Bernoulli&#x27;s equation problem help

You've got Bernoulli-

p + \frac{1}{2}\rho{v^2} + \Phi = const

for some constant. Using that, you can get the equation

v_2^2 - v_1^2 = something

Now you have to consider the mass flow rate: mass flow in the first section equals the second. That gives you

v_2(v_1)

. That should lead you to a solution.

The more formal equation to use I would think is mass continuity

\partial_t\rho + \nabla\cdot{\rho\vec{v}} = 0

but that's just the same as eyeballing the mass flow rate, anyhow. (you'd take it into the integral form for that.)

(edited 2 years ago)

Reply 2

OP

Original post by Callicious

You've got Bernoulli-

p + \frac{1}{2}\rho{v^2} + \Phi = const

for some constant. Using that, you can get the equation

v_2^2 - v_1^2 = something

Now you have to consider the mass flow rate: mass flow in the first section equals the second. That gives you

v_2(v_1)

. That should lead you to a solution.

The more formal equation to use I would think is mass continuity

\partial_t\rho + \nabla\cdot{\rho\vec{v}} = 0

but that's just the same as eyeballing the mass flow rate, anyhow. (you'd take it into the integral form for that.)

Are you suggesting that I rearrange the mass flow rate equation to make V2 the subject and then sub that back into V2^2-V1^2?

Reply 3

Original post by Soul Wavel3ngth

Are you suggesting that I rearrange the mass flow rate equation to make V2 the subject and then sub that back into V2^2-V1^2?

I'm likely wrong mind you- I've never had to actually deal with this kind of stuff. Try giving it a go though? (I've tried but haven't gotten the right answer, so yeah. Will try again and feed back)

Reply 4

OP

Original post by Callicious

I'm likely wrong mind you- I've never had to actually deal with this kind of stuff. Try giving it a go though? (I've tried but haven't gotten the right answer, so yeah. Will try again and feed back)

Ah ok. Btw I forgot to mention that the value of G in this sheet is 9.807

Reply 5

Original post by Soul Wavel3ngth

Ah ok. Btw I forgot to mention that the value of G in this sheet is 9.807

Okay well.. the only experience I have with this is a one-shot back in 1st year. Anyhow... I really do think that this should give the right answer (though I can't manage it for some reason-) The reason I got it wrong was I used a circular pipe- I always forget that Bernoulli eqs like scaling things purely along one axis :lol:

Unparseable latex formula:

\displaystyle \frac{1}{2}\rho{v_2^2} + p_1 = \frac{1}{2}\rho{v_1^2} + p _2 \\[br]p_1 - p_2 = \frac{1}{2}\rho\left(v_2^2 - v_1^2\right) \\ [br]\partial_t\rho + \nabla\cdot\rho\vec{v} = 0 \\ [br]0 + \oint \rho\vec{v}\cdot{d\vec{s}} = 0 \\ [br]\rho{v_1}{A_1} = \rho{v_2}{A_2} \\ [br]v_1\left(\frac{d_1}{d_2}\right)^2 = v_2 \\ [br]\therefore p_1 - p_2 = \frac{1}{2}\rho{v_1^2}\left(\left(\frac{d_1}{d_2}\right)^4 - 1\right)[br]

That's for a circular pipe. I'm an idiot! :lol:

If instead of using an area proportional to the diameter squared, and instead use it just proportional to diameter,

p_1 - p_2 = \frac{1}{2}\rho{v_1^2}\left(\left(\frac{d_1}{d_2}\right)^2 - 1\right)

I do get the right answer.

You don't need g at all for this: this is a horizontal pipe. For more info, see https://www.researchgate.net/publication/280717530_First_Theoretical_Constructions_to_the_Fluid_Mechanics_Problem_of_the_Discharge.

(edited 2 years ago)

Reply 6

OP

Original post by Callicious

Okay well.. the only experience I have with this is a one-shot back in 1st year. Anyhow... I really do think that this should give the right answer (though I can't manage it for some reason-) The reason I got it wrong was I used a circular pipe- I always forget that Bernoulli eqs like scaling things purely along one axis :lol:

Unparseable latex formula:

\displaystyle \frac{1}{2}\rho{v_2^2} + p_1 = \frac{1}{2}\rho{v_1^2} + p _2 \\[br]p_1 - p_2 = \frac{1}{2}\rho\left(v_2^2 - v_1^2\right) \\ [br]\partial_t\rho + \nabla\cdot\rho\vec{v} = 0 \\ [br]0 + \oint \rho\vec{v}\cdot{d\vec{s}} = 0 \\ [br]\rho{v_1}{A_1} = \rho{v_2}{A_2} \\ [br]v_1\left(\frac{d_1}{d_2}\right)^2 = v_2 \\ [br]\therefore p_1 - p_2 = \frac{1}{2}\rho{v_1^2}\left(\left(\frac{d_1}{d_2}\right)^4 - 1\right)[br]

That's for a circular pipe. I'm an idiot! :lol:

If instead of using an area proportional to the diameter squared, and instead use it just proportional to diameter,

p_1 - p_2 = \frac{1}{2}\rho{v_1^2}\left(\left(\frac{d_1}{d_2}\right)^2 - 1\right)

I do get the right answer.

