Answer:
A) centerline velocity = 1.894 m/s
B) flow rate = 7.44 x 10^(-3) m³/s
Explanation:
A) The flow velocity intensity for the input radial coordinate "r" is given by;
U(r) = (Δp•D²/16μL) [1 - (2r/D)²]
Velocity at the centre of the tube can be expressed as;
V_c = (Δp•D²/16μL)
Thus,
U(r) = (V_c)[1 - (2r/D)²]
From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m
But we are to find the velocity at the centre of the tube, thus;
We will use the radius across the horizontal distance which will be;
0.05 - 0.012 = 0.038m
Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.
Thus;
U(r) = (V_c)[1 - (2r/D)²]
0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]
0.8 = (V_c)[1 - (0.76)²]
V_c = 0.8/0.4224 = 1.894 m/s
B) flow rate is given by;
ΔV = Average Velocity x Area
Now, average velocity = V_c/2
Thus, average velocity = 1.894/2 = 0.947 m/s
Area(A) = πr² = π x 0.05² = 0.007854 m²
So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s
