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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answered: 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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answered: 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

answered: Guest

there are no correct statementd on the list of choices you provided.

answered: Guest

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)

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The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in...