High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > Fluid Mechanics

Pressure

Stability and BalancePrevious Section
Pascal's PrincipleNext Section
Popular Courses
Find My Course
0:00 / 0:00

Density of a Fluid


Density is defined as the mass per unit volume.
ρ=mV\boxed{\rho=\dfrac{m}{V}}

The unit for density is kg/m3kg/m^3

Exam Tip
  • You can use following change in unit of Volume to find other density units:

1m3=106cm3=106mL=1000L1m^3=10^6cm^3=10^6mL=1000L

  • The density of water is worth knowing:1g/cm3=1000kg/m31g/cm^3=1000kg/m^3. Technically it only takes this value at 4 degrees Celsius, but at other "everyday" temperatures, this is close enough for calculations.


Wize Concept
The specific gravity of an object is its density divided by the density of water (1000 kg/m3).


0:00 / 0:00

Mass and Density

The Mass of an object measures the amount of material of the object. The Density of an object measures the mass per volume of the object.

Density

For an object with mass MMand volume VV, having constant density ρ\rho

ρ=MV\boxed{\displaystyle \rho=\frac{M}{V}}

Mass Integrals

For an object with cross-sectional area A(x)A(x)and density ρ(x)\rho(x)on the interval [a,b][a,b], the mass MMis
M=∫abρ(x)A(x)dx\boxed{\displaystyle M=\int_a^b\rho(x)A(x)dx}

Wize Concept
For 3-D solids, A(x)A(x) can be determined using the methods of volume of revolution by cross sections.

0:00 / 0:00

Pressure


Pressure is defined as the normal force per unit of area of surface:


P=FA\boxed{P=\frac{F}{A}}


Pressure is a scalar quantity. It involves the molecules or atoms of the fluid striking the container walls.

The SI unit of pressure is the Pascal: 1 Pa = 1 Nm21\ Pa\ =\ 1\ \frac{N}{m^2}


Hydrostatic Pressure

Hydrostatic pressure is due to the weight of a column of fluid in stationary fluids:


P=mgA=[ρ(h×A)]gA=ρghP=\dfrac{mg}{A}=\dfrac{[\rho\left(h\times A\right)]g}{A}=\rho gh

whereAA is the cross sectional area of the column, hh is the height of column and ρ\rhois the density of the fluid.


Wize Concept
  • Fluids (gases and liquids) assume the shape of their containers.
  • For this course, we assume that liquids are incompressible (I.e. their density does not depend on pressure).


The variation of pressure with depth for an incompressible liquid is given by:


P=Po+ρgh\boxed{P=P_o+\rho gh}
  • P0P_0 is the pressure at the surface.
  • hh is the depth of the liquid column between PP and P0P_0.

Air Pressure

Air pressure on the surface of the earth is assumed to be constant and is defined to be equal to one atmosphere, 1 atm1\ atm(defined at sea level):

Pair=1atm=101325Pa=101.325 kPaP_{air}=1atm=101325Pa=101.325\ kPa

Exam Tip
For many exam problems it is enough to consider 1atm1 atm to be 105Pa10^5 Pa.

Gauge pressure is defined as the absolute pressure subtracted by the air pressure.

Pgauge=Pabsolute−Pair\boxed{P_{gauge}=P_{absolute}-P_{air}}














Example: Gauge Pressure


At depth hh within a saltwater solution, the gauge pressure is 33 times the atmospheric pressure. At what depth will the total pressure be 66 times the atmospheric pressure?


At depth hh, we have:

Pgauge=ρgh=3PatmP_{gauge}=\rho gh=3P_{atm}


Therefore the atmospheric pressure can be expressed in terms of the depth as follows:

Patm=ρgh3\bct {P_{atm}=\dfrac{\rho gh}{3}}


At the unknown depth h′h', we have:

Ptotal=Patm+PgaugeP_{total}=P_{atm}+P_{gauge}

6Patm=Patm+ρgh′6P_{atm}=P_{atm}+\rho gh'

5Patm=ρgh′5P_{atm}=\rho gh'

5⋅ρgh3=ρgh′5\cdot\bct {\dfrac{\rho gh}{3}}=\rho gh'

53 h=h′\dfrac{5}{3} \ h=h'

Therefore, the new depth has to be 53\dfrac{5}{3} of the original depth.


Practice: Vacuum Lifter

A vacuum lifter consists of five suction pads that are attached to a frame and a vacuum pump that keeps the pressure under each pad at 0.250.25 atm. The diameter of each pad is 1010 cm. What is the heaviest piece of glass that can be lifted using this vacuum lifter?