Can somebody please explain how the product rule was used here?

[aur_khan]

Hello,
In one of my lecturer's slides, he states he has used the product rule to go from one line to the next. But i am not sure how he has applied it as my understanding of the product rule is that if we were to differentiate (uv) with respect to x, we would get u(dv/dx) + v(du/dx). However the lecturer gets a coefficient of "1/2" and i think he has applied the product rule backwards or something? Can somebody please explain what he has done and how he has done it?

I am attaching a picture of the problem below.

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. . . . .\(\displaystyle U_s\, \dfrac{\partial U_s}{\partial s}\, =\, -\dfrac{1}{\rho}\, \dfrac{\partial P}{\partial s}\, +\, g_s\)

Use product rule and, as g_s = –g in the vertical direction:

. . . . .\(\displaystyle \dfrac{1}{2}\, \dfrac{\partial (U_s^2)}{\partial s}\, +\, \dfrac{1}{\rho}\, \dfrac{\partial P}{\partial s}\, +\, g\,\dfrac{\partial z}{\partial s}\)

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I do not understand how the instructor goes from the first line above to the next.

My understanding of the product-rule formula is as follows:

. . . . .\(\displaystyle \dfrac{d}{dx}(uv)\, =\, v\, \cdot\, \dfrac{du}{dx}\, +\, u\, \cdot\, \dfrac{dv}{dx}\)

In this case, how does the first term go from this:

. . . . .\(\displaystyle U_s\, \dfrac{\partial U_s}{\partial s}\)

...to this:

. . . . .\(\displaystyle \dfrac{1}{2}\, \dfrac{\partial (U_s^2)}{\partial s}\)

Thank you very much in advance for your help and explanation.

Regards,

Ahsan

 

[stapel]

In one of my lecturer's slides, he states...

---

. . . . .\(\displaystyle U_s\, \dfrac{\partial U_s}{\partial s}\, =\, -\dfrac{1}{\rho}\, \dfrac{\partial P}{\partial s}\, +\, g_s\)

Use product rule and, as g_s = –g in the vertical direction:

. . . . .\(\displaystyle \dfrac{1}{2}\, \dfrac{\partial (U_s^2)}{\partial s}\, +\, \dfrac{1}{\rho}\, \dfrac{\partial P}{\partial s}\, +\, g\,\dfrac{\partial z}{\partial s}\)

---

I do not understand how the instructor goes from the first line above to the next.

My understanding of the product-rule formula is as follows:

. . . . .\(\displaystyle \dfrac{d}{dx}(uv)\, =\, v\, \cdot\, \dfrac{du}{dx}\, +\, u\, \cdot\, \dfrac{dv}{dx}\)

In this case, how does the first term go from this:

. . . . .\(\displaystyle U_s\, \dfrac{\partial U_s}{\partial s}\)

...to this:

. . . . .\(\displaystyle \dfrac{1}{2}\, \dfrac{\partial (U_s^2)}{\partial s}\)

I have a suspicion that at least part of the answer will depend upon the specifics of g, U, s, z, and rho. Can you tell us what lead up to the portion with which you're having difficulty?

Thank you!

 

[aur_khan]

Apply RANS to Frictionless Flow

I have a suspicion that at least part of the answer will depend upon the specifics of g, U, s, z, and rho. Can you tell us what lead up to the portion with which you're having difficulty?

Thank you!

That is actually the first and only slide related to that matter. The slides previous to that are about application of Reynolds Averaged Navier-Stokes equations to a static fluid, which is different.

However the specifics of the symbols according to my understanding are as follows:

g is gravity per unit mass
For steady flow along a streamline, velocity U_s is a function of position s on the curve but is constant with time
g_s = -g in the vertical direction
rho = density of the fluid per unit mass
z = one of the three axes (a fluid is considered to have velocity components in the x, y and z planes)


I have attached a photograph of the teacher's slide below as well FYI. Please let me know if you have any clue how he applied the product rule to go from the first line to the second.

Thank you.

Regards,

Ahsan

 

[Deleted member 4993]

That is actually the first and only slide related to that matter. The slides previous to that are about application of Reynolds Averaged Navier-Stokes equations to a static fluid, which is different.

However the specifics of the symbols according to my understanding are as follows:

g is gravity per unit mass
For steady flow along a streamline, velocity U_s is a function of position s on the curve but is constant with time
g_s = -g in the vertical direction
rho = density of the fluid per unit mass
z = one of the three axes (a fluid is considered to have velocity components in the x, y and z planes)


I have attached a photograph of the teacher's slide below as well FYI. Please let me know if you have any clue how he applied the product rule to go from the first line to the second.

Thank you.

Regards,

Ahsan

View attachment 7974

Do you agree that:

\(\displaystyle \frac{d}{dt}(v^2) \ = \ (2v)\frac{dv}{dt}\)

 

[aur_khan]

It makes sense now

Do you agree that:

\(\displaystyle \frac{d}{dt}(v^2) \ = \ (2v)\frac{dv}{dt}\)

Thank you Subhotosh it makes sense now. So basically like when we differentiate y^2 with respect to dt we get 2y.dy/dt so in a similar fashion the teacher is basically re writing that expression in a different way and multiplying it by a half as the net result will be the same as before applying the product rule.
thanks very much !!!