Momentum equation and open channel flow

In summary, the conversation discusses using the momentum principle to determine y2 (ft) and v2 (ft/s) in a fluid dynamics problem. The speaker mentions a past fluids class and a professor's lecture, and points out a potential error in the professor's provided equation. They also discuss integrating the pressure over area and the difference between previous cases and the current problem. The conversation ends with the speaker expressing gratitude for the help provided.
  • #1
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Homework Statement


Determine y2 (ft)and v2 (ft/s) using the momentum principle assume widths are the same at 1 and 2
Screen Shot 2018-09-08 at 2.52.08 PM.png

Homework Equations


from what i remember in my past fluids class, the momentum equation is:
Force = Qvout - Qvin assuming same density and direction

However in my lecture the professor provided another equation

Screen Shot 2018-09-08 at 2.58.06 PM.png

but when i work out the units in this equation left side yields Force and right side is pressure (force/area)

did my professor forget to multiply each pressure with its respective Area? P1A1?


The Attempt at a Solution


using ##\rho####\beta##QV2 - ##\rho####\beta##QV1 = P1A1-P2A2
assume both ##\beta##'s = 1
assume width = 1
using continuity equation Q = V1A1 = V2A2 = (4.55)(1)(8.55)=y2(1)V2

##\rho##(4.55)(1)(8.55)(V2 - V1)= P1(4.55)(1)-P2y2(1)

##\rho##38.9(V2 - 8.55)= P14.55-P2y2

I am stuck
 

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  • #2
fayan77 said:

Homework Statement


Determine y2 (ft)and v2 (ft/s) using the momentum principle assume widths are the same at 1 and 2
View attachment 230419

Homework Equations


from what i remember in my past fluids class, the momentum equation is:
Force = Qvout - Qvin assuming same density and direction

However in my lecture the professor provided another equation

View attachment 230421
but when i work out the units in this equation left side yields Force and right side is pressure (force/area)

did my professor forget to multiply each pressure with its respective Area? P1A1?


The Attempt at a Solution


using ##\rho####\beta##QV2 - ##\rho####\beta##QV1 = P1A1-P2A2
assume both ##\beta##'s = 1
assume width = 1
using continuity equation Q = V1A1 = V2A2 = (4.55)(1)(8.55)=y2(1)V2

##\rho##(4.55)(1)(8.55)(V2 - V1)= P1(4.55)(1)-P2y2(1)

##\rho##38.9(V2 - 8.55)= P14.55-P2y2

I am stuck
You are correct about needing to get the pressures times the areas, but the pressure is varying with depth, so you need to integrate the pressure (which is hydrostatic in the y direction) over the area. Also note that, from the continuity equation, y2 is a function of v2.

What are you doing about the friction term?
 
  • #3
Solved it, my professor's notes are filled with errors as he calls P1,P2 a pressure when really it is a Force, also he has some typos on the equations. Where I was confused was, I thought (from my fluid dynamics class) when we look at a controlled volume we immediately add forces on the cut which where (Pressure) X (Cross sectional Area) where as here we take the integral of ##\rho##gydA from 0 to h where dA = (width)dy

What is the difference?
 
  • #4
fayan77 said:
Solved it, my professor's notes are filled with errors as he calls P1,P2 a pressure when really it is a Force, also he has some typos on the equations. Where I was confused was, I thought (from my fluid dynamics class) when we look at a controlled volume we immediately add forces on the cut which where (Pressure) X (Cross sectional Area) where as here we take the integral of ##\rho##gydA from 0 to h where dA = (width)dy

What is the difference?
What is what difference?
 
  • #5
last semester, in a different course when we analyzed a controlled volume the pressure force was simply P X (Cross sectional Area).

Now we integrate with respect to depth geometry

Was it for simplicity's sake that we did P X A?
 
  • #6
In the cases you looked at, the channel was closed, and there was no free surface, and the average pressure was much higher than the hydrostatic difference in pressure between the top of the channel and the bottom of the channel. Here, the pressure is zero at the top, and twice the average at the bottom.
 
  • #7
True, thank you for your help, you a real one.
 

What is the momentum equation and how is it used in open channel flow?

The momentum equation is a fundamental equation in fluid mechanics that relates the forces acting on a fluid to its acceleration and mass. In open channel flow, it is used to calculate the forces acting on a fluid in an open channel, such as a river or canal, and predict its behavior.

What are the key terms and variables in the momentum equation?

The key terms in the momentum equation are force, acceleration, and mass. The variables include the fluid velocity, density, and cross-sectional area of the channel.

How does the momentum equation differ from the continuity equation?

The continuity equation is used to describe the conservation of mass in a fluid, while the momentum equation describes the conservation of momentum. In open channel flow, both equations are used together to analyze the behavior of the fluid.

What are the assumptions made in the momentum equation for open channel flow?

The momentum equation assumes steady, uniform flow with no changes in velocity or channel geometry. It also assumes that the flow is frictionless and there are no external forces acting on the fluid.

How is the momentum equation applied in practical engineering applications?

The momentum equation is used in various engineering applications, such as designing irrigation systems, analyzing flood control measures, and predicting the behavior of ships and other watercraft. It is also used in the design of hydraulic structures, such as dams and spillways.

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