The Reynold number(N) for a liquid flowing through a pipe depends upon diameter(D) of pipe, speed (v), density (d) and coefficient of viscosity (n) of liquid. Derive the expression for Reynold number.
Answers
Reynold number , R_é = rho vd)μ. It's a dimensionless constant showing the ratio between the inertial force and viscous force on a body .
Here, we have to derive an expression for the number.
N ∝ D
N ∝ V
N ∝rho(density)
N ∝ n( coefficient of viscosity)
[ D ] = l
[ V ] = LT^-1
[ n ] = ML^-1T^-1
[ rho ] = ML^-3
N = [D]^a . [ V]^b. [ n ]^c . [ rho]^d
> N = l^a . [ LT^-1]^b . [ ML^-1T^-1]^c. [ML^-3]^d
> N = l^a. l^b . t^-b . m^c . t^-c. m^d . l^-3d
But N = l^0 . m^0 . t^0 ( As reynold's number is a dimensionless constant mentioned above)
> l^a+b-3d = l^0
> m^c+d = m^0
> t^-b-c = t^0
a + b - 3d = 0
c + d = 0
-b -c = 0
>> a + b = 3d ; c + d = 0, b = -c
d = -c (2nd eqn) and b = -c. So b = d
a + d = 3d
>> a = 2d = 2b .
This is the required expression .
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Answer:
☆Answered by Rohith kumar. R.
☆Renoyld number.It is a dimensionless constant showing the ratio between the inertial force and viscous force on the object.
▪N is proportional to D
▪N is proportional to V
▪N is proportional to density
▪N is proportional ton (viscosity)
▪[D]=1
▪[V]=LT^-1
▪[n]=ML^-1T^-1
▪[rho]=ML^-3
▪N=1^a×1^b.t^-b.m^c.t^-c.m^d
▪But we know that
▪N= 1^0.t^0.m^0
=1^a+b-3d=1^0
=m^c+d=m^0
=t^-c-c=t^0
=a+b-3d=0
=c+d=0
=-b-c=0
=a+b=3d,c+d=0,b=-c
=d=-c.and b=-c
▪so we can tell that b=d
▪Therefore a=2d=2b.
Hope it helps u mate.
Thank you.