Water flows through the headrace channel of a hydroelectric

Water flows through the headrace channel of a hydroelectric project, i.e. the powerhouse is at the downstream end of the channel. The powerhouse is equipped with five small turbines. Water is flowing at a rate of 800 cfs through the headrace channel. The channel is rectangular and 10 feet wide. The water depth is 8 feet. The channel walls are 12 feet high. Answer the following questions: What is the flow regime in the headrace channel? If all turbines are suddenly shut down, will the resulting surge overtop the walls? If the discharge is reduced by 40% by shutting down two turbines, what will be the height and speed of the surge in the headrace channel?

Solution

a) The flow regime in the channel is Laminar flow with least viscous fluid flow and in a no slip condition.

b) The discharge rate of the flow is 800cfs which accounts only a portion of the channel cross sectional area, i.i 8\' x 10\'. During a surge the discharge volume spreads out across the entire channel at 800fcs. SInce channel area is greater than it cross section the 800cfs discharge does not overflow. (assuming the velocity of the flow is the same).

c) The 40 % decrease in the discharge of the flow is retained in the channel resulting an increase in its height.

The decrease in volume of flow is 0.4 x 800 = 320 cfs, velocity of flow v = 320 / (62.4 x 10 x 8 x 0.0069) =9.29 ft /s

This discharge equals increase in volume across the channel,

320 = density x A channel x v = 62.4 x 10 x length of the channel x (h1-h)

increased height is h1 = 320 (62.4 x 10 x L) +h

The speed of surge is given by increase in pressure head,

p2-p1 = density x g x (h1 - h)