Chapter 5, Problem 5.40P

To determine

The given relation in a dimensionless form using Ipsen’s method.

Answer to Problem 5.40P

$t_d\frac{\mu }{\rho d^2}=f\left(\frac{v_0}{d^3},\frac{h_0}{d}\right)$

Explanation of Solution

Given information:

The time $t_d$ takes to drain the liquid completely from a tank depends on:

Hole diameter $d$

Initial volume $v_o$

Initial depth $h_0$

Density $\rho$

Viscosity $\mu$

Ipsen’s method is a way to find all pi products at once.

We have to use either multiplication or division to eliminate each dimension in a function.

Usually we eliminate mass $M$ in a function that contains dimensions $M,L,T$

Then $L$ and $T$ respectively

By doing this, finally we get a dimensionless relation between the variables.

Calculation:

Write the function with respective dimensions,

$\begin{array}{l}{t}_d=f\left(d,v_0,h_0,\rho ,\mu \right)\\ \left\{T\right\}=\left\{L\right\}\left\{L^3\right\}\left\{L\right\}\left\{M{L}^{-3}\right\}\left\{M{L}^{-1}{T}^{-1}\right\}\end{array}$

First of all,

Eliminate mass $M$

For that, select density $\rho$ and divide it with other variables which contain mass

Therefore, we can rewrite the equation as,

$\begin{array}{l}{t}_d=f\left(d,v_0,h_0,\cancel{\rho },\frac{\mu }{\rho }\right)\\ \left\{T\right\}=\left\{L\right\}\left\{L^3\right\}\left\{L\right\}\left\{M{L}^{-3}\right\}\left\{L^2{T}^{-1}\right\}\end{array}$

Now, discard $\rho$ and eliminate time $T$

Select $\frac{\mu }{\rho }$, to eliminate dimension $T$

Therefore, we can rewrite the equation as,

$\begin{array}{l}{t}_d\frac{\mu }{\rho }=f\left(d,v_0,h_0,\cancel{\frac{\mu }{\rho }}\right)\\ \left\{L^2\right\}=\left\{L\right\}\left\{L^3\right\}\left\{L\right\}\left\{L^2{T}^{-1}\right\}\end{array}$

Now, finally eliminate $L$ by dividing by appropriate powers of $L$

$\begin{array}{l}{t}_d\frac{\mu }{\rho d^2}=f\left(\cancel{d},\frac{v_0}{d^3},\frac{h_0}{d}\right)\\ \left\{1\right\}=\left\{L\right\}\left\{1\right\}\left\{1\right\}\end{array}$

Now discard $d$ and write the final relation,

$t_d\frac{\mu }{\rho d^2}=f\left(\frac{v_0}{d^3},\frac{h_0}{d}\right)$

Conclusion:

The relation between the given variables can be written as,

$t_d\frac{\mu }{\rho d^2}=f\left(\frac{v_0}{d^3},\frac{h_0}{d}\right)$.