# 11.6 Gauge pressure, absolute pressure, and pressure measurement  (Page 3/7)

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## Systolic pressure

Systolic pressure is the maximum blood pressure.

## Diastolic pressure

Diastolic pressure is the minimum blood pressure.

## Calculating height of iv bag: blood pressure and intravenous infusions

Intravenous infusions are usually made with the help of the gravitational force. Assuming that the density of the fluid being administered is 1.00 g/ml, at what height should the IV bag be placed above the entry point so that the fluid just enters the vein if the blood pressure in the vein is 18 mm Hg above atmospheric pressure? Assume that the IV bag is collapsible.

Strategy for (a)

For the fluid to just enter the vein, its pressure at entry must exceed the blood pressure in the vein (18 mm Hg above atmospheric pressure). We therefore need to find the height of fluid that corresponds to this gauge pressure.

Solution

We first need to convert the pressure into SI units. Since $1.0 mm Hg=\text{133 Pa}$ ,

$P=\text{18 mm Hg}×\frac{\text{133 Pa}}{1.0 mm Hg}=\text{2400 Pa}\text{.}$

Rearranging ${P}_{\text{g}}=\mathrm{h\rho g}$ for $h$ gives $h=\frac{{P}_{\text{g}}}{\mathrm{\rho g}}$ . Substituting known values into this equation gives

$\begin{array}{lll}h& =& \frac{\text{2400 N}{\text{/m}}^{2}}{\left(1\text{.}0×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)}\\ & =& \text{0.24 m.}\end{array}$

Discussion

The IV bag must be placed at 0.24 m above the entry point into the arm for the fluid to just enter the arm. Generally, IV bags are placed higher than this. You may have noticed that the bags used for blood collection are placed below the donor to allow blood to flow easily from the arm to the bag, which is the opposite direction of flow than required in the example presented here.

A barometer is a device that measures atmospheric pressure. A mercury barometer is shown in [link] . This device measures atmospheric pressure, rather than gauge pressure, because there is a nearly pure vacuum above the mercury in the tube. The height of the mercury is such that $\mathrm{h\rho g}={P}_{\text{atm}}$ . When atmospheric pressure varies, the mercury rises or falls, giving important clues to weather forecasters. The barometer can also be used as an altimeter, since average atmospheric pressure varies with altitude. Mercury barometers and manometers are so common that units of mm Hg are often quoted for atmospheric pressure and blood pressures. [link] gives conversion factors for some of the more commonly used units of pressure.

Conversion factors for various pressure units
Conversion to N/m 2 (Pa) Conversion from atm
$1.0 atm=1\text{.}\text{013}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1.0 atm=1\text{.}\text{013}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$
$1.0\phantom{\rule{0.25em}{0ex}}{\text{dyne/cm}}^{2}=0\text{.}\text{10}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{atm}=1\text{.}\text{013}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{dyne/cm}}^{2}$
$1\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{kg/cm}}^{2}=9\text{.}8×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{atm}=1\text{.}\text{013}\phantom{\rule{0.25em}{0ex}}{\text{kg/cm}}^{2}$
$1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{lb/in}{\text{.}}^{2}=6\text{.}\text{90}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{atm}=\text{14}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{lb/in}{\text{.}}^{2}$
$1.0 mm Hg=\text{133}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{atm}=\text{760 mm Hg}$
$1\text{.}0 cm Hg=1\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{atm}=\text{76}\text{.}0 cm Hg$
$1\text{.}0 cm water=\text{98}\text{.}1\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{atm}=1\text{.}\text{03}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{cm water}$
$1.0 bar=1\text{.}\text{000}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1\text{.}0\phantom{\rule{0.25em}{0ex}}\text{atm}=1.013 bar$
$1.0 millibar=1\text{.}\text{000}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ $1.0 atm=\text{1013 millibar}$

## Section summary

• Gauge pressure is the pressure relative to atmospheric pressure.
• Absolute pressure is the sum of gauge pressure and atmospheric pressure.
• Aneroid gauge measures pressure using a bellows-and-spring arrangement connected to the pointer of a calibrated scale.
• Open-tube manometers have U-shaped tubes and one end is always open. It is used to measure pressure.
• A mercury barometer is a device that measures atmospheric pressure.

## Conceptual questions

Explain why the fluid reaches equal levels on either side of a manometer if both sides are open to the atmosphere, even if the tubes are of different diameters.

[link] shows how a common measurement of arterial blood pressure is made. Is there any effect on the measured pressure if the manometer is lowered? What is the effect of raising the arm above the shoulder? What is the effect of placing the cuff on the upper leg with the person standing? Explain your answers in terms of pressure created by the weight of a fluid.

Considering the magnitude of typical arterial blood pressures, why are mercury rather than water manometers used for these measurements?

## Problems&Exercises

Find the gauge and absolute pressures in the balloon and peanut jar shown in [link] , assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking $h=0\text{.}\text{0500 m}$ for each.

Balloon:

$\begin{array}{lll}{P}_{\text{g}}& =& 5.00 cm\phantom{\rule{0.25em}{0ex}}{\text{H}}_{2}\text{O,}\\ {P}_{\text{abs}}& =& 1.035×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{cm}\phantom{\rule{0.25em}{0ex}}{\text{H}}_{2}\text{O.}\end{array}$

Jar:

$\begin{array}{lll}{P}_{\text{g}}& =& -\text{50.0 mm Hg}\text{,}\\ {P}_{\text{abs}}& =& \text{710 mm Hg.}\end{array}$

(a) Convert normal blood pressure readings of 120 over 80 mm Hg to newtons per meter squared using the relationship for pressure due to the weight of a fluid $\left(P=\mathrm{h\rho g}\right)$ rather than a conversion factor. (b) Discuss why blood pressures for an infant could be smaller than those for an adult. Specifically, consider the smaller height to which blood must be pumped.

