Chapter 7, Problem 116P

Repeal Prob. 7-112, but with the distance r from the sound source as an additional independent parameter.

To determine

(a)

A dimensionless relationship for I as a function of the other parameters by using the method of repeating variables in mass-based primary dimensions by using distance r from the sound source as an additional independent parameter.

Answer to Problem 116P

Dimensionless relationship between sound intensity and remaining parameter.

  $\frac{I}{\rho c^3}=f(\frac{I}{\rho c^2})$

Explanation of Solution

Given Information:

  $\begin{array}{l}Soundpressure=p\hfill \\ Density=\rho \hfill \\ Soundspeed=c\hfill \\ Soundintencity=I\hfill \\ Distance=r\hfill \end{array}$

Concept used:

The mass-based primary dimension will be used in this question. In this system all the possible variables are replaced by the mass. Mass, time length dimensions are represented as, [M],[L] and [t].

Concept of Buckingham's Pi method will also be used. It is represented as-

  $k=n-j$

Where,

n= number of physical variables

k =independent physical quantities

n = total number of variable parameters

Calculation:

  $I=F(p,c,\rho ,r)$

Primary dimensions of each parameter,

  $\begin{array}{l}\text{velocity of}\text{sound }=c=\left\{\frac{L}{t}\right\}=\left\{L^1{t}^{-1}\right\}\hfill \\ Density=\rho =\left\{\frac{mass}{volume}\right\}=\left\{\frac{m}{L^3}\right\}=\left\{m^1{L}^{-3}\right\}\hfill \\ \begin{array}{l}pressure=p=\left\{\frac{force}{area}\right\}=\left\{\frac{mL{t}^{-2}}{L^2}\right\}=\left\{m^1{L}^{-1}{t}^{-2}\right\}\\ Distance=r=\left\{L\right\}\\ Soundintencity=I=\left\{\frac{power}{area}\right\}=\left\{\frac{m{L}^2{t}^{-3}}{L^2}\right\}=\left\{m{t}^{-3}\right\}\\ \end{array}\hfill \\ \hfill \\ \hfill \end{array}$

The total mass of parameters, n = 5

No. Of primary dimensions, j = 3

Expected no of $\pi$ term, k = n-j =2

Dependant $\pi$ by combining repeating parameter with remaining parameters.

  ${(\pi )}_1=(I{r}^{a1}{\rho }^{b1}{c}^{c1})$

Rewriting,

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

Mass,

  $\begin{array}{l}\left\{M^0\right\}=\left\{M^1{M}^{b1}\right\}\\ 0=1+b_1\\ b_1=-1\end{array}$

Time,

  $\begin{array}{l}\left\{T^0\right\}=\left\{T^{-3}{T}^{-c_1}\right\}\\ 0=-3-c_1\\ c_1=-3\end{array}$

Length,

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

Putting value in $\pi$ the term.

  $\begin{array}{l}{\pi }_1=I{r}^0{\rho }^{-1}{c}^{-3}\\ {\pi }_1=\frac{I}{\rho c^3}\end{array}$

Dependant Pi using independent variable P.

  ${\pi }_2=\mathrm{P}{r}^{a_2}{\rho }^{b_2}{c}^{c_2}$

  $\left\{M^0{L}^0{T}^0\right\}=\{(M{L}^{-1}{T}^{-2}){(L)}^{a2}{(M{L}^{-3})}^{b2}{(L^1{T}^{-1})}^{c_2}\}$

Mass,

  $\begin{array}{l}\left\{M^0\right\}=\left\{M^1{M}^{b_2}\right\}\\ 0=1+b_2\\ b_2=-1\end{array}$

Time,

  $\begin{array}{l}\left\{T^0\right\}=\left\{T^{-2}{T}^{-c_2}\right\}\\ 0=-2-c_2\\ c_2=-2\end{array}$

Length,

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

Putting values in $\pi$ the term,

  $\begin{array}{l}{\pi }_2={\mathrm{Pr}}^0{\rho }^{-1}{c}^{-2}\\ =\frac{P}{\rho c^2}\end{array}$

Thus, from the equation of ${\pi }_1$ and ${\pi }_2$

  $\frac{I}{\rho c^3}=(\frac{P}{\rho c^2})f(\frac{I}{\rho c^2})$

Conclusion:

In this way, we are able to produce a dimensionless relationship for sound intensity using an independent parameter.

  $\frac{I}{\rho c^3}=f(\frac{I}{\rho c^2})$

To determine

(b)

The expression for dimensionless relationship of I by using the force-based system of repeating variables by using distance r from the sound source as an additional independent parameter.

Answer to Problem 116P

Intensity by using force-based primary dimension system$I=\left[F{L}^{-1}{T}^{-1}\right]$

For three repeating variables,

  $\frac{1}{\rho c^3}=f\left(\frac{1}{\rho c^2}\right)$

Explanation of Solution

Given:

  $\begin{array}{l}\begin{array}{l}Soundpressure=p\hfill \\ Density=\rho \hfill \\ Soundspeed=c\hfill \\ Soundintencity=I\hfill \\ Distance=r\hfill \end{array}\\ Force=f\end{array}$.

