H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈0.3164Re−1/4 is reasonably accurate for Reynolds numbers between 4000 and 105. Use this information for the following scenario. Water at 20 °C is to flow through a 3‐cm I. D. plastic pipe at the rate of 0.001 m3/s. Find the incline angle of the pipe needed to make the static pressure constant along the pipe

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm  for D

v=  0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine  the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g=  acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle  the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

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