Momentum equation and open channel flow

[fayan77]

Homework Statement

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

Homework Equations

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

However in my lecture the professor provided another equation

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

 

[Chestermiller]

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 = Qv_out - Qv_in 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?

 

[fayan77]

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?

 

[Chestermiller]

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?

 

[fayan77]

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?

 

[Chestermiller]

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.

 

[fayan77]

True, thank you for your help, you a real one.