Chapter 7, Problem 53P

Consider developing Couette flow-the same flow as Prob. 7-2 except that the flow is not yet steady-state, but is developing with time. In other words, time t is an additional parameter in the problem. Generate a dimensionless relationship between all the variables.

To determine

The functional relationship between given parameters.

Answer to Problem 53P

The functional relationship between given parameters is $\frac{u}{V}=f\left(\mathrm{Re},\frac{y}{h},\frac{tV}{h}\right)$.

Explanation of Solution

Given information:

The viscosity of the fluid is $\mu$, the distance is $y$, the velocity of the fluid is $u$, the speed of the top plate is $V$, the distance of the top plate is $h$ and the time is $t$.

Write the expression of function of fluid velocity.

$u=f\left(\rho ,\mu ,V,h,y,t\right)$  ....(I)

Write the expression for the number of the pi terms.

$N=n-j$  ....(II)

Here, the total number of the repeating variable is $j$ and the total number of the variable is $n$.

The total number of the variable are 7 $\left(u,\rho ,\mu ,V,h,y,t\right)$ and the number of the repeating variables are $3$

$\left(\rho ,V,h,\right)$.

Substitute $4$ for $n$ and $3$ for $j$ in equation (II).

$\begin{array}{c}N=7-3\\ =4\end{array}$

Write the expression for the first pi-terms.

${\prod }_1=u{h}^a{\rho }^b{V}^c$   ....(III)

Here, the constants are $a,b$, and $c$.

Write the expression for the second pi-terms.

${\prod }_2=\mu h^a{\rho }^b{V}^c$   ....(IV)

Write the expression for the third pi-terms.

${\prod }_3=y{h}^a{\rho }^b{V}^c$   ....(V)

Write the expression for the forth pi-terms.

${\prod }_4=t{h}^a{\rho }^b{V}^c$   ....(VI)

Write the expression for the relation between the pi terms.

${\Pi }_1=f\left({\Pi }_2,{\Pi }_3,{\Pi }_4\right)$  ....(VII)

Write the dimensional expression for the fluid velocity.

$u=\left[L{t}^{-1}\right]$   ....(VIII)

Here, the length is $L$ and the time is $t$.

Write the dimensional expression for the viscosity of the fluid.

$\mu =\left[m{L}^{-1}{t}^{-1}\right]$   ....(IX)

Here, the mass is $m$.

Write the dimensional expression for the density.

$\rho =\left[m{L}^{-3}\right]$  ....(X)

Write the dimensional expression for the length.

$L=\left[L\right]$   ....(XI)

Write the dimensional expression for the speed of the plate.

$V=\left[L{t}^{-1}\right]$   ....(XII)

Write the dimensional expression for the time.

$t=\left[t\right]$  ....(XIII)

Calculation:

Substitute $\left[m^0{L}^0{t}^0\right]$ for ${\Pi }_1$, $\left[L{t}^{-1}\right]$ for $u$, $\left[m{L}^{-3}\right]$ for $\rho$, $\left[L{t}^{-1}\right]$ for $V$ and $\left[L\right]$ for $h$ in Equation (III)

$\begin{array}{l}\left[m^0{L}^0{t}^0\right]=\left[L{t}^{-1}\right]{\left[L\right]}^a{\left[m{L}^{-3}\right]}^b{\left[L{t}^{-1}\right]}^c\\ \left[m^0{L}^0{t}^0\right]={\left[m\right]}^b{\left[L\right]}^{1+a-3b+c}{\left[t\right]}^{-1-c}\end{array}$  ....(XIV)

Compare the power of the dimensional terms of $m$ in Equation (XIV).

$b=0$

Compare the power of the dimensional terms of $t$ in Equation (XIV).

$\begin{array}{l}-1-c=0\\ c=-1\end{array}$

Compare the power of the dimensional terms of $L$ in Equation (XIV).

$\begin{array}{l}1+a-3b+c=0\\ 1+a-3\left(0\right)+\left(-1\right)=0\\ a=0\end{array}$

Substitute $0$ for $a$, $0$ for $b$ and $-1$ for $c$ in Equation (III).

$\begin{array}{c}{\prod }_1=u{h}^0{\rho }^0{V}^{-1}\\ =\frac{u}{V}\end{array}$  ....(XV)

Substitute $\left[m^0{L}^0{t}^0\right]$ for ${\Pi }_1$, $\left[m{L}^{-1}{t}^{-1}\right]$ for $\mu$, $\left[m{L}^{-3}\right]$ for $\rho$, $\left[L{t}^{-1}\right]$ for $V$ and $\left[L\right]$ for $h$ in Equation (IV)

$\begin{array}{l}\left[m^0{L}^0{t}^0\right]=\left[m{L}^{-1}{t}^{-1}\right]{\left[L\right]}^a{\left[m{L}^{-3}\right]}^b{\left[L{t}^{-1}\right]}^c\\ \left[m^0{L}^0{t}^0\right]={\left[m\right]}^{1+b}{\left[L\right]}^{-1+a-3b+c}{\left[t\right]}^{-1-c}\end{array}$   ....(XVI)

Compare the power of the dimensional terms of $m$ in Equation (XVI).

