Chapter 7, Problem 107P

To determine

Expression showing relationship x-component of fluid velocity.

Answer to Problem 107P

Dimensionless relationship for x-component of fluid velocity is

  $\frac{u}{V_{top}-V_{bottom}}=f(\mathrm{Re},\frac{y}{h})$

Explanation of Solution

Given Information:

Top plate speed, $V_{top}$

Bottom plate speed, $V_{bottom}$

Steady flow, incompressible, 2 dimensional.

Concept used:

Buckinghams Pi theorem.

Calculation:

Fluid velocity − u

Distance between parallel plate − h

Distance − y

Fluid density - $\rho$

Fluid viscosity - $\mu$

Plate speeds − V

No. Pf parameters, n = 6

The relative velocity of the top and bottom plate.

  $V=V_{top}-V_{bottom}$

Fluid velocity u is the function of remaining five parameters

  $u=f(\rho ,\mu ,(V_{top}-V_{bottom}),h,y)$

The primary dimension of each parameter.

Velocity,

  $\left\{u\right\}=\left\{\frac{L}{T}\right\}=\left\{L^1{T}^{-1}\right\}$

Density,

  $\left\{\rho \right\}=\left\{\frac{mass}{volume}\right\}=\left\{\frac{M}{L^3}\right\}=\left\{M^1{L}^{-3}\right\}$

Viscosity,

  $\left\{\mu \right\}=\left\{\frac{M}{LT}\right\}=\left\{M^1{L}^{-1}{T}^{-1}\right\}$

Distance y,

  $\left\{y\right\}=\left\{L^1\right\}$

Distance h,

  $\left\{h\right\}=\left\{L^1\right\}$

Plate speeds,

  $\{V_{top}-V_{bottom}\}=\left\{\frac{L}{T}\right\}=\left\{L^1{T}^{-1}\right\}$

No. Of primary dimensions used in the problem is 3. (M,L,T)

J=3

K=n-j=6-3=3

As per Buckinghams Pi theorem, no. of this $\pi \text{'}s$ is = 3

Since j=3, we need to select 3 repeating parameters.

According to the guideline, we should select $\rho ,h$ and $(V_{top}-V_{bottom})$ as a repeating parameter.

(1) Now creating $\pi \text{'}s$ by combing repeating parameters with the remaining parameters.

Dependant $\pi$

  $({\pi }_1)=u{h}^{a1}{\rho }^{b1}{(V_{top}-V_{bottom})}^{c1}$

The primary dimension of the above term is

  $\left\{\pi \right\}=\left\{M^0{L}^0{T}^0\right\}$

  $=\left\{u{h}^{a1}{\rho }^{b1}{(V_{top}-V_{bottom})}^{c1}\right\}=\{(L^1{T}^{-1})(L^1){(M{L}^{-3})}^{b1}{(L^1{T}^{-1})}^{c1}\}$

Now equation becomes,

  $\left\{m^0{L}^0{t}^0\right\}=\{(L^1{T}^{-1}){(L^1)}^{a1}{(M{L}^{-3})}^{b1}{(L^1{T}^{-1})}^{c1}\}$

Equating each primary dimension to solve $a_{1,}{b}_{1,}{c}_1$,

Mass,

  $\begin{array}{l}\left\{M^0\right\}=\left\{M^{b1}\right\}\\ 0=b_1\\ b_1=0\end{array}$

Length,

  $\begin{array}{l}\left\{L^0\right\}=\left\{L^1{L}^{a1}{L}^{-3b1}{L}^{c1}\right\}\\ 0=1+a_1-3b_1+c_1\\ a_1=0\end{array}$

Time,

  $\begin{array}{l}\left\{T^0\right\}=\left(T^{-1}{T}^{-c1}\right)\\ 0=-1-c_1\\ c_1=-1\end{array}$

Putting values $a_{1,}{b}_{1,}{c}_1$ in the equation,

  ${\pi }_1=u{h}^0{\rho }^0{(V_{top}-V_{bottom})}^{-1}$

${\pi }_1=\frac{u}{(V_{top}-V_{bottom})}$ --------------(1)

