Calculate Thickness of Oak Layer to Limit Heat Loss to 740kJ/hr in Door

In summary, the question asks for the thickness of a layer of oak wood that would limit the heat loss of a door, which has two steel layers with a thickness of 0.47 mm each and dimensions of 725 mm by 1800 mm, to 740 kJ per hour. The inside temperature is 18C and the outside temperature is -20C. The equation used is H=area(temp1-temp2)/((thickness1/k1)+(thickness2/k2)), where H is the heat loss, A is the area, temp1 and temp2 are the inside and outside temperatures respectively, and k1 and k2 are the thermal conductivity of the two materials. The unknown variable
  • #1
Tonyuguccioni
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Warning! All queries in the Homework section must use the Posting Template provided when a new thread is started.
The information I am given is : a door has two steel layers both are .47 mm thick, the door itself is 725 mm by 1800mm. The question asks, how thick of a layer of wood (oak) would have to be put in the door to limit the heat loss to 740kJ per hour? Temp inside is 18C and outside is -20C

All work is done with only three significant digits.

I have been working with the equation ;
H=area(temp1-temp2)/((thickness1/k1)+(thickness2/k2))
Where; k is the thermal conductivity of the material
X is the unknown

I have run threw the equation several times but every time I either can't isolate the variable or end up with a wrong answer.

0.205kJ/sec= ((1.30m^2)(38C))/((9.40*10^-5m/43W/mC)+(Xm/0.17W/mC))

Is this the right equation and/or have I made a simple error with conversions ?

Please don't post full answer I want to find it but tips or suggestions would be awesome !
 
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  • #3
thanks but it is still escaping me i feel like i am just one step away from getting it... very frustrating but it i will prevail ! :P
 
  • #4
Did you try simply solving for X in your equation? (Note that it is usually easier not to make mistakes if you solve the algebraic expression first and insert numbers when this is done)
 
  • #5
whats the next step? and what is the equation you started with was it H= -kA (dt/dx)
 

FAQ: Calculate Thickness of Oak Layer to Limit Heat Loss to 740kJ/hr in Door

1. How do I calculate the thickness of the oak layer?

To calculate the thickness of the oak layer, you will need to use the formula for heat conduction. This formula takes into account the thermal conductivity of oak, the surface area of the door, and the desired heat loss of 740kJ/hr. By manipulating this formula, you can solve for the thickness of the oak layer.

2. What is the thermal conductivity of oak?

The thermal conductivity of oak can vary depending on the type and density of the wood. On average, the thermal conductivity of oak is around 0.17 Watts per meter Kelvin (W/mK). However, it is always best to use the specific thermal conductivity value for the type of oak you are using in your calculations.

3. How do I determine the surface area of the door?

The surface area of the door can be calculated by multiplying the width and height of the door. If the door has any additional panels or features that may affect heat loss, these should also be taken into account when calculating the surface area.

4. Can I use a different type of wood besides oak?

Yes, you can use a different type of wood besides oak. However, the thermal conductivity value will differ, and you will need to recalculate the thickness of the wood layer to achieve the desired heat loss. It is recommended to use a wood with a similar thermal conductivity to ensure accurate calculations.

5. Is there a limit to the thickness of the oak layer?

Yes, there is a limit to the thickness of the oak layer. While a thicker layer may provide better insulation, there will come a point where the thickness will not significantly affect the heat loss. It is important to balance the desired heat loss with the practicality and feasibility of a thicker oak layer.

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