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Question 5 a) As shown in Figure Q5 (a), crude oil at 60°C with ρ-860 kg/m3 and μ-3.91 x 10 kg s/m (about four times of visc

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Answer #1

Answer:

Re = \rho * v * D / \mu

= 860 * 2.83 * 1.22 / ( 3.91 * 10^-4)

= 7.6 * 10^6

so turbulent flow

first we need to calculate the head loss by Darcy's equation

head = h = f * L * (Q/A)^2 / ( 2*g*D)

area = (\pi/4)*D^2

= (\pi/4)*1.22^2

= 1.17 m2

Q/A = 3.31 / 1.17 = 2.83 m/s

h = 0.0125 * 1286 * 2.83^2 / ( 2*9.81*1.22)

= 5.38 m

Power = q*h*g*\rho

q(m3 /h) = Q*3600

= 3.31 * 3600

= 11916 m3/h

Power = P = 11916 * 5.38 * 9.81 * 860

= 540854227.7 W

= 725005.6672 HP

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Question 5 a) As shown in Figure Q5 (a), crude oil at 60°C with ρ-860 kg/m3 and μ-3.91 x 10" kg s/m (about four times of viscosity of water) is pumped across from City A to City B pipeline, a...
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