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Chapter:
The fluid of density `rho` and viscosity `mu` flows through a pipe of diameter `d`. Show by Rayleighs method, resistance per unit area is,
`F=rhoV^2 phi(R_e)`
where V is mean velocity of fluid and `R_e` is Reynolds number.
SOLUTION:
The resistance per unit area depends upon:
Density `rho`
Viscosity `mu`
Velocity `V`
Diameter `D`
Mathematically,
`F=f(rho,V,D,mu)`
Or, `F=K*rho^a*V^b*D^c*mu^d`
Where `K` is the dimensionless parameter and `a,b,c,d` are the exponents to be determined.
The equation for dimension is,
`ML^-1T^-2=K*[ML^-3]^a*[LT^-1]^b*[L]^c*[ML^-1 T^-1]^d`...(I)
Now, for dimensional homogeneity, the exponents of each dimension on both sides must be equal.
`1=a+d`..(i)
`-1=-3a+b+c-d`...(ii)
`-2=-b-d`...(iii)
Here we have four unknowns but only three equation. therefore, it is not possible to find the values of variables `a,b,c,d`. However, three of them can be expressed in terms of fourth variable.
Thus,
`a=1-d`
`b=2-d`
and `-3+3d+2-d+c-d=-1`
or,`c=-d`
Thus,
`F=K*rho....
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