Chapter 4, Problem 4.40P

To determine

The temperature distribution $T\left(y\right)$ for a constant wall temperature $t_w$.

Answer to Problem 4.40P

The temperature distribution $T\left(y\right)$ for a constant wall temperature $t_w$ is $T_w+\frac{\mu U^2}{3k}\left(1-\frac{y^4}{h^4}\right)$.

Explanation of Solution

Given information:

The velocity component of laminar flow between parallel plate in $x$ direction, $y$ direction and $z$ direction is $u=U\left(1-\frac{y^2}{h^2}\right)$, $v=0$, $w=0$ respectively Here, the velocity in $x$ direction is $u$, the velocity in $y$ direction is $v$, the velocity in $z$ direction is $w$, the distance between the parallel plate is $h$, the distance from the centreline is $y$.

Write the expression for energy equation for the flow.

$\rho c_{v}u\frac{\partial T}{\partial t}=k\frac{{\partial }^{2}T}{\partial y^2}+\mu {\left(\frac{\partial u}{\partial y}\right)}^2$        …… (I)

Here, the specific volume of the fluid is $c_v$, thermal conductivity is $k$ and dynamic viscosity of the fluid is $\mu$, density of the fluid is $\rho$, the rate of change of velocity component $u$ with respect to distance from the centre line is $\frac{\partial u}{\partial y}$.

Write the expression for change of temperature with respect to time

$\frac{\partial T}{\partial t}=\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}+w\frac{\partial T}{\partial z}$        …… (II)

Here, the rate of change of temperature with respect to $x$ is $u\frac{\partial T}{\partial x}$, the rate of change of temperature with respect to $y$ is $v\frac{\partial T}{\partial y}$, the rate of change of temperature with respect to $z$ is $w\frac{\partial T}{\partial z}$, the rate of change of temperature with respect to $t$ is $\frac{\partial T}{\partial t}$.

Since there is no change in temperature in y and z Substitute $0$ for $v\frac{\partial T}{\partial y}$, $0$ for $w\frac{\partial T}{\partial z}$ in Equation (II).

$\begin{array}{l}\frac{\partial T}{\partial t}=\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}+0+0\\ \frac{\partial T}{\partial t}=\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}\end{array}$

Substitute $\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}$ for $\frac{\partial T}{\partial t}$ in Equation (I).

$\rho c_{v}u\left(\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}\right)=k\frac{{\partial }^{2}T}{\partial y^2}+\mu {\left(\frac{\partial u}{\partial y}\right)}^2$        …… (III)

Substitute $0$ for $\frac{\partial T}{\partial t}$ Equation (III).

$\rho c_{v}u\left(u\frac{\partial T}{\partial x}\right)=k\frac{{\partial }^{2}T}{\partial y^2}+\mu {\left(\frac{\partial u}{\partial y}\right)}^2$        …… (IV)

Since temperature only varies in $y$ axis, Substitute $0$ for $\frac{\partial T}{\partial x}$ in Equation (IV)

$\begin{array}{l}\rho c_{v}u\left(u\left(0\right)\right)=k\frac{{\partial }^{2}T}{\partial y^2}+\mu {\left(\frac{\partial u}{\partial y}\right)}^2\\ 0=k\frac{{\partial }^{2}T}{\partial y^2}+\mu {\left(\frac{\partial u}{\partial y}\right)}^2\\ k\frac{{\partial }^{2}T}{\partial y^2}=-\mu {\left(\frac{\partial u}{\partial y}\right)}^2\end{array}$        …… (V)

Calculation:

Substitute $U\left(1-\frac{y^2}{h^2}\right)$ for $u$ in Equation (V).

$\begin{array}{c}k\frac{{\partial }^{2}T}{\partial y^2}=-\mu {\left(\frac{\partial \left[U\left(1-\frac{y^2}{h^2}\right)\right]}{\partial y}\right)}^2\\ k\frac{{\partial }^{2}T}{\partial y^2}=-\mu {\left(U\frac{\partial }{\partial y}\left(1-\frac{y^2}{h^2}\right)\right)}^2\\ k\frac{{\partial }^{2}T}{\partial y^2}=-\mu {\left(U\left(\frac{2y}{h^2}\right)\right)}^2\\ k\frac{{\partial }^{2}T}{\partial y^2}=-\frac{4\mu U^2{y}^2}{h^4}\end{array}$

$\frac{{\partial }^{2}T}{\partial y^2}=\frac{4\mu U^2{y}^2}{k{h}^4}$

Further integrating the equation.

$\begin{array}{c}\frac{\partial T}{\partial y}=-{\displaystyle \int \left(\frac{4\mu U^2{y}^2}{k{h}^4}\right)}dy\\ =-{\displaystyle \int \frac{4\mu U^2}{h^{4}k}\left(y^2\right)}dy\\ =-\frac{4\mu U^2{y}^3}{3h^{4}k}+c\end{array}$        …… (VI)

Substitute the boundary equation $0$ for $y$, $0$ for $\frac{\partial T}{\partial y}$ in Equation (VI).

$\begin{array}{l}0=-\frac{4\mu U^2\left(0\right)}{3h^{4}k}+c\\ 0=c\end{array}$

Substitute $0$ for $c$ in Equation (VI)

$\begin{array}{l}\frac{\partial T}{\partial y}=-\frac{4\mu U^2{y}^3}{3h^{4}k}+\left(0\right)\\ \frac{\partial T}{\partial y}=-\frac{4\mu U^2{y}^3}{3h^{4}k}\end{array}$

Further integrating the equation.

$\begin{array}{l}{\displaystyle \int \frac{\partial T}{\partial y}}=-{\displaystyle \int \frac{4\mu U^2{y}^3}{3h^{4}k}}dy\\ T\left(y\right)=-\frac{4\mu U^2}{3h^{4}k}{\displaystyle \int y^3}dy\\ T\left(y\right)=-\frac{4\mu U^2}{3h^{4}k}\left[\frac{y^4}{4}\right]+c\\ T\left(y\right)=-\frac{\mu U^2{y}^4}{3h^{4}k}+c\end{array}$        …… (VII)

Substitute the boundary equation $T_w$ for $T$, $h$ for $y$ in Equation (VII).

$\begin{array}{l}{T}_w=-\frac{\mu U^2{h}^4}{3h^{4}k}+c\\ T_w=-\frac{\mu U^2}{3k}+c\\ T_w+\frac{\mu U^2}{3k}=c\end{array}$

Substitute $T_w+\frac{\mu U^2}{3k}$ for $c$ in Equation (VII).

$\begin{array}{c}T\left(y\right)=-\frac{\mu U^2{y}^4}{3h^{4}k}+\left(T_w+\frac{\mu U^2}{3k}\right)\\ =T_w+\frac{\mu U^2}{3k}\left(1-\frac{y^4}{h^4}\right)\end{array}$

Conclusion:

The temperature distribution $T\left(y\right)$ for a constant wall temperature $t_w$ is $T_w+\frac{\mu U^2}{3k}\left(1-\frac{y^4}{h^4}\right)$.