Experiments show that the pressure drop for flow through an orifice plate of diameter d mounted in a length of pipe of diameter D may be expressed as Δp=p1−p2 =f ðrho, μ, V , d, DÞ. You are asked to organize some experimental data. Obtain the resulting dimensionless parameters.

The question is not clear and the complete question says;

Experiments show that the pressure drop for flow through an orifice plate of diameter d mounted in a length of pipe of diameter D may be expressed as Δp = p1 − p2 =f (ρ, μ, V, d, D). You are asked to organize some experimental data. Obtain the resulting dimensionless parameters.

The set of dimensionless parameters is; (Δp•d)/Vµ = Φ((D/d), (ρ•d•V/µ))

Explanation:

First of all, let's write the functional equation that lists all the variables in the question ;

Δp = f(d, D, V, ρ, µ)

Now, since the question said we should express as a suitable set of dimensionless parameters, thus, let's write all these terms using the FLT (Force Length Time) system of units expression.

Thus;

Δp = Force/Area = F/L²

d = Diameter = L

D = Diameter = L

V = Velocity = L/T

ρ = Density = kg/m³ = (F/LT^(-2)) ÷ L³ = FT²/L⁴

µ = viscosity = N.s/m² = FT/L²

From the above, we see that all three basic dimensions F,L & T are required to define the six variables.

Thus, from the Buckingham pi theorem, k - r = 6 - 3 = 3.

Thus, 3 pi terms will be needed.

Let's now try to select 3 repeating variables.

From the derivations we got, it's clear that d, D, V and µ are dimensionally independent since each one contains a basic dimension not included in the others. But in this case, let's pick 3 and I'll pick d, V and µ as the 3 repeating variables.

Thus:

π1 = Δp•d^(a)•V^(b)•µ^(c)

Now, let's put their respective units in FLT system

π1 = F/L²•L^(a)•(L/T)^(b)•(FT/L²)^(c)

For π1 to be dimensionless,

π1 = F^(0)•L^(0)•T^(0)

Thus;

F/L²•L^(a)•(L/T)^(b)•(FT/L²)^(c) = F^(0)•L^(0)•T^(0)

By inspection,

For F,

1 + c = 0 and c= - 1

For L; -2 + a + b - 2c = 0

For T; -b + c = 0 and since c=-1

-b - 1 = 0 ; b= -1

For L, -2 + a - 1 - 2(-1) = 0 ; a=1

So,a = 1 ; b = -1; c = -1

Thus, plugging in these values, we have;

π1 = Δp•d^(1)•V^(-1)•µ^(-1)

π1 = (Δp•d)/Vµ

Let's now repeat the procedure for the second non-repeating variable D2.

π2 = D•d^(a)•V^(b)•µ^(c)

Now, let's put their respective units in FLT system

π1 = L•L^(a)•(L/T)^(b)•(FT/L²)^(c)

For π2 to be dimensionless,

π2 = F^(0)•L^(0)•T^(0)

Thus;

L•L^(a)•(L/T)^(b)•(FT/L²)^(c) = F^(0)•L^(0)•T^(0)

By inspection,

For F;

-2c = 0 and so, c=0

For L;

1 + a + b - 2c = 0

For T;

-b + c = 0

Since c =0 then b =0

For, L;

1 + a + 0 - 0 = 0 so, a = -1

Thus, plugging in these values, we have;

π2 = D•d^(-1)•V^(0)•µ^(0)

π2 = D/d

Let's now repeat the procedure for the third non-repeating variable ρ.

π3 = ρ•d^(a)•V^(b)•µ^(c)

Now, let's put their respective units in FLT system

π3 = F/T²L⁴•L^(a)•(L/T)^(b)•(FT/L²)^(c)

For π4 to be dimensionless,

π3 = F^(0)•L^(0)•T^(0)

Thus;

FT²/L⁴•L^(a)•(L/T)^(b)•(FT/L²)^(c) = F^(0)•L^(0)•T^(0)

By inspection,

For F;

1 + c = 0 and so, c=-1

For L;

-4 + a + b - 2c = 0

For T;

2 - b + c = 0

Since c =-1 then b = 1

For, L;

-4 + a + 1 +2 = 0 ;so, a = 1

Thus, plugging in these values, we have;

π3 = ρ•d^(1)•V^(1)•µ^(-1)

π3 = ρ•d•V/µ

Now, let's express the results of the dimensionless analysis in the form of;

π1 = Φ(π2, π3)

Thus;

(Δp•d)/Vµ = Φ((D/d), (ρ•d•V/µ))

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