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8.1.3Fluids

Fluid Pressure Equations — AP PHYSICS 1 Revision Notes & Topical Questions

5 min read 9 parts Lesson 5 of 14 1 quiz

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Lesson summary

8.1.3 Fluid Pressure Equations — AP PHYSICS 1 notes: Use P = P0 + ρgh for the pressure at depth in a static fluid and apply it to forces, manometers and hydraulic systems.Origin — a column of area A and height h has weight ρAhg, so the extra pressure at its base is ρgh.Gauge pres…

  • Gauge pressure
  • Quantities
  • Why P0 counts
  • Linear in depth
  • Same depth, same pressure
  • Independent of volume or width

Section 1 of 9Pressure and depth

Origin
  • Origin a column of area A and height h has weight ρAhg, so the extra pressure at its base is ρgh.
Gauge pressure
  • Pgauge = ρgh, from the weight of the fluid column above the point.
P
P0 + ρgh
  • absolute pressure at depth h, with P0 the pressure at the surface (atmospheric, about 1.0 × 105 Pa for an open container).
Quantities
  • ρ in kg/m3, g = 9.8 N/kg, h in m.
Why P0 counts
  • Why P0 counts an incompressible fluid transmits any surface pressure to submerged objects.

Section 2 of 9Consequences

1Linear in depth
  • Linear in depth a graph of P against h is a straight line with gradient ρg and intercept P0.
2Same depth, same pressure
  • Same depth, same pressure in a connected static fluid, whatever the container shape.
3Independent of volume or width
  • Independent of volume or width a thin tube and a wide lake of equal depth give the same bottom pressure.
4Independent of the object
  • Independent of the object a cardboard box and a steel cube at the same depth feel the same pressure.
5All directions
  • All directions pressure at a point is the same in every direction.

Section 3 of 9Pascal's principle and hydraulics

Pascal's principle
  • Pascal's principle a pressure change in an enclosed incompressible fluid is transmitted undiminished to every point and to the container walls.
Hydraulic lift
  • F1/A1 = F2/A2, so force is multiplied by the area ratio.
Energy
  • Energy the small piston moves farther, F1d1 = F2d2, so no energy is created.

Section 4 of 9Manometers and barometers

Pressure difference
  • ΔP = ρgΔh between two depths in the same fluid.
U-tube manometer
  • U-tube manometer the height difference between the arms is proportional to the pressure difference.
Mercury barometer
  • Patm = ρHggh, so h = 1.01 × 105/(13 600 × 9.8) = 0.76 m.
Water barometer
  • Water barometer would need to be about 10 m tall.

Section 5 of 9Layers of fluids

Immiscible layers
  • P = P0 + ρ1gh1 + ρ2gh2.
Example
  • 0.10 m oil (800 kg/m3) on 0.30 m water gives 784 + 2940 = 3.7 × 103 Pa gauge at the bottom.
Arrangement
  • Arrangement the dense fluid sits at the bottom and each interface is a horizontal surface of constant pressure.
U-tube with two liquids
  • U-tube with two liquids equate the pressures at the interface level down each arm.

Section 6 of 9Applying the equation

1
Decide whether absolute pressure (include P0) or gauge pressure (only ρgh) is needed.
2
Measure h vertically from the free surface, not along a slanted path.
3
For a force use F = PA, with the average pressure or centre pressure for a vertical surface.
4
Check units: kg/m3 × N/kg × m = N/m2 = Pa.

Section 7 of 9More from the specification

More from the specification
Continuity equation in incompressible fluids
  • Continuity equation in incompressible fluids applying mass conservation across flowing fluids provides a reliable analytical means to calculate laminar stream velocity.
Ideal fluid
  • Incompressible fluid with no viscosity.
Density
  • Mass per unit volume.
Perpendicular
  • At right angles.
Displacement
  • Change in position.
Component
  • Part of a vector along one axis.

Section 8 of 9Key terms

TermMeaning
Hydrostatic pressurePressure in a static fluid from the weight above, ρgh.
Gauge pressurePressure relative to atmospheric, ρgh.
Pascal's principlePressure changes in an enclosed fluid are transmitted undiminished.
Hydraulic liftDevice using Pascal's principle to multiply force.
ManometerDevice measuring pressure differences with a liquid column.

Section 9 of 9Exam technique

1
Forgetting to add atmospheric pressure when absolute pressure is asked.
2
Using the length of a tilted tube instead of the vertical depth.
3
Thinking pressure depends on volume or container shape.
4
Using the area ratio the wrong way round in hydraulic problems.
5
Thinking a hydraulic lift gives free energy.

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