I have a question regarding Reynolds number and equation and hydrostatics. I know that from the book and the content and lectures that the value of 20,000 Pa is generally the threshold from whether we determine that the flow of an incompressible fluid will become turbulent instead of laminar. As far as the equation goes, is this something we should just go ahead and memorize or is that something that is lower yield?
My other question is about hydrostatics with gauge pressure. In the Fluids 2 content review video, the parts where we are talking about a buildup of pressure from above when it is not the atmosphere with air pockets in the cave for example, why is it in those scenarios that the P above cannot be simply 1atm? And also with regards to the gauge pressure with distance, when do we use the value given from the bottom versus the top of the given scenario?
Thank you very much for your help.
 Shaun
reynolds number

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Re: reynolds number
It's not too likely that you'd be asked to use that equation. It's more likely, if that topic were to be tested at all, that a conceptual understanding would be required rather than a specific equation like that memorized. It'd be good to know the relationships and the general trends that the equation predicts/reflects, since that'd be enough to answer the question even if they did ask something specific (you could use the answer choices and your conceptual understanding to determine which is correct).
The pressure can't be Patm in this case because the two pressures must be equal (otherwise the trapped air pocket would be bigger or smaller). So, P1=P2. Then, we can expand those values:
P1above (atm pressure) + rho(g)z1 = P2above + rho(g)z2
Then, we can see that z2=0 since that point is 0 meters below the surface. Thus,
P2above = P1above + rho(g)z1
This means that P2above must be larger than P1above, since we're adding something to P1above to get P2 above!
Gauge pressure is basically the pressure value without atmospheric pressure included. You'd use the distance from the surface of the fluid to calculate pressure.
I hope this helps!
The pressure can't be Patm in this case because the two pressures must be equal (otherwise the trapped air pocket would be bigger or smaller). So, P1=P2. Then, we can expand those values:
P1above (atm pressure) + rho(g)z1 = P2above + rho(g)z2
Then, we can see that z2=0 since that point is 0 meters below the surface. Thus,
P2above = P1above + rho(g)z1
This means that P2above must be larger than P1above, since we're adding something to P1above to get P2 above!
Gauge pressure is basically the pressure value without atmospheric pressure included. You'd use the distance from the surface of the fluid to calculate pressure.
I hope this helps!