An Example on Heat Conduction

Question:

A room has a wall constructed of a ½ inch wooden wall. Find the rate of heat loss in Btu/h if the wall surface temperature is 10^oF outside the room and 70^oF inside the room. The thermal conductivity of the wood wall is 0.062 Btu/(h * ft * ^oF).

Answer:

To find the heat loss we will apply the heat conduction equation

Q = -kA(D T)/(D x)

Q is the heat we are solving for,

k is the thermal conductivity = 0.062 Btu/(h * ft * ^oF)

A is the area of the wall

D T is the temperature difference = 70 ^oF - 10 ^oF = 60 ^oF

D x is the thickness of the wall = ½ inch

To make our units correct, we will need to deal with D x in feet, not inches. To do that, we will do the following unit conversion:

D x = ( ½ inch) * (1 ft / 12 in) = 1/24 ft

Remember: 1 Foot = 12 Inches

Before plugging the above values into our equation, we also need to determine an area. The heat flow does not depend on the area of the wall. Because of this, we will assume that the thermal conductivity, k, given is "per unit area." This means that we can assume area = 1. This can only be done when the heat flow does not depend on area.

Let’s plug the numbers into the equation and solve for heat loss.

The value of Q is negative because we are losing heat. Let’s think about that. We discussed that heat always flows from hot to cold. Therefore, the heat must be flowing from the inside to the outside, like on a cold Illinois winter day. Because the room is losing heat, or getting colder, the value of Q is negative.

In a real home, there is not 89 Btu/h of heat lost to the outside. Why do you think that is? It is because there is insulation in the walls to protect against heat loss. The insulation has a high resistance to heat loss, in other words it does not allow heat to escape. Let's do an example of a wall with insulation.

A cold storage room has walls constructed of a 4-inch layer of corkboard insulation contained between double wooden walls, each ½ inch thick. Find the rate of heat loss in Btu/h if the wall surface temperature is 10^oF outside the room and 70^oF inside the room. The thermal conductivity of corkboard in 0.025 Btu/(h * ft * ^oF) and the thermal conductivity of the wood walls is 0.062 Btu/(h * ft * ^oF).

Answer:

To find the total heat loss, we will find the heat loss through each of the layers of wall, and then add them all together.

Because we do not know the temperatures between every layer, we will define a new term called the thermal resistance. Like discussed before, this is the resistance of that layer to loosing heat. We would expect the insulating corkboard to have a higher resistance than the wood walls.

Thermal resistance = D x / (kA)

We will do the problem the same as we did the last one, but this time we will use thermal resistances since there is more than one material being used. If we find the thermal resistance of each layer we can find the total heat loss by the following equation

Let's try it:

1. Thermal resistance of first wall = D x / (kA)

D x = ½ inch = ( ½ inch) * (1 foot / 12 inches) = 1/24 ft

(Note: 12 inches = 1 foot)

kA = 0.062 Btu/(h * ft * ^oF)

Thermal Resistance of first wall = D x/(kA) = (1/24 ft) / [0.062 Btu/(h * ft * ^oF)]

= 0.67 (h * ^oF) / Btu

2. Thermal Resistance of corkboard = D x / (kA)

D x = 4 inches = ( 4 inches) * (1 foot / 12 inches) = 1/3 ft

(Note: 12 inches = 1 foot)

kA = 0.024 Btu/(h * ft * ^oF)

Thermal Resistance of corkboard = D x/(kA) = (1/3 ft) / [0.024 Btu/(h * ft * ^oF)]

= 13.9 (h * ^oF) / Btu

3. Thermal Resistance of second wall D x / (kA)

D x = ½ inch = ( ½ inch) * (1 foot / 12 inches) = 1/24 ft

(Note: 12 inches = 1 foot)

kA = 0.062 Btu/(h * ft * ^oF)

Thermal Resistance of second wall = D x/(kA) = (1/24 ft)/[0.062 Btu/(h * ft * ^oF)]= 0.67 (h * ^oF) / Btu

Putting 1-3 together with the equation labeled "equation 1," we can solve for Q.

Q = - (T₂ – T₁) / sum of the thermal resistances

We know the following:

T₂ = 70 ^oF

T₁ = 10 ^oF

Sum of thermal resistances = 0.67 + 13.9 + 0.67 = 15.24 (h * ^oF) / Btu

Therefore we can find Q:

Q = - (70 ^oF – 10 ^oF) / [15.24 (h * ^oF) / Btu]

We can see that we are still loosing heat to the outside, but this time we are only loosing 3.9 Btu/h as opposed to almost 90 Btu/h. Therefore we can see the use of insulation in our homes.