Saturday, August 14, 2021

A smooth pipe of diameter 400mm and length 800m carries water at the rate of 0.04m3/s0.04m3/s. Determine the head lost due to friction, wall shear stress, center-line velocity and thickness of laminar sub-layer. Take the kinematics viscosity of water as 0.018 stokes.

 A smooth pipe of diameter 400mm and length 800m carries water at the rate of 

0.04m3/s. Determine the head lost due to friction, wall shear stress, center-line velocity and thickness of laminar sub-layer. Take the kinematics viscosity of water as 0.018 stokes.


Solution:-

 Given:-

Diameter of pipe, D=400mm=0.40m

Radius, R=D2=0.42=0.2m

Length of pipe, L=800m

Discharge, Q=0.04m3/s

Kinematics viscosity, v=0.018stokes

v=0.018cm2/s=0.018×104m2/s

Average velocity, V¯=QArea=0.04π4(0.4)2=0.318m/s


Reynolds number

Re=V×Dv

Re=V¯×Dv

Re=0.3183×0.40.018×104=7.073×104

The flow is turbulent.


Step 1: The coefficient viscosity of friction ‘f’ is obtained from

f=0.0791(Re)1/4

f=0.0791(7.073×104)1/4

f=0.0791(16.30)=0.00485


Step 2: Head lost due to friction,

hf=4fLV2D×2g

hf=4fLV¯2D×2g

hf=4×0.00485×800×(0.3183)20.40×2×9.81=0.2m


Step 3: Wall shear stress (τ0)

τ0=fρV22

τ0=fρV¯22

τ0=0.00485×1000×(0.3184)22

τ0=0.245Nm2


Step 4: Center-line viscosity (umax)for smooth pipe,

umaxu=5.75log10uRv+5.55

[put u=umax at y=R]

u=τ0ρ

u=0.2451000=0.0156m/s


Substituting the value of u, R and v in the above equation,

umax0.0156=5.75log100.0156×0.200.018×104+5.55

umax0.0156=24.173

umax=0.0156×24.173=0.377m/s

Step 5: Thickness of laminar sublayer (δ)

δ=11.6×vu

δ=11.6×0.018×1040.0156=0.001338m

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