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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 exp
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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.
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2021-08-12T21:16:39+00:00
2021-08-12T21:16:39+00:00 1 Answers
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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.
Answer:
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))