Heat Transfer Problem Conduction and Convection

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Discussion Overview

The discussion revolves around a heat transfer problem involving conduction and convection in a tent designed for cold-weather conditions. Participants explore the calculations necessary to determine the inside temperature of the tent, as well as the surface temperatures of the insulation material, given specific thermal properties and conditions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant presents a homework problem involving a tent's thermal properties, including conductivity, surface area, and heat transfer coefficients.
  • Another participant suggests that assuming steady state allows for analysis of heat flux values across the tent's surfaces.
  • Some participants express confusion regarding the concept of heat flux and its relation to the problem, seeking clarification on definitions and equations.
  • There is a discussion about the implications of steady state, with one participant noting that the energy of the tent does not change under these conditions.
  • Participants attempt to derive equations for heat transfer through convection and conduction, leading to calculations of surface and air temperatures.
  • One participant questions the signs used in temperature differences, prompting a discussion about expected temperature gradients from the heater to the outside.
  • Subsequent calculations yield different temperature values for the inside air and surface temperatures, with participants reflecting on the reasonableness of their results.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the final temperature values calculated by participants, as some express uncertainty about their results and the assumptions made during calculations. The discussion includes multiple viewpoints on the interpretation of temperature differences and the application of heat transfer principles.

Contextual Notes

Participants express uncertainty regarding the definitions and application of heat flux, steady state, and the correct setup of temperature differences in their equations. There are unresolved questions about the assumptions made in the calculations and the implications of the results.

Who May Find This Useful

This discussion may be useful for students studying heat transfer principles, particularly in the context of thermal insulation and convection in engineering applications.

charan1
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Homework Statement


A tent being considered for cold-weather conditions has insulation on the top and sides with a conductivity, k = 0.040 W/(m*K) and a thickness, l = 0.50 cm. The tent (and surrounding medium) has the following conditions: a heater (inside the tent) producing heat at a rate of 1500 W, the total surface area of the tent is 33 m2, an average heat transfer coefficient between the inside air and inside surface of hi = 7 W/(m2*K), an average heat transfer coefficient between the outside air and outside surface of ho = 10 W/(m2*K), and an outside temperature of tao = 5*C. Determine the tent temperature, tai, with the insulated tent. Also find the inside surface temperature of the tent insulation material, tsi, and the outside surface temperature, tso.

Thermal Conductivity of tent material (k) = .040 W/m*k
Surface Area of Tent (A) = 33m2
Outside air Temp (tao) = 5*C + 273.15 = 278.15K
Convective heat transfer coefficient inside tent(hi) = 7 W/(m2*K)
Convective heat transfer coefficient outside tent(ho) = 10W/(m2*K)
q inside tent = 1500W
Tent Thickness (\Deltax) = .005m
Inside air Temp (tai) = ?
Inside Tent Surface Temp tsi = ?
Outside Tent Surface Temp tso = ?

Homework Equations


Convection:
q=h*A*\DeltaT

Conduction:
q=-k*A*\DeltaT / \Deltax

The Attempt at a Solution



For Outside Air to outside tent Surface (Which is convection) so \DeltaT = tso - tao :
q=(10W/m2*K)(33m2)(tso - 278.15K)

For the Heat transfer between the inner and outer surface of the tent (Which is Conduction):
q= -(.04W/m*k)(33m2)(tso - tsi) / .005m

For Heat transfer between the inside air and inner tent surface (which is convection) so \DeltaT = tsi - tai :
1500W=(7W/m2*k)(33m2)(\DeltaT)
\DeltaT=6.4935k
6.4935k=tsi - tai

I got this far, I know I have to set some equations equal to each other but every time I attempted that I end up having about 2-3 unknowns in one equation...please point me in the right direction to solve this.
Thank you
 
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If you assume steady state, can you say anything about all three heat flux values?
 
I'm not exactly sure what that is...we have never gone over anything that has to do with heat flux this is a basic Engineering Sciences 101 homework. We only covered convection and conduction and have these formulas.
 
The heat flux is q, and it's measured in J/s, or W... Know what I'm getting at?
 
Oh I see so, delta U - Work = Q

Delta U would be the temperature change? And W is equal to 0? because there is no mechanical energy?

What would I set Delta U equal to with the information I have?
 
Not quite... the energy of the tent doesn't change if it's at steady state. More like the heater puts out 1500 W, so this is the heat flux through all the layers. This leaves you with an equation with only one unknown.
 
Last edited:
What is the definition of steady state? Since that explains that the Q is 1500Watts throughout...
 
Steady state exists when none of the system's state variables (like temperature) change with time.
 
For Outside Air to outside tent Surface (Which is convection) so \DeltaT = tso - tao :
1500W=(10W/m2*K)(33m2)(tso - 5*C)
tso=9.595*C

For the Heat transfer between the inner and outer surface of the tent (Which is Conduction):
1500W= -(.04W/m*k)(33m2)(tsi - 9.595*C) / .005m
tsi=3.913*C

For Heat transfer between the inside air and inner tent surface (which is convection) so \DeltaT = tsi - tai :
1500W=(7W/m2*k)(33m2)(3.913*C-tai)
tai=-2.58*C

these are the numbers I got, I'm not sure they are correct because it doesn't really make sense to me...

I think the problem is something with the \DeltaT's...The way I subtracted them...I was unsure about which value actually goes first in the delta...for instance the surface temp or the air temp...
 
  • #10
Well, let's look at it this way: wouldn't you expect the temperature to get colder as you transition from the heater to the outside? Does that help get the signs right?
 
  • #11
For Outside Air to outside tent Surface (Which is convection) so \DeltaT = tso - tao :
1500W=(10W/m2*K)(33m2)(tso - 5*C)
tso=9.595*C

For the heat transfer between the inner and outer surface of the tent (Which is Conduction):
1500W= -(.04W/m*k)(33m2)(9.595*C - tsi) / .005m
tsi=15.27*C

For Heat transfer between the inside air and inner tent surface (which is convection) so \Delta T = tsi - tai :
1500W=(7W/m2*k)(33m2)(tai - 15.27*C)
tai=21.77*C

This made more sense...what do you think?
 
  • #12
Much more sense!
 
  • #13
Thank you!
 

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