Calculate Pressure from Height & Specific Gravity

[TyErd]

Homework Statement

question is in the diagram.
basically I am not sure how to use the specific gravity and convert it into pressure.

Homework Equations

P=p(rho)gh

The Attempt at a Solution

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[tiny-tim]

Hi TyErd!

(have a rho: ρ )

Use the specific gravity to find ρ, the density …

ρ-of-oil = ρ-of-water times specific-gravity-of-oil

(so specific gravity of water = 1)

 

[TyErd]

ohhh! so the density of any liquid is density of water x specific gravity of the liquid?

 

[TyErd]

in knowing that my working out is where density of water is 1000...: 80 + 1000(0.3)(9.81) - 1000(13.6)(9.81)(h) - 1000(0.72)(9.81)(0.75) = P_atmospheric. now what, i have two unknowns or do i assume P_atm is 101.325?

 

[TyErd]

even if i do use 101.325 the answer is still wrong... i don't know why.

 

[tiny-tim]

Hi TyErd!

(just got up :zzz: …)

ohhh! so the density of any liquid is density of water x specific gravity of the liquid?

Yup!

in knowing that my working out is where density of water is 1000...: 80 + 1000(0.3)(9.81) - 1000(13.6)(9.81)(h) - 1000(0.72)(9.81)(0.75) = P_atmospheric. now what, i have two unknowns or do i assume P_atm is 101.325?

ah, the question says the gage pressure (or gauge pressure) is 80 KPa …

gauge pressure is defined as actual pressure minus atmospheric pressure …

so you can ignore the atmospheric pressure! ​

 

[TyErd]

oh yeah so then what does all of that equal? it should equal the pressure at the top of the oil column so would that just be the 0 because we can assume the mass of air negligible?

 

[tiny-tim]

oh yeah so then what does all of that equal? it should equal the pressure at the top of the oil column so would that just be the 0 …

yes

the sum of the mgh's has to equal the difference between the outside air and the measurer (80 KPa) …

you can think of them as being an initial and final pressure, much like initial and final energy in a "solids" equation

… because we can assume the mass of air negligible?

no, the mass of air is very heavy, it's because the 80 KPa is given as the difference from that

 

[TyErd]

okay so i did this: 80 + 1000(0.3)(9.81) - 1000(13.6)(9.81)(h) - 1000(0.72)(9.81)(0.75) =0, h=-0.017047 which is incorrect. where have i gone wrong?

 

[tiny-tim]

80,000 ?

 

[TyErd]

even then its incorrect. the answer is h=0.582m

 

[tiny-tim]

even then its incorrect. the answer is h=0.582m

I do get 0.582

Please show your full calculations.