Flow Rate in Horizontal Rectangular Channel: Solve Now

In summary: B]In summary, the problem involves a horizontal rectangular open channel with a width of 20 m and a water depth of 9 m. When a 1.5 m high hump is introduced in the channel floor, the water surface drops by 1 m. The flow rate, neglecting energy losses, is needed to be determined. It is also proposed to place a pier at the center of the channel on the hump, and the maximum width of the pier needs to be determined to avoid backwater effects. The equations used for the solution include E=h+(v^2/2g), h+(v^2/2g)=constant, hc=(q^2/g)^1/3, Emin=1
  • #1
10
0

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.
[/B]

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 = 6.5+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, Emin=1.5hc given Emin=9.75

I 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

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




[/B]
 
Physics news on Phys.org
  • #2
I'm not sure I'm interpreting the problem correctly. I have attached a figure. I assume the hump goes across the entire width of the channel.

Ragnar1995 said:

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 = 6.5+v2^2/2g

and i got to v1-v2=7m/s, wasnt sure what to do from there,

[/B]

When using the formula h + v2/(2g) = const. you have apparently chosen two points in the fluid flow to apply the equation. Can you describe the location of these two points? Where are you choosing h = 0? Have you measured the height h for the two points from the same reference level (h = 0)?

Also note that ##\sqrt{v_2^2 - v_1^2} \neq v_2 - v_1 ##.
 

Attachments

  • Channel with bump.png
    Channel with bump.png
    1.8 KB · Views: 417
Last edited:
  • #3
Ragnar1995 said:
9+v1^2/2g = 6.5+v2^2/2g
Think again about that 6.5m. Remember that this term represents the lost PE. How does the height of the bump come into that?
 

What is flow rate in a horizontal rectangular channel?

Flow rate in a horizontal rectangular channel refers to the volume of fluid passing through the channel per unit time. It is typically measured in cubic meters per second or cubic feet per second.

How is flow rate calculated in a horizontal rectangular channel?

Flow rate in a horizontal rectangular channel can be calculated using the formula Q = V * A, where Q is the flow rate, V is the average velocity of the fluid, and A is the cross-sectional area of the channel.

What factors affect flow rate in a horizontal rectangular channel?

The flow rate in a horizontal rectangular channel can be affected by several factors, including the size and shape of the channel, the viscosity of the fluid, and the slope of the channel.

Why is it important to determine flow rate in a horizontal rectangular channel?

Determining flow rate in a horizontal rectangular channel is important for various reasons, such as designing efficient irrigation systems, predicting flood risks, and optimizing industrial processes that involve fluid flow.

How can flow rate be increased in a horizontal rectangular channel?

Flow rate in a horizontal rectangular channel can be increased by increasing the slope of the channel, decreasing the channel width, or using pumps to increase the velocity of the fluid.

Suggested for: Flow Rate in Horizontal Rectangular Channel: Solve Now

Replies
19
Views
1K
Replies
10
Views
886
Replies
6
Views
786
Replies
4
Views
707
Replies
16
Views
1K
Replies
6
Views
1K
Replies
8
Views
1K
Back
Top