How Do You Calculate Heat Transfer in a Truncated Cone Using Fourier's Law?

[JSBeckton]

A truncated cone 30cm high is made of Aluminum. The dia at the top is 7.5cm, and 12.5cm at the bottom. The lower surface is maintained at 93 deg C, the upper surface at 540 deg C. the other surface is insulated. Assuming 1 dimensional heat flow, calculate the rate of heat transfer in watts.

Ok, my book is seriously lacking in the example department, and we have only gone through derivations in class, no examples yet.

I understand that this is a conduction problem so I am to use fouriers law

q=-kA dT
dx

k= 202 W/mC for Al

Now the tricky part, I know that the problem that I face is that the area changes along the axis. Am I just to integrate:

pi r^2 from 3.25cm to 6.25cm, then multiply that by (540-93)/.3?

Thanks

 

[nilesh_pat]

in advance for any help!Yes, that is the correct approach. You will need to calculate the area of a circle for each radius (using pi*r^2) and then use this in your integration. The result of the integration should be multiplied by the temperature difference (540-93) and the thermal conductivity of aluminum (202 W/mC).

 

[piercas]

for your question. You are on the right track with using Fourier's law to solve this problem. In order to calculate the rate of heat transfer, we need to first calculate the temperature gradient along the cone. This can be done by taking the derivative of the temperature with respect to the distance along the cone (x-axis).

dT/dx = (540-93)/30 = 14.9 °C/cm

Next, we need to calculate the area at each point along the cone. We can do this by using the formula for the surface area of a truncated cone:

A = π(r1 + r2)√(h^2 + (r1-r2)^2)

Where r1 is the radius at the top (3.75cm) and r2 is the radius at the bottom (6.25cm), and h is the height of the cone (30cm).

Plugging in these values, we get the following areas at each point along the cone:

x = 0cm (bottom of cone): A = π(6.25+6.25)√(30^2 + (6.25-6.25)^2) = 1885.8 cm^2

x = 30cm (top of cone): A = π(3.75+3.75)√(30^2 + (3.75-3.75)^2) = 705.5 cm^2

Now, we can use these values to calculate the rate of heat transfer at each point along the cone using Fourier's law:

q = -kA(dT/dx)

At x = 0cm (bottom of cone):

q = -202 W/mC * 1885.8 cm^2 * (14.9 °C/cm) = -5,739,196.8 W

At x = 30cm (top of cone):

q = -202 W/mC * 705.5 cm^2 * (14.9 °C/cm) = -2,017,359.8 W

Therefore, the total rate of heat transfer from the bottom to the top of the cone is:

q = -5,739,196.8 W - (-2,017,359.8 W) = -3,721,837 W

This means that the heat is transferring from the lower surface (at 93 °C) to the upper surface (