Use the method outlined in Problem 76 to show that the...

Use the method outlined in Problem 76 to show that the steady heat-flow rate in the direction of the axis of a truncated cone with conductivity $k,$ faces of radii $R_{1}$ and $R_{2},$ and length $L$ is $H=\pi k R_{1} R_{2}\left(T_{1}-T_{2}\right) / L .$ Here, $T_{1}$ and $T_{2}$ are the temperatures on the two faces, and insulation prevents any heat flow out the sides (Fig. 16.17 ). (FIGURE CANNOT COPY)

Key Concepts

Fourier's Law of Heat Conduction

This principle states that the heat flux is proportional to the negative gradient of temperature. In essence, it provides the basic equation for conduction heat transfer in materials, linking the rate of heat transfer through a material to its thermal conductivity and temperature difference over a distance.

Steady-State Heat Conduction

In steady-state conduction, the temperature distribution within the material does not change with time. This implies that the heat flow is constant along any cross-section, and the energy input equals the energy output. This concept is key in setting up and solving problems where the heat flow rate is independent of time.

Integration over a Variable Cross-sectional Area

When dealing with objects whose cross-sectional area changes along the direction of heat flow, such as a truncated cone, the local heat conduction must be expressed in differential form. By integrating the differential forms that account for the continuously changing area, one can obtain the total heat flow across the body.

Truncated Cone Geometry

This involves understanding the spatial dimensions and how the radius, and consequently the cross-sectional area, varies linearly along the length of the cone. The mathematical description of this variation is essential for setting up the integration in problems of heat conduction through non-uniform bodies.

Transcript

00:01 Okay, so the question wanted us to derive the equation and to prove that the expression, the final expression, match, the expression that is given, final question.

00:13 Okay, so we know heat loss rate can be equal to negative k -a times d t over the r.

00:20 Okay, k is the thermoc conductivity, a is the surface area.

00:23 T and r here, t is the temperature, and r is the radius.

00:27 So since we assume that the size of a corn are perfectly insolid, therefore the heat flow is only along the lens of the comb.

00:38 Okay, so rx is the radius function of x.

00:44 Okay, i'm not sorry, r by way, it's r r r.

00:48 It's the radius function of r.

00:49 So we have r2 minus r1 over l times r plus r1.

00:53 Therefore, the function for the area, for the surface area, will be pi r square or pi times r function square.

01:03 Then we have pi times r2 minus r1 over l times r plus r1 to a power of 2.

01:09 If we plug in, we'll have h is equal negative k pi, r2 minus r1l times r plus r1 square, dt over the r.

01:20 Then we move all the variables that is associated with r2, and derivative r on the left side, we have hdr over r2 minus r1 over l times r plus r1 square is equal to negative k pi d t.

01:38 Now if we put integral on both sides, we'll have since the range for r is from 0 to l, okay, l is the length of a corn, and the range for d t is from t1 to t2 then we'll have eventually we have hl over r2 minus r1 times 1 over r2 r2 minus r1 is equal to k pi t2 minus t1 you guys may wonder where's the negative sign well i can tell you guys is that if you guys do the calculations the negative sign on each side because the integral for the d r part also will appear a negative sign up to the anti -derivative.

02:44 So since there are negative signs on the left side and there are a negative sign on the right side, so eventually the negative sign cancels off...

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