Equations governing cavitation, flow speed, and delta vapor pressure...

[ellenb899]

Homework Statement

I am looking for equations governing cavitation, flow speed, and delta vapor pressure, local pressure please!

Relevant Equations

Bernoulli's equations...

Looking for the flow speed for cavitation to occur

 

[Borek]

30 second googling suggests good start could be something called "cavitation number".

 

[ellenb899]

Yes I have found that. It is the rearranging for velocity as I do not have a value for cavitation number.

 

[erobz]

In a flow, if the pressure drops below the vapor pressure of the fluid cavitation can occur. There is no “cavitation velocity” as far as I remember.

 

[ellenb899]

Flow velocity of the fluid is what I am talking about

 

[erobz]

Flow velocity of the fluid is what I am talking about

but it depends on pressure, not velocity. There is no single “cavitation velocity”.

 

[ellenb899]

The question gives the 2 pressures, and asks which flow speed would cavitation be possible in the liquid.

 

[erobz]

The question gives the 2 pressures, and asks which flow speed would cavitation be possible in the liquid.

Post the whole question please. Is there a diagram?

 

[ellenb899]

No diagram. Question is as follows:
The vapor pressure of water is 23.3hPa and the total pressure is 30hPa. At which flow speed would cavitation be possible in the liquid?

 

[erobz]

No diagram. Question is as follows:
The vapor pressure of water is 23.3hPa and the total pressure is 30hPa. At which flow speed would cavitation be possible in the liquid?

Do you have a relationship for the total pressure?

 

[Lnewqban]

No diagram. Question is as follows:
The vapor pressure of water is 23.3hPa and the total pressure is 30hPa. At which flow speed would cavitation be possible in the liquid?

Please, see:
https://www.grc.nasa.gov/www/k-12/airplane/bern.html

As the flow velocity increases, so does the dynamic pressure at expense of the static pressure, if the total pressure remains the same (as the problem seems to imply).
As the velocity of the flow increases, cavitation will start when the decreasing static pressure reaches the value of the vapor pressure.

 

[ellenb899]

Do you have a relationship for the total pressure?

No I dont

 

[erobz]

No I dont

But you have in the relevant equations section "Bernoulli's"?

$$ P_s + \rho g h + \rho \frac{V^2}{2} = \text{constant} = P_{total} $$

 

[ellenb899]

That was an attempt. Referring to your first answer it is cavitation number equation that is used.

 

[erobz]

That was an attempt. Referring to your first answer it is cavitation number equation that is used.

But that is the relevant equation... Imagining a single location of a flow having total pressure $P_{total} = 30 \rm{hPa}$. The Static Pressure $P_s$ cannot drop below the vapor pressure of the liquid ( i.e. $P_{static} > P_{vapor}$), or cavitation can occur. Since we are examining at a single location, the elevation $h$ can be taken as the $0$ datum without consequence.

What are you left with using all the information?

 

[ellenb899]

$$ v = \sqrt \frac 2P p $$

 

[erobz]

$$ v = \sqrt \frac 2P p $$

Lets take this one step at a time:

Start with:

$$ P_{static} > P_{vapor} $$

The left hand side $P_{static}$ can be represented in terms of $P_{total}$ and $V$ using the equation for $P_{total}$ in post #13. What is that result?

P.S. thank you for using $LaTeX$

 

[ellenb899]

$$ \sqrt{\frac{2p}{p}} = \frac{p_{\text{static}}}{v} $$

 

[erobz]

$$ \sqrt{\frac{2p}{p}} = \frac{p_{\text{static}}}{v} $$

It's an inequality. We need to end with $V < \text{something}$. And, that is not one step...

Substitute into the inequality for $P_{static}$ using the equation in post#13 ( it labled $P_s$ in that post) . We are trying to get rid of $P_{static}$ in the inequality by replacing it with variables we know $P_{total}$ and variables we want to know $V$.

Please just write the resulting inequality of that step.

$$ \cdots > P_{vapor}$$

fill in the blank $\cdots$ on the left hand side (LHS).

 

[ellenb899]

$$ \frac{p_{\text{static}}}{v} > P_{\text{vapour}} $$

But if looking for v value, it cannot be an inequality?

 

[erobz]

$$ \frac{p_{\text{static}}}{v} > P_{\text{vapour}} $$

Incorrect. There should be nothing on the lefthand side other than the variables $P_{total}$ and $V$. try again.

But if looking for v value, it cannot be an inequality?

You have to think a bit here about what things mean. You're just ( rather poorly - without care) throwing variables around...its a waste of time.

 

[ellenb899]

$$ \frac{p_{\text{total}}}{V} > P_{\text{vapour}} $$

 

[erobz]

$$ \frac{p_{\text{total}}}{V} > P_{\text{vapour}} $$

No. use the equation in post #13... This step needs to look like:

$$ \overbrace{\text{something} \pm \text{something}}^{ = P_{static}} > P_{vapor} $$