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Reynold number $\left( R \right)$ determines whether or not the flow of a liquid through a pipe is streamlined. This constant is a combination of velocity $\left( v \right)$, density $\left( \rho \right)$, coefficient of viscosity $\left( n \right)$ and diameter $D$ of the pipe. Deriving the formula for $R$ using the method of dimensions, write the power of $D$ , on which the Reynold number depends.

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Last updated date: 20th Jun 2024
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Answer
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Hint The Reynolds number can be said as the ratio of the inertial force to that of the viscous force in a fluid. It can be derived from the given parameters by writing its dimensions and equating it with the unit less Reynolds number and simplifying it.

Complete step by step solution
It is given that the Reynold number depends on the velocity, density, coefficient of viscosity and the diameter. Hence this relation is also written as
$R = K{v^a}{\rho ^b}{\eta ^c}{D^1}$
The constant $K$ is added because there may be some change in the value even though it is derived from the given relation. Let us write the dimension of the above parameters to derive the formula. We know that the dimension of the velocity is $m{s^{ - 1}}$ , the dimension of the density is $Kg{m^{ - 3}}$ , coefficient of viscosity is $sc{m^{ - 1}}$ and for the diameter is $m$.
${M^0}{L^0}{T^0}$= ${\left[ {L{T^{ - 1}}} \right]^a}{\left[ {M{L^{ - 3}}} \right]^b}{\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^c}{L^1}$
In the above step, the Reynold number is substituted with no unit.
${M^0}{L^0}{T^0}$= ${M^{b + c}}{L^{a - 3b - c + 1}}{T^{a - c}}$
By equating the corresponding dimensions and the values. First let us equate the values of the mass.
$b + c = 0$
$b = - c$ --------------(1)
Then the parameters of length are considered.
$a - 3b - c + 1 = 0$
Substitute the value of the (1) in the above equation,
$a + 3c - c = - 1$
$a + 2c = - 1$ ---------------(2)
Then the parameter of time is considered.
$ - a - c = 0$
$a = - c$
Substituting this value in the equation (2),
$ - c + 2c = - 1$
$c = - 1$
$a = b = 1$
Substituting these in the dimension equation, we get
$R = K{v^1}{\rho ^1}{\eta ^{ - 1}}{D^1}$
Hence the above equation may also written as
$R = K\dfrac{{v\rho D}}{\eta }$
The above formula holds for the Reynolds number.

Note The Reynolds number determines the characteristics of the fluid that either it is streamlined or not. If the Reynolds number is less than $2000$ , it is laminar flow and if the Reynolds number is between $2000$ to $4000$ , it is turbulent flow.