Pressure in a Fluid

The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged

prev | up | next

Discussion

theory

Pressure in a uniform fluid.

The absolute pressure in a uniform fluid at a particular depth is given by …

P = P0 + ρgh

The absolute pressure in a uniform or nonuniform fluid at a particular depth h measured along the vertical or z-axis is given by …

  h  
 P = P0 +  ⌠
⌡		 ρ(z)g(z)dz
  0  

barometer

barometer, manometer, Hare's apparatus

physiology

blood pressure

Circulatory Pressures (mm Hg)
location systolic diastolic mean
aorta 120 80 100
left ventricle 120 8 –
left atrium 7 10 4
pulmonay artery 15 7 12
right ventricle 15 2 –
right atrium 4 4 0
pulmonary capillary wedge 7 10 4
Source: Physics of the Body

ear pressure in the middle ear: eardrum at end of outer ear connected to smaller oval window at beginning of inner ear. 15-30 times greater pressure. combination of difference in membrane diameters and lever effects of middle ear bones.

eye pressure and glaucoma

pascal's principle

Pascal's principle: Pressure changes applied to the surface of an enclosed fluid are transmitted evenly throughout the fluid.

hydraulics

Summary

Problems

practice

  1. The first measurements of blood pressure were made in 1726 by the English botanist, physiologist, and clergyman, Stephen Hales. Hales performed several experments on horses deemed "unfit for service". You must recall that at the time horses were primarily used as working animals. Those that were seriously injured, chronically ill, or otherwize unable to perform their duties were routinely slaughtered and eaten. Read Hales' description of one such experiment.

    In December I laid a common Field Gate on the Ground, with some Straw upon it, on which a white Mare was cast on her right Side, and in that Posture bound fast to the Gate; she was fourteen Hands and three Inches high [150 cm]; lean, tho' not to a great Degree, and about ten or twelve Years old. This and the above-mentioned Horse and Mare were to have been killed, as being unfit for Service ….

    Then laying bare the left Carotid Artery, I fixed to it towards the Heart the Brass Pipe, and to that the Wind-Pipe of a Goose; to the other End of which a Glass Tube was fixed, which was twelve Feet nine Inches long [388 cm]. The Design of using the Wind-Pipe was by its Pliancy to prevent the Inconveniencies [sic] that might happen when the Mare struggled; if the Tube had been immediately fixed to the Artery, without the Intervention of this pliant Pipe.

    There had been lost before the Tube was fixed to the Artery, about seventy cubick Inches of Blood [1.15 L]. The Blood rose in the Tube in the same manner as in the Case of the two former Horses, till it reached to nine Feet six Inches Height [290 cm]. I then took away the Tube from the Artery, and let out by Measure sixty cubick Inches of Blood [0.98 L], and then immediately replaced the Tube to see how high the Blood would rise in it after each Evacuation; this was repeated several times, till the Mare expired ….

    Given that blood has a density of about 1035 kg/m3 determine the blood pressure of this poor, unfortunate horse.
    Solution …
    • Use the formula for the gauge pressure in a uniform fluid, take the maximum height of the column of blood, and solve.
       
      ΔP = ρgΔh = (1035 kg/m3)(9.8 m/s2)(3.88 m) = 39,354.84 Pa
      ΔP = 39 kPa
       
      Compared to the typical human values of 10 to 16 kPa for arterial blood pressure, this result seems reasonable. Horses are much bigger than people and thus need a generally higher arterial pressure to squeeze the blood to every distant nook and cranny. Also, blood pressure is generally higher when an animal is under stress. Slowly bleeding to death is definitely a stressful situation. This method of determining blood pressure is called invasive catheterization and is almost never used. Blood pressure is now routinely determined by much less deadly means.
  2. Determine the maximum height that a lift pump can raise water from a well.
    Solution …
    • A lift pump works by reducing the pressure above of a column of water. The greatest difference possible would be atmospheric pressure at the bottom and vacuum at the top. Set this pressure difference equal to the pressure difference within the column of water and solve for height.
             
      P0 =  ρgh
      h =  P0  =  (101,325 Pa)
      ρg (1000 kg/m3)(9.80 m/s2)
      h =  10.3 m
             
  3. When the human body is accelerated vertically, blood pressure in the brain will drop. Determine the maximum vertical acceleration that a human can withstand before losing consciousness; that is, determine the acceleration that would reduce the blood pressure in the brain to zero. Assume a typical systolic pressure of 16 kPa and that the base of the brain is 20 cm above the top of the heart. (Blood has a density of about 1035 kg/m3).
    Solution …
    • In this problem we are given pressure difference, height difference, and density and are asked to find acceleration.
         
