A gravity-feed piping system between two large tanks is used to transport a 20%

Question

A gravity-feed piping system between two large tanks is used to transport a 20% magnesium chloride solution (MgC12 under brines in EES). The magnesium chloride is at 27 degree C and the tanks are open to the atmosphere. All fittings and valves have inner diameters of 60 mm and wall roughness of 0.046 mm. The total straight length of pipe in the system is 20 m. The elevation difference between the free surfaces of the tanks is 5 m. Determine the volumetric flow rate (I min^-1) between the two tanks. Account for minor losses in the standard elbows and close return bends.

Explanation / Answer

Solution:-

Diameter of Pipe, D = 60 mm

Length of straight portion of pipe, L = 20 m

Wall roughness, f = 0.046

Elevation difference between free surfaces of both tank, H = 5 m

Number of 90 degree bend = 4, Number of close return bend = 1

Total head loss = Head loss in straight pipe + head loss in bends

Head loss in straight pipe = Hf , Head loss in bends = hf

Hf = 4fLV2 / 2gD where V = velocity of flow, L = length of pipe, D = diameter of pipe, f = surface roughness

hf = kV2 / 2g where k = minor loss coefficent k = 0.3 for 90 degree bend & k = 0.2 for close return bend

So, putting the above formula and values,

V2 / 2g = H - Hf - hf = H - 4fLV2 / 2gD - kV2 / 2g

=> V2 / 2g = 5 - 4 x 0.046 x 20 V2 / 2 x 9.81 x 0.06 - (4 x 0.3 + 0.2) V2 / 2 x 9.81

=> 3.2483 V2 = 5

=> V = (5 / 3.2483)1/2   = 1.24 m/s

Volumetric Flow Rate = 22/7 x D2 /4 x V = 22/7 x 0.062/4 x 1.24 = 3.507 x 10-3 m3/s

Volumetric Flow Rate = 3.507 x 10-3 x 103 x 60 = 210.42 liter / min (Ans)

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