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## Homework Statement

In a horizontal rectangular open channel 20 m wide the water depth is 9 m. When a

smooth hump 1.5 m high is introduced in the channel floor, a drop of 1 m is produced in

the water surface. What is the flow rate, neglecting energy losses? It is proposed to place

a pier at the centre of this channel on the hump. Determine the maximum width of this

pier if it is not to cause any backwater effects.

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## Homework Equations

all the equations i can think of are :

E=h+(v^2/2g) also

**h+(v^2/2g)=constant**

E=h+(q^2/2gh^2)

hc=(q^2/g)^1/3

Emin=1.5hc

Q=qB

E1=y+E2

v=q/h

Where E=specific energy

h=depth of flow

hc=critical depth

q=flow rate per unit width

Q=flow rate[/B]

## The Attempt at a Solution

First of i tried to find velocities at each point using :

**h+(v^2/2g)=constant**

9+v1^2/2g = 6.59+v1^2/2g = 6.5

**+v2^2/2g**

and i got to v1-v2=7m/s, wasnt sure what to do from there, so i tried a different method...

I assumed the critical depth would be 6.5,and i got to v1-v2=7m/s, wasnt sure what to do from there, so i tried a different method...

I assumed the critical depth would be 6.5,

**Emin=1.5hc given Emin=9.75**

I then subbed that into the equationI then subbed that into the equation

**E=h+(v^2/2g) to give v=7.99m/s**

i then subbed into eq v=q/h to give q=51.935

i then subbed into eq v=q/h to give q=51.935

then

**Q=qB**

= 51.935 x 20

=1038.7 m^3/s

however answer is

832.52 m^3

/s

any help would be appreciated. Thanks

= 51.935 x 20

=1038.7 m^3/s

however answer is

832.52 m^3

/s

any help would be appreciated. Thanks

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