PLEASE SHOW WORK MMEE Viscous Effects Example 3 Consider the

PLEASE SHOW WORK

MMEE Viscous Effects Example 3 Consider the following smooth pipe network. Water at 283 K enters the system at a flow rate of 142 L/s. What power must the pump deliver to the fluid if half of the total flow is to be carried by each branch of the network? Neglect viscous losses in pipe tees. p 1,000 kg/m u 1.31 x 103 kg/ m*s) Pump 7.5 cm ID 15 m 30 m 142 L/s 1 15 cm ID

Solution

Flow rate in each branch Q = 142 / 2 = 71 L/s = 0.071 m³/s

Cross-section area of 15 cm ID pipe = 3.14 / 4 * 15² = 176.625 cm² = 0.0177 m²

Velocity in 15 cm ID pipe = 0.071 / 0.0177 = 4.02 m/s

Reynolds number in 15 cm ID pipe, = rho*V*D / u = 1000*4.02 * 0.15 / (1.31*10^-3) = 4.6*10⁵

From Moody diagram, for Re 4.6*10⁵ and for smooth pipe, we get Friction factor f = 0.0135

For 15 cm ID pipe, head loss = f*L/D*V² / 2g = 0.0135 * (30 / 0.15) * 4.02² / (2*9.81) = 2.223 m

Cross-section area of 7.5 cm ID pipe = 3.14 / 4 * 7.5² = 44.156 cm² = 0.0044 m²

Velocity in 7.5 cm ID pipe = 0.071 / 0.0044 = 16.08 m/s

Reynolds number in 7.5 cm ID pipe, = rho*V*D / u = 1000*16.08 * 0.075 / (1.31*10^-3) = 9.2*10⁵

From Moody diagram, for Re 9.2*10⁵ and for smooth pipe, we get Friction factor f = 0.012

For 7.5 cm ID pipe, head loss = f*L/D*V² / 2g = 0.012 * ((30+15+15) / 0.075) * 16.08² / (2*9.81) = 126.516 m

For 15 cm pipe, P1 - P2 = 2.223 m

For 7.5 cm pipe, P1 - P2 = 126.516 - H..........where H is the head added by pump.

Equating both, 126.516 - H = 2.223

H = 124.293 m

Power delivered by pump = rho*g*Q*H

= 1000*9.81*0.071*124.293

= 86.571 kW