and outer wall temperature is kept at $$T_2$$ (temperature of the surroundings), after equilibrium has been reached. The material of the cyllinder has heat transfer coefficient $$H$$. Also, $$T_2<T_1$$. (It also asks explicitly if the result depends on $$H$$.)

Any tips or hints to get me started?

Thermal Conductivity coefficient is what you're looking for. The equation to use is:
$$q = -(kA)\frac{dt}{dx}$$

where

$$q$$ is heat flow rate is watts.

$$k$$ is thermal conductivity in $$\frac{W}{m \Dot K}$$

$$A$$ is cross sectional area normal to flow in $$m^2$$

$$\frac{dt}{dx}$$ is the temperature gradient in $$\frac{K}{m}$$

Regards,

First of all, thanks for the help Nenad.
I guess I should've mentioned the text was in Norwegian, and heat transfer coefficient seemed like the most direct translation of the wording in the text. Another thing I think i mis-translated was that it said stationary conditions, not equillibrium.

I think I got it right thanks to your help though:
First setting $$A=2 \pi rL$$, and then using your equation: $$\frac{dQ}{dt}=-(HA)\frac{dT}{dr}=C$$ , where I set the heat flow rate to a constant since there are stationary conditions (but not zero since that would be absurd as long as we are using energy to maintain $$T_1$$ on the inner wall, right?).
For simplicity, I then set $$C'=-\frac{C}{H2 \pi L}$$, and the resulting differential equation is:
$$dT=\frac{C'}{r}dr$$
which gives:
$$T(r)=C'\ln(r)+T_0$$
solving for $$C'$$ and $$T_0$$ using given initial conditions ($$T(r_1)=T_1$$ and $$T(r_2)=T_2$$) gives:
$$T(r)=\frac{T_2-T_1}{\ln(\frac{r_2}{r_1})}\ln(\frac{r}{r_1})+T_1$$
Thus $$T(r)$$ doesn't depend on $$H$$.

Last edited:

## 1. What is the heat transfer coefficient?

The heat transfer coefficient is a measure of the ability of a material or substance to transfer heat. It is defined as the amount of heat transferred per unit area per unit time per unit temperature difference.

## 2. How is the heat transfer coefficient calculated?

The heat transfer coefficient can be calculated using the formula h = q / (A * ΔT), where h is the heat transfer coefficient, q is the heat transfer rate, A is the surface area, and ΔT is the temperature difference between the two surfaces.

## 3. What factors affect the heat transfer coefficient?

The heat transfer coefficient is affected by various factors such as the thermal conductivity of the material, the surface roughness, the flow rate of the fluid, and the temperature difference between the two surfaces.

## 4. How does the heat transfer coefficient influence heat transfer?

The heat transfer coefficient plays a crucial role in heat transfer as it determines the rate at which heat is transferred between two surfaces. A higher heat transfer coefficient means that heat can be transferred more quickly, while a lower heat transfer coefficient means that heat transfer will be slower.

## 5. How can the heat transfer coefficient be optimized?

The heat transfer coefficient can be optimized by using materials with high thermal conductivity, increasing the surface area for heat transfer, and maximizing the temperature difference between the two surfaces. Additionally, improving the flow rate of the fluid and reducing surface roughness can also help to increase the heat transfer coefficient.

• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
Replies
1
Views
657
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Engineering and Comp Sci Homework Help
Replies
22
Views
1K
• Thermodynamics
Replies
1
Views
753
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
654