Chapter 7, Problem 73P

To determine

The dimensionless relationship between given parameters.

Answer to Problem 73P

The functional relationship between given parameters is $\frac{h}{R}=f\left(\text{Fr,}\omega t,\mathrm{Re}\right)$.

Explanation of Solution

The angular velocity is $\omega$, the fluid density is $\rho$, the acceleration due to gravity id $g$, the radius is $R$, the viscosity is $\mu$, the time is $t$ and the elevation difference is $h$.

Write the expression for the elevation difference.

$h=f\left(\omega ,\rho ,g,R,t,\mu \right)$   ....(I)

Write the expression for the number of the pi terms.

$k=n-j$  ....(II)

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

Substitute $7$ for $n$ and $3$ for $j$ in Equation (II).

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

Write the expression for the first pi term.

${\Pi }_1=h{\omega }^{a_1}{\rho }^{b_1}{R}^{c_1}$   ....(III)

Write the expression for the independent pi term.

${\Pi }_2=g{\omega }^{a_2}{\rho }^{b_2}{R}^{c_2}$   ....(IV)

Write the expression for third pi term.

${\Pi }_3=t{\omega }^{a_3}{\rho }^{b_3}{R}^{c_3}$   ....(V)

Write the expression for fourth pi term.

${\Pi }_4=\mu {\omega }^{a_4}{\rho }^{b_4}{R}^{c_4}$   ....(VI)

Write the expression for Reynold's number.

$\mathrm{Re}=\frac{\rho \omega R^2}{\mu }$

Write the dimensional formula for $\Pi$.

$\Pi =\left[M^0{L}^0{T}^0\right]$

Here, the dimension of mass is $\left[M\right]$, the dimension of length is $\left[L\right]$ and the dimension of time is $\left[T\right]$.

Write the dimensional formula for elevation difference.

$h=\left[L^1\right]$

Write the dimensional formula for density.

$\rho =\left[M^1{L}^{-3}\right]$

Write the dimensional formula for radius.

$R=\left[L^1\right]$

Write the dimensional formula for angular velocity.

$\omega =\left[T^{-1}\right]$

Write the dimensional formula for acceleration due to gravity.

$g=\left[L^1{T}^{-2}\right]$

Write the dimensional formula for time.

$t=\left[T^1\right]$

Write the dimensional formula for viscosity.

$\mu =\left[M^1{L}^{-1}{T}^{-1}\right]$

Calculation:

Substitute $[M^0{L}^0{T}^0]$ for ${\Pi }_1$, $[L^1]$ for $h$, $[L^1]$ for $R$, $[T^{-1}]$ for $\omega$ and $[M^1{L}^{-3}]$ for $\rho$ in Equation (III).

$\begin{array}{l}\left[M^0{L}^0{T}^0\right]=\left[L^1\right]{\left[T^{-1}\right]}^{a_1}{\left[M^1{L}^{-3}\right]}^{b_1}\left[L^{c_1}\right]\\ \left[M^0{L}^0{T}^0\right]={\left[M\right]}^{b_1}{\left[L\right]}^{-3b_1+c_1+1}{\left[T\right]}^{-a_1}\end{array}$   ....(VII)

Compare the power of the dimensional terms of $M$ in Equation (VII).

$b_1=0$

Compare the power of the dimensional terms of $T$ in Equation (VII).

$a_1=0$

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

$\begin{array}{l}-3b_1+c_1+1=0\\ 0+c_1+1=0\\ c_1=-1\end{array}$

Substitute $0$ for $a_1$, $-1$ for $c_1$ and $0$ for $b_1$ in Equation (III).

$\begin{array}{c}{\Pi }_1=h{\omega }^0{\rho }^0{R}^{-1}\\ =\frac{h}{R}\end{array}$   ....(VIII)

Substitute $\left[M^0{L}^0{T}^0\right]$ for ${\Pi }_2$, $\left[L^1{T}^{-2}\right]$ for $g$, $\left[L^1\right]$ for $R$, $\left[T^{-1}\right]$ for $\omega$ and $\left[M^1{L}^{-3}\right]$ for $\rho$ in Equation (IV).

$\begin{array}{l}\left[M^0{L}^0{T}^0\right]=\left[L^1{T}^{-2}\right]{\left[T^{-1}\right]}^{a_2}{\left[M^1{L}^{-3}\right]}^{b_2}{\left[L^1\right]}^{c_2}\\ \left[M^0{L}^0{T}^0\right]=\left[L^1{T}^{-2}\right]\left[T^{-a_2}\right]\left[M^{b_2}{L}^{-3b_2}\right]\left[L^{c_2}\right]\\ \left[M^0{L}^0{T}^0\right]=\left[M^{b_2}{L}^{-3b_2+c_2+1}{T}^{-2-a_2}\right]\end{array}$.  ....(IX)

Compare the power of the dimensional terms of $M$ in Equation (IX).

$b_2=0$

Compare the power of the dimensional terms of $T$ in Equation (IX).

