00:00:00
hi so for today we're going to talk
00:00:01
about fluid mechanics and the basics of
00:00:04
fluid mechanics hey this is an
00:00:07
introductory subject for civil
00:00:09
engineering or mechanical engineering
00:00:10
their major course fluid mechanics okay
00:00:15
so we'll go to discuss in this video I'm
00:00:17
going to discuss the basics on the other
00:00:19
with mechanics so let us start with some
00:00:22
definition of terms what is fluid
00:00:25
mechanics fluid mechanics is basically
00:00:27
either branch on science that which
00:00:30
deals with the study or end behavior of
00:00:33
fluids such as gases or liquids
00:00:36
misconception all about fluids that when
00:00:39
we talk about fluids the first thing
00:00:41
that comes to our mind is that fluid is
00:00:45
limited to a liquid form but in the case
00:00:48
of the study of fluid mechanics fluid is
00:00:51
not only in liquid form but basically
00:00:54
fluid is any substance that can flow
00:00:57
either a liquid or a gas meaning to say
00:01:00
even a gas can be considered as fluid
00:01:03
why because it can flow okay so in this
00:01:07
kind of fluid that is the basic
00:01:08
definition of fluid mechanics so we have
00:01:11
two branches of fluid mechanics and we
00:01:14
have the fluid statics with this which
00:01:15
is the study of fluids at rest meaning
00:01:18
the fluid is not moving Ray's just
00:01:22
resting all right so the magnitude of
00:01:25
forces acting along that fluid is zero
00:01:27
because it is at rest rest in
00:01:30
equilibrium situations and we have the
00:01:33
fluid dynamics once this fluids starts
00:01:36
to move then we are talking about fluid
00:01:38
dynamics here and this fluid dynamics is
00:01:41
much more complicated or complex than
00:01:45
the fluid statics because now the fluid
00:01:47
is now moving right so here is a sample
00:01:52
important quantities when talking about
00:01:54
flowing fluid mechanics so we have
00:01:57
worked on the density of the material so
00:01:59
what's this density basically density is
00:02:02
defined as the mass per unit volume okay
00:02:06
so it is basically represented by a
00:02:09
Greek alphabet Rho and that is
00:02:11
equivalent to mass of this
00:02:14
the volume of the substance they were in
00:02:17
our unit for density is SI unit is
00:02:20
kilogram per cubic meter because all it
00:02:23
is cubic per meter okay a homogeneous
00:02:26
material that is a material that has the
00:02:30
same material composition of material
00:02:36
has the same density in other words for
00:02:39
example we have the same steel range
00:02:44
here and steel nail they are both still
00:02:46
though they have different shape and
00:02:50
different mass okay their density is the
00:02:53
same because they are a homogeneous
00:02:56
material they are composed of the same
00:02:58
material so different mass same density
00:03:01
world steel and still they regardless of
00:03:04
their size and regardless of their mass
00:03:07
so because the wrench in made are both
00:03:09
made of steel they have the same density
00:03:11
must per unit input and again density is
00:03:15
a important parameter or whether we the
00:03:19
object would actually sink or float away
00:03:22
when submerged into a liquid okay so
00:03:25
density that is basically density so
00:03:28
here are some densities of some common
00:03:30
substances we have the most common is
00:03:34
water wherein the density of water is 1
00:03:37
times 10 raised to 3 kilogram per cubic
00:03:40
meter so that is the density of water
00:03:43
for every element we have the density in
00:03:47
the constant density all right so do we
00:03:51
need to memorize all of this not
00:03:52
necessarily but we have to take look
00:03:54
what is the density of water because
00:03:56
sometimes in the problem it's not given
00:03:58
so that is basically 1 times 10 raised
00:04:01
to 3 kilogram per cubic meter and we
00:04:05
have the air 1.2 kilograms per cubic
00:04:07
meter
00:04:07
you have the ethanol the benzene the
