Pressure

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Discussion

definition

Pressure is the ratio of force applied per area covered …

P =  F
A

The unit of pressure is the pascal

⎡
⎣		 Pa =  N  =  kg m/s2  = kg ⎤
⎦
m2 m2 m s2

The pascal is also a unit of stress and the topics of pressure and stress are connected.

gauge vs. absolute

spring loaded pressure gauge

Selected Gauge Pressures (black - positive, red - negative)
atm kPa device, event, phenomena, process
200 20,000 pressurized breathing apparatus
150 15,000 milk homogenization
110 11,000 rupture compression strength of vertebral disks
7 - 14 700 - 1400 puffed cereal manufacture
9 900 espresso machine
4 - 7 400 - 700 bicycle tire
2.7 - 4.1 275 - 415 champagne at serving temperature (10 °C)
2.7 275 carbonated soft drinks
2.0 - 2.5 200 - 250 car tire
> 4 > 400 blast wave, 100% lethality
2.3 - 4.0 230 - 400 blast wave, 50% lethality
1.6 - 2.3 160 - 230 blast wave, 1% lethality
1.02 103 typical household pressure cooker
1 101.325 one standard atmosphere over environment
  47 bottom of feet while standing
  20 lungs, extreme exhalation
  17 sustained pressure, eardrum ruptures
  8 sustained pressure, eardrum senses pain
  7 - 14 aircraft shock wave
  13 - 19 blood pressure, arterial, systolic (during a heartbeat)
  8 - 12 blood pressure, arterial, diastolic (between heartbeats)
  8.8 blowing your nose
  11 eye, severe glaucoma
  1.6 - 3.0 eye, normal
  4.0 blood pressure, capillary, arterial end
  1.3 blood pressure, capillary, venous end
  15 bladder, voiding, maximum
  3 bladder, micturition reflex ("gotta go urge")
  2 - 4 bladder, voiding, sustained
  1.3 - 2.6 gastrointestinal tract
  0.6 - 1.6 cerebrospinal fluid
  0.4 - 0.9 blood pressure, venous
  0.6 - 0.8 interstitial fluid (osmotic pressure)
  2 acoustic pressure, eardrum ruptures (160 dB)
  0.02 acoustic pressure, eardrum senses pain (120 dB)
  2 × 10−8 acoustic pressure, threshold of hearing (0 dB)
0 0 environmental pressure
  -1.3 lungs, resting
  -1.5 lungs, drinking through a 15 cm straw
  -25 lungs, extreme inhalation
-1 -101.325 one standard atmosphere below environment
a perfect vacuum in a standard atmosphere environment

 

Selected Absolute Pressures
atm kPa device, event, location, phenomena, process
3.4 × 1011   center of the sun
????   center of Jupiter
3.6 × 106   center of earth
1070   Marianas Trench, Pacific Ocean (-10,924 m)
160   Lake Baikal, Asia (-1620 m)
140   Lake Tanganyika, Africa (-1470 m)
40   Lake Superior, North America (-406 m)
????   record dive by a human
90   surface of Venus
26   helium freezes at about 1 K
  108.38 record high, altitude adjusted (Siberia, 1968)
  106 dead sea (-400 m)
1 101.325 sea level, standard atmosphere
  90 atmospheric pressure at 1000 m, interior of concorde
  88.8 record low, altitude adjusted (Hurricane Gilbert, 1988)
  80 atmospheric pressure at 2000 m, interior of typical jet aircraft
  52 La Paz, Bolivia (5200 m)
~ ½ 51 maximum altitude of permanent human habitation (5500 m)
  ~ 40 vertical limit of human survivability (~7000 m)
~ ⅓ 31 Mount Everest (8848 m)
~ ⅕ 19 altitude of typical commercial jet aircraft (12,000 m)
> 0.033  > 3.3  "low vacuum"
< 0.033  < 3.3  "medium vacuum"
0.025 2.2 altitude of reconnaissance plane (SR-71, 26,000 m)
0.011 1.1 altitude of highest skydive (1960, 31,330 m)
0.006 0.6 altitude of highest manned balloon flight (1961, 34,668 m)
0.007   surface of Mars
~ 10−5   surface of Pluto, maximum (late 1990s)
< 10−6    "high vacuum"
< 10−9    "very high vacuum"
~ 10−13   surface of the moon, daytime
< 10−12    "ultra high vacuum"
~ 10−15   surface of the moon, nighttime
< 10−15    "extreme ultrahigh vacuum"
~ 10−17   I am told that below this value all vacuum equipment leaks.

the atmosphere

Standard Atmospheric Tables

Chemical Composition of the Atmosphere
gas formula molecular weight
(g/mol)		 fraction
nitrogen N2 028.0134000 0.78084
oxygen O2 031.9988000 0.209476
argon Ar 039.9480000 0.00934
carbon dioxide CO2 044.0099500 0.000314
neon Ne 020.1830000 0.00001818
helium He 004.0026000 0.00000524
methane CH4 016.0430300 0.000002
krypton Kr 083.8000000 0.00000114
hydrogen H2 002.0159400 0.0000005
xenon Xe 131.3000000 0.000000087
overall – 028.9644253 0.999997147
Source: US Standard Atmosphere (1976)

Summary

Problems

practice

  1. Determine the mass of the earth's atmosphere.
    Solution …
    • Since pressure is force divided by area, the force of the atmosphere pressing on the surface of the earth can be found by multiplying standard atmospheric pressure by the surface area of the earth. Given that the force of an object's weight is its mass times the acceleration due to gravity, the mass of the earth's atmosphere is the force it exerts divided by the acceleration due to gravity. Or symbolically …
                 
      F = PA = P(4πr2) = W = mg
      m =  W  =  P(4πr2)  =  (101,325 Pa)(4π)(6.37 × 106)2
      g g 9.8 m/s2
      m =  5.27 × 1018 kg
                 
  2. Write something.
    • Answer it.
  3. Write something.
    • Answer it.
  4. Write something completely different.
    • Answer it.

conceptual

  1. A question about American football. How is it that a weekend carpenter can apply a relatively small amount of force on a 5 pound hammer and drive a nail though a board, but a 300 pound tackle running at top speed will never be able to pierce the helmet of a quarterback?

numerical

  1. A car can be weighed by measuring the "footprint" of each tire and multiplying by that tire's gauge pressure. Determine the weight of this car (in English units) using the data in the following table. (Adapted from: Beakman's World, Episode 401 "Sweat, Beakmania and Weighing a Car" ca. 1995-1996.)
     
    Determine the Weight of This Car
    tire length (in.) width (in.) pressure (p.s.i.) weight (lbs)
    front right 6 4 24
    front left 6 4 24
    rear right 5½ 4 22
    rear left 5½ 4 22
    total → → →
     

investigative

  1. Obtain a pen-sized tire pressure gauge for a car or bicycle — the kind with a cylindrical body and sliding, calibrated rod. When the gauge is connected to the tire valve, the air inside the tire pushes a piston attached to a spring and the calibrated rod. The whole apparatus moves until the force of the air escaping from the tire equals the force of the spring pushing back. (For an in-depth description with illustrations see How a Tire Pressure Gauge Works.)
    1. Measure the diameter of the bore.
    2. Choose a convenient pressure value and measure the length of the stem from that value to the zero.
    3. Calculate the spring coefficient from these two measurements and the pressure value you chose.

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