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- Thread starter MrMechanic
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- #2

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So, now calculate the pressures at 33m in each, then at 30m in each, etc.

- #3

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I'll start with the left tank

P = Patm + (0.82)(9.81)(38-30)

is it correct?

- #4

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If by Patm you mean the pressure of the air in that tank, that term is ok. But watch out for the units in the other term. What does "specific gravity" mean exactly?

I'll start with the left tank

P = Patm + (0.82)(9.81)(38-30)

is it correct?

- #5

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yes If Patm =14.7 psi = 101.325kPa = 101.325kN/m^2

and

P = 101.325 + (0.82)(9.81(38-30)

P = 36.9714 kPa

What i don't know is how to do I form an equation where i can get the height (h)

- #6

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OK, but you need to specify the reference substance. Did you meanSpecific gravity is the ratio of the density of a substance to the density. that's why it has no unit

Specific gravity is the ratio of the density of a substance to the density **of water**.

?The density of that substance should therefore figure in the formula. As it stands you have (spec grav) * (acceleration) * ( height), which has dimension L

If you continue to work out the pressure at each height, working down from the top, and put in an unknown h for the height you need, you eventually get to two expressions for the pressure right at at the bottom. They must be equal, of course.What i don't know is how to do I form an equation where i can get the height (h)

- #7

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(-40) + (0.82)(9.81)(38-30) + (1.5) (9.81) (h) - (9.81)(h+3) - 20.6 = P1

and i'll perform another on the other side? or should I set P1 = 0?

- #8

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When I perform P1 = 0

I get h = 5.235

I get h = 5.235

- #9

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Two problems.

(-40) + (0.82)(9.81)(38-30) + (1.5) (9.81) (h) - (9.81)(h+3) - 20.6 = P1

and i'll perform another on the other side? or should I set P1 = 0?

1. I had been ignoring the minus sign in "-30cm Hg". I don't understand how an absolute air pressure can be negative. These are sealed containers, so it shouldn't be relative to ambient pressure. Do you have any explanation for that?

2. You are ignoring my comment about density and units. If the s.g. is 0.82 relative to water, what is the density in kg/m

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The density of oil is not given. And also gasoline. In order to get their density you have to multiply their specific gravity to the density of water which is 9.81kg/m^3

- #11

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No, that's where you are going wrong. The 9.81 is gravitational acceleration. That factor converts mass to weight. The density of water is much larger.to get their density you have to multiply their specific gravity to the density of water which is 9.81kg/m^3

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Oh yeah sorry. It's not kg/m^3 ... It's 9.81 kN/m^3 forgot to change that sorry

- #13

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Right, but answer my other question; what is the density of water in these units?Oh yeah sorry. It's not kg/m^3 ... It's 9.81 kN/m^3 forgot to change that sorry

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