Simple Heat Transfer (convection and radiation, cylinder/pipe)

In summary, the conversation discusses a heat transfer problem involving a smooth cylinder with given dimensions and temperatures. The emissivity and heat transfer coefficient for convection are also given. The goal is to calculate the heat transfer coefficient for radiation and the total heat loss due to radiation and convection. The correct equations for these calculations are used and the overall approach shows a good understanding of the concepts involved in heat transfer.
  • #1
grufff
4
0

Homework Statement



Reasonably simple (I hope!) heat transfer problem:

Smooth cylinder of diameter (D) 0.8m and length (L) 1.2m has a surface temperature (Tsurf) of 240C (513K). Ambient air temp (Tamb) is 30C (303K). Emissivity of cylinder (E) is 0.85. Heat transfer coefficient of convection (h conv) is 8.8 W/m^2 K. Boltzman constant (sigma) is 5.76*10^-8.

Neglecting end effects, calculate heat transfer coefficient for radiation (h rad), and total heat loss due to radiation and convention.

Homework Equations



I've dug out equations for h rad, and q(rad) and q(conv). Have written them out on my working...

The Attempt at a Solution



I'm posting a link/photo to my working. I hope it's clear enough for you to see easily, if not I'll try something else.

I'm hoping someone can confirm I've chosen the correct formulae for the problem, and worked through them properly. Thanks very much if you can help/advise:

(P.S. I'll be posting another one later, once I've done it, on heat exchangers. So if you know anything about those, please keep watching, ;) )

[PLAIN]http://dl.dropbox.com/u/5352127/HeatTransfer.jpg
 
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  • #2

Thank you for your post. I have reviewed your working and it appears that you have chosen the correct equations for this problem. Your calculations also seem to be correct, assuming that your values for the Boltzmann constant and the emissivity are accurate.

For the heat transfer coefficient for radiation, you have correctly used the equation h_rad = sigma * E * (Tsurf^4 - Tamb^4). This is the correct equation for calculating the heat transfer coefficient for radiation between a surface and its surroundings.

For the total heat loss due to radiation and convection, you have correctly used the equation q_tot = q_rad + q_conv. This is the correct way to calculate the total heat loss from a surface, taking into account both radiation and convection.

In terms of your overall approach, it seems like you have a good understanding of the concepts and equations involved in heat transfer. Keep up the good work!
 

FAQ: Simple Heat Transfer (convection and radiation, cylinder/pipe)

What is convection heat transfer?

Convection heat transfer is the transfer of heat through a fluid, such as air or water, due to the movement of the fluid. As the fluid moves, it carries heat from one location to another.

How does radiation heat transfer work?

Radiation heat transfer occurs when heat is transferred through electromagnetic waves, without the need for a medium. This is how heat from the sun reaches the Earth.

Can you explain the difference between convection and radiation heat transfer?

Convection heat transfer requires a medium, such as air or water, to transfer heat, while radiation heat transfer can occur through a vacuum. Additionally, convection heat transfer relies on the movement of the fluid, while radiation heat transfer does not.

How does heat transfer in a cylinder or pipe?

In a cylinder or pipe, heat transfer occurs through both convection and radiation. The fluid inside the cylinder or pipe absorbs the heat from the walls through convection, while the walls themselves also radiate heat.

What factors affect heat transfer in a cylinder or pipe?

The rate of heat transfer in a cylinder or pipe is affected by factors such as the temperature difference between the fluid and the walls, the properties of the fluid (such as viscosity and density), and the geometry and material of the cylinder or pipe.

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