00:04:09
eyes and different elements here okay so
00:04:13
we have the densities of some common
00:04:17
materials okay so let's try to solve a
00:04:21
problem okay
00:04:22
I've written the problem here so for our
00:04:26
problem number one
00:04:30
find the mass and weight of the air at
00:04:33
20 degrees Celsius that is in Celsius in
00:04:36
a living room with the four meters by 5
00:04:38
meters in a ceiling of 3 meters high
00:04:41
so basically we're asked to find the
00:04:44
mass of the air and the weight of the
00:04:48
air inside the room so what is the mass
00:04:51
of the air what is the weight of that
00:04:54
air all right
00:04:56
using the formula density is equal to
00:04:59
mass over volume we can calculate the
00:05:02
air the mass of the air inside the room
00:05:06
so of course inside the air inside the
00:05:10
room we have the density of air is
00:05:12
equals to the mass of air divided by the
00:05:16
volume of the room ok so what are we
00:05:21
going to do is to get the mass of here
00:05:23
see we cross multiply that a mass of the
00:05:27
air is simply density of the air and the
00:05:31
volume of the room okay so the density
00:05:34
of air is shown I wanna go 1.2 kilograms
00:05:38
per cubic metre and the volume of the
00:05:43
room is that that is a cube so because
00:05:46
we have not a tube necessarily so we
00:05:48
have a horror meter by 5 meter by 3
00:05:51
meter so length times width times height
00:05:53
of the boom so we have 4 meters
00:05:56
multiplied by 5 meters multiplied by 3
00:06:00
meters all right and this should give us
00:06:03
this meter cubic meter will cancel
00:06:07
nobody and that should give us 72
00:06:12
kilograms if I'm not mistaken so let's
00:06:16
try to verify so 72 kilograms that is
00:06:20
the mass of the air inside that room how
00:06:23
about the weight of the air well
00:06:24
actually we know that the weight of the
00:06:26
air is simply the mass times the
00:06:30
acceleration due to gravity that is 72
00:06:33
kilograms multiplied by 9.81 meters per
00:06:39
second square
00:06:40
and weight should be neutral so that is
00:06:43
75 times point eighty one that should be
00:06:46
706 706 point 32 that should be our and
00:06:55
so for the weight and the mass of e here
00:07:00
hey
00:07:01
so this is our computation thing we have
00:07:06
the same answer okay so how about
00:07:09
pressure pressure is also an important
00:07:11
quantity when we are talking about fluid
00:07:14
mechanics and pressures basically
00:07:16
defined as the amount of force exerted
00:07:19
on a given area this force is absolutely
00:07:24
perpendicular to the normal or the
00:07:27
surface area of the object right so and
00:07:34
it's SI unit is given in Pascal and
00:07:39
Pascal is denoted by E a and of course
00:07:43
the unit of pressure is denoted by one
00:07:48
Newton meter per meter squared is
00:07:50
absolutely equal to one Pascal so in
00:07:53
other words if you're going to write
00:07:55
oops for going to write the pressure
00:08:01
equation so pressure is defined as the
00:08:03
force per unit area and the unit of
00:08:07
force is actually one Newton divided by
00:08:10
this SI unit of area is meters fear so
00:08:13
we say that the unit of pressure is that
00:08:16
say one Pascal denoted by 1 p.m. and
00:08:20
that is basically equivalent to 1 Newton
00:08:24
per meter squared okay and we say that
00:08:28
this force is actually 1 it's actually
00:08:31
always perpendicular to the surface to
00:08:35
which it acts for example we have this
00:08:38
plate of a object with an area a so the
00:08:42
force acting should be always
00:08:45
perpendicular to the normal or the
00:08:48
surface of the object so that is the
00:08:51
pressure so a
00:08:53
in our equation if you are going to know
00:08:55
what this equation if we increase the
00:08:58
force acting per unit area well of
00:09:01
course if we if the pressure here is
00:09:04
also the force over area what if we
00:09:06
don't want the force like what will
00:09:10
happen is that since this is the scene
