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Question #593 from EK Physics:
This is a tricky problem. I really need some guidance here.
Alright so before solving anything, I realized that there is a pressure gradient between both tanks. This pressure gradient results in a forward velocity from Tank B (higher gauge pressure) to Tank A.
But the issue I'm having is figuring out how to express this quantitatively:
Using Bernoulli's Equation which basically assumes conservation of energy, I can solve for this velocity.
P(a) + pgh(a) + 1/2pv(a)^2 = P(b) + pgh(b) + 1/2pv(b)^2
Alright so I'm not really sure what the P's stand for here. (Is it static pressure?) Well, I assuming they're unknown so I'll leave them in the equation.
Relative to the connecting tube, both tanks have potential energy, so I'll leave both of those terms.
For both tanks, I'll assume that they begin in static equilibrium and take velocity as 0 m/s, which cancels out both terms on either side of the equation.
So solving for the unknowns, we have P(a)-P(b) = pgh(b) - pgh(a) or
Pressure Difference = pg[h(b) - h(a)]
Pressure Difference = pg[5m]
To be honest, at this point I'd have no idea what to do next. But according to the back of the book, which I don't completely understand why, we take this Pressure Differential and equate it 1/2pv^2. Then we solve for Velocity.
The unknown density cancels out resulting in velocity equating the square root of (10 x 5 x 2 ) or 10 m/s.
-------------------------
Can someone please tell me if my logic sounds right in solving this problem?
Why do we equate the pressure differential to 1/2pv^2? Quantitatively, is there anyway someone can explain this. I think I have some idea intuitively.
This is a tricky problem. I really need some guidance here.
Alright so before solving anything, I realized that there is a pressure gradient between both tanks. This pressure gradient results in a forward velocity from Tank B (higher gauge pressure) to Tank A.
But the issue I'm having is figuring out how to express this quantitatively:
Using Bernoulli's Equation which basically assumes conservation of energy, I can solve for this velocity.
P(a) + pgh(a) + 1/2pv(a)^2 = P(b) + pgh(b) + 1/2pv(b)^2
Alright so I'm not really sure what the P's stand for here. (Is it static pressure?) Well, I assuming they're unknown so I'll leave them in the equation.
Relative to the connecting tube, both tanks have potential energy, so I'll leave both of those terms.
For both tanks, I'll assume that they begin in static equilibrium and take velocity as 0 m/s, which cancels out both terms on either side of the equation.
So solving for the unknowns, we have P(a)-P(b) = pgh(b) - pgh(a) or
Pressure Difference = pg[h(b) - h(a)]
Pressure Difference = pg[5m]
To be honest, at this point I'd have no idea what to do next. But according to the back of the book, which I don't completely understand why, we take this Pressure Differential and equate it 1/2pv^2. Then we solve for Velocity.
The unknown density cancels out resulting in velocity equating the square root of (10 x 5 x 2 ) or 10 m/s.
-------------------------
Can someone please tell me if my logic sounds right in solving this problem?
Why do we equate the pressure differential to 1/2pv^2? Quantitatively, is there anyway someone can explain this. I think I have some idea intuitively.
