Chapter 7, Problem 115P

Sound intensity I is defined as the acoustic power per unit area emanating from a sound source. We know that I is a function of sound pressure level P (dimensions of pressure) and fluid properties $\partial$ (density) and speed of sound c. (a) Use the method of repeating variables in mass-based primary dimension to generate a dimensionless relationship for I as a function of the other parameters. Show all your work. What happens if you choose three repeating variables? Discuss (b) Repeat pall (a). but use the force-based primary dimension system Discuss.

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.

Answer to Problem 115P

Dimensionless quantity of I is

$I=\left(\frac{c^2\rho }{P}\right)$

The dimension of I in mass- based primary function is

$I=\left[M{T}^{-3}\right]$

For three repeating variables

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

Explanation of Solution

Given:

Acoustic power per unit area is called as the sound intensity which is releasing from any sound source. In the given question sound intensity I is represented as a function of P called as pressure level, c called as the speed of sound and the ρ which is a density of fluid properties.

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 Buckinghams 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:

As per the given details, the sound intensity has four parameters which are-

I = sound intensity

P = pressure level

c = speed of sound

ρ = density

This is also represented as

$I=f(P,c,\rho )$

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

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

For density = $\left[\frac{mass}{volume}\right]=\left[\frac{M}{L^3}\right]$

For pressure level = $\left[\frac{force}{area}\right]$

Dimension for force is = $\left[ML{T}^{-2}\right]$

Therefore, P = $\left[\frac{ML{T}^{-2}}{L^2}\right]=\left[M{L}^{-1}{T}^{-2}\right]$

For the primary dimension of intensity, the formula of intensity is

$I=\left[\frac{power}{area}\right]$

Power is given as = $\left[\frac{work}{time}\right]$

And work is given as = $force\times distance$

Therefore, $I=\left[\frac{\frac{work}{time}}{area}\right]$ or $I=\left[\frac{\frac{force\times distance}{time}}{area}\right]$

$I=\left[\frac{\frac{ML{T}^{-2}\times L}{T}}{L^2}\right]$

$I=\left[M{T}^{-3}\right]$

Next case is to discuss about when all three variables are chosen.

The total number of primary dimension is (m,L,t) 3. Then the j =3

According to Buckinghams pi theorem,

Total number of ps = n-j

Where n=4 and j=3

Total number of ps = 4-3 =1

There are only three parameters present at the right side of intensity formula which are P,ρ and c as, j=3. Therefore these three parameters are repeating parameters.

Dimensionless quantity by these three variables is

$\left(\frac{c^2\rho }{P}\right)$

Now a number of ps are calculated by combining given three parameters with other parameters.

Dependent p is calculated by using the I dependent variable.

Therefore,

${\Pi }_1=I{P}^{a1}{\rho }^{b1}{c}^{c1}$............ equ (1)

By using primary dimensions-

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

For, $I{P}^{a1}{\rho }^{b1}{c}^{c1}$ is-

  $\left[I{P}^{a1}{\rho }^{b1}{c}^{c1}\right]=\left[\left(M{T}^{-3}\right){\left(M{L}^{-1}{T}^{-2}\right)}^{a1}{\left(M{L}^{-3}\right)}^{b1}{\left(L^{-1}{T}^{-1}\right)}^{c1}\right]$

From equ(1) and equ (2)

  $\left[M^0{L}^0{T}^0\right]=\left[\left(M{T}^{-3}\right){\left(M{L}^{-1}{T}^{-2}\right)}^{a1}{\left(M{L}^{-3}\right)}^{b1}{\left(L^{-1}{T}^{-1}\right)}^{c1}\right]$

Equating the exponents of both sides,

For mass:

$\left[M^0\right]=\left[M^{a1},M^{b1},M^{c1}\right]$

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

For time:

$\left[T^0\right]=\left[T^{-3},T^{2a1},T^{c1}\right]$

$\begin{array}{l}0=-3-2a_1+c_1\\ c_1=-3-2a_1\end{array}$

For length:

  $\left[L^0\right]=\left[L^{-a1},L^{-3b1},L^{c1}\right]$

$\begin{array}{l}0=-a_1-3b_1+c_1\\ c_1=a_1+3b_1\end{array}$

If the values of a₁ and b₁ are known from the obtained values, then we can solve for other values. Then there is no unique solution. There must be a mistake in choosing in thenumber of repeating variables because for variable a₁ and b₁, there isan infinite number of p₁s.

For this condition reduce the number of j by one.

So, j = 3-1 =2.

