Saturday, August 14, 2021

A smooth pipe line of 100 mm diameter carries 2.27m32.27m3 per minute of water at 20∘C20∘C with kinematic viscosity of 0.0098 Stokes. Calculate the friction factor maximum velocity as well as shear stress at the boundary.

  A smooth pipe line of 100 mm diameter carries 2.27m32.27m3 per minute of water at 

20C with kinematic viscosity of 0.0098 Stokes. Calculate the friction factor maximum velocity as well as shear stress at the boundary.


Solution:- 

Given:-

Diameter of pipe, D=100mm=0.1m

Radius, R=0.12=0.05m

Discharge, Q=2.27m3/min=2.2760=0.0378m3/s

Kinematic viscosity, v=0.0098stokes

v=0.0098×104m2/s


Therefore Average velocity,

U¯=QA=0.0378π4(0.1)2=4.817m/s

Therefore Reynolds number is given by,

Re=U¯×Dv

Re=4.817×0.10.0098×104=4.9154×105

The flow is turbulent and Re is more than 105.

Hence for smooth pipe, the coefficient of friction ‘f’ is obtained by,

14=2.0log10(Re4f)0.8

14f=2.0log10(4.9154×1054f)0.8

14f=2.0[log10(4.9154×105)+log104f]0.8

14f=2.0[5.6915+log104f]0.8

14f=2×5.6915+2log104f0.8

14f=11.3830+log10(4f)20.8

14f=11.383+log10(4f)0.8

14flog10(4f)=11.3830.8=10.583.(1)


Step 1: Friction factor

(f*)=4× coefficient of friction=4f

substitute the value in equation (1),

1flog10(f)=10.583(2)


The above equation is solved by hit and trial method,

Therefore, let f*=0.1, then LHS of equation (2), becomes

LHS=10.1log10(0.1)=3.16(1.0)=4.16

Let f*=0.01, then LHS becomes,

LHS=10.01log10(0.01)=10(2)=12


But for exact solution, LHS should be (10.583). Hence value of f* lies between 0.1 and 0.01.

Let f*=0.013, then LHS becomes,

LHS=10.013log10(0.013)=8.77(1.886)=10.656

which is approximately (10.583).

Hence the value of (f*=0.013).

Therefore friction factor, f*=0.013.


Step 2: Maximum velocity (umax)

We know, f*=4f

f=f4

f=0.0134=0.00325

Now,

u=U¯f2

u=4.817×0.003252

u=4.817×0.04303=0.194

For smooth,

umaxu=5.75log10uyv+5.55


The velocity will be maximum at the centre of pipe.

Where, y=R=0.05,

umaxu=5.75log10uRv+5.55

umax0.194=5.75log100.194×0.050.0098×104+5.55

umax=0.194[22.974+5.55]=5.528m/s


Step 3: Shear stress at boundary (τ0)

u=τ0ρ

τ0=ρu2

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