Crude oil is to be transported from reservoir A to reservoir B at a volumetric f

Question

Crude oil is to be transported from reservoir A to reservoir B at a volumetric flow rate of 0.02 m3/s. The pipeline has a diameter of 0.25 m and a length of 25000 m. The power required to drive the pump having an efficiency of 75 percent is 10 kW. Assuming that the minor head losses are 50 percent of the head losses in the straight pipe sections, determine the elevation difference between reservoirs A and B. The density and absolute viscosity of the crude oil are 900 kg in and 0.07 Pa s. respectively. (Ans. 19.53 m)

Explanation / Answer

Let the surface of the higher reservoir be section 1, and that of the lower reservoir be section 2.

Let Z1 - Z2 = h. Obviously, p1 = p2 = patm, V1 = V2 0.

The fluid velocity in the pipe is

V = Q/A = 0.02/(*0.252/4) = 0.4075 m/s

Re = Vd/ = 900*0.4075*.25/0.07 = 1309.6   <2300 Laminar flow

f = 64/Re = 0.04887

hf = f(L/d)[V2/(2g)] = 0.048*(25000/0.25)*[0.40752/(2*9.81)] = 41.36 m

Minor loss      hm = 50%*hf= 20.68 m

Pump head is (P is the power, is the efficiency)

hp= P/(gQ) = 75%*10000/(900*9.81*0.02) = 42.47 m

The energy equation for this problem is

p1/ + V12/(2g)+ z1 = p2/ + V22/(2g)+ z2 +hf   + hm -hp           =>

patm/ + 02/(2g)+ h = patm/ + 02/(2g)+41.36+20.68-42.47      =>

h = 19.57 m

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