You don't need g at all for this: this is a horizontal pipe. For more info, see https://www.researchgate.net/publication/280717530_First_Theoretical_Constructions_to_the_Fluid_Mechanics_Problem_of_the_Discharge.

Ah I see thanks, Idk why my professor included stuff in a worksheet that's intended to only cover the stuff he's been through with us lmao

Reply 7

Original post by Soul Wavel3ngth

Ah I see thanks, Idk why my professor included stuff in a worksheet that's intended to only cover the stuff he's been through with us lmao

What's he covered? If you're dealing with Bernoulli then you will definitely have covered continuity...

Reply 8

OP

Original post by Callicious

What's he covered? If you're dealing with Bernoulli then you will definitely have covered continuity...

I don't recall us ever going over the horizontal pipe vs circular pipe stuff lol, I'll re-check some of the lectures to see if I missed out anything in my notes

Reply 9

Original post by Soul Wavel3ngth

I don't recall us ever going over the horizontal pipe vs circular pipe stuff lol, I'll re-check some of the lectures to see if I missed out anything in my notes

The circular pipe thing was all me-

Now that I recall, we never covered circular pipes- we only scaled them in one axis (i.e. the diameter doubled or halved with the same height of the pipe, etc.) I just didn't remember and assumed a circular pipe for my solution, which wasn't what the question wanted (they assumed the normal single-axis scaling.)

Reply 10

OP

Original post by Callicious

Okay well.. the only experience I have with this is a one-shot back in 1st year. Anyhow... I really do think that this should give the right answer (though I can't manage it for some reason-) The reason I got it wrong was I used a circular pipe- I always forget that Bernoulli eqs like scaling things purely along one axis :lol:

Unparseable latex formula:

\displaystyle \frac{1}{2}\rho{v_2^2} + p_1 = \frac{1}{2}\rho{v_1^2} + p _2 \\[br]p_1 - p_2 = \frac{1}{2}\rho\left(v_2^2 - v_1^2\right) \\ [br]\partial_t\rho + \nabla\cdot\rho\vec{v} = 0 \\ [br]0 + \oint \rho\vec{v}\cdot{d\vec{s}} = 0 \\ [br]\rho{v_1}{A_1} = \rho{v_2}{A_2} \\ [br]v_1\left(\frac{d_1}{d_2}\right)^2 = v_2 \\ [br]\therefore p_1 - p_2 = \frac{1}{2}\rho{v_1^2}\left(\left(\frac{d_1}{d_2}\right)^4 - 1\right)[br]

That's for a circular pipe. I'm an idiot! :lol:

If instead of using an area proportional to the diameter squared, and instead use it just proportional to diameter,

p_1 - p_2 = \frac{1}{2}\rho{v_1^2}\left(\left(\frac{d_1}{d_2}\right)^2 - 1\right)

I do get the right answer.

You don't need g at all for this: this is a horizontal pipe. For more info, see https://www.researchgate.net/publication/280717530_First_Theoretical_Constructions_to_the_Fluid_Mechanics_Problem_of_the_Discharge.

I was just looking back at this thread and wanted to know why the pgz(1) and pgz(2) were omitted from the bernoulli's equation you set up in step 1 of your calc?

Reply 11

Original post by Soul Wavel3ngth

I was just looking back at this thread and wanted to know why the pgz(1) and pgz(2) were omitted from the bernoulli's equation you set up in step 1 of your calc?

Because gravitational potential energy matters not with this problem. Bernoulli is energy conservation- if we don't use gravity, it needn't be kept in the equation since it all cancels out.

Reply 12

OP

Original post by Callicious

Because gravitational potential energy matters not with this problem. Bernoulli is energy conservation- if we don't use gravity, it needn't be kept in the equation since it all cancels out.

Ah ok thanks for the clarification, I'm assuming this would be the opposite when it comes to inclined pipes? Also when it comes to area/diameter stuff for inclined pipes, should you treat it as being the same as circular or horizontal pipes?

Reply 13