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Fluid mechanicsbuckingham pi theoremQ5. In ~ a sudden contraction in a pipeline the diameter alters from D1 to D2. The pressure drop, Δp , which


Fluid mechanicsbuckingham pi theoremQ5. In ~ a sudden contraction in a pipeline the diameter transforms from D1 come D2. The push drop, Δp , i beg your pardon develops throughout the contraction is a function of D1 and D2, and the velocity, V, in the larger pipe, and the liquid density, rho, and viscosity, µ. Use D1, V, and also µ together repeating variables to determine a suitable collection of dimensionless parameters. Why would certainly it it is in incorrect to encompass the velocity in the smaller pipe as secondary variable?Thank you!!!




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Answer:

A) collection of dimensionless parameters is; (Δp•D1)/Vµ = Φ((D2/D1), (ρ•D1•V/µ))

B) Yes, it will be untrue to encompass the velocity in the smaller sized pipe as second variable.

Explanation:

First of all, let’s create the sensible equation the lists every the variables in the concern ;

Δp = f(D1, D2, V, ρ, µ)

Now, since the inquiry said we have to express as a suitable set of dimensionless parameters, thus, let’s create all this terms using the FLT (Force length Time) device of systems expression.

Thus;

Δp = Force/Area = F/L²

D1 = Diameter = L

D2 = Diameter = L

V = Velocity = L/T

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

µ = viscosity = N.s/m² = FT/L²

From the above, we watch that all three simple dimensions F,L & T are compelled to specify the six variables.

Thus, native the Buckingham pi theorem, k – r = 6 – 3 = 3.

Thus, 3 pi terms will be needed.

Let’s now shot to choose 3 repeating variables.

From the derivations us got, it’s clear that D1, D2, V and µ space dimensionally independent due to the fact that each one consists of a basic dimension not consisted of in the others. Yet in this case, let’s choose 3 and also I’ll choose D1, V and also µ together the 3 repeating variables.

Thus:

π1 = Δp•D1^(a)•V^(b)•µ^(c)

Now, let’s placed their particular units in FLT device

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 also c= – 1

For L; -2 + a + b – 2c = 0

For T; -b + c = 0 and also 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•D1^(1)•V^(-1)•µ^(-1)

π1 = (Δp•D1)/Vµ

Let’s now repeat the procedure because that the second non-repeating change D2.

π2 = D2•D1^(a)•V^(b)•µ^(c)

Now, let’s placed their corresponding 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 also 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 this values, we have;

π2 = D2•D1^(-1)•V^(0)•µ^(0)

π2 = D2/D1

Let’s currently repeat the procedure because that the 3rd non-repeating variable ρ.

π3 = ρ•D1^(a)•V^(b)•µ^(c)

Now, let’s put their corresponding units in FLT device

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 also 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 this values, us have;

π3 = ρ•D1^(1)•V^(1)•µ^(-1)

π3 = ρ•D1•V/µ

Now, let’s refer the outcomes of the dimensionless analysis in the type of;

π1 = Φ(π2, π3)

Thus;

(Δp•D1)/Vµ = Φ((D2/D1), (ρ•D1•V/µ))

B) In continuous equation,

Q1 = Q2

Thus,

ρV1A1 = ρV2A2

And so,

V1A1 = V2A2

V2 = (V1A1)/A2

Since area = πD²/4

Thus,V2 = (V1D1)/D2

Now, V2 is the velocity of the smaller and is dependency on V1, D1 and also D2.

Thus, it will certainly be not correct to incorporate the velocity in the smaller pipe as secondary variable.