Chapter 7, Problem 50P

To determine

The dimensionless relationship.

Answer to Problem 50P

The dimensionless relationship is $\frac{\dot{W}}{\rho {\omega }^3{D}^5}=f\left(\frac{\mu }{\omega \rho D^2},\frac{h_{\text{tank}}}{D},\frac{D_{\text{tank}}}{D}\right)$.

Explanation of Solution

Given information:

The shaft power is $\dot{W}$, the angular velocity is $\omega$, the fluid density is $\rho$, the fluid dynamic viscosity is $\mu$, the stirrer diameter is $D$, the diameter of the tank is $D_{\text{tank}}$ and the average liquid depth is $h_{\text{tank}}$.

Write the expression of function of shaft power.

$\dot{W}=f\left(\omega ,\rho ,\mu ,D,D_{\text{tank}},h_{\text{tank}}\right)$   ....(I)

Write the dimension of density in $MLT$ unit system.

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

Here, the dimension of mass is $M$ and the dimension of the length is $L$.

Write the dimension of the dynamic viscosity.

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

Here, the dimension of time is $T$.

Write the dimension of diameter in $MLT$ unit system.

$D=\left[L\right]$  ....(IV)

Write the expression of angular velocity.

$\omega =\left(\frac{\theta }{t}\right)$   ....(VI)

Here, the angle is $\theta$ and the time is $t$.

Write the dimension of time in $MLT$ unit system.

$t=\left[T\right]$

Write the dimension of angle in $MLT$ unit system.

$\theta =1$.

Substitute $T$ for $t$ and $1$ for $\theta$ in Equation ( VI).

$\begin{array}{c}\omega =\left(\frac{1}{T}\right)\\ =\left[T^{-1}\right]\end{array}$  ....(VII)

Write the expression of power.

$\dot{W}=\frac{E}{t}$  ....(VIII)

Here, the energy is $E$.

Write the expression of Energy.

$E=m\times a_c\times d$   ....(IX)

Here, the mass is $m$, the acceleration is $a_c$ and the displacement is $d$.

Write the dimension of mass in $MLT$ unit system.

$m=\left[M\right]$

Write the dimension of acceleration in $MLT$ unit system.

$a_c=\left[ML{T}^{-2}\right]$

Write the dimension of displacement in $MLT$ unit system.

$d=\left[L\right]$

Substitute $\left[M\right]$ for $m$, $\left[ML{T}^{-2}\right]$ for $a_c$ and $\left[L\right]$ for $d$ in Equation (IX).

$\begin{array}{c}E=\left[M\right]\times \left[L{T}^{-2}\right]\times \left[L\right]\\ =\left[M{L}^2{T}^{-2}\right]\end{array}$

Substitute $\left[M^2{L}^2{T}^{-2}\right]$ for $E$ and $\left[T\right]$ for $t$ in Equation (VIII)

$\begin{array}{c}\dot{W}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[T\right]}\\ =\left[M{L}^2{T}^{-3}\right]\end{array}$

Write the dimension of tank diameter in $MLT$ unit system.

$D_{\text{tank}}=\left[L\right]$  ....(X)

Write the dimension of average liquid depth in $MLT$ unit system.

$h_{\text{tank}}=\left[L\right]$  ....(XI)

Here, the number of variable is $7$ these are $\dot{W},\omega ,\rho ,\mu ,D,D_{\text{tank}},\text{and}{h}_{\text{tank}}$ and the number of dimension is $3$ these are $M,L\text{and}T$.

Write the expression of number of pi-terms.

$\text{pi-terms}=n-j$   ....(XII)

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

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

$\begin{array}{c}\text{pi-terms}=7-3\\ =4\end{array}$

Here, four pi-terms is present.

Here, the basic variable is $\omega$, $\rho$ and $D$.

Write the expression of first pi-terms.

${\prod }_1=\dot{W}{\omega }^a{\rho }^b{D}^c$   ....(XIII)

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

Write the expression of second pi-terms.

${\prod }_2=\mu {\omega }^a{\rho }^b{D}^c$   ....(XIV)

Write the expression of third pi-terms.

${\prod }_3=h_{\text{tank}}{\omega }^a{\rho }^b{D}^c$   ....(XV)

Write the expression of fourth pi-terms.

${\prod }_3=D_{\text{tank}}{\omega }^a{\rho }^b{D}^c$   ....(XVI)

Write the dimension of first pi-term.

${\prod }_1=\left[M^0{L}^0{T}^0\right]$

Write the dimension of second pi-term.

${\prod }_2=\left[M^0{L}^0{T}^0\right]$

Write the dimension of third pi-term.

${\prod }_3=\left[M^0{L}^0{T}^0\right]$

Write the dimension of fourth pi-term.

${\prod }_4=\left[M^0{L}^0{T}^0\right]$

Substitute $\left[M^0{L}^0{T}^0\right]$ for ${\prod }_1$, $\left[M{L}^2{T}^{-3}\right]$ for $\dot{W}$, $\left[T^{-1}\right]$ for $\omega$, $\left[M{L}^{-3}\right]$ for $\rho$ and $\left[L\right]$ for $D$ in Equation (XIII).