Concept Used:

The force-based primary dimension will be used in this question. In this system all the possible variables are replaced by the mass. Force, time, length dimensions are represented as, [F], [L] and [t].

Concept of Buckingham's Pi method will also be used. It is represented as,

  $k=n-j$

Where,

n= number of physical variables

k =independent physical quantities

n = total number of variable parameters

Calculation:

  $I=f(P,c,\rho )$

Now, the primary dimensions of all parameters are given below:

speed of sound = $c=\left[\frac{L}{T}\right]$

density = $\left[\frac{\frac{force}{accerlation}}{volume}\right]=\left[\frac{\frac{F}{\frac{L}{T^2}}}{L^3}\right]=\left[\frac{F{L}^{-1}{T}^2}{L^3}\right]$ = $\left[F^1{L}^{-4}{T}^2\right]$

pressure level = $\left[\frac{force}{area}\right]$ = $\left[\frac{F}{L^2}\right]=\left[F{L}^{-2}\right]$

Sound Intensity, $I=\left[\frac{power}{area}\right]=\left[\frac{\frac{F\times L}{T}}{L^2}\right]=\left[F{L}^{-1}{T}^{-1}\right]$

The total mass of parameters, n = 5

No. Of primary dimensions, j = 3

Expected no of $\pi$ term, k = n-j =2

Dependent p is calculated by using the I dependent variable.

Therefore,

  ${\Pi }_1=I{\rho }^{a1}{c}^{b1}$............equ (1)

By using primary dimensions,

  ${\Pi }_1=\left[F^0{L}^0{T}^0\right]$...............equ(2)

For, $I{\rho }^{a1}{c}^{b1}$ is-

  $\left[I{\rho }^{a1}{c}^{b1}\right]=\left[\left(F^1{L}^{-1}{T}^{-1}\right){\left(F^1{L}^{-4}{T}^2\right)}^{a1}{\left(L^{-1}{T}^{-1}\right)}^{b1}\right]$

From equ(1) and equ (2),

  $\left[F^0{L}^0{T}^0\right]=\left[\left(F^1{L}^{-1}{T}^{-1}\right){\left(F^1{L}^{-4}{T}^2\right)}^{a1}{\left(L^{-1}{T}^{-1}\right)}^{b1}\right]$

Equating the exponents of both sides,

For force:

  $\left[F^0\right]=\left[F^1,F^{a1}\right]$

  $\begin{array}{l}0=1+a_1\\ a_1=-1\end{array}$

For time:

  $\left[T^0\right]=\left[T^{-1},T^{2a1}{T}^{-b1}\right]$

  $\begin{array}{l}0=-1+2a_1-b_1\\ 0=-1-2-b_1\\ b_1=-3\end{array}$

Putting the values of a₁ and b₁ in equation (1),

  ${\Pi }_1=I{\rho }^{-1}{c}^{-3}$

  ${\Pi }_1=\frac{1}{\rho c^3}$..................... (a)

Dependent p is calculated by using P independent variable.

Therefore,

  ${\Pi }_2=P{\rho }^{a_2}{c}^{b_2}$............equ (3)

By using primary dimensions,

  ${\Pi }_2=\left[F^0{L}^0{T}^0\right]$...............equ(4)

For, $P{\rho }^{a_2}{c}^{b_2}$ is

  $\left[P{\rho }^{a_2}{c}^{b_2}\right]=\left[\left(F{L}^{-2}\right){\left(F^1{L}^{-4}T\right)}^{{}^{a_2}}{\left(L^{-1}{T}^{-1}\right)}^{b_2}\right]$

From equ(3) and equ (4),

  $\left[F^0{L}^0{T}^0\right]=\left[\left(F{L}^{-2}\right){\left(F^1{L}^{-4}T\right)}^{{}^{a_2}}{\left(L^{-1}{T}^{-1}\right)}^{b_2}\right]$

Equating the exponents of both sides,

For mass:

  $\left[F^0\right]=\left[F^1,F^{a_2}\right]$

  $\begin{array}{l}0=1+a_2\\ a_2=-1\end{array}$

For time:

  $\left[T^0\right]=\left[T^{-2a_2},T^{-b_2}\right]$

  $\begin{array}{l}0=-2a_2-b_2\\ b_2=-2\end{array}$

Putting the values of a₂ and b₂ in equation (3),

  ${\Pi }_2=P{\rho }^{-1}{c}^{-2}$

  ${\Pi }_2=\frac{P}{\rho c^2}$..................... (b)

From equation (a) and (b),

  $\frac{1}{\rho c^3}=f\left(\frac{1}{\rho c^2}\right)$

Conclusion:

Intensity by using force-based primary dimension system.

  $I=\left[F{L}^{-1}{T}^{-1}\right]$

For three repeating variables,

  $\frac{1}{\rho c^3}=f\left(\frac{1}{\rho c^2}\right)$