(2) Independent $\pi$,

  $({\pi }_2)=\mu h^{a2}{\rho }^{b2}{(V_{top}-V_{bottom})}^{c2}$

Primary dimensions of the above term

  $\left\{{\pi }_2\right\}=\left\{M^0{L}^0{T}^0\right\}$

  $\left\{\mu h^{a2}{\rho }^{b2}{(V_{top}-V_{bottom})}^{c2}\right\}=\{(M^1{L}^{-1}{T}^{-1}){(L^1)}^{a2}{(M{L}^{-3})}^{b2}{(L^1{T}^{-1})}^{c2}\}$

Now equation becomes,

  $\left\{m^0{L}^0{t}^0\right\}=\{(M^1{L}^{-1}{T}^{-1}){(L^1)}^{a2}{(M{L}^{-3})}^{b2}{(L^1{T}^{-1})}^{c2}\}$

Equating each primary dimension to solve $a_{2,}{b}_{2,}{c}_2$

Mass,

  $\begin{array}{l}\left\{m^0\right\}=\left\{M^1{M}^{b2}\right\}\\ 0=1+b_1\\ b_1=-1\end{array}$

Length,

  $\begin{array}{l}\left\{L^0\right\}=\left\{L^1{L}^{a2}{L}^{-3b2}{L}^{c2}\right\}\\ 0=-1+a_2-3b_2+c_2\\ a_2=-1\end{array}$

Time,

  $\begin{array}{l}\left\{t^0\right\}=\left(T^{-2}{T}^{-c2}\right)\\ 0=-1-c_2\\ c_2=-1\end{array}$

Putting values $a_{2,}{b}_{2,}{c}_2$ in the equation

  ${\pi }_2=\mu h^{-1}{\rho }^{-1}{(V_{top}-V_{bottom})}^{-1}$

  ${\pi }_2=\frac{\mu }{h\rho (V_{top}-V_{bottom})}$

Equation of ${\pi }_2$ is reciprocal of Reynolds number.

$\frac{\mu }{h\rho (V_{top}-V_{bottom})}=\frac{1}{\mathrm{Re}}$ -----------------(2)

(3) Independent $\pi$ using y variable.

  ${\pi }_3=y{h}^{a3}{\rho }^{b3}{v}^{c3}$

The primary dimension of above term.

  $\left\{{\pi }_3\right\}=\left(M^0{L}^0{T}^0\right)$

  $\left\{y{h}^{a3}{\rho }^{b3}{v}^{c3}\right\}=\{(L^1){(L^1)}^{a3}{(M{L}^{-3})}^{b3}{(L^1{T}^{-1})}^{c3}\}$

Now equation becomes,

  $\left\{m^0{L}^0{t}^0\right\}=\{(L^1){(L^1)}^{a3}{(M{L}^{-3})}^{b3}{(L^1{T}^{-1})}^{c3}\}$

Equating each primary dimension to solve $a_{3,}{b}_{3,}{c}_3$

Mass,

  $\begin{array}{l}\left\{M^0\right\}=\left\{M^{b3}\right\}\\ 0=b_3\\ b_3=0\end{array}$

Time,

  $\begin{array}{l}\left\{T^0\right\}=\left\{T^{-c3}\right\}\\ 0=-c_3\\ c_3=0\end{array}$

Length,

  $\begin{array}{l}\left\{L^0\right\}=\left\{L^1{L}^{a3}{L}^{-3b3}{L}^{c3}\right\}\\ 0=1+a_3-3b_3+c_3\\ a_3=-1\end{array}$

Putting all the values in the equation,

  $\begin{array}{l}{\pi }_3=y{h}^{-1}{\rho }^0{(}^{V_{top}}\\ {\pi }_3=\frac{y}{h}\end{array}$ --------------------(3)

Comparing equation (1)(2)(3)

  $\frac{u}{h\rho (V_{top}-V_{bottom})}=f(\mathrm{Re},\frac{y}{h})$

Conclusion:

Thus, we can develop a dimensionless relationship for x-component of fluid velocity which is a function of fluid viscosity $\mu$, plate speeds, distance h, and distance y, density $\rho$ by using Buckingham Pi theorem.