A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If the flow is laminar, determine the centerline velocity and the flowrate.

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

answer:

a

explanation:

answer:

no man

explanation:

no i dont want it