How Much Natural Gas Is Needed to Heat a House to 28°C?

Click For Summary

Homework Help Overview

The problem involves calculating the amount of natural gas required to heat a house to a specific temperature, given the thermal properties of the house's walls and roof, as well as the outside temperature. The subject area includes thermodynamics and heat transfer principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of power using thermal conductivity and question the method of applying it to find heat flux. There are attempts to compute the total heat required based on the temperature difference and the dimensions of the house.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the initial calculations and the correct application of thermal conductivity. Some guidance has been offered regarding the use of thermal conductivity in relation to heat flux and surface area.

Contextual Notes

Participants are working under the assumption that radiation and heat loss through the ground can be disregarded, focusing solely on the heat transfer through the walls and roof.

blue88
Messages
8
Reaction score
0
The average thermal conductivity of the walls (including windows) and roof of a house in Figure P11.32 is 4.8 10-4 kW/m°C, and their average thickness is 20.0 cm. The house is heated with natural gas, with a heat of combustion (energy given off per cubic meter of gas burned) of 9300 kcal/m3. How many cubic meters of gas must be burned each day to maintain an inside temperature of 28.0°C if the outside temperature is 0.0°C? Disregard radiation and loss by heat through the ground.

The house's dimensions are 10.0m x 8.0m x 5.0m


I tried this problem, and I keep getting the answer completely wrong... if anyone could show me how to do this one, that would be awesome.

Thanks.
 
Physics news on Phys.org
What have you tried so far? We can only help you if you tell us what you've done up to this point.
 
I did P=(4.8E-4)(.2)(28)

P= 2.688W


Q=Pnet(t)
Q=(2.688)(86400)
Q= 232243.2 J of heat


but i don't think that I am even starting it right.
 
Why did you multiply the thermal conductivity by length when computing the emitted power? How do you use the thermal conductivity to calculate the heat flux through a surface of area A and thickness L with a temperature difference \Delta T?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
12K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
5
Views
5K
  • · Replies 12 ·
Replies
12
Views
37K
Replies
2
Views
4K
Replies
1
Views
4K