$\begin{array}{l}\left[M^0{L}^0{T}^0\right]=\left[M{L}^2{T}^{-3}\right]{\left[T^{-1}\right]}^a{\left[M{L}^{-3}\right]}^b{\left[L\right]}^c\\ \left[M^0{L}^0{T}^0\right]=\left[M^{1+b}{L}^{2-3b+c}{T}^{-3-a}\right]\end{array}$   ....(XVII)

Compare the power of $M$ on both sides of Equation (XVII).

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

Compare the power of $T$ on both sides of Equation (XVII).

$\begin{array}{l}0=-3-a\\ a=-3\end{array}$

Compare the power of $L$ on both sides of Equation (XVII) and substitute $-1$ for $b$ and $-3$ for $a$.

$\begin{array}{l}0=2-3\left(-1\right)+c\\ 0=2+3+c\\ c=-5\end{array}$

Substitute $-3$ for $a$, $-1$ for $b$ and $-5$ for $c$ in Equation (XIII).

$\begin{array}{c}{\prod }_1=\dot{W}{\omega }^{-3}{\rho }^{-1}{D}^{-5}\\ =\frac{\dot{W}}{\rho {\omega }^3{D}^5}\end{array}$

Substitute $\left[M^0{L}^0{T}^0\right]$ for ${\prod }_2$, $\left[M{L}^{-1}{T}^{-1}\right]$ for $\mu$, $\left[T^{-1}\right]$ for $\omega$, $\left[M{L}^{-3}\right]$ for $\rho$ and $\left[L\right]$ for $D$ in Equation (XIV).

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

Compare the power of $M$ on both sides of Equation (XVIII).

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

Compare the power of $T$ on both sides of Equation (XVIII).

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

Compare the power of $L$ on both sides of Equation (XIV) and substitute $-1$ for $b$ and $-1$ for $a$.

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

Substitute $-1$ for $a$, $-1$ for $b$ and $-2$ for $c$ in Equation (XII).

$\begin{array}{c}{\prod }_2=\mu {\omega }^{-1}{\rho }^{-1}{D}^{-2}\\ =\frac{\mu }{\omega \rho D^2}\end{array}$

Substitute $\left[M^0{L}^0{T}^0\right]$ for ${\prod }_2$, $\left[L\right]$ for $h_{\text{tank}}$, $\left[T^{-1}\right]$ for $\omega$, $\left[M{L}^{-3}\right]$ for $\rho$ and $\left[L\right]$ for $D$ in Equation (XV).

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

Compare the power of $M$ on both sides of Equation (XIX).

$b=0$

Compare the power of $T$ on both sides of Equation (XIX).

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

Compare the power of $L$ on both sides of Equation (XIX) and substitute $0$ for $b$ and $0$ for $a$.

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

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

$\begin{array}{c}{\prod }_3=h_{\text{tank}}{\omega }^0{\rho }^0{D}^{-1}\\ =\frac{h_{\text{tank}}}{D}\end{array}$

Substitute $\left[M^0{L}^0{T}^0\right]$ for ${\prod }_2$, $\left[L\right]$ for $D_{\text{tank}}$, $\left[T^{-1}\right]$ for $\omega$, $\left[M{L}^{-3}\right]$ for $\rho$ and $\left[L\right]$ for $D$ in Equation (XVI).

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

Compare the power of $M$ on both sides of Equation (XX).

$b=0$

Compare the power of $T$ on both sides of Equation (XX).

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

Compare the power of $L$ on both sides of Equation (XX) and substitute $0$ for $b$ and $0$ for $a$.

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

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

$\begin{array}{c}{\prod }_4=h_{\text{tank}}{\omega }^0{\rho }^0{D}^{-1}\\ =\frac{D_{\text{tank}}}{D}\end{array}$

According to Buckingham pi-theorem the first pi-term is the function of rest of all another pi-term.

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

Substitute $\frac{\dot{W}}{\rho {\omega }^3{D}^5}$ for ${\prod }_1$, $\frac{\mu }{\omega \rho D^2}$ for ${\prod }_2$, $\frac{h_{\text{tank}}}{D}$ for ${\prod }_3$ and $\frac{D_{\text{tank}}}{D}$ for ${\prod }_4$ in Equation (XXI).

$\frac{\dot{W}}{\rho {\omega }^3{D}^5}=f\left(\frac{\mu }{\omega \rho D^2},\frac{h_{\text{tank}}}{D},\frac{D_{\text{tank}}}{D}\right)$

Conclusion:

The dimensionless relationship is $\frac{\dot{W}}{\rho {\omega }^3{D}^5}=f\left(\frac{\mu }{\omega \rho D^2},\frac{h_{\text{tank}}}{D},\frac{D_{\text{tank}}}{D}\right)$.