00:09:12
sports is directly proportional to the
00:09:13
pressure so we will be having pores our
00:09:17
pressure is equal to twice the F over a
00:09:20
so our pressure will increase our
00:09:22
actually by two also because of the
00:09:25
increase in the force so what will
00:09:28
happen if if we try to apply a force to
00:09:34
a very small area of a much smaller than
00:09:38
this so what will happen let's say we
00:09:40
have here an object okay it is acted by
00:09:43
a force this is an area a prime which is
00:09:47
basically smaller than this so we have
00:09:49
the same force of life so the pressure
00:09:52
here is absolutely much greater than the
00:09:55
pressure here because the area is now
00:09:59
smaller and we if we decrease the area
00:10:02
to which the force is acting what will
00:10:04
happen to the pressure it would increase
00:10:06
alright so because this has the area and
00:10:10
the pressure has a indirectly
00:10:12
proportional or indirectly proportional
00:10:13
to each other or inversely proportional
00:10:16
to each other so that if we're going to
00:10:18
decrease the area the pressure will
00:10:20
increase and of course increasing the
00:10:22
area to which the force is acting upon
00:10:24
well will also decrease the pressure
00:10:27
right so that's the basic concept of the
00:10:30
pressure so in other words if the
00:10:32
applied force is acting on a small area
00:10:35
then the pressure will be large and vice
00:10:38
versa all right so take note that
00:10:42
pressure has many lives its SI units
00:10:44
only given by Pascal but we can convert
00:10:49
pressure in different types of units
00:10:52
such a store mmHg
00:10:54
bar over a atm we have many units for
00:10:58
pressure but for now we're going to talk
00:11:00
about the facts cos okay again the area
00:11:04
here is the prof sectional
00:11:06
the object to which the voice is up so
00:11:10
something to think about when we are
00:11:12
dealing with pressure for ladies do you
00:11:15
feel pain or pressure whenever you wear
00:11:18
your hiatus for a long time
00:11:20
well of course some of our friends was
00:11:23
going to party though those ladies are
00:11:25
wearing high heels maybe five inches so
00:11:30
what happens the tendency is that
00:11:33
whenever they wear that for a very long
00:11:37
time they will feel pain in their feet
00:11:41
high or even in relaxed why because
00:11:44
there is the presence of pressure
00:11:45
if you're going to recall from our
00:11:47
equation a while ago that the pressure
00:11:50
is inversely proportional to the area
00:11:53
cross sectional area so for example your
00:11:56
friend is wearing a high heels so the
00:11:58
tendency is that the weight now become
00:12:01
becomes the force because weight is also
00:12:04
a force so it is now acting on the very
00:12:07
small cross sectional area of the hills
00:12:11
imagine force if I'm going to draw that
00:12:15
so let's just make this as a as an
00:12:18
exaggeration okay suppose this is the
00:12:21
heels of your friend and this is the
00:12:23
force acting away on this hill of the
00:12:28
shoes of the five-inch shoes of your
00:12:31
heels of your friend so this force is
00:12:35
after the per week okay and it let's say
00:12:39
that if the weight is of the user of
00:12:42
your friend is really larger than in
00:12:44
other words it's a little chubby
00:12:46
what will happen is that the the
00:12:49
pressure that your feet will feel is
00:12:54
very large because the cross section
00:12:56
here is an S is actually a very small
00:13:01
squared away area of a sphere so very
00:13:06
small area so that's why we cannot
00:13:09
actually be R or ladies you're not
00:13:12
absolutely bear wearing that highest for
00:13:16
a very long time because the
00:13:18
area which daily acting is acting is on
00:13:22
a very very small area my compared to
00:13:26
when we wear shoes the normal shoes the
00:13:29
area of our shoes is really really large
00:13:36
compared to the area when wearing high
00:13:38
