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

In summary, the student is trying to solve for Tw using two different Js and then combining them to eliminate the term.
  • #1
Donna14
27
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...
 

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  • #2
You get two heat flow equations, one for the styrofoam and one for the bricks. Using both together, you can elimitate Tw.
 
  • #3
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?
 
  • #4
With the standard ways to manipulate equations.

Why did you use two different J?
 
  • #5
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?
 
  • #6
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
 
  • #7
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!
 
  • #8
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
 
  • #9
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:

1. How does heat escape from a house?

Heat can escape from a house through three main processes: conduction, convection, and radiation. Conduction is the transfer of heat through solid materials, such as the walls and roof of a house. Convection is the movement of heat through liquids and gases, such as air currents in a room. Radiation is the transfer of heat through electromagnetic waves, such as sunlight entering through windows.

2. What factors contribute to heat loss from a house?

There are several factors that can contribute to heat loss from a house. These include the type and thickness of insulation, the quality of windows and doors, the presence of air leaks, and the efficiency of the heating system. Climate and weather conditions also play a role in heat loss.

3. How can I reduce heat loss from my house?

There are several steps you can take to reduce heat loss from your house. These include adding insulation to walls and attics, sealing air leaks around windows and doors, upgrading to energy-efficient windows, and maintaining your heating system. You can also consider using alternative heating sources, such as a fireplace or space heater, during colder months.

4. Does the size of a house affect heat loss?

Yes, the size of a house can affect heat loss. Generally, larger houses have more surface area, which means there are more opportunities for heat to escape through walls, windows, and doors. However, the efficiency of a house's insulation and heating system also play a significant role in heat loss, so it is not the only determining factor.

5. Can landscaping affect heat loss from a house?

Yes, landscaping can affect heat loss from a house. Trees and shrubs can act as natural barriers against wind and can help block out sunlight, reducing heat transfer through windows. However, if landscaping is too close to the house, it can also block out natural light and airflow, making the house colder and more prone to heat loss. It is important to carefully plan and maintain landscaping to optimize its effects on heat loss.

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