The power required by an agitator in a tank is a function of the following four variables: (a) diameter of impeller, (b) number of rotations of the impeller per unit time, (c) viscosity of liquid, (d) density of liquid. From a dimensional analysis, obtain a relation between the power and the four

Question

The power required by an agitator in a tank is a function of the following four variables:
(a) diameter of impeller,
(b) number of rotations of the impeller per unit time,
(c) viscosity of liquid,
(d) density of liquid.
From a dimensional analysis, obtain a relation between the power and the four variables.
The power consumption is found, experimentally, to be proportional to the square of the speed of rotation. By what factor would the power be expected to increase if the impeller diameter were doubled?

Accepted Answer

If the power [latex]P=\phi(D N ho \mu)[/latex], then a typical form of the function is [latex]P=k D^a N^b ho^c \mu^d[/latex], where k is a constant. The dimensions of each parameter in terms of M, L, and T are: power, [latex]P= M L ^2 / T ^3[/latex], density, [latex] ho= M / L ^3[/latex], diameter, D = L, viscosity, [latex]\mu= M / L T[/latex], and speed of rotation, [latex]N= T ^{-1}[/latex]

Equating dimensions:

M : [latex]1=c+d[/latex]
L : [latex]2=a-3 c-d[/latex]
T : [latex]-3=-b-d[/latex]
Solving in terms of d : [latex]a=(5-2 d), b=(3-d), c=(1-d)[/latex]

∴ [latex]P=k\left(\frac{D^5}{D^{2 d}} \frac{N^3}{N^d} \frac{ ho}{ ho^d} \mu^d ight)[/latex]

or: [latex]P / D^5 N^3 ho=k\left(D^2 N ho / \mu ight)^{-d}[/latex]

that is: [latex]N_P=k R e^m[/latex]

Thus the power number is a function of the Reynolds number to the power m. In fact [latex]N_P[/latex] is also a function of the Froude number, DN²/g. The previous equation may be written as:

[latex]P / D^5 N^3 ho=k\left(D^2 N ho / \mu ight)^m[/latex]

Experimentally: [latex]P \propto N^2[/latex]

From the equation, [latex]P \propto N^m N^3[/latex], that is m + 3 = 2 and m = 1

Thus for the same fluid, that is the same viscosity and density:

[latex]\left(P_2 / P_1 ight)\left(D_1^5 N_1^3 / D_2^5 N_2^3 ight)=\left(D_1^2 N_1 / D_2^2 N_2 ight)^{-1} ext { or: }\left(P_2 / P_1 ight)=\left(N_2^2 D_2^3 ight) /\left(N_1^2 D_1^3 ight)[/latex]

In this case, [latex]N_1=N_2[/latex] and [latex]D_2=2 D_1[/latex].

∴ [latex]\left(P_2 / P_1 ight)=8 D_1^3 / D_1^3=\underline{\underline{8}}[/latex]

A similar solution may be obtained using the Recurring Set method as follows:

[latex]P=\phi(D, N, ho, \mu), f(P, D, N, ho, \mu)=0[/latex]

Using M, L and T as fundamentals, there are five variables and three fundamentals and therefore by Buckingham’s theorem, there will be two dimensionless groups.
Choosing D, N and ρ as the recurring set, dimensionally:

[latex]\left.\begin{array}{l}D \equiv L \N \equiv T ^{-1} \ ho \equiv M L ^{-3}\end{array} ight] \quad ext { Thus: }\quad\left[\begin{array}{l} L \equiv D \T \equiv N^{-1} \M \equiv ho L ^3= ho D^3\end{array} ight.[/latex]

First group, [latex]\pi_1[/latex], is [latex]P\left( M L ^2 T ^{-3} ight)^{-1} \equiv P\left( ho D^3 D^2 N^3 ight)^{-1} \equiv \frac{P}{ ho D^5 N^3}[/latex]

Second group, [latex]\pi_2[/latex], is [latex]\mu\left( M L ^{-1} T ^{-1} ight)^{-1} \equiv \mu\left( ho D^3 D^{-1} N ight)^{-1} \equiv \frac{\mu}{ ho D^2 N}[/latex]

Thus: [latex]f \left(\frac{P}{ ho D^5 N^3}, \frac{\mu}{ ho D^2 N} ight)=0[/latex]

Although there is little to be gained by using this method for simple problems, there is considerable advantage when a large number of groups is involved.

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