What is the derivation of the second equation for area velocity?

In summary, the first equation is derived from the fact that the particle moves by the distance traveled multiplied by the altitude of the triangle. The second equation is derived from the fact that the particle moves in the (x, y, 0) plane.
  • #1
Shreya
188
65
Homework Statement
I want to understand the second equation for Areal velocity as given below.
Relevant Equations
Refer below.
Screenshot from 2023-01-19 20-29-43.png
Screenshot from 2023-01-19 20-30-05.png

I think I understand how the first equation comes about.

Screenshot from 2023-01-19 20-43-16.png

In ##dt## the particle travels by ##dr##, I considered it as a triangle with altitude ##r## and base ##dr##. On dividing the area travelled in ##dt## by ##dt## we get the above equation.
A similar argument can be applied to ##\frac 1 2 \rho^2 \frac {d\phi} {dt}## as ##\rho## is same as r and ##\rho \frac {d\phi} {dt}## is same as ##\frac {dr} {dt}##

But, I am not able to understand the 2nd equation. I can provide a similar argument for ##xv_y## but can,t seem to reason any further.
Please be kind to help.
 

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  • #2
The unsigned area swept out in time [itex]\delta t[/itex] is approximately [tex]|\delta S| \approx \tfrac12
\|\mathbf{r} \times (\mathbf{r} + \mathbf{v}\delta t)\| =
\tfrac12 \|\mathbf{r} \times \mathbf{v}\| |\delta t|.[/tex] This is because the magnitude of the cross-product gives the area of the parallelogram bounded by [itex]\mathbf{r}[/itex] and [itex]\mathbf{r} + \mathbf{v} \delta t[/itex], and the area swept out is approximately the area of a triangle which is one half of the parallelogram. Dividing by [itex]\delta t[/itex] and taking the limit [itex]\delta t \to 0[/itex] gives [tex]
\left|\frac{dS}{dt}\right| = \tfrac12 \|\mathbf{r} \times \mathbf{v}\|.[/tex] This is essentially the derivation of the first formula.

To derive the second, in the case of motion in the [itex](x,y,0)[/itex] plane we can calculate [itex]\frac12\|\mathbf{r} \times \mathbf{v}\| [/itex] either in cartesians or in plane polars using the standard formulae [tex]\begin{split}
\mathbf{r} &= r\mathbf{e}_r(\theta) \\
\mathbf{v} &= \dot r \mathbf{e}_r(\theta) + r\dot \theta \mathbf{e}_{\theta}(\theta) \end{split}[/tex] where [tex]
\mathbf{e}_r(\theta) \times \mathbf{e}_\theta(\theta) \equiv \mathbf{e}_z.[/tex] We can then adopt the convention that [itex]S[/itex] increases in the direction of increasing [itex]\theta[/itex] to remove the absolute value signs.
 
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Likes Shreya
  • #3
pasmith said:
To derive the second, in the case of motion in the (x,y,0) plane we can calculate 12‖r×v‖ either in cartesians or in plane polars using the standard formulae r=rer(θ)v=r˙er(θ)+rθ˙eθ(θ) where er(θ)×eθ(θ)≡ez. We can then adopt the convention that S increases in the direction of increasing θ to remove the absolute value signs.
Thank You so much @pasmith !
##\vec r = x \hat i + y \hat j##
##\vec v = v_x \hat i + v_y \hat j##

##\vec r \times \vec v = (x v_y - y v_x) \hat k##
##\frac {dS} {dt} = \frac 1 2 (xv_y-yv_x)##
Is the above right?
 

1. What is the second equation for area velocity?

The second equation for area velocity is the product of the cross-sectional area and the average velocity of a fluid flow. It is represented as A x V = Q, where A is the cross-sectional area, V is the average velocity, and Q is the volumetric flow rate.

2. How is the second equation for area velocity derived?

The second equation for area velocity is derived from the continuity equation, which states that the mass flow rate into a control volume must equal the mass flow rate out of the control volume. By rearranging the continuity equation and substituting the definition of volumetric flow rate, we arrive at the equation A x V = Q.

3. What are the units of the second equation for area velocity?

The units of the second equation for area velocity depend on the units used for each variable. The cross-sectional area is typically measured in square meters (m²), the average velocity in meters per second (m/s), and the volumetric flow rate in cubic meters per second (m³/s).

4. Can the second equation for area velocity be applied to all types of fluid flow?

Yes, the second equation for area velocity can be applied to all types of fluid flow, including liquids and gases. However, it assumes that the flow is steady, incompressible, and has a constant cross-sectional area throughout the control volume.

5. How is the second equation for area velocity used in practical applications?

The second equation for area velocity is commonly used in various engineering and scientific fields, such as fluid mechanics, hydraulics, and environmental studies. It is used to calculate the flow rate of fluids in pipes, channels, and rivers, and is also used in the design of pumps, turbines, and other hydraulic systems.

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