Calculating Pressure Difference Between Sections 1 and 2

timothy1875
Messages
9
Reaction score
0
Trying to calculate pressure difference between sections 1 and 2.
1=32mm 2=20mm
density = 1000 kg/m^3

I have done P1-P2 = 1/2 density x (vi^2 - vj^2)

Is this right?

How do I now find head loss if difference between 1 and 2 were 45mm ?

45/density x gravity ?

Thanks,
 
Physics news on Phys.org
Your equation for P1 - P2 looks OK (in the absence of friction and assuming the discharge coefficient = 1).

The relation between static pressure difference and fluid column height ("head loss") is:

P2 - P1 = density X g X h1 - h2. Is your head loss 45 mm or was that just an example?
 
Ye 45mm is the head loss but it's theoretical and not related to the initial equation. Would the formula be h1 - h2 = p2 - p1 / density x g ?

Thanks for your help
 
timothy1875 said:
Ye 45mm is the head loss but it's theoretical and not related to the initial equation. Would the formula be h1 - h2 = p2 - p1 / density x g ?

Thanks for your help
Yes.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top