Water at 16 degrees Celsius rho10011 kgm3 and mew113103 kgms

Water at 16 degrees Celsius (rho=1001.1 kg/m^3 and mew=1.13*10^-3 kg/m*s) is pumped steadily at 10 L/s through a 55 m long and 15 cm diameter horizontal pipe connected to a 4.5 cm diameter, 69 m long pipe by a 45 degree reducer. Both sections are stainless steel with a friction factor of 0.0159, and both ends are open to the atmosphere.

Determine the head in meters the pump must supply

Take the acceleration of gravity to be 9.81 m/s^2

Solution

For square reduction in pipe diameter, Loss coeff K = (0.6 + 0.48*f1) (D1 / D2)² [(D1 / D2)² - 1]

= (0.6 + 0.48*0.0159) (15 / 4.5)² [(15 / 4.5)² - 1]

= 68.26

For 45 deg taper reduction this value of K should be multiplied by 0.61.

Ktaper = 0.61*68.26 = 41.64

Cross-section area of first pipe, A1 = 3.14 /4 * 0.15² = 0.0176625 m²

Cross-section area of second pipe, A2 = 3.14 /4 * 0.045² = 0.00159 m²

Velocity in first pipe V1 = Q / A1 = 10*10^-3 / 0.0176625 = 0.566 m/s

Velocity in first pipe V2 = Q / A2 = 10*10^-3 / 0.00159 = 6.29 m/s

Head loss in pipe = f*L1 / D1 * (V1² / (2g)) + f*L2 / D2 * (V2² / (2g)) + Ktaper*V1² / (2g)

= 0.0159*55 / 0.15 * (0.566² / (2*9.81)) + 0.0159*69 / 0.045 * (6.29² / (2*9.81)) + 41.64*(0.566² / (2*9.81)

= 49.93 m