$\begin{array}{l}1+b=0\\ b=-1\end{array}$

Compare the power of the dimensional terms of $t$ in Equation (XVI).

$\begin{array}{l}-1-c=0\\ c=-1\end{array}$

Compare the power of the dimensional terms of $L$ in Equation (XVI).

$\begin{array}{l}-1+a-3b+c=0\\ -1+a-3\left(-1\right)+\left(-1\right)=0\\ a=-1\end{array}$

Substitute $-1$ for $a$, $-1$ for $b$ and $-1$ for $c$ in Equation (IV).

$\begin{array}{c}{\prod }_2=\mu h^{-1}{\rho }^{-1}{V}^{-1}\\ =\frac{\mu }{h\rho V}\end{array}$  ....(XVII)

Since, the Equation (IX) is the reciprocal of the Reynolds number hence $\frac{1}{\mathrm{Re}}=\frac{\mu }{h\rho V}$.

Substitute $\mathrm{Re}$ for $\frac{h\rho V}{\mu }$ in Equation (XVII).

${\prod }_2=\mathrm{Re}$   ....(XVIII)

Substitute $\left[m^0{L}^0{t}^0\right]$ for ${\Pi }_1$, $\left[L\right]$ for $y$, $\left[m{L}^{-3}\right]$ for $\rho$, $\left[L{t}^{-1}\right]$ for $V$ and $\left[L\right]$ for $h$ in Equation (V)

$\begin{array}{l}\left[m^0{L}^0{t}^0\right]=\left[L\right]{\left[L\right]}^a{\left[m{L}^{-3}\right]}^b{\left[L{t}^{-1}\right]}^c\\ \left[m^0{L}^0{t}^0\right]={\left[m\right]}^b{\left[L\right]}^{1+a-3b+c}{\left[t\right]}^{-c}\end{array}$  ....(XIX)

Compare the power of the dimensional terms of $m$ in Equation (XIX).

$b=0$

Compare the power of the dimensional terms of $t$ in Equation (XIX).

$\begin{array}{l}-c=0\\ c=0\end{array}$

Compare the power of the dimensional terms of $L$ in Equation (XIX).

$\begin{array}{l}1+a-3b+c=0\\ 1+a-3\left(0\right)+\left(0\right)=0\\ a=-1\end{array}$

Substitute $-1$ for $a$, $0$ for $b$ and $0$ for $c$ in Equation (V).

$\begin{array}{c}{\prod }_3=y{h}^{-1}{\rho }^0{V}^0\\ =\frac{y}{h}\end{array}$  ....(XX)

Substitute $\left[m^0{L}^0{t}^0\right]$ for ${\Pi }_1$, $\left[t\right]$ for $t$, $\left[m{L}^{-3}\right]$ for $\rho$, $\left[L{t}^{-1}\right]$ for $V$ and $\left[L\right]$ for $h$ in Equation (VI)

$\begin{array}{l}\left[m^0{L}^0{t}^0\right]=\left[t\right]{\left[L\right]}^a{\left[m{L}^{-3}\right]}^b{\left[L{t}^{-1}\right]}^c\\ \left[m^0{L}^0{t}^0\right]={\left[m\right]}^b{\left[L\right]}^{a-3b+c}{\left[t\right]}^{1-c}\end{array}$   ....(XXI)

Compare the power of the dimensional terms of $m$ in Equation (XXI).

$b=0$

Compare the power of the dimensional terms of $t$ in Equation (XXI).

$\begin{array}{l}1-c=0\\ c=1\end{array}$

Compare the power of the dimensional terms of $L$ in Equation (XXI).

$\begin{array}{l}a-3b+c=0\\ a-3\left(0\right)+\left(1\right)=0\\ a=-1\end{array}$

Substitute $-1$ for $a$, $0$ for $b$ and $1$ for $c$ in Equation (VI).

$\begin{array}{c}{\prod }_4=t{h}^{-1}{\rho }^0{V}^1\\ =\frac{tV}{h}\end{array}$  ....(XXII)

Substitute $\frac{tV}{h}$ for ${\Pi }_4$, $\frac{y}{h}$ for ${\Pi }_3$, $\mathrm{Re}$ for ${\Pi }_2$ and $\frac{u}{V}$ for ${\Pi }_1$ in Equation (VII).

$\frac{u}{V}=f\left(\mathrm{Re},\frac{y}{h},\frac{tV}{h}\right)$

Conclusion:

The functional relationship between given parameters is $\frac{u}{V}=f\left(\mathrm{Re},\frac{y}{h},\frac{tV}{h}\right)$.