heels okay so the point is that there is
00:13:42
a great pressure in the high heels of
00:13:45
your friend who is going to wear this
00:13:48
idea sorry so number two do you feel
00:13:50
friendship when you are at the drugs
00:13:52
average sometimes and then use the
00:13:53
elevator to go to a much higher level in
00:13:56
a building all right so when of course
00:13:59
sometimes we are tough at the ground
00:14:01
level and we try to ride the elevator
00:14:03
and from ground to xxx let's say that is
00:14:07
constantly going up okay so what happens
00:14:11
is that the tendency is we feel pressure
00:14:13
in our ears
00:14:14
I mean sometimes there is a thickening
00:14:17
sound in our ears because of the
00:14:21
pressure okay so not only because the
00:14:25
pressure is limited by this formula but
00:14:29
also the pressure is absolutely has
00:14:32
another formula of which the pressure to
00:14:36
which our body feels is absolutely equal
00:14:39
also into what we call the Rho gh and it
00:14:42
is a function of height so Rho gh is
00:14:45
actually the Rho is the density G is the
00:14:49
gravitational force
00:14:52
alright levitational force observation
00:14:57
due to gravity and this is now the
00:14:59
height so whenever we go to a higher
00:15:03
elevation we experience pressure for
00:15:06
example we write the flight from Manila
00:15:10
kadavul were actually riding a an
00:15:13
airplane so at the very high height know
00:15:19
we feel pressure in our ears because
00:15:21
pressure is also a function of height so
00:15:25
we have this following formula pressure
00:15:26
is also equal to Rho G each so number
00:15:29
three do you feel pressure when you have
00:15:31
to swim at the very bottom of the
00:15:33
swimming pool well actually we always
00:15:35
feared that whenever we try to swim at
00:15:38
the very bottom of the swimming hole if
00:15:40
you don't know how to swim good for you
00:15:43
you don't feel any pressure but but most
00:15:46
of us maybe have experienced this that
00:15:49
whenever we go deeper okay in the into
00:15:53
the deep of the swimming pool what
00:15:54
happens is that the tendencies we feel
00:15:56
pressure in our nose okay maybe in our
00:16:00
ears and even in our head because of the
00:16:03
pressure and that is all because of the
00:16:07
formula Rosie eats so suppose we have a
00:16:10
pan filled with a fluid density ro that
00:16:13
just like what I would say the wine
00:16:14
about that French rule is equals to Rho
00:16:17
gh okay so at the very bottom of this
00:16:21
pan right there is a pressure and that
00:16:24
pressure is actually Rho G H they not
00:16:27
including the what we call t atmospheric
00:16:30
pressure but what later on we'll be
00:16:32
discussing that
00:16:33
so this equation helps us understand why
00:16:36
the deeper we go underwater
00:16:37
the greater the pressure our body fills
00:16:40
think so atmospheric pressure there's
00:16:44
also what we call the atmospheric
00:16:46
pressure and that is basically defined
00:16:47
as the P sub atm
00:16:50
alright and when do we use that constant
00:16:55
alright for atmospheric pressure well of
00:16:57
course if we're going to use that for an
00:17:00
open tank we should include the pressure
00:17:02
outside the plan
00:17:03
pressing on the top of the surface of
00:17:05
the fluid or in other words whenever we
00:17:08
have an open of a an open Pam and it is
00:17:13
actually adapted by the atmospheric
00:17:16
pressure and if we're going to make sure
00:17:18
the very pressure of the very bottom of
00:17:21
that pan we should improve the
00:17:23
atmospheric pressure for you to analyze
00:17:25
what I'm trying to say is that whenever
00:17:27
we have an open path and this time for
00:17:31
example is actually filled with water
00:17:34
right up to this level to a height H if
00:17:40
we want to measure the pressure at this
00:17:43
point at the very
00:17:44
bottom of the of the time what is the
00:17:47
pressure there so there is what we call
00:17:50
the atmospheric pressure and for every
00:17:53
open and we must include that in our
00:17:56
[Music]
00:17:58
competition so the pressure at the very
00:18:01
bottom of this time is the atmospheric
00:18:04
pressure of a multiplied by the pressure
00:18:08
due to regimes so the total pressure at
00:18:11
the very bottom for an open and acted by
00:18:14
the pressure or the air is actually the
00:18:17
atmospheric pressure plus the what we
00:18:21
call the Rho G H okay so atmospheric
00:18:25
pressure is actually equal to 101.3 kilo
00:18:29
Pascal's and that is equivalent to 760
00:18:33
Torr equal to 760 mmHg and that is
00:18:37
basically equal to 1 atm all of these
00:18:40
are units of pressure raised 11.3
00:18:45
kilopascal or in other word one one
00:18:47
other words one one point three to five
00:18:50
kilo Pascal's for accurate combinations
00:18:54
right so for this problem the swimming
00:18:59
pool fresh or saltwater so compare the
00:19:01
total pressure at the bottom of a
00:19:03
swimming pool of a depth three meters
00:19:05
for about 99 feet fi or eight feet if it
00:19:09
is filled with fresh water and sea water
00:19:11
so first we're going to compute the
00:19:15
pressure when we are when we go we are
00:19:19
going to use the fresh water all right
00:19:22
so let's calculate the pressure so for
00:19:25
example here is our pool okay here is
00:19:30
our pool this pool is filled with what
00:19:35
we call the fresh water a fresh water
00:19:37
first and it's high array is actually
00:19:44
three meters so that is for a three
00:19:49
meters height so we're going to compute
00:19:51
now the total pressure at the very
00:19:54
bottom of the spoon so the total
00:19:58
pressure
00:19:58
is also the people that is P atm because
00:20:03
this is an open swimming pool so we have
00:20:05
to include the pressure due to
00:20:09
atmospheric pressure oops
00:20:11
so that is added to the pressure Rho G H
00:20:19
so we're going to include pressure
00:20:21
atmospheric pressure this for number
00:20:23
three
00:20:24
so 101.3 oh five or 11.3 let's be
00:20:30
consistent
00:20:31
then the pressure atmospheric pressure
00:20:34
is 101 point three kilo Pascal's Plus D
00:20:39
Rho what is this the density of water is
00:20:42
1 times 10 raised to 3 kilogram per
00:20:45
cubic meter multiplied by 9.81 m/s^2 and
00:20:51
that should be multiplied by 3 meters so
00:20:56
what are we going to get here is
00:20:58
actually 11.3 Giga Pascal's 1 times 10
00:21:02
this would be okay let me just conclude
00:21:06
9.81 they're three meters so one 30
00:21:11
point seventy three okay so let's check
00:21:17
by or one point 31 times 10 raised to 5
00:21:21
Pascal's one point 31 times 10 raised to
00:21:25
5 Pascal's or engineering in engineering
00:21:29
notation 130 point 73 kilo Pascal's so
00:21:40
what will happen if we're going to
00:21:41
replace this fresh water this fresh
00:21:45
water by soft water no for soft water
00:21:50
what will happen is that the same
00:21:53
competition we have to include the
00:21:56
atmospheric pressure plus the role of
00:21:59
now the sea or the soft water times the
00:22:03
G times the height well of course the
00:22:06
only thing that changed here is the
00:22:09
density of
00:22:11
the object so from fresh water you see
00:22:15
water so we have total pressure is
00:22:19
absolutely equal to 101.3 Ito Pascal's
00:22:23
plus the density of water or sea water
00:22:26
is one point zero three times ten days
00:22:30
to three kilogram per cubic meter
00:22:37
multiplied by G 9.81 times the height
00:22:41
the same that is three so the total
00:22:43
pressure we're going to compute for da
00:22:46
is 101.3 a dope-ass cos plus one times
00:22:51
one point zero three times 10 raised to
00:22:54
3 9.81 times three and that is one point
00:23:00
what 1.30 to one point 32 times 10
00:23:07
raised to 5 fast palace or engine in
00:23:10
engineering notation we have what 30 1.1
00:23:17
1.6 1 to 9 Pascal's or kilo Pascal's and
00:23:24
slowly kilo Pascal's this is in terms of
00:23:30
