Physics, 25.03.2020 23:26 Ashtree519

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.

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Answer from: kanga06

For this velocity profile $V_{max}=4\ m/sec,\ \ \ V_{avg}=2\ m/sec,\ and\ \dot Q= 0.00251428\ m^3/sec$ .

Given to us:

Inner radius, r = 2 cm = 0.02 m

Velocity profile,

$u(r)=4(1-\dfrac{r^2}{R^2 } )$

a.) To find out maximum velocity,

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

$\dfrac{du}{dr}=0$

$\dfrac{d4(1-\dfrac{r^2}{R^2 } )}{dr}=0\\\\$

On differentiating,

$\dfrac{d4(1-\dfrac{r^2}{R^2 } )}{dr}=0\\\\4(0-\dfrac{2r}{R^2 } )=0\\\\\\\dfrac{-8r}{R^2}=0$

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

$V_{max}= 4(1-\dfrac{r^2}{R^2 } )\\\\V_{max}= 4(1-\dfrac{0^2}{R^2 } )\\\\V_{max}= 4\ m/sec$

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

$V_{avg}=\dfrac{V_{max}}{2}\\ \\V_{avg}=\dfrac{4}{2}\\\\V_{avg}=2\ m/sec$

c.) To find out volume flow rate $\bold {(\dot Q)}$ ,

volume flow rate,

$\dot Q= A\cdot V_{avg}\\\\\dot Q= (\pi r^2)\cdot V_{avg}\\\\\dot Q= (\pi\times 0.02^2)\cdot 2\\\\\dot Q= 0.00251428\ m^3/sec$

Hence, for this velocity profile $V_{max}=4\ m/sec,\ \ \ V_{avg}=2\ m/sec,\ and\ \dot Q= 0.00251428\ m^3/sec$ .

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Answer from: sankarjgoutham

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

Answer from: Quest

real, smaller and upside-down. so the answer should be b.

Answer from: Quest

d. air pressure decreases as altitude increases.

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