Bernoulli equation, question regarding flow speed EK

[samiam99]

Here is the question:

it's from examkracker :
7. If both ends of the pipe were opened, all of the following would decrease significantly at point B as the unkown fluid drained from the pipe EXCEPT:

a) Volume flow rate

b) Fluid Velocity**** I'm stuck on this. I know of the equation v=√2gD- but this only applies when we are looking at a hole in say a tank, where the pressure at that point is equal to the atm pressure.
-I'm now sure where to go from here.

c) Fluid density- this is the answer, which makes sense

d) Fluid pressure- this would decrease according the equation P=density(depth)g- since as the fluid level drops the depth is also dropping

thanks for the help!

The pipe
7.

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[type12]

Your reasoning for D is the reason for B: imagine popping a hole into a full container of water. First, it's a burst of fluid, but as the height of the fluid goes down, it turns into a slow drip. v=2gh, and h is decreasing as fluid is being drained.

Normally, when you are using v=2gh, you are looking at a moment, because the height continually drops.

 

[samiam99]

Your reasoning for D is the reason for B: imagine popping a hole into a full container of water. First, it's a burst of fluid, but has the height of the fluid goes down, it turns into a slow drip. v=2gh, and h is decreasing as fluid is being drained.

Normally, when you are using v=2gh, you are looking at a moment, because the height continually drops.

thanks! so basically speed of flow is driven by height?

 

[type12]

thanks! so basically speed of flow is driven by height?

Yes, this stems from Bernoulli's equation: the energy density of a system cannot change (as far as the MCAT is concerned):
K (of state 1) = K (of state 2)
--> K is made of P + pgh + 1/2pv^2
(P1 + pgh1 + 1/2pv1^2) = (P2 + pgh2 + 1/2pv2^2)
--> normally, you look at an instantaneous state change (popping a hole), so P1 has not had a chance to change, and is equal to P2
pgh1 + 1/2pv1^1/2 = pgh2 + 1/2pv2^2
--> Bernoulli's principle actually states that a change in velocity happens at the cost of potential energy (and pressure, but that's beyond the MCAT), and in our first state, the fluid wasn't moving (before be popped the hole
pgh1 + 0 = 0 + 1/2pv2^2

And now you can see where v=rad(2gh) comes from.

 

[samiam99]

Yes, this stems from Bernoulli's equation: the energy density of a system cannot change (as far as the MCAT is concerned):
K (of state 1) = K (of state 2)
--> K is made of P + pgh + 1/2pv^2
(P1 + pgh1 + 1/2pv1^2) = (P2 + pgh2 + 1/2pv2^2)
--> normally, you look at an instantaneous state change (popping a hole), so P1 has not had a chance to change, and is equal to P2
pgh1 + 1/2pv1^1/2 = pgh2 + 1/2pv2^2
--> Bernoulli's principle actually states that a change in velocity happens at the cost of potential energy (and pressure, but that's beyond the MCAT), and in our first state, the fluid wasn't moving (before be popped the hole
pgh1 + 0 = 0 + 1/2pv2^2

And now you can see where v=rad(2gh) comes from.

that makes sense. I was looking at it this way.

(P1 + pgh1 + 1/2pv1^2) = (P2 + pgh2 + 1/2pv2^2)

I looked at P1 as the pressure at the surface which would be athmospheric pressure. And P2 as the pressure at point B which is also the reference level y2=0.
and P2 as the pressure at point B which be Patm+Pgauge.

can assume v1= zero because the drop in liquid level is slow
(P1 + pgh1 ) = (P2 + 1/2pv2^2)
p1+pgh1-p2=1/2pv2^2

from this I figure that pgh1 is decreasing as is p2 , therefore v2 is decreasing.

doesn't v=rad(2gh) only hold true, for the speed of flow of a stream that is coming out of a hole, because it that case the pressure in the stream would be =to athmospheric pressure.