How tall must a water-filled manometer be to measure blood pressures as high as 300 mm Hg?

4.08 m

Pressure cookers have been around for more than 300 years, although their use has strongly declined in recent years (early models had a nasty habit of exploding). How much force must the latches holding the lid onto a pressure cooker be able to withstand if the circular lid is $\text{25.0 cm}$ in diameter and the gauge pressure inside is 300 atm? Neglect the weight of the lid.

Suppose you measure a standing person’s blood pressure by placing the cuff on his leg 0.500 m below the heart. Calculate the pressure you would observe (in units of mm Hg) if the pressure at the heart were 120 over 80 mm Hg. Assume that there is no loss of pressure due to resistance in the circulatory system (a reasonable assumption, since major arteries are large).

$\begin{array}{}\Delta P=\text{38.7 mm Hg,}\\ \text{Leg blood pressure}=\frac{\text{159}}{\text{119}}\text{.}\end{array}$

A submarine is stranded on the bottom of the ocean with its hatch 25.0 m below the surface. Calculate the force needed to open the hatch from the inside, given it is circular and 0.450 m in diameter. Air pressure inside the submarine is 1.00 atm.

Assuming bicycle tires are perfectly flexible and support the weight of bicycle and rider by pressure alone, calculate the total area of the tires in contact with the ground. The bicycle plus rider has a mass of 80.0 kg, and the gauge pressure in the tires is $3\text{.}\text{50}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$ .

$\text{22}\text{.}4\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}$

can some one tell me how v=RW is dimensionally correct?
ms-1 = m X Hz
babar
What is displacement
shortest distance b/w two points
bilal
distance+direction
A.d
explain distanace+direction
bilal
the change of postion from one point to another with direction
A.d
if we change thrle direction then displacement is destroy?
bilal
change the direction then?
bilal
what do u mean by i didnt understand bro
A.d
displacement is one dimension...?
bilal
displacement is the total length an object cover from initial to the final with respect to direction as Well as time.
mohammed
thanks
bilal
displacement is the ratio of speed with respect to particular time
Bhautik
shortest distance travel from initial point to final point
ankit
is straight shortest line that connect initial pt with final pt.
Zeleke
what are the differences between vector and scalar quantity
vector is assigned to those physical quantity that has both direction and magnitude! example velocity ,scalar just has magnitude example Mass of an object. hope it helps
Mudang
velocity is produce in fan...?
how many electrons are there in 5 microcouloumb
can a given total amount of mechanical energy be totally converted into heat energy..if so give example
human running
Emmanuel
what is the fumula for calculating specific heat capacity, fusion,fission and vaporization?
Q=cm(∆t)
Emmanuel
Q=cm∆T
what is difference b/w vaporization and evaporation
evaporation is the process of extracting moisture while vaporization is process of becoming a vapor or gas
Emmanuel
From a molecular standpoint they are both cooling processes. Also, you may want to explore states of matter😊 #myTwoCents ~Shi~
Shii
cooling is a similarlity in both process I am confused in difference
1- Evaporation is a process where a liquid change to gas without reaching its boiling point. 2- Vaporization is a process where a liquid change to gas after reaching its boiling point. 3- Sublimation is a process where a solid changes into vapour without passing through a liquid state
Victor
I see. Evaporation is a type of vaporization, that occurs on the surface of a liquid as it changes into the gaseous phase before reaching its boiling point. hope that aids
Shii
vaporisation is cooling process while vaporization is heating process
Emmanuel
I mean to write evaporation is an heating process while vaporization is cooling process
Emmanuel
Yea here are two applications. 1- your wet washed clothes dry under the sun, the water EVAPORATES 2- when u are cooking, it reaches a point where u need to add more water because the water you added previously is getting dried. this is VAPORIZATION. Am not sure which is a cooling or heating process
Victor
vaporization occur only when the evaporation get to level where the above cloud is been (saturated) so cooling take place and started to change to liquid (eg rain fall)
Emmanuel
They are both properties of the same process so they're both cooling
Shii
what about sublimation? cooling or heating process?
Victor
exact
evaporation is the increase in kinetic energy of the liquid which can be gone by adding heat
Emmanuel
so its an heating process
Emmanuel
sublimation is when a solid change to gas
Emmanuel
evaporation is very definitely a cooling process. respectfully@Emmanuel when liquid turns to gas it requires more energy from its surroundings, this energy is in the form of heat, and when heat energy leaves the evaporating liquid it leaves it cooler. Thus, cooling process.
Shii
.
Shii
evaporation is very definitely a cooling process. respectfully@Emmanuel
Shii
kk
Emmanuel
You're right @Shi. I get your point
Victor
eascape velocity on the surface of Earth is 11.2 kms-1 the escape velocity on the surface of another planet of same mass as that of Earth but of 1/4 times of radius of Earth is a5.6kms-1 b11.2 kms-1 c22.4kms-1 d5.6ms-1
Emm.. is that a question? or..
Victor
it is McQ
a)5.6km/s
Alvis
c= Q/cm◇T
A.d
units...
Shii
vital
Shii
the time period of the artificial satellite is given by ?
raza
Why is there no 2nd harmonic in the classical electron orbit?
how to reform magnet after been demagneted
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Issac Newton devised a genius way to calculate changing quantities...
Shii
what is the error during taking work done of a body..
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas