Answer:
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'