According to Buckingham Pis theorem,

$\begin{array}{l}k=n-j\\ k=4-2=2\end{array}$

There are only two repeating variables which are ρ and c.

Dependent p is calculated by using the I dependent variable.

Therefore,

${\Pi }_1=I{\rho }^{a1}{c}^{b1}$............equ (3)

By using primary dimensions,

${\Pi }_1=\left[M^0{L}^0{T}^0\right]$...............equ(4)

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

  $\left[I{\rho }^{a1}{c}^{b1}\right]=\left[\left(M{T}^{-3}\right){\left(M{L}^{-3}\right)}^{a1}{\left(L^{-1}{T}^{-1}\right)}^{b1}\right]$

From equ(3) and equ (4),

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

Equating the exponents of both sides,

For mass:

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

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

For time:

$\left[T^0\right]=\left[T^{-3},T^{-b1}\right]$

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

Putting the values of a₁ and b₁ in equation (3),

  ${\Pi }_1=I{\rho }^{a1}{c}^{b1}$

  ${\Pi }_1=I{\rho }^{-1}{c}^{-3}$

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

Dependent p is calculated by using the Pindependent variable.

Therefore,

${\Pi }_2=P{\rho }^{a_2}{c}^{b_2}$............equ (5)

By using primary dimensions,

${\Pi }_2=\left[M^0{L}^0{T}^0\right]$...............equ(6)

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

  $\left[P{\rho }^{a_2}{c}^{b_2}\right]=\left[\left(M{L}^{-1}{T}^{-2}\right){\left(M{L}^{-3}\right)}^{{}^{a_2}}{\left(L^{-1}{T}^{-1}\right)}^{b_2}\right]$

From equ(5) and equ (6),

  $\left[M^0{L}^0{T}^0\right]=\left[\left(M{L}^{-1}{T}^{-2}\right){\left(M{L}^{-3}\right)}^{{}^{a_2}}{\left(L^{-1}{T}^{-1}\right)}^{b_2}\right]$

Equating the exponents of both sides,

For mass:

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

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

For time:

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

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

Putting the values of a₂ and b₂ in equation (5),

  ${\Pi }_2=P{\rho }^{a_2}{c}^{b_2}$

  ${\Pi }_2=\frac{P}{\rho c^2}$

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

From equation (a) and (b),

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

Conclusion:

Dimensionless quantity of I is

$I=\left(\frac{c^2\rho }{P}\right)$

For three repeating variables,

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

To determine

(b)

The expression for dimensionless relationship of I by using the force-based system of repeating variables.

Answer to Problem 115P

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:

Acoustic power per unit area is called as the sound intensity which is releasing from any sound source. In the given question sound intensity I is represented as a function of P called as pressure level, c called as the speed of sound and the ρ which is a density of fluid properties.

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 Buckinghams 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:

As per the given details, the sound intensity has four parameters which are

I = sound intensity

P = pressure level

c = speed of sound

ρ = density

This is also represented as

$I=f(P,c,\rho )$

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

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

For density = $\left[\frac{mass}{volume}\right]=\left[\frac{M}{L^3}\right]$

$mass=\frac{force}{accerlation}$

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

For density = $\left[F^1{L}^{-4}{T}^2\right]$

For pressure level = $\left[\frac{force}{area}\right]$

Therefore, P = $\left[\frac{F}{L^2}\right]=\left[F{L}^{-2}\right]$

For the primary dimension of intensity, the formula of intensity is

$I=\left[\frac{power}{area}\right]$

Power is given as = $\left[\frac{work}{time}\right]$

And work is given as = $force\times distance$

Therefore, $I=\left[\frac{\frac{work}{time}}{area}\right]$ or $I=\left[\frac{\frac{force\times distance}{time}}{area}\right]$

$I=\left[\frac{\frac{F\times L}{T}}{L^2}\right]$

$I=\left[F{L}^{-1}{T}^{-1}\right]$

Next case is to discuss about when all three variables are chosen,

The total number of primary dimension is (F,L,t) 3. Then j =3.

According to Buckinghams pi theorem,

Total number of ps = n-j

Where, n=4 and j=3

Total number of ps k= 4-3 =1

There are only three parameters present at the right side of intensity formula which are P,ρ and c as, j=3. Therefore these three parameters are repeating parameters.

But if these three variables are chosen as repeating variable, they will form dimensionless quantity therefore reduce the value of j.

For this condition reduce the number of j by one.

So, j = 3-1 =2.

According to Buckingham Pis theorem,

$\begin{array}{l}k=n-j\\ k=4-2=2\end{array}$

There are only two repeating variables which are ρ and c.

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 Pindependent 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,

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