Respuesta :
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.
Answer:
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
For this velocity profile [tex]V_{max}=4\ m/sec,\ \ \ V_{avg}=2\ m/sec,\ and\ \dot Q= 0.00251428\ m^3/sec[/tex].
Given to us:
Inner radius, r = 2 cm = 0.02 m
Velocity profile,
[tex]u(r)=4(1-\dfrac{r^2}{R^2 } )[/tex]
a.) To find out maximum velocity,
To obtain the maximum velocity, Put the maximum condition of a single variable laminar flow. Therefore,
[tex]\dfrac{du}{dr}=0[/tex]
[tex]\dfrac{d4(1-\dfrac{r^2}{R^2 } )}{dr}=0\\\\[/tex]
On differentiating,
[tex]\dfrac{d4(1-\dfrac{r^2}{R^2 } )}{dr}=0\\\\4(0-\dfrac{2r}{R^2 } )=0\\\\\\\dfrac{-8r}{R^2}=0[/tex]
Hence, on putting r = 0. The velocity will be maximum at the center of the pipe.
[tex]V_{max}= 4(1-\dfrac{r^2}{R^2 } )\\\\V_{max}= 4(1-\dfrac{0^2}{R^2 } )\\\\V_{max}= 4\ m/sec[/tex]
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
[tex]V_{avg}=\dfrac{V_{max}}{2}\\ \\V_{avg}=\dfrac{4}{2}\\\\V_{avg}=2\ m/sec[/tex]
c.) To find out volume flow rate[tex]\bold {(\dot Q)}[/tex],
volume flow rate,
[tex]\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[/tex]
Hence, for this velocity profile [tex]V_{max}=4\ m/sec,\ \ \ V_{avg}=2\ m/sec,\ and\ \dot Q= 0.00251428\ m^3/sec[/tex].
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