How can I eliminate Tw in the equation for heat loss from a house?

Click For Summary

Homework Help Overview

The discussion revolves around the heat loss from a house, specifically focusing on the equations governing heat flow through different materials, such as styrofoam and brick. The original poster is attempting to understand how to eliminate the variable Tw, which represents the temperature at the interface between the two materials.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of heat flow equations for different materials and the need to eliminate Tw. There are attempts to manipulate equations and clarify the relationship between heat flow through the styrofoam and brick walls.

Discussion Status

Some participants have provided guidance on the concept of heat flow being consistent through both materials, drawing analogies to electrical circuits. However, there remains uncertainty about how to combine the equations effectively to eliminate Tw, with ongoing requests for hints and clarification.

Contextual Notes

There is a mention of different thermal conductivities for the materials involved, which raises questions about the assumptions made regarding heat flow. The original poster expresses confusion about the process and seeks further assistance.

Donna14
Messages
27
Reaction score
0

Homework Statement




I'm struggling to understand question 1.C
How do I incorporate Tw and later eliminate this?


Homework Equations



-k*A*(ΔT/ΔX)

The Attempt at a Solution



For 1.a I used this equation: -k*A*((Tinside-Toutside)/thickness wall) I thought to put T inside first as they ask for the inward heat flow and that will be negative (heat loss)...

For 1.c I'm really stuck...
 

Attachments

  • image.jpg
    image.jpg
    58 KB · Views: 495
Physics news on Phys.org
You get two heat flow equations, one for the styrofoam and one for the bricks. Using both together, you can elimitate Tw.
 
Thanks for your response.


Js=-ks*A*((Tinside-Tw)/thickness styrofoam)
Jb=-kb*A*((Tw-Toutside)/thickness brick wall)

How do I now eliminate Tw?
 
With the standard ways to manipulate equations.

Why did you use two different J?
 
I thought that I had two different J as they both represent a different number...
Im really not sure how to continue from here... I am sorry... Can you give me a hint?
 
Donna14 said:
I thought that I had two different J as they both represent a different number...
Im really not sure how to continue from here... I am sorry... Can you give me a hint?

J is the same for both because the heat goes through both walls in succession. The walls are in series.

Chet
 
Chestermiller said:
J is the same for both because the heat goes through both walls in succession. The walls are in series.

Chet

But won't the flow through styrofoam be different from the flow through brick as they have a different k? And one Goes from inside to the brick wall and the other from the brick wall to outside.
I understand that there is also a 'total' flow from into outside.

And how do I combine these 2 equations. I might be very dumb, but really trying to see it!
 
Donna14 said:
But won't the flow through styrofoam be different from the flow through brick as they have a different k? And one Goes from inside to the brick wall and the other from the brick wall to outside.
I understand that there is also a 'total' flow from into outside.

And how do I combine these 2 equations. I might be very dumb, but really trying to see it!
This is very much analogous to an electric circuit with two unequal resistors in series. The overall voltage drop is analogous to the overall temperature difference, and the current is analogous to the heat flow. The current flow through each of the resistors is the same. The temperature at the interface between the styrofoam and the brick is analogous to the voltage in the wire between the two resistors. So:

I = ΔV1/R1

and

I = ΔV2/R2

ΔV=ΔV1+ΔV2=I (R1 + R2)

I = ΔV/(R1 + R2)

Chet
 
Chestermiller said:
This is very much analogous to an electric circuit with two unequal resistors in series. The overall voltage drop is analogous to the overall temperature difference, and the current is analogous to the heat flow. The current flow through each of the resistors is the same. The temperature at the interface between the styrofoam and the brick is analogous to the voltage in the wire between the two resistors. So:

I = ΔV1/R1

and

I = ΔV2/R2

ΔV=ΔV1+ΔV2=I (R1 + R2)

I = ΔV/(R1 + R2)

Chet


I really appreciate your effort!
I understand now that J is the same, but still don't know what to do with the to equations... Sorry!
 
  • #10
Donna14 said:
I really appreciate your effort!
I understand now that J is the same, but still don't know what to do with the to equations... Sorry!
Do the same thing we did here with the two voltage drops. Solve for the two temperature differences, and then add them to eliminate Tw.

Chet
 
  • #11
Chestermiller said:
Do the same thing we did here with the two voltage drops. Solve for the two temperature differences, and then add them to eliminate Tw.

Chet

I got it!
Thank you so much!
 
Last edited:

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
12
Views
3K