Shear Stress Distribution in Hollow Circular Section Formula

Shear Stress distribution in a cylindrical tube - Suppose that there is a viscous fluid steady flow in a horizontal tube. Whose density is constant and a disk of r radius and dL length is concentric with the axis of the speed element tube. 

Let the fluid pressure on the upstream and downstream faces of the disk be P and (P + dL), doing work. The flowing fluid is a viscous fluid, a shea force (τ) will be generated on the rim of the disc in the direction opposite to the flow of the fluid. 

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Form Principle of Torque 

∑F = Pa.Sa - Pb.Sb - Fs - Fg ..........(i)

Where, 

Pa and Pb = Inlet and outlet pressure 

Sa and Sb = Inlet and outlet cross-section

Fs = Shear force

Fg = Gravitational force

Here, Pa = P 

Pb = (P + dL) 

Sa and Sb = πr²

Pa. Sa = Pπr² 

Pb. Sb = (P + dP). πr² = Pπr² + dPπr² 

Fg = 0, Fs = 2πr.d.L.τ 

Putting these all values in equation (i). 

∑F = Pπr² - Pπr² - dPπr² - 0 - 2πr.d.L.τ 

After solving this equation, We have 

∑F = 0 

0 = - dPπr² - 2πr.d.L.τ 

dP/dL = - 2πr.d.L.τ/πr² 

dP/dL = - 2τ / r 

dP/dL + 2τ / r = 0 

Pressure remains constant at any cross-section of a tube for all fow in steady laminar or turbulent.

dP/dL, r  is independent of 

Take us, τ = τw and r = rw 

Where τw = Shear stress at the wall 

rw = radius of the tube 

dP/dL + 2τw/rw = 0 

The upper equation is shear Stress Distribution in a cylindrical tube flow Calculation formula or equation. 

Velocity Distribution for Newtonian Fluids 

According to Newton's law viscosity equation, there is the following relation between shear stress and velocity gradient. 

μ = τ / (du/dr) ..........(i) 

sing in the above equation is because in a pipe μ (velocity) decreases with an increase in r ( radius) with an increase in r there is a decrease in μ. 

du/dr = - τ/μ ..........(ii) 

We know that

τw/rw = τ/μ 

Putting this value in equation (ii) 

du/dr = - τw/(rw.μ).r 

du = - τw/(rw.μ).rdr .......(iii) 

On integration 

∫limt 0 to u du = τw/(rw.μ) ∫limt rw to r rdr 

μ = τw/(2rw.μ) [r²w - r²] ........(iv) 

The maximum value of the local velocity (μ max)is located at the center of the pipe.

At the center of the pipe, r = 0 

μmax = τw.rw/2μ .........(v) 

Dividing from equations (iv) and (v)

μ/μmax = 1- (r/rw)² 

It is a parabolic equation it is shows that viscosity distribution with respect for is a parabola.

The formula for Shear Stress

Shear stress is a type of stress that occurs when a force is applied parallel to the cross-sectional area of ​​an object. It is defined as the force per unit area required to deform a material in a direction parallel to it.

τ = F / A

Where

• τ represents shear stress, measured in Pascals (Pa)
• F represents the applied force, measured in Newtons (N)
• A represents the cross-sectional area, measured in square meters (m^2)

The formula states that shear stress is equal to the applied force divided by the cross-sectional area.

The formula for this assumes that the force is being applied parallel to the surface of the object. If the force is applied at an angle, the formula for shear stress becomes:

τ = (F sin θ) / A

Where:

• θ represents the angle between the applied force and the object's surface.

In this case, the formula takes into account the fact that only the component of the force parallel to the surface of the object contributes to the shear stress.