$\begin{array}{l}-2-a_2=0\\ a_2=-2\end{array}$

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

$\begin{array}{l}-3b_2+c_2+1=0\\ 0+c_2+1=0\\ c_2=-1\end{array}$

Substitute $-2$ for $a_2$, $-1$ for $c_2$ and $0$ for $b_2$ in Equation (IV).

$\begin{array}{c}{\Pi }_2=g{\omega }^{-2}{\rho }^0{R}^{-1}\\ =\frac{g}{{\omega }^{2}R}\end{array}$

Write the relation between the two pi terms.

${\Pi }_1=f\left({\Pi }_2\right)$  ....(X)

Write the relation between Froude Number and second pi term.

$\text{Fr}=f\left({\Pi }_2^{\frac{-1}{2}}\right)$   ....(XI)

Compare Equation (X) and Equation (XI).

${\Pi }_1=f\left(\text{Fr}\right)$   ....(XII)

Substitute $\frac{h}{R}$ for ${\Pi }_1$ in Equation (XII).

$\frac{h}{R}=f\left(\text{Fr}\right)$

Substitute $\left[M^0{L}^0{T}^0\right]$ for ${\Pi }_3$, $\left[T^1\right]$ for $t$, $\left[L^1\right]$ for $R$, $\left[T^{-1}\right]$ for $\omega$ and $\left[M^1{L}^{-3}\right]$ for $\rho$ in Equation (V).

$\begin{array}{l}\left[M^0{L}^0{T}^0\right]=\left[T^1\right]{\left[T^{-1}\right]}^{a_3}{\left[M^1{L}^{-3}\right]}^{b_3}{\left[L^1\right]}^{c_3}\\ \left[M^0{L}^0{T}^0\right]=\left[M^{b_3}\right]\left[L^{-3b_3+c_3}\right]\left[T^{1-a_3}\right]\end{array}$   ....(XIII)

Compare the power of the dimensional terms of $M$ in Equation (XIII).

$b_3=0$

Compare the power of the dimensional terms of $T$ in Equation (XIII).

$\begin{array}{l}1-a_3=0\\ a_3=1\end{array}$

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

$\begin{array}{l}-3b_3+c_3+1=0\\ 0+c_3=0\\ c_3=0\end{array}$

Substitute $1$ for $a_3$, $0$ for $c_3$ and $0$ for $b_3$ in Equation (V)

$\begin{array}{c}{\Pi }_3=t{\omega }^1{\rho }^0{R}^0\\ =t\omega \end{array}$

Substitute $\left[M^0{L}^0{T}^0\right]$ for ${\Pi }_4$, $\left[M^1{L}^{-1}{T}^{-1}\right]$ for $\mu$, $\left[L^1\right]$ for $R$, $\left[T^{-1}\right]$ for $\omega$ and $\left[M^1{L}^{-3}\right]$ for $\rho$ in Equation (VI).

$\begin{array}{l}\left[M^0{L}^0{T}^0\right]=\left[M^1{L}^{-1}{T}^{-1}\right]{\left[T^{-1}\right]}^{a_4}{\left[M^1{L}^{-3}\right]}^{b_4}{\left[L^1\right]}^{c_4}\\ \left[M^0{L}^0{T}^0\right]=\left[M^{1+b_4}\right]\left[L^{-1-3b_4+c_4}\right]\left[T^{-1-a_4}\right]\end{array}$   ....(XIV)

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

$\begin{array}{l}1+b_4=0\\ b_4=-1\end{array}$

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

$\begin{array}{l}-1-a_4=0\\ a_4=-1\end{array}$

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

$\begin{array}{l}-1-3b_4+c_4=0\\ -1-3\left(-1\right)+c_4=0\\ c_4=-2\end{array}$

Substitute $-1$ for $a_4$, $-2$ for $c_4$ and $-1$ for $b_4$ in Equation (VI)

$\begin{array}{c}{\Pi }_4=\mu {\omega }^{-1}{\rho }^{-1}{R}^{-2}\\ =\frac{\mu }{\omega \rho R^2}\end{array}$

Equation (XI) is the reciprocal of Reynold's Number.

Modify the Equation (XI).

${\Pi }_4=\frac{\rho \omega R^2}{\mu }$   ....(XV)

Substitute $\mathrm{Re}$ for $\frac{\rho \omega R^2}{\mu }$ in Equation (XV).

${\Pi }_4=\mathrm{Re}$

Write the expression for relation between pi terms.

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

Substitute $\frac{h}{R}$ for ${\Pi }_1$, $\text{Fr}$ for ${\Pi }_2$, $\omega t$ for ${\Pi }_3$ and $\mathrm{Re}$ for ${\Pi }_4$ in Equation (XVI).

$\frac{h}{R}=f\left(\text{Fr,}\omega t,\mathrm{Re}\right)$

Conclusion:

The functional relationship between given parameters is $\frac{h}{R}=f\left(\text{Fr,}\omega t,\mathrm{Re}\right)$.