TyErd
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Calculate Pressure from Height & Specific Gravity
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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)
(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
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ohhh! so the density of any liquid is density of water x specific gravity of the liquid?
TyErd
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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?
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TyErd
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even if i do use 101.325 the answer is still wrong... i don't know why.
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Hi TyErd!
(just got up :zzz: …)
Yup!
ah, the question says the gage pressure (or gauge pressure) is 80 KPa …
gauge pressure is defined as actual pressure minus atmospheric pressure …
(just got up :zzz: …)
TyErd said:
Yup!
TyErd said:
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
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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?
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TyErd said:
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
no, the mass of air is very heavy, it's because the 80 KPa is given as the difference from that
TyErd
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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?
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80,000 ? 
TyErd
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even then its incorrect. the answer is h=0.582m
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TyErd said:
I do get 0.582
Please show your full calculations.