      ΔP =  ρg′Δh
      g′ =  ΔP  =  (16,000 Pa)  = 77.294 … m/s2
      ρΔh (1035 kg/m3)(0.20 m)
      g′ ≈  8 g
         
      The value g′ is the apparent acceleration due to gravity. Since gravity is already pulling us down with 1 g, the absolute acceleration that a human could withstand is on the order 7 g. Since the height difference was measured from the bottom of the brain, 7 g would be the acceleration at which the brain was entirely emptied of blood. The actual acceleration that would induce unconsciousness would be somewhat lower and would be preceded by a period of greyout and then blackout as the visual cortex was drained of blood. With training and special clothing, it is possible to remain conscious at accelerations greater than what we just calculated. At the 2002 Ilopango Airshow in El Salvador, aerobatic pilot Greg Poe pulled a maximum of 11.4 g apparent acceleration for a second or two at the start of a very rapid ascent. This is the current record for a civilian pilot and may be an overall record. Since most air forces keep this kind of information classified, we can't be sure.
  4. Astronomical Pressures.
    1. Derive an expression for the pressure in a spherical, astronomical body with uniform density.
    2. Use this formula to estimate the pressure at the center of …
      1. the earth
      2. the sun
    Solutions …
    1. We simplify things a bit by assuming a constant density …
             
      ρ =  m  =  3m
      V 4πr3
             
      but we can't do the same for gravity. On astronomical scales, gravity varies considerably. This means it's time to reach for a calculus-based solution. We'll begin by determining just how gravity varies. Start with Newton's law of universal gravitation …
         
      g(r) =  Gm(r)
      r2
         
      On the surface of the earth we'd use the whole mass of the earth in this equation, but inside the earth we use only the fraction that's at a greater depth; that is, at a distance r from the center of the earth smaller than the the radius R of the whole earth. The mass of this portion can be found by multiplying density and volume …
                 
      m(r) = ρV(r) =  3m   4πr3  =  mr3
      4πR3 3 R3
                 
      which makes sense. Mass varies as the cube of length, so the fraction should be some sort of ratio of the cubes. Now, substitute and simplify.
                 
      g(r) =  G   mr3  =  Gmr
      r2 R3 R3
                 
      Pressure in a fluid (yes, I know the earth is mostly solid, but the equation works) is the weight of the fluid above a surface divided by the area of the surface. The surface can have any area and, through the magic of algebra, disappears from the equation so that we are left with the product of density (ρ), gravity (g), and height (h in swimming pools and blood vessels, r in astronomical situations like this). Now for the calculus. You can't assign a value for gravity in this situation. It varies from 9.8 m/s2 on the surface to zero at the center. We can reduce the amount of variation if we examine just a part of this total distance (Δr). We can reduce it even more if we examine an even smaller part. And we can reduce the variation to nothing if we examine an infinitesimal part (dr). Now the product of density (ρ), gravity (g), and height (dr) works again. All we have to do is add up the contributions to the pressure made by the infinite number of infinitesimal parts from the surface of the earth down to its center. The process of adding infinitesimals is called integration.
        R     R               R
      P =  ⌠
      ⌡      		 ρg(r)dr  =  ⌠
      ⌡      		 3m   Gmr  dr =  3Gm2 ⎡
      ⎣      		 r2 ⎤
      ⎦
      4πR3 R3 4πR6 2
        r     r               r
      and here's our equation …
           
      P =  3Gm2  (R2 − r2)
      8πR6
           
      which reduces to …
         
      P0 =  3Gm2
      8πR4
         
      at the center where r = 0.
    2. Let's do it.
      1. For the earth …
               
         P0 =  3Gm2  =  3(6.67 × 10−11 Nm2/kg2)(5.98 × 1024 kg)2
        8πR4 8π(6.34 × 106 m)4
               
         P0 =  1.7 × 1011 Pa = 170 GPa = 1.7 million atmospheres 
               
        The actual value is closer to 360 GPa or about twice the value calculated above, which is annoyingly big, but at least we got the right order of magnitude. To do this correctly, we'd have to account for variations in density with depth. The density of the earth starts at about 2300 kg/m3 at the crust, increases (nonuniformly) with depth in the mantle, jumps drastically at the outer core where it nearly doubles, and keeps increasing (nonuniformly) hitting a maximum of 12,580 kg/m3 at the center.
      2. For the sun …
               
         P0 =  3Gm2  =  3(6.67 × 10−11 Nm2/kg2)(1.99 × 1030 kg)2
        8πR4 8π(6.96 × 108 m)4
               
         P0 =  1.3 × 1014 Pa = 130 TPa = 1.3 billion atmospheres 
               
        The actual value is on the order of 100 to 300 billion atmospheres. I wonder what's going on here?

numerical

  1. Determine the pressure under the thickest part of the Antarctic ice cap (4776 m) in kPa and atm.

calculus

  1. The data in the text file earth.txt gives the density and gravitational field strength of the earth at various depths below the surface. Using data analysis software (preferably something that can do numerical integration) generate a data column for the the pressure at various depths below the surface. The value in the center of the core will be on the order of 360 GPa, so you can ignore the contribution of atmospheric pressure in your calculations.

Resources

prev | up | next

Another quality webpage by

Glenn Elert		 eglobe logo home | contact

bent | chaos | eworld | facts | physics