Calculate Pressure from Height & Specific Gravity

In summary, the homework statement asks how to convert specific gravity into pressure. The student attempted to solve the equation using the specific gravity of oil and the density of water to calculate pressure, but ran into difficulty. The student is not sure why the answer is incorrect.
  • #1
TyErd
299
0

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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  • #2
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
ohhh! so the density of any liquid is density of water x specific gravity of the liquid?
 
  • #4
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
even if i do use 101.325 the answer is still wrong... i don't know why.
 
  • #6
Hi TyErd! :smile:

(just got up :zzz: …)
TyErd said:
ohhh! so the density of any liquid is density of water x specific gravity of the liquid?

Yup! :biggrin:
TyErd said:
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
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
TyErd said:
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
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
80,000 ? o:)
 
  • #11
even then its incorrect. the answer is h=0.582m
 
  • #12
TyErd said:
even then its incorrect. the answer is h=0.582m

I do get 0.582 :confused:

Please show your full calculations.
 

1. How is pressure calculated from height and specific gravity?

Pressure can be calculated using the formula P = hρg, where P is pressure, h is height, ρ is density (specific gravity), and g is the acceleration due to gravity.

2. What is the unit of measurement for pressure in this calculation?

The unit of measurement for pressure in this calculation is typically expressed in pascals (Pa) or newtons per square meter (N/m²).

3. Can this calculation be used for any type of fluid?

Yes, this calculation can be used for any type of fluid as long as the density (specific gravity) and acceleration due to gravity are known.

4. How does height and specific gravity affect pressure?

The height and specific gravity both play a role in determining the pressure of a fluid. As the height increases, the pressure also increases. Similarly, as the specific gravity increases, the pressure also increases.

5. How accurate is this calculation for determining pressure?

This calculation can provide a reasonably accurate estimate of pressure, but it may not take into account certain factors such as changes in temperature or the presence of other substances in the fluid. For more precise measurements, a more comprehensive equation or experimental data may be needed.

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