Help with TBR Fluids

[hibni]

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For q.7 my reasoning was that as the hot air balloon rises higher the upward buoyant force increases. Since P=F/A if F increases then the pressure should also increase. I am not understanding why this reasoning is not correct?

For q. 28. water will flow from a high pressure to a low pressure. If water is flowing for input to output then the output end has a lower pressure. How is II correct then?

 

[rabbott1971]

Be careful with the difference between physics "pressure" and our normal idea of "pressure." In physics the larger diameter gives more pressure, whereas I always think of a pressure washer with the skinny metal spray wand giving lots of pressure, when actually this is flow velocity. Maybe that's just me, but I thought I'd mention it. Here's a page about it.

Pressure

As far as the balloon, their answer makes sense, and you don't really have to do physics. The balloon goes up, less air is stacking up on top of it, so it experiences less external pressure. Don't get formula-lock on these things, instead picture the system and what's going on. Just friendly advice, good luck.

 

[hibni]

Be careful with the difference between physics "pressure" and our normal idea of "pressure." In physics the larger diameter gives more pressure, whereas I always think of a pressure washer with the skinny metal spray wand giving lots of pressure, when actually this is flow velocity. Maybe that's just me, but I thought I'd mention it. Here's a page about it.

Pressure

As far as the balloon, their answer makes sense, and you don't really have to do physics. The balloon goes up, less air is stacking up on top of it, so it experiences less external pressure. Don't get formula-lock on these things, instead picture the system and what's going on. Just friendly advice, good luck.

That makes sense intuitively. I'm just not understanding in which situations we can use P=F/A correctly.

 

[rabbott1971]

Found this little explanation, it might help. The rest is from the link, with a couple of underlined bracketed comments by me.

Relation between pressure, velocity and area

So your problem is with P=F/A. Well, there's no problem with it. What is really wrong with your thinking is that you're not paying attention to the equation: the force changes too.

Let's recap what happens in your situation:

[referring to the diagram below]

1. There's a change in cross-sectional area: A2 < A1;
2. Thanks to conservation of mass, (1) implies v2 > v1
3. Thanks to Bernoulli, (2) implies p2 < p1

The dark blue rectangle on the left is what we call an element. Like the rest of the flow in the bigger section, it flows with velocity v1. It is delimited left and right by faces with area A1. Note that, since the liquid left and right of it has pressure p1, this element is compressed by forces F1=p1*A1 on each side. [notice the rearrangement of P = F/A]

Now to the element on the smaller section, which flows faster. Its cross-sectional area is smaller. The pressure left and right of it is also smaller. As a result, the forces compressing it, F2=p2*A2, are also smaller.

So, P=F/A still holds. Yes, when the situation changes, A is smaller, which by itself would make P bigger. However, as we saw above, then new F is smaller than the old one too, which by itself would make P smaller. The net effect of p2 > p1 (which we know beforehand from Bernoulli) means, therefore, simply that F has diminished more than A did.

 

[hibni]

That makes so much sense. Thanks so much!!