Water flows through a pipe system as shown with a velocity p

Water flows through a pipe system as shown with a velocity profile:

u = 19.1(1-r²/R²)

The relative roughness of the pipes, ?/D=0.001. For these conditions find:

a) The volumetric flow rate, Q, through the system.

Now, assuming the mean velocity, V = 9.55 m/s calculate:

b) The friction factor, f, for the pipe using 1 iteration of the Colebrook equation.

c) The total headloss between points A and B,h_L.

d) The total headloss between A and B,h_L, if a second pipe is added as shown in the figure to reduce headloss. Assume the volumetric flow rate stays the same, and the friction factor is constant (same as part b) and equal between the two pipes. In this case, two of the elbows are replaced with tees.

Proposed 100 m 15 cm dia _ u-19.1 (1-r/R2) m/s 10 m, 20 cm dia 10 m, 20 cm dia 100 m, 20 cm dia Elbow Tee K 1.3 0.9

Solution

a)

u = 19.1(1 - r^2 / R^2)

Max velocity Umax will occur at centreline where r = 0. Umax = 19.1*(1 - 0) = 19.1

Mean velocity V = Umax/2 = 19.1 / 2 = 9.55 m/s

Pipe cross section area A = 3.14/4 * 0.2^2 = 0.0314 m^2

Flow rate Q = A*V = 9.55*0.0314 = 0.3 m^3 /s

b)

For water, kinematic viscosity v = 1.004*10^-6 m^2/s

Re = V*D/v

= 9.55*0.2 / (1.004*10^-6)

Re = 1,902,390

Colebrook equation: f^(-0.5) = -2 log [2.51 / (Re*f^0.5) + e/3.72]

Assuming f = 0.02 we get

Left Hand side = 0.02^(-0.5) = 7.071

Right Hand side = -2 log [2.51 / (1902390*0.02^0.5) + 0.001/3.72] = 10.06

LHS is not close to RHS.

In next iteration, assuming f = 0.01 we get

Left Hand side = 0.01^(-0.5) = 10

Right Hand side = -2 log [2.51 / (1902390*0.01^0.5) + 0.001/3.72] = 9.76

Now, LHS is close to RHS and hence f = 0.01

c)

Head loss = (fL/D + K) * V^2 / (2g)

= [0.01*(10 + 100 + 10) / 0.2 + 1.3 + 1.3 + 1.3 + 1.3]* (9.55^2) / (2*9.81)

= 52.06 m

d)

Head loss between A and B via long route = [0.01*(10 + 100 + 10) / 0.2 + 1.3 + 1.3 + 0.9 + 0.9]* (V_long^2) / (2*9.81)

= 10.4*V_long^2 / (2*9.81)

Head loss between A and B via short route = [0.01*100 / 0.2 + 0.9 + 0.9]* (V_short^2) / (2*9.81)

= 6.8*V_short^2 / (2*9.81)

Equating both the expressions,

10.4*V_long^2 / (2*9.81) = 6.8*V_short^2 / (2*9.81)

Vshort / Vlong = 1.2367

By mass conservation, A*V = A*Vshort + A*Vlong

9.55 = Vshort + Vlong

9.55 = 1.2367*Vlong + Vlong

Vlong = 4.27 m/s

Vshort = 5.28 m/s

Head loss = = 6.8*5.28^2 / (2*9.81) = 9.66 m