# 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 $3\text{.}\text{50}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$ .

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

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?
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
what's the big bang?
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
Julius
great Mudang
Kossi
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
ancient greek language physis = nature
isidor
what is phyacs
if i am going to start studying physics where should i start?
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
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suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir