Solve Integral of dp^2 | Lower Boundary Ps, Upper P0

[hermano]

intgral of dp^2??

Hi,

I have the following equation:

\dot{m} = c \frac{dp^{2}}{dx}

Integration gives:

\dot{m} \int dx = c \int dp^{2}

How can I solve \int dp^{2}??

The lower boundary is Ps and the upper boundary is P0 for the integral!

 

[tiny-tim]

hi hermano!

do you mean dq/dx where q = p² ?

just integrate over q from q = (p_s)² to q = (p₀)²

 

[hermano]

Hi tiny-tim ;-)

I want to integrate dp^2! So you take some parameter q and say that this parameter q is equal to p^2 and then solve the integral of dq instead of dp^2. This gives q again and thus p^2! Then fill in the upper and lower boundary for p: P0^2 - Ps^2

Am I right?

 

[tiny-tim]

if I'm correctly understanding your original equation, yes

 

[hermano]

My original question is how to solve this equation:

\dot{m} \int dx = c \int dp^{2}

My problem is solving the right hand side of the equation namely:

c \int dp^{2}

Your answer is: replace p^{2} by q and solve this 'normal' integral c \int dq! This gives c q, then replace q by p^{2} again so the solution is c p^{2}. Then I only have to fill in the boundaries for p. Right?

 

[tiny-tim]

I think so, yes.

 

[K^2]

dp²/dx is an infinitesimal quantity. On the left side of equation, you have a finite quantity. Something is wrong.

You either have (dp/dx)² or d²p/dx². In either case, review your equation. Where do you get it from, anyways?

 

[hermano]

Hi K^2,

This equation comes from Darcy's law namely \dot{m} = \frac{p Q}{R T}

Fill Q (volumetric flow rate form Darcy's law) in in the above equation gives:

\dot{m} = c p \frac{dp}{dx}

With c all the constants in the equation.

p \frac{dp}{dx} is equal to \frac{1}{2} \frac{dp^2}{dx}

And thus finally I got \dot{m} = \frac{c}{2} \frac{dp^2}{dx}

Solving this:

\dot{m} \int dx = \frac{c}{2} \int dp^{2}

Therefore my question: How to solve \int dp^{2} ??

 

[tiny-tim]

ah, now i see where it comes from!

yes, the whole point of changing pdp/dx to 1/2 d(p²)/dx

(and btw, it's better to write it that way, with brackets)

is because that is a perfect derivative, and you can immediately integrate it to p_f² - p_i².

Incidentally, the same trick enables us to go from F = ma (in mechanics) via a = vdv/dx to ∆(1/2 mv²) ​

 

[hermano]

Tiny-tim,

Do I make a fault against the mathematics by saying that \dot{m} = \frac{c}{2} \frac{dp^2}{dx} can be solved through \dot{m} \int dx = \frac{c}{2} \int dp^{2}??

Then still my question is: how to solve the right hand site of this equation namely the integral of \int dp^{2} ??

 

[K^2]

Yeah, you shouldn't write it as dp²/dx, because that means (dp)²/dx, and that just isn't right. d(p²)/dx, on the other hand, does make sense, and d(p²) can be integrated over using variable substitution.

\int f(u)du = \int f(x)\frac{du}{dx}dx

Which in this case, turns integral over d(p²) into integral over 2p dp.

 

[tiny-tim]

hi hermano!

(btw, we say "a fault in the mathematics" … though you can say "a crime against mathematics"! )

Do I make a fault against the mathematics by saying that \dot{m} = \frac{c}{2} \frac{dp^2}{dx} can be solved through \dot{m} \int dx = \frac{c}{2} \int dp^{2}??

No, we often treat a derivative exactly like a fraction, and multiply both sides by dx (and then integrate) …

that's fine ​

Then still my question is: how to solve the right hand site of this equation namely the integral of \int dp^{2} ??

∫ d(p²) = [p²]

 

[hermano]

Thanks, now it is clear for me!

Maybe you both can help me with another small issue I'm facing with this equation! For the integral of my pressure, I have to fill in the upper and lower boundary for p^2! However, I doubt of the input pressure is my upper or lower boundary, and also the same for the output pressure. What tells me which pressure (input - output) is the upper and lower boundary for my integral?

The equation gives the mass flow through a pipe (Poiseuille) from the high input pressure to the lower ouput pressure.