# Homework Help: What am I doing wrong in this simple pressure calc?

1. Sep 13, 2015

### Bluestribute

I'm trying to find gauge pressure at multiple points in a "mixed" container. Which, I thought would be equal to ρghliquid 1 + ρghliquid 2 etc. Well, that's not right? And I don't know why?

So for point A, I did: (55.1 * 32.1 * 4) + (62.4 *32.1 * 4). Then I divided by 144 to get it into psi. I got 105. That's not right . . .

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2. Sep 13, 2015

### Bystander

What's "32.1?"

3. Sep 13, 2015

### paisiello2

Clearly it's g.

Does the extra weight of the column of water @ B add pressure to the column of water @ A?

4. Sep 13, 2015

### Bystander

Clearly. And 62.4?

5. Sep 13, 2015

### SteamKing

Staff Emeritus
For the figures in the diagram for water and oil, γ = ρg, so you don't need to multiply γw = 62.4 lbf / ft3 by g ...

6. Sep 13, 2015

### Bluestribute

Wait, that's what that letter means? So it's just ϒh + ϒh?

EDIT: Jeez that would have been helpful to know. Or know to infer . . . Yes, just do that to get psf and convert to psi to get the right answer . . . Wow. Thanks. But what about B? The only thing above B is air . . . and they don't give any constants for air. Should that just be known (because it isn't negligible . . . I tried 0 psi with no luck).
LAST EDIT: Work backwards from A and subtract. Got it!

Last edited: Sep 13, 2015
7. Sep 13, 2015

### SteamKing

Staff Emeritus
That's one of the things about working in Imperial versus working in SI. In Imperial, you get accustomed to working with weight instead of mass, as in SI.

Fresh water weighs 62.4 lbf / ft3. You can work back to find the mass density in slugs / ft3, which is approximately 2.

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