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Topic 4 (Fluid Density & Pressure , and mass Flow & Energy)

Where we deal with stuff that is spread out

Quiz 4 was Thr.Mar.29 . . . here is my grading key as *.jpg image

vector quantities (directional arrows, like momentum or Force) spread over a Surface Area (2-d)
the important Area is perpendicular to the vector arrows
. . . Force arrows that pierce an Area contribute to the Force density, that is spread over that (2 dimensional) Area
      Force density is usually called "Pressure" because it presses the Surface
. . . momentum that goes thru a surface, is spread over that surface Area as a momentum density
      momentum density is changed (time-wise) by Pressure exerted for that time duration
      - but the stuff was moving, so the new momentum density is at a different place by then.

Massive objects occupy Volume

=> m_object   =   ρ_material · V_object   . . .   ρ = "mass density"

condensed matter occupies Volume because their atoms touch one another
      in solids, each is stuck to its neighbors, so they retain their relative positions
      in liquids, each molecule can slide past its neighbors in any direction with little resistance ... they can also rotate (spin)
      in gasses, the molecules are usually far from one another, with empty space between them, so seldom touch (bounce)
. . . look up mass densities of elements and compounds in Tables
      dry air 1.3 kg/m³ . . . gasoline 680 kg/m³ . . . liquid nitrogen 808 kg/m³ . . .oil 900 kg/m³
      ice 920 kg/m³ . . .tapwater 1000 kg/m³ . . . seawater 1030 kg/m³ . . . liquid oxygen 1140 kg/m³
      graphite 2100 kg/m³ . . . NaCl 2160 kg/m³ . . . Al 2700 kg/m³ . . . Ca 1550 kg/m³
      Fe 7800 kg/m³ . . . Cu 8900 kg/m³ . . . Pb 11300 kg/m³ . . . Hg 13500 kg/m³ . . . Au 19300 kg/m³ . . . Pt 21400 kg/m³

different kinds of atoms are not exactly the same size . . . using V = m / ρ .
V_Pb compared to V_O . . . = (m_Pb / m_O ) / ( ρ_Pb / ρ_O )
      = ( 207 / 16 ) / ( 11300 / 1140 )   =   12.9 / 9.9   =   1.3 × as large in Volume .
      diameter goes as the cube root of Volume . . . diameter of lead atom is 1.09 × the diameter of an Oxygen atom .
V_C / V_O = (m_C / m_O ) / ( ρ_C / ρ_O ).
      = ( 12 / 16 ) / ( 2100 / 1140 )   =   0.75 / 1.84   =   0.41 × .
      Carbon atom diameters are only ³√(.4) = 3/4 as large as an Oxygen atom.
. . . even these extreme cases are not tremendously different in diameter . . .
=> atoms with more "atomic mass" (more protons & neutrons) make objects with higher mass densities . . . almost proportional.

Pressure is Force Spread over an Area ; Force is aggregate Pressure thru that Area

=> P = F / A   . . . or . . . F = P · A

the Force pierces the Area that the Pressure presses on.
      from the subject, F arrows go thru the entire contact Area, into the object.
. . . the full Force that the subject applies comes from adding up the Force thru each square inch of contact.
      (or whatever Area units your Pressure is in terms of ... they're on the bottom!)
. . . the fluid presses outward in all directions . . . the object gets squeezed inward by the fluid from all directions

in the US, pressure is reported in pounds per square inch ... the metric unit (Newtons/meter) is also called a "Pascal"
. . . "standard atmospheric pressure" is 14.7 lb/in² (psi) . . . 101,325 N/m² . . . 1.013 "bar" . . . 760 Torr . . . 29.9 "inches of mercury"
. . . typical house water in its pipes is 30 - 60 psi more intense than the atmospheric pressure outside the pipes.
      this internal overpressure is called "gauge" pressure ... the Pressure your tire gauge reads, also
=> P_gauge = P_inside − P_outside . . . P_inside   plumbing ≈ 45 - 75 psi.

Pressure-caused Force is added (just like all other applied Forces) to yield the total (net) Force
an adult male African elephant's mass is about 6000 kg ; each foot is elliptical, about .4 m wide × .5 m long.
. . . the ground pushes up through 4 of these   (π/4 · .4 m × .5 m =)   0.16 m² foot-pad Areas => 0.63 m² total (on all fours)
      total support Force from the ground must average opposite the elephant's weight   mg = 60,000 N .
=> P = 60,000 N / 0.63 m² = 95,500 N/m² . . . less than an atmosphere .

Fluid Pressure accumulates with depth in the fluid
consider a jar of water ... the Force by the jar's bottom needs to hold up (against gravity) the weight of all the water
. . . jar's F = P·A = m g(↑) = ρ·V g(↑) . . . = ρ·A×d g(↑) . . . where d is water depth .
=> water's P = ρ·d g . . . (F by water to jar is opposite the jar's Pressure applied to the water ... Newton's 3rd Law)
      . . . this of course presumes that the fluid is not accelerating !

