The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1 2 r2/R2). Determine the average and maximum velocities in the pipe and the volume flow rate.

For this velocity profile .

Given to us:

Inner radius, r = 2 cm = 0.02 m

Velocity profile,

a.) To find out maximum velocity,

To obtain the maximum velocity, Put the maximum condition of a single variable laminar flow. Therefore,

On differentiating,

Hence, on putting r = 0. The velocity will be maximum at the center of the pipe.

b.) To find out Average velocity,

As we know the average velocity for a fully developed laminar flow in a circular pipe is always half of the maximum velocity, Therefore

c.) To find out volume flow rate,

volume flow rate,

Hence, for this velocity profile .

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The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

increases it

80cm/s.

24cm/s.

53.3cm/s.