What am I doing wrong in this simple pressure calc?

[Bluestribute]

I'm trying to find gauge pressure at multiple points in a "mixed" container. Which, I thought would be equal to ρgh_{liquid 1} + ρgh_{liquid 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 . . .

 

[Bystander]

What's "32.1?"

 

[paisiello2]

Clearly it's g.

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

 

[Bystander]

Clearly it's g.

Clearly. And 62.4?

 

[SteamKing]

I'm trying to find gauge pressure at multiple points in a "mixed" container. Which, I thought would be equal to ρgh_{liquid 1} + ρgh_{liquid 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 . . .

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

 

[Bluestribute]

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

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!

 

[SteamKing]

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!

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.