Determine the maximum discharge obtainable in a 35ftdiameter

Determine the maximum discharge obtainable in a 3.5-ft-diameter commercial steel penstock that carries water from a mountain reservoir to a hydroelectric power plant. The penstock entrance is squared-edged and is located 100-ft below the reservoir\'s water surface. It is 1, 500 ft long, contains a globe valve, and discharges water into the atmosphere at an elevation 750 ft below the entrance.

Solution

For square-edged pipe entrance, Loss coeff K = 0.5

For fully-open Globe valve, K = 6.4

For pipe exit, K = 1

Total K = 0.5 + 6.4 + 1 = 7.9

For commercial steel, roughness, e = 150*10^-6 ft

Relative roughness, e/d = 150*10^-6 / 3.5 = 42.857*10^-6

Pressure head at entrance P1 = 100 ft

At exit, since pressure is atmospheric, hence P2 = 0

Since pipe area is constant, hence V1 = V2

Also, it is given that z1 - z2 = 750 ft

We have P1 + V1² / 2g + z1 = P2 + V2² / 2g + z2 + (fL/D + K)V² / 2g

100 + 750 = (f*1500/3.5 + 7.9)*V² / 2g

(f*1500/3.5 + 7.9)*V² = 850*2*32.2

(f*1500/3.5 + 7.9)*V² = 54740.............eqn1

Assume Re = 1.8E7.

Using Colebrook equation, we get f = 0.01046

(0.01046*1500/3.5 + 7.9)*V² = 54740

V = 66.5 ft/s

Re = V*D/neu

= 66.5*3.5 / (0.926E-5)

= 2.51E7

This is not close to what we had assumed (Re = 1.8E7).

Assume Re = 2.2E7.

Using Colebrook equation, we get f = 0.01043

(0.01043*1500/3.5 + 7.9)*V² = 54740

V = 66.52 ft/s

Re = V*D/neu

= 66.52*3.5 / (0.926E-5)

= 2.514E7

This is not close to what we had assumed (Re = 2.2E7).

Assume Re = 2.5E7.

Using Colebrook equation, we get f = 0.010409

(0.010409*1500/3.5 + 7.9)*V² = 54740

V = 66.54 ft/s

Re = V*D/neu

= 66.54*3.5 / (0.926E-5)

= 2.515E7

This is close to what we had assumed (Re = 2.5E7).

Hence, Re ~ 2.5E7

and V = 66.54 ft/s

Flow rate Q = pi/4*D^2 * V

= 3.14/4*3.5^2 * 66.54

= 639.87 ft^3/s