1 Torr is the Pressure caused by a 1 mm deep column of mercury
Pressure .001 m deep in mercury should be   (using static Pressure formula above)
. . . P = ρ·d g = 13500 kg/m³ · .001 m · 9.8 N/kg = 132 N/m² . . . not very much pressure
      760 mm deep in Hg would have   P = ρ·d g = 13500 kg/m³ · .760 m · 9.8 N/kg = 102600 N/m²   . . . = 1 atmosphere
. . . human blood pressure (at heart level) should be about 120 Torr when the heart pumps (systolic)
      blood pressure at your feet will be ≈ 1060 kg/m³ · 1.3 m · 9.8 N/kg = 13,500 N/m² ≈ 100 Torr more intense than this (when standing)

example: depth of Earth's atmosphere
. . . atmospheric Pressure is caused by the weight of Earth's atmosphere . . . 101,000 N above each square meter of surface
. . . air has a noticeable density here at the surface due to being compressed by this weight . . . 1.3 kg in each cubic meter.
. . . we can find out how deep the atmosphere is (or would be, if it had sea-level density all the way up):
      P = ρ·d g = 101,000 N/m² = 1.3 kg/m³ · d · 9.8 N/kg  
=> d = 7930 m   . . . only 5 miles deep !   if its density was uniform ... (but gas density is less at high altitude where P less intense)

Gas Pressure is proportional to density
. . . so, half-way to the top of the atmosphere, the Pressure would be half of the Pressure at sea level,
      so the density there is only half the density at sea level
. . . so that ½-way point is HIGHER than 4 km up ... turns out that 8 km up still has 1/2.72 of atmospheric Pressure
      even 16 km up has 1/(2.72)² = 13% of sea-level's Pressure and density.

Suppose you have a huge cubic balloon 1 m × 1 m × 1 m , fully submerged in water.
the Pressure at the balloon bottom is ρ·d g = 1000 kg/m³ · 1 m · 9.8 N/kg   = 9800 N/m²   more than at the top.
      => F_buoyant = 9800 N/m² · 1 m² = 9800 N (↑).
. . . this is, of course the weight of the cubic meter of water that the ballon has displaced (pushed out of the way)
      since the balloon itself only weighs 1.2 kg/m³ · 1 m³ · 9.8 N/kg   = 11.76 N (↓)
. . . the balloon will accelerate upward rapidly unless held down by some VERY strong Force

how deep does the balloon rest, in the water, while it is floating?
when floating normally , F_buoyant = − m_balloon g = ρ_water (A × d_bottom) g . . . d_bottom = (1.2 kg) / (1000 kg/m³ 1 m²) = .0012 m .
=> the fraction of Volume that is submerged ... is equal to the ratio of densities .

Fluid Pressure can do Work , so is a Potential Energy density

=> P·ΔV = W , Work done by Pressure = "Pressure Potential Energy"

Hole in a water bottle ... hi Pressure inside pushes water toward low pressure outside
      the stronger Force   (P·A_hole) does positive Work -
      the weaker Force does less negative Work
. . . the total Work done by the Pressure difference, as the water moves distance Δx ,
      becomes Kinetic Energy in the water as it spurts out the hole.
      the deeper the water above the hole, the faster the water will emerge from the hole.
. . . partly block the end of a garden hose with your thumb, causing less water to come out;
      less water flow implies less friction, so higher Pressure behind the thumb
      higher Pressure in the hose makes water come out at faster speed.
      (but less water per second, because the Area is smaller ... mass/second = ρ v·A)
=> P_inside − P_outside = ½ ρ v ²   . . . Kinetic Energy density.

Bernoulli equation is Energy conservation for fluids
the starting KE density + PE density , has Work density done to it by flowing to less intense Pressure , to obtain its KE + PE density there .
½ ρ v ² + ρ g h + P = ½ ρ v ² + ρ g h + P

cautions: . . . If some effect is caused by Bernoulli's KE density, it cannot depend on momentum's direction
      . . . most fluid flow is dominated by viscosity (wet friction), so Mechanical Energy becomes Thermal Energy

for gases,   P V = N k_B T  
the Energy density called "Pressure", is multiplied by Volume , to be an Energy.
. . . it is the potential Work that the gas might be able to do, if it were released.
What's Temperature doing there? . . . The right-hand side is 2/3 × the random KE of the gas atoms.
. . . (2/3 is the fraction that hit the 2-dimensional surface Area; 3/3 would be all the KE, in the 3-d Volume)
k_B is "Boltzmann's konstant", which tells how much Energy each atom has, at that Temperature
. . .   = 1.38E−23 Joule/Kelvin ... ( a room-Temperature gas atom actually has 3/2 k_B T = 8.14 E−21 Joule, at T = 293 K )
If you double the number of molecules   N , held in the same Volume   V,
      the Pressure will double since there are twice as many molecules bounce off the walls (recall Δp / Δt )
If you double the Temperature, the molecules must be moving faster (KE → 2 KE means v → 1.414v)
      so each molecule bounces off the walls 1.414× as often, and it exchanges 1.414× as much momentum (with the wall)
. . . so double Temperature leads to   (1.414×1.414 = ) 2× the Force from each molecule (F = Δp / Δt ).
If you squeeze the Volume down to ½ as much, with a piston (say) , its Pressure would double also (density doubles)
      unless the Work your piston does to the gas also increases its Temperature (which it probably will) - but that's Topic 5.

Mass is conserved . . . so usually the same amount of mass flows into a pipe section, as flows out of it

=> Δ m / Δ t = ρ v · A   . . .   mass per Volume times Volume flow rate  

slow flow can "almost" have constant Energy density . . . Bernoulli eq'n

=> ½ ρ v ² + ρ g h + Pressure . . . is constant or uniform