Question

Water to a residential area is transported at a rate of 1.5 m³/s via 70-cm-internal-diameter concrete pipes with a surface roughness of 3 mm and a total length of 1520 m. In order to reduce pumping power requirements, it is proposed to line the interior surfaces of the concrete pipe with 2-cm thick petroleum-based lining that has a surface roughness thickness of 0.04 mm. There is a concern that the reduction of pipe diameter to 66 cm and the increase in average velocity may offset any gains. Taking ρ = 1000 kg/m³ and v = 1 × 10^–6 m²/s for water, determine the percent increase or decrease in the pumping power requirements due to pipe frictional losses as a result of lining the concrete pipes. (Round the final answer to one decimal place.)

Answer 1

Area of pipe before lining,

A = (pi / 4) * d^2 = (pi / 4)*0.7^2

A = 0.3848 m^2

Average velocity of flow, V = Q / A

V = 1.50 / 0.3848 = 3.898 m/s

Reyynolds number, Re = Vd / v

Re = 3.898*0.70 / 1*10^(-6) = 2.728*10^6

Re > 4000. so flow is turbulance.

Relative roughness = e / d = 0.003 / 0.70 = 0.00428

Now we have values of Re and revative roughness.

From moody's chart, Friction factor for turbulance flow,

f1 = 0.029

Head loss due to friction,

h = f1*LV^2 / 2gd

h = 0.029*1520*(3.898)^2 / 2*9.8*0.70

h = 48.76 m

New diameter after internal lining,

d' = d - 2*r

d' = 0.70 - 2*0.02 = 0.66 m

New average velocity, V' = Q / A'

V' = 1.5 / (pi / 4)*0.66^2 = 4.384 m/s

Reynolds number, Re = 4.384*0.70 / 1*10^(-6)

Re' = 2.893*10^6

Relative roughness = 0.04*10^(-3) / 0.66 = 0.0606*10^(-3)

Now we have values of Re and revative roughness.

From moody's chart, Friction factor for turbulance flow,

f1' = 0.0117

Head loss due to friction,

h' = f1*LV^2 / 2gd

h' = 0.011*1520*(4.384)^2 / 2*9.81*0.66

h' = 24.81 m

Percentage power reduction,

% = (h' / h) * 100

% = (24.81 / 48.76) * 100

Power reduction = 50.89 %