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Pressure and height

  • Thread starter TyErd
  • Start date
  • #1
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Homework Statement


question is in the diagram.
basically im 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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Answers and Replies

  • #2
tiny-tim
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Hi TyErd! :smile:

(have a rho: ρ :wink:)

Use the specific gravity to find ρ, the density …

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

(so specific gravity of water = 1)
 
  • #3
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ohhh! so the density of any liquid is density of water x specific gravity of the liquid?
 
  • #4
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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?
 
Last edited:
  • #5
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even if i do use 101.325 the answer is still wrong... i don't know why.
 
  • #6
tiny-tim
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Hi TyErd! :smile:

(just got up :zzz: …)
ohhh! so the density of any liquid is density of water x specific gravity of the liquid?
Yup! :biggrin:
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! :wink:
 
  • #7
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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?
 
  • #8
tiny-tim
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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 …
yes :smile:

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 :wink:
… 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 :wink:
 
  • #9
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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?
 
  • #10
tiny-tim
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80,000 ? o:)
 
  • #11
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even then its incorrect. the answer is h=0.582m
 
  • #12
tiny-tim
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even then its incorrect. the answer is h=0.582m
I do get 0.582 :confused:

Please show your full calculations.
 

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