Question

please do it correctly ! last person gave me wrong data ! please help !

Problem 3: An old pipe 2 m in diameter has a roughness of -30 mm. A 12-mm thick coating would reduce the pipe roughness to & 1 mm. 0 0 l km-pipe a) How much the friction head loss would be reduced per kilometer of pipe, for water at 20 °C at a flow rate of 6 m3/s? b) How much pumping costs would be reduced per kilometer of pipe? The pumps are 75% efficient and the cost of energy is $1 per 70 MJ (megajoule). Notice that new pipe diameter after coating would be reduced to (D-2t), where t is the coating thickness

Answer 1

Ans) Given,

Diameter of pipe (D) = 2 m

Thickness of coating = 12 mm or 0.012 m

a) We know , friction head loss (H_f) = f L V² / D(2g)

where, 'f' = Darcy friction factor

V = flow velocity

L = Length of pipe

V = Q / A

Given, Q = 6 m³/s

Also, A = (/4)D² = 0.785 x 2 x 2 = 3.14 m²

71

=> V = 6 / 3.14

V = 1.910 m/s

To determine Darcy friction factor 'f' , we have to calculate Reynolds number(Re),

Re = $\rho$ V D/$\mu$

where, $\rho$ = density of water

  $\mu$ = dynamic viscosity of water ( 10^-3 Ns/m² at 20^oC)

=> Re = 1000 x 1.91 x 2 / 10^-3

=> Re = 3.82 x 10⁶

According to Moody's diagram, for e/D = 30 / 2000 = 0.015 and Re = 3.82 x 10⁶ , f = 0.048

Therefore, head loss per km(1000m) = 0.048 x 1000 x 1.91²/ (2 x 9.81 x 2)

=> H_f = 4.4625 m

Now, after coating , new diameter = 2 - 2(0.012) = 1.976

V' = Q/A = 6 / (0.785 x 1.976²)

V' = 1.957 m/s   

Re will approximately same as above = 3.82 x 10⁶

  According to Moody's diagram, for e/D = 1 / 2000 = 0.0005 and Re = 3.82 x 10⁶ , f = 0.018

  Therefore, head loss per km(1000m) = 0.018 x 1000 x 1.957²/ (2 x 9.81 x 2)

=> H_f = 1.757 m

Therefore, headloss reduced = 4.4625 - 1.757 = 2.7055 m

% headloss reduced = (2.7055/4.4625) x 100 = 60.62 %

Ans b) Power consumed by pump per hour = $\rho$gQH_p x 1 hr

Pump head = loss due to friction

P = 1000 x 9.81 x 6 x 4.4625 x 1

P = 262.66 kWh or 945.576 MJ

Since efficiency of pump = 75 %, energy actually consumed = 945.576/0.75 = 1260.768 MJ

Pumping cost for 70 MJ = 1$

Pump cost for 945.576 MJ = 1260.768/70

= 18.01 $

Now, after coating, Power consumed = 1000 x 9.81 x 6 x 1.757 x 1

P = 103.417 kWh or 372.3 MJ

Actual energy consumed = 342.3/0/0.75 = 496.4 MJ

New operating cost = 496.4/70 = 7.1$

Pumping cost reduced = 18.01 - 7.10 = 10.91$

% Cost reduced = 10.91/18.01 = 60.57 %