Question

A hydroelectric turbine passes 1.7 million gal/min through its blades. If the average velocity of the flow in the circular cross-sectional conduit leading to the turbine is not to exceed 34 ft/s, determine the minimum allowable diameter of the conduit.

Answer 1

11.87 ft

Step-by-step explanation:

Volumetric flow rate of the turbine Q = 1.7 million gal/min

Q = 1.7 x
`10^(6)` gal/min

1 gal = 0.00378541 m^3

1 min = 60 sec

1.7 x
`10^(6)` gal/min = (1.7 x
`10^(6)` x 0.00378541)/60 = 107.25 m^3

average velocity of flow through turbine = 34 ft/s

1 ft = 0.3048 m

34 ft/s = 34 x 0.3048 = 10.36 m/s

According to continuity equation, Q = AV

where Q = volumetric flow rate

A = Area of conduit

V = velocity of flow through turbine

A = Q/V = 107.25/10.36 = 10.35 m^2

Area of conduit =
`\pi r ^(2)`

radius r =
`\sqrt{(area)/(\pi ) }` =
`\sqrt{(10.35)/(3.142) }` = 1.81 m

diameter = 2 x radius = 2 x 1.81 = 3.62 m

diameter = 3.62 m = 3.62 x 3.28084 = 11.87 ft