Derivation of Reynolds number is based on various definitions which are explained below.
What is Reynolds number?
Reynolds number which referred to as R_{e} is a dimensionless quantity which is used in fluid mechanics to predict the flow patterns in different fluid flow conditions. It is defined as the ratio of inertial force to viscous force within the fluid. The mathematical representation is given as:
\(R_{e}=\frac{\rho uL}{\mu }=\frac{uL}{\nu }\)Where,
⍴: density of the fluid in kg.m^{-3}
𝜇: dynamic viscosity of the fluid in N.s.m^{-2}
𝜈: kinematic viscosity of the fluid in m^{2}.s^{-1}
u: velocity of fluid with respect to the object in m.s^{-1}
L: characteristic linear dimension in m
Using Reynolds number it becomes easy to determine whether the fluid flow is laminar or turbulent. Following is the boundary values of a circular pipe that can be used to determine the flow pattern:
- If Re > 2300, then the flow is said to be laminar.
- If 2300 < Re < 4000, then the flow is said to be transient.
- If Re > 4000, then the flow is said to be turbulent.
Reynolds number derivation
- The mathematical form of the Reynolds number can be derived as follows:
R_{e}=inertial force/viscous force
\(\frac{ma}{\tau A}\) \(\frac{\rho V.\frac{du}{dt}}{\mu \frac{du}{dy}.A}\propto \frac{\rho L^{3}\frac{du}{dt}}{\mu \frac{du}{dy}L^{2}}\) \(\frac{\rho L.\frac{dy}{dt}}{\mu}\propto \frac{\rho u_{0}L}{\mu }\) \(\frac{u_{0}L}{\nu }\)Where,
t: time
y: cross-sectional position
\(u=\frac{dx}{dt}\) flow speedτ: shear stress in Pa
A: cross-sectional area of the flow
V: volume of the fluid element
L: characteristic linear dimension in m
u_{0}: maximum speed of the object relative to the fluid in m.s^{-1}
⍴: density of the fluid in kg.m^{-3}
𝜇: dynamic viscosity of the fluid in N.s.m^{-2}
𝜈: kinematic viscosity of the fluid in m^{2}.s^{-1}
- Following is the Reynolds number representation for the flow in a pipe: \(R_{e}=\frac{\rho uD_{H}}{\mu }\) \(=\frac{uD_{H}}{\nu }\) \(=\frac{QD_{H}}{\nu A }\)
Where,
D_{H}: hydraulic diameter of the pipe in m
Q: volumetric flow rate in m^{3}.s^{-1}
A: pipe’s cross-sectional area in m^{2}
u: mean velocity of the fluid in m.s^{-1}
⍴: density of the fluid in kg.m^{-3}
𝜇: dynamic viscosity of the fluid in N.s.m^{-2}
𝜈: kinematic viscosity of the fluid in m^{2}.s^{-1}
Critical Reynolds number and hydraulic diameter
Critical Reynold number of any fluid flow is defined when the flow pattern changes from laminar to turbulent. It is referred as Re_{cr}. The critical Reynold number is not fixed as it varies from geometry to geometry and also is dependent on flow conditions.
For a circular pipe, the critical Reynolds number is given as Re_{cr}=2300 and for non-circular pipes it depends on the hydraulic diameter D_{h} which is defined as
\(D_{h}=\frac{4A_{c}}{P}\)Where,
A_{c}: cross-sectional area of the pipe
P: wetted perimeter of the pipe
Hydraulic radius is defined as the cross-sectional area of the channel divided by the wetted perimeter and referred as R_{H}.
\(R_{H}=\frac{A_{c}}{P}\)Following is the table of hydraulic diameter of non-circular tubes and channels:
Geometry |
Hydraulic diameter |
Circular tube |
\(D_{H}=\frac{4.\frac{\pi D^{2}}{4}}{\pi D}=D\) |
Annulus |
\(D_{H}=\frac{4.\frac{\pi (D_{out}^{2}-D_{in}^{2})}{4}}{\pi (D_{out}+D_{in})}=D_{out}-D_{in}\) |
Square duct |
\(D_{H}=\frac{4a^{2}}{4a}=a\) |
Rectangular duct |
\(D_{H}=\frac{4ab}{2(a+b)}=\frac{2ab}{a+b}\) |
Partially filled rectangular duct |
\(D_{H}=\frac{4ab}{2a+b}\) |
Reynolds number examples
Example 1: A Newtonian fluid with a dynamic viscosity of 0.40n.s.m^{-2} and a specific gravity of 0.93 flows through a 28mm diameter pipe with a velocity of 2.9m.s^{-1}.
Answer: The density can be calculated using the specific gravity like
⍴=0.93(1000kg.m^{-3})
=930kg.m^{-3}
The Reynolds number can be calculated using the formula as
\(R_{e}=\frac{\rho uL}{\mu }=\frac{uL}{\nu }\)R_{e}=(930kg.m^{-3}) (2.9m.s^{-1}) (28mm) (10^{-3}m.mm^{-1})/(0.38Ns.m^{-2})
= 198 (kg.m.s^{-2})/N
Since R_{e}=198 the flow is laminar.
Example 2: A person turns on the water tap in his kitchen. Water flows out from a copper pipe with 7.00mm diameter at a velocity of 1.00m.s^{-1}. The density of water in the pipe is 1000kg.m^{-3} and viscosity 0.00135 Pa.s^{-1}. What is the Reynolds number and what is the type of flow in the pipe?
Answer: The diameter of the pipe=7.00mm=0.007m
Using Reynolds number formula:
\(R_{e}=\frac{\rho uL}{\mu }=\frac{uL}{\nu }\)R_{e}=(1000kg.m^{-3}) (1.00m.s^{-1}) (0.007m)/0.00135 Pa.s^{-1}
=(1000kg.m^{-3}) (1.00m.s^{-1}) (0.007m)/0.00135 kg.m^{-1}.s^{-1}
=5185
Since R_{e}=5185 the flow is turbulent.
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