A 60 aqueous solution of sucrose is to be pumped through a

A 60 % aqueous solution of sucrose is to be pumped through a pipe. An engineer is interested in comparing pumping cost under two conditions. Solution is pumped through the inner cylinder (its radius is Ri=1.26 cm). Solution is pumped in the annulus region, between inner and outer cylinder (the radius of outer cylinder is Ro=2.8 cm). At 20 degree C, the solution density is 1286 kg/m^3 and the viscosity is 0.057 Pa.sec. The pipe is 8.2 m in length. The pressure difference impressed through is 37162 Pa. Which pumping configuration can you suggest? Please verify your answer with calculations.

Solution

1)

Cross-section area A = 3.14 * 1.26² = 4.985 cm²

Reynolds number Re = rho*V*D / meu

= 1286 * V * (2*1.26*10^-2) / 0.057

= 568.54*V

Assuming laminar flow, friction factor f = 64 / Re

= 64 / (568.54*V)

= 0.1126 / V

Pressure drop = f*L/D * (1/2)*rho*V²

37162 = (0.1126 / V) * 8.2 / (2*1.26*10^-2) * (1/2)*1286*V²

V = 1.578 m/s

For validation, Re = 568.54*1.578 = 897 which is less than critical value of 2200. Hence, our laminar flow assumption was correct.

Flow rate Q = A*V

= (4.985*10^-4) * 1.578

= 7.866*10^-4 m³/s

2)

Cross-section area A = 3.14 * (2.8² - 1.26²) = 19.6325 cm²

Wetted perimeter P = 2*3.14* (2.8 + 1.26) = 25.4968 cm

Equivalent dia D_e = 4A/P

= 4*19.6325 / 25.4968

D_e = 3.08 cm

Reynolds number Re = rho * V * D_e / meu

= 1286 * V * (3.08*10^-2) / 0.057

= 694.89*V

Aspect ratio R_o / R_i = 2.8 / 1.26 = 2.22

From tables for an annular pipe, with aspect ratio = 2.22, we get C = 95........where friction factor f = C / Re

Hence, f = 95 / (694.89*V)

f = 0.1367 / V

Pressure drop = f*L/D_e * (1/2)*rho*V²

37162 = (0.1367 / V) * 8.2 / (3.08*10^-2) * (1/2)*1286*V²

V = 1.588 m/s

Flow rate Q = A*V

= 19.6325*10^-4 * 1.588

= 31.174*10^-4 m³/s

Clearly, flow rate in annulur pipe is higher than the circular pipe. Hence, annular pipe is suggested.