How Does Heat Transfer Through Welded Metal Bars Differ?

Click For Summary

Homework Help Overview

The problem involves heat transfer through two welded metal bars, one made of steel and the other of copper, with specified lengths, temperatures, and thermal conductivities. The objective is to find the temperature at the junction and the total rate of heat flow.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the heat conduction equation and the appropriate area to use in calculations, debating between cross-sectional and surface area. There is confusion regarding the correct area to apply in the context of insulated bars.

Discussion Status

Participants are actively engaging in clarifying the correct approach to the problem. Some have identified a mistake in the original reasoning regarding area calculations, while others are reflecting on their understanding of the concepts involved.

Contextual Notes

There is mention of pre-exam nerves affecting one participant's confidence, indicating a potential pressure to perform well in upcoming assessments.

Pomico
Messages
24
Reaction score
0

Homework Statement



A steel bar 10cm long is welded end-to-end to a copper bar 20cm long. Both bars are insulated perfectly along their sides. Each bar has a radius of 2.0cm. The free end of the steel bar is maintained at 100 degrees C and the free end of the copper bar is maintained at 0 degrees C. Find the temperature at the junction between the two bars and the total rate of flow of heat.

k(steel) = 50.2 Wm[tex]^{-1}[/tex]K[tex]^{-1}[/tex]
k(copper) = 385.0 Wm[tex]^{-1}[/tex]K[tex]^{-1}[/tex]

Homework Equations



H = kA(T1-T2)/L

The Attempt at a Solution



I substituted the given values into the above equation for copper and again for steel and set them equal to each other (with the temperature in Kelvin):

kA(T1-T2)/L (steel) = kA(T2-T1)/L (copper)

and I ended up with an answer of 284.5K

My tutor, however, ended up with an answer of 293.7K
I used the different surface areas of the steel and copper parts of the tube in each side of the equation whereas my tutor used the area of the full length of the bar (i.e. an area of 2pi*rL with L = 30cm.
Besides that our workings are the same.

I don't understand why the the total area of the bar should be used instead of the area of each respective material.
I hope that makes sense...

Please help!
 
Last edited:
Physics news on Phys.org
Pomico said:
(i.e. an area of 2pi*rL with L = 30cm.
That isn't the correct area. Surely [itex]A = \pi r^2 = 0.0004\pi[/itex].
 
Last edited:
Are you sure it isn't an area?
2pi*rL in units is cm*cm = cm^2.

As the bars are cylindrical wouldn't your way give the volume when you introduce the length of the bar?

Aren't I looking for surface area as opposed to cross-sectional area?
 
Last edited:
Pomico said:
Are you sure it isn't an area?
2pi*rL in units is cm*cm = cm^2.

As the bars are cylindrical wouldn't your way give the volume when you introduce the length of the bar?

Aren't I looking for surface area as opposed to cross-sectional area?
Why are you looking for the surface area when the sides are insulating? Surely the heat conduction would occur through the cross-sectional area?
 
Ooh hang on, it is cross-sectional area isn't it!
So that would be why my tutor's method cancels out the areas =p

Sorry about that!
 
Yeah sorry, normally I'm ok at these. My exam's coming up and I think it's pre-exam nerves =/

Sorry again!
 
Pomico said:
Yeah sorry, normally I'm ok at these. My exam's coming up and I think it's pre-exam nerves =/

Sorry again!
Hey don't worry about it, you spotted the mistake so you know what your talking about. Listen if your revising, you don't need to be stressing, it those that don't revise that should be worried. Good luck in your exam :biggrin:
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 52 ·
2
Replies
52
Views
7K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
2
Views
3K
Replies
1
Views
7K
Replies
1
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K