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
