Question

A reservoir maintains the water surface at an elevation of 380 feet. An 8-inch diameter pipe connects to the reservoir at an elevation of 270 ft and runs at a slope to an exit nozzle at an elevation of 210 ft. The nozzle is 4 inches in diameter. If the head lost through the pipe and nozzle is 36 feet , calculate the flow rate.

Answer 1

Known :

z1 = 380 ft

z2 = 210 ft

D1 = 8 in

D2 = 4 in

hL = 36 ft

Solution :

Continuity Equation

Q1 = Q2

A1 • V1 = A2 • V2

(πD1²/4) • V1 = (πD2²/4) • V2

D1² • V1 = D2² • V2

8² • V1 = 4² • V2

V2 = 4V1 ... (i)

Energy Equation :

P1/γ + V1²/2g + z1 = P2/γ + V2²/2g + z2 + hL

Since P1 = P2, then

V1²/2g + z1 = V2²/2g + z2 + hL

V1²/2(32.2) + 380 = V2²/2(32.2) + 210 + 36

V2² - V1² = 8.63 × 10³ ... (ii)

Subtitute (i) into (ii)

(4V1)² - V1² = 8.63 × 10³

15V1² = 8.63 × 10³

V1 = 24 ft/s

Q = A1 • V1

Q = [π(8/12)² / 4] • 24

Q = 8.377 cfs