Euler's equation pressure difference

In summary, the problem involves finding the directional derivative of p from point C to point A without breaking it into components. The equations used are ##\frac{\partial}{\partial l}## (p+##\gamma##l) = ##\rho##a and ##\nabla p## = ##\rho##a. The solution involves using the unit vector in the direction from C to A and dotting it with the equations to get the directional derivative. Another method involves resolving the equation into components to get the partial derivatives in the x and y directions, but it may not work for all problems.
  • #1
fayan77
84
0

Homework Statement



Screen Shot 2018-02-15 at 2.45.13 PM.png


I am after PC - PA
However I must do so without breaking into components. My problem has different values

L=3
H=4
SG=1.2
downward a = 1.5g
horizontal a = 0.9g
and my coordinate is conventional positive y up and positive x to the right
cos##\theta## = 3/5
sin##\theta## = 4/5

Homework Equations



- ##\frac {\partial} {\partial l}## (p+##\gamma##l) = ##\rho##a

- ##\nabla p## = ##\rho##a

The Attempt at a Solution


a = 0.9g##\hat i## -1.5g##\hat j##
therefore,
- ##\frac {dp} {dl}##-##\gamma## = ##\rho##(0.9g##\hat i## -1.5g##\hat j##)

where dp = PC - PA
dl = 5 because of the 3,4,5 triangle
and in order to get rid of (i hat and j hat) i use triangle relationship and project 0.9g##\hat i## to l direction using 0.9g(3/5) and -1.5g(4/5)
therefore,
- (PA - PC) / 5 = ##\rho##(0.9g(3/5)-1.5g(4/5)) + ##\gamma##
PC - PA= -5##\rho##g(.9(3/5)-1.5(4/5)+1)
PC - PA= not correct answer
 

Attachments

  • Screen Shot 2018-02-15 at 2.45.13 PM.png
    Screen Shot 2018-02-15 at 2.45.13 PM.png
    9.8 KB · Views: 945
Last edited by a moderator:
Physics news on Phys.org
  • #2
This doesn't seem to have been set up correctly. You have vectors on one side of the equations and scalars on the other. That's a no-no.

The starting equation should be:
$$-\nabla{p}-\gamma\hat{j}=\rho\hat{a}$$
To get the derivative of p in the direction from C to A, you need to dot this equation with a unit vector in that direction. What is the equation for the unit vector in the direction from C to A?
 
  • #3
(3/5)##\hat i## + (4/5)##\hat j##
 
  • #4
fayan77 said:
(3/5)##\hat i## + (4/5)##\hat j##
When you dot ##\nabla p## with this unit vector, you get the "directional derivative" ##\frac{\partial p}{\partial l}##, where l is measured along the diagonal from C to A. What do you get when you dot the other terms in the equation with this unit vector?
 
  • #5
let (3/5)##\hat i## + (4/5)##\hat j## = W
where ##\gamma## = ##\rho##g

-##\frac{\partial p}{\partial l}## -##\rho##g##\hat{j}## DOT W =##\rho\hat{a}## DOT W
-##\frac{\partial p}{\partial l}## = [##\rho##g##\hat{j}## DOT W] DOT [##\rho\hat{a}## DOT W]
-##\frac{\partial p}{\partial l}## = ##\rho##g[.9##\left( \frac 3 5 \right) \hat i## - .5##\left( \frac 4 5 \right) \hat j##]
 
  • #6
fayan77 said:
let (3/5)##\hat i## + (4/5)##\hat j## = W
where ##\gamma## = ##\rho##g

-##\frac{\partial p}{\partial l}## -##\rho##g##\hat{j}## DOT W =##\rho\hat{a}## DOT W
-##\frac{\partial p}{\partial l}## = [##\rho##g##\hat{j}## DOT W] DOT [##\rho\hat{a}## DOT W]
-##\frac{\partial p}{\partial l}## = ##\rho##g[.9##\left( \frac 3 5 \right) \hat i## - .5##\left( \frac 4 5 \right) \hat j##]
That’s not done correctly. The dot product of two vectors is a scalar.
 
  • #7
got it,
-##\frac{\partial p}{\partial l}## = ##\rho##g(.14)

-(PA - PC) / 5 = ##\rho##g(.14)

PC - PA = 5 (1.2)(1.94)(32.2)(.14) = 52.4

So what we did here was just projecting everything along CA?
Is there another way of doing this? I saw a solution on the internet solved differently. Do you mind explaining the other solution?
Screen Shot 2018-02-15 at 7.29.06 PM.png
 

Attachments

  • Screen Shot 2018-02-15 at 7.29.06 PM.png
    Screen Shot 2018-02-15 at 7.29.06 PM.png
    26.6 KB · Views: 1,085
  • #8
fayan77 said:
got it,
-##\frac{\partial p}{\partial l}## = ##\rho##g(.14)

-(PA - PC) / 5 = ##\rho##g(.14)

PC - PA = 5 (1.2)(1.94)(32.2)(.14) = 52.4

So what we did here was just projecting everything along CA?

Sure. What they wanted you to realize was that you can get the directional derivative by dotting the equation with a unit vector in the desired direction.
Is there another way of doing this? I saw a solution on the internet solved differently. Do you mind explaining the other solution?
View attachment 220408
This is just basically equivalent to resolving the equation into components, to get the partial derivatives in the x and y directions. I really don't like what they've done here. The only reason this method works is that, for this particular problem, the partial derivatives of p with respect to both x and y are constant, independent of x and y.
 
Last edited:
  • #9
okay, and thank you for your help.
 

1. What is Euler's equation for pressure difference?

Euler's equation for pressure difference is a mathematical expression that describes the relationship between fluid velocity, pressure, and fluid density in a moving fluid. It is commonly used in fluid mechanics to calculate the pressure difference between two points in a fluid flow.

2. How is Euler's equation for pressure difference derived?

Euler's equation for pressure difference is derived from the Navier-Stokes equations, which are a set of equations that describe the motion of a fluid. It is derived by applying the principles of conservation of mass, momentum, and energy, along with the assumption of a steady and inviscid flow.

3. What are the units of Euler's equation for pressure difference?

The units of Euler's equation for pressure difference depend on the units used for velocity, pressure, and density. In SI units, the equation is typically expressed in terms of meters per second (m/s) for velocity, Newtons per square meter (N/m²) for pressure, and kilograms per cubic meter (kg/m³) for density.

4. What is the significance of Euler's equation for pressure difference in fluid mechanics?

Euler's equation for pressure difference is a fundamental equation in fluid mechanics, as it allows for the calculation of pressure differences in fluid flows. It is used in a variety of applications, such as designing pumps and turbines, analyzing airfoil performance, and understanding the behavior of fluids in pipes and channels.

5. Can Euler's equation for pressure difference be used for compressible fluids?

Yes, Euler's equation for pressure difference can be used for both incompressible and compressible fluids. In the case of compressible fluids, the equation is modified to take into account the change in density with pressure. This modified equation is known as the compressible Euler equation and is commonly used in aerodynamics and gas dynamics.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
210
  • Quantum Physics
Replies
1
Views
630
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
973
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Quantum Interpretations and Foundations
2
Replies
54
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
818
  • Engineering and Comp Sci Homework Help
Replies
1
Views
650
  • Advanced Physics Homework Help
Replies
20
Views
2K
Back
Top