Physics Asked by Adeel on December 24, 2020

Refer to this image showing the temperature of a metal pipe at the inlet and the outlet:

The temperature $T(z,t)$ is a function of the length $z$ and time $t$. Let the average temperature be

$$T_mathrm{avg}(t)=frac{1}{2}big(T(a,t)+T(b,t)big).$$

Integrating the partial derivative of $T$ with respect to $t$, $frac{partial T}{partial t}$ over the entire length of the pipe (from $z=a$ to $z=b$) and applying Leibniz’s rule, we should get

$$int_{z=a}^{z=b}frac{partial}{partial t}T(z,t)dz=frac{d}{dt}int_{z=a}^{z=b}T(z,t)dz-T(b,t)frac{db}{dt}+T(a,t)frac{da}{at}.$$

In this paper (eqs. (21) and (22)), the result is reported as follows:

$$int_{z=a}^{z=b}frac{partial}{partial t}T(z,t)dz=(b-a)frac{d}{dt}T_mathrm{avg}(t)+big(T(a,t)-T_mathrm{avg}(t)big)frac{da}{dt}+big(T_mathrm{avg}(t)-T(b,t)big)frac{db}{dt}.$$

Are the two equivalent? How to correctly solve this integral? Thanks!

**Note:** The integral in this question is only a portion of a larger integration problem in the energy balance, not reported for brevity. Equations (21) and (22) in the reference describe the complete energy balance.

**This answer is based on the original answer here**. Some supplemental details relevant to the problem are added.

Referring to the original problem, let $T(z,t)$ vary linearly with $z$ as shown in the following figure:

As proposed in the original solution, integrating integral of the partial derivative of $T$ from $a$ to $b$ and applying Leibniz's Rule: begin{align} int_{a}^{b}frac{partial}{partial t}T(z,t)dz&=int_{a}^{b}frac{partial}{partial t}big(T(z,t)-T_mathrm{avg}(t)+T_mathrm{avg}(t)big)dz\ &=int_{a}^{b}frac{partial}{partial t}T_mathrm{avg}(t)dz + underbrace{int_{a}^{b}frac{partial}{partial t}big(T(z,t)-T_mathrm{avg}(t)big)dz}_text{apply Leibniz Rule}\ &=(b-a)frac{d}{dt}T_mathrm{avg}(t)\ &quad+underbrace{frac{d}{dt}int_{a}^{b}big(T(z,t)-T_mathrm{avg}(t)big)dz+big(T(a,t)-T_mathrm{avg}(t)big)frac{da}{dt}+big(T_mathrm{avg}(t)-T(b,t)big)frac{db}{dt}.}_text{Leibniz's Rule applied} end{align}

The area under the curve in the figure (shaded green) can be found using the integral $int_{a}^{b}T(z,t)dz$. Approximating the curve with a straight line, we can write

begin{align} int_{a}^{b}T(z,t)dz&approxunderbrace{frac{1}{2}(b-a)big(T(b,t)-T(a,t)big)}_text{area of top triangle}+underbrace{T(a,t)(b-a)}_text{area of bottom rectangle}\ &=frac{1}{2}big(T(a,t)+T(b,t)big)(b-a)=(b-a)T_mathrm{avg}(t). end{align}

Thus, the second term in the third equality of the first equation vanishes, and the desired result is obtained: $$int_{a}^{b}frac{partial}{partial t}T(z,t)dzapprox(b-a)frac{d}{dt}T_mathrm{avg}(t)+big(T(a,t)-T_mathrm{avg}(t)big)frac{da}{dt}+big(T_mathrm{avg}(t)-T(b,t)big)frac{db}{dt}.$$

Correct answer by Adeel on December 24, 2020

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