Help Understanding Volume Flow Rate / Bernoulli's

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Understanding the relationship between volume flow rate and Bernoulli's equation involves recognizing that the cross-sectional area (A) and velocity (v) are inversely proportional; if A2 is 1/4 of A1, then v2 will be 4 times v1. The term A in the equations refers to the cross-sectional area of the pipe, which can be calculated as A = πr² for circular pipes. When applying Bernoulli's equation, the ratio A1/A2 may not appear explicitly because the equation focuses on energy conservation rather than flow continuity. Volumetric flow rate is defined as volume per unit time, and it can be expressed as the product of cross-sectional area and fluid velocity. Clarifying these concepts can help resolve confusion in applying them to specific problems.
KendrickLamar
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Hey guys, just have a question because I'm having a little trouble understanding like how exactly to use the volume flow rate with bernoulli's equation. (wasnt sure which section to put this in but is more of a concept question using homework examples)

Like the volume flow rate i understand A1v1=A2v2
for example i was doing this problem : https://www.physicsforums.com/showthread.php?t=80584

Well its inversely proportional right but (im just being stupid probably because I'm really tired) but if A2 is 1/4 the size of A1 for example, then v2 is going to be 4 x the size of V1? like

A1v1 = 1/4A2v2 so 4A1/v1 = A2v2 then v2 = 4(A1/A2)(v1), well like in the problem i linked, what happens to that A1/A2? why does it disappear when being plugged into bernoulli's equation?

Also another question is the A = cross sectional area right? what's that even mean like is that the same as surface area, or regular area, or what? because in a problem like this https://www.physicsforums.com/showthread.php?t=65810 (now I am not sure if the member solved it 100% correct) but the person here puts Volume Flow Rate = pir^2v1 for example, what is the volume flow rate in this case i don't understand and is A just the area of a circle on the tube , even tho its open its not used like a cylinder's area?

thanks if anyone can help me out I am like brain dead at the moment.
 
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If A_2 is 1/4 A_1 then yes, V_2 will be 4 times V_1.

I'm not really sure what you mean when you say it "disappears?"

Cross sectional area is the area of a cross section. Say you have a piece of pipe. Stand the pipe on it's end on top of a piece of paper, then trace the inside of the pipe onto the paper. Find the area of that circle (A=\pi r^2) and that's the cross sectional area.

Volumetric flow rate is a volume per unit time (e.g., \frac{m^3}{sec}) If you do unit analysis you'll see that cross sectional area (e.g., m^2) times fluid velocity (e.g., \frac{m}{sec}) gives you volumetric flow rate.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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