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Derivation of ∆p = ρg∆h (CIE A Level Physics)

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Leander

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Leander

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Derivation of ∆p = ρg∆h

  • Hydrostatic pressure is the pressure at any given point within a fluid, that is exerted by the weight of the fluid above that point 
  • If the fluid is at rest, then all the points within the fluid are in equilibrium
  • Therefore, the pressure acts in all directions at each point

Pressure on a point in a fluid

4-3-3-pressure-at-a-point-within-a-fluid-cie-new

The hydrostatic pressure on area A is due to the weight, W, of the volume of fluid above it

  • The weight, W, of the fluid above area A is given by:

W space equals space m g

  • Using the density equation, mass can be given as:

rho space equals space m over V

m space equals space rho V space equals space rho A h

  • Substituting this expression for mass into the weight equation gives:

W space equals space rho A h g

  • Therefore, the pressure exerted on area A can be given as:

p space equals space F over A space equals space W over A space equals space fraction numerator rho A h g over denominator A end fraction space equals space rho h g

  • Within the volume of the cube V, there is a change in pressure between the top and bottom surfaces

Change in pressure through a volume of fluid

4-3-3-change-in-pressure-through-volume-of-fluid-cie-new

The pressure at the bottom of the cube with volume V is greater than the area A at the top of the cube, because there is an increasing amount of fluid above, which increases the force of weight W acting upon it

  • The change in pressure can be found by considering the change in height of the volume of fluid above the lower surface
  • This gives the equation for hydrostatic pressure:

increment p space equals space rho increment h g

  • Where:
    • Δp = change in pressure in pascals (Pa)
    • ρ (Greek letter rho) = density of fluid in kilograms per metre cubed (kg m-3)
    • Δh = change in height in metres (m)
    • g = gravitational field strength in newtons per kg (N kg-1)

Examiner Tip

You will be expected to remember all the steps for this derivation for an exam question.

You can recap the equations used in this derivations in the topics Density and Pressure

Using the equation for hydrostatic pressure

  • When an object is immersed in a liquid, the liquid will exert a pressure which acts in all directions, squeezing the object
  • The size of this pressure depends upon:
    • The density, ρ, of the liquid
    • The depth, h, of the object
    • The gravitational field strength, g

  • The total pressure acting on the object considers both the weight of the fluid above the object, and the weight of the air above the object
  • When asked about the total pressure, the atmospheric pressure must also be included

Total pressure = Hydrostatic pressure + Atmospheric pressure

  • Atmospheric pressure (also known as barometric pressure) is 101 325 Pa

U-tube manometer

  • A manometer is an instrument to measure pressure and density of two liquids

A U-tube manometer

U-tube diagrams, downloadable AS & A Level Physics revision notes

A U-tube manometer shows changes in pressure as the liquid moves in the tube

  • In Figure 1: The level of liquid is equal because the atmospheric pressure (Patm) is the same
  • In Figure 2: If the pressure on one side rises, the liquid will be forced down making the liquid in the other limb rise. The difference between the two levels gives the pressure difference between the two ends of the tube
  • In Figure 3: The U-tube now has two different liquids. The density of the blue one is greater than that of the orange one. The pressure at each point is due to the atmospheric pressure plus the weight of the liquid above it

Worked example

Atmospheric pressure at sea level has a value of 100 kPa. The density of sea water is 1020 kg m-3.

At what depth in the sea would the total pressure be 250 kPa?

A. 20 m               B. 9.5 m               C. 18 m          D. 15 m

Answer: 

Step 1: List the known quantities and convert to SI units:

  • Atmospheric pressure = 100 × 103 Pa
  • Total pressure = 250 × 103 Pa
  • Density of sea water, ρ = 1020 kg m-3
  • Gravitational field strength, g = 9.81 N kg-1

Step 2: Recall the equation for total pressure

total pressure = hydrostatic pressure + atmospheric pressure

Step 3: Substitute in the hydrostatic pressure equation

total pressure = rho h g + atmospheric pressure

Step 4: Rearrange for change in height 

h space equals space fraction numerator total space pressure space minus atmospheric space pressure over denominator rho g end fraction

Step 5: Substitute in the known values to calculate

h space equals space fraction numerator open parentheses 250 cross times 10 cubed close parentheses space minus space open parentheses 100 cross times 10 cubed close parentheses over denominator 1020 space cross times space 9.81 end fraction

h space equals space 15 space straight m

Examiner Tip

The values for pressure given in exam questions can vary widely and depend on metric prefixes such as kPa or MPa.

It is always safest to convert all the pressure values to SI units (Pa) before you begin the calculation, to minimise the chance of making a mistake.

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Leander

Author: Leander

Expertise: Physics

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.