email passwords all right so there is a
00:23:35
big difference in the pressure at the
00:23:38
very bottom of the spirit for the VTC
00:23:40
fresh water and so far as you can see
00:23:42
here the depressor the very bottom of
00:23:44
the swimming pool if it is fresh water
00:23:46
is given by 1 point 31 times 10 raised
00:23:49
to 5 Pascal we're asked for salt water
00:23:51
that is 1 point 32 times 10 raised to 5
00:23:53
Pascal there is a great difference so
00:23:56
that the determining difference here is
00:23:58
actually because of the tens in the
00:24:00
difference of their days so that's why
00:24:03
we have to include that whenever we
00:24:05
screen we try to stream at the very
00:24:07
bottom of a fresh water the pressure is
00:24:10
not really that great but when you try
00:24:13
to swim the very bottom of in salt water
00:24:16
forcing water what will happen is that
00:24:19
the tendency is that you will feel a
00:24:22
greater pressure why because
00:24:23
the density again of the seawater is
00:24:25
much greater than the density or in
00:24:28
freshwater okay let's go back to our
00:24:32
discussion so conversion of conversion
00:24:36
of pressure units we have different
00:24:38
conversion of pressure units because we
00:24:40
have many units of fractures or Pascal
00:24:45
denoted by P a1 e is equals to 100 per
00:24:48
square meter and we have what people do
00:24:50
bar bar is a unit of pressure also in
00:24:54
one part is equal to 0.1 mega Pascal's
00:24:58
that's a huge amount of pressure so 0.1
00:25:02
mega Pascal is absolutely equal to 0.1
00:25:05
times 10 raised to six Oscars we have
00:25:09
the water column meter way in that is
00:25:11
the equivalent conversion units honestly
00:25:14
I don't know that way for the irony
00:25:18
haven't encountered that during my time
00:25:21
in my college but I hope someday we'll
00:25:25
know that that in your major subjects so
00:25:28
we have the atmospheric pressure what 1
00:25:30
atm ATM 118 is basically equal to this
00:25:34
the pressure atmospheric pressure is 101
00:25:38
3 to 5 Pascal's or what we have record a
00:25:42
while ago that is one one point three
00:25:44
times ten days to one one point three
00:25:49
kilo Pascal's so the most are more
00:25:52
accolade is 101.325 kilo Pascal's or 1 1
00:25:57
3 to 5 Pascal's so we have the mercury
00:26:00
kilometer given by this conversion and
00:26:03
we have the third 37 the pressure in
00:26:06
that one 4 is equal to 1 millimeter HD
00:26:09
of pressure okay
00:26:11
that's basic the basic conversion of
00:26:14
pressure units so let's go to Pascal's
00:26:17
principle so Pascal principle states
00:26:19
that a change in the pressure applied to
00:26:23
improve in I start smitten undiminished
00:26:26
to every point in the fluid and to the
00:26:29
walls of the container in other words
00:26:30
Pascal's principle whenever we have a
00:26:33
fluid inside and we apply force the
00:26:37
Oh pressure on the left side the
00:26:39
pressure on the left side must be always
00:26:41
equal to the pressure on the right side
00:26:44
of the container okay
00:26:46
the pressure is transmitted undiminished
00:26:49
to every point in the fluid so in the
00:26:53
absence of the gravity the pressure is
00:26:54
the same everywhere in the vessel so
00:26:57
remember the pressure is a scalar
00:26:59
quantity but the force is actually a
00:27:02
vector so but the pressure is scalar
00:27:06
even though the force is vector and the
00:27:09
fall is always perpendicular to the
00:27:10
cross sectional area so what will happen
00:27:14
here is that if we have for example one
00:27:19
application of Pascal's principle is
00:27:22
actually what we call the hydraulic jaw
00:27:25
this hydraulic or hydraulic press or
00:27:28
hydraulic jump this hydraulic presses
00:27:30
actually I used to give first in a hard
00:27:32
lifting service station or in a company
00:27:37
where cars are being manufactured
00:27:41
okay so consider a hydraulic press or
00:27:44
hydraulic job this hydraulic job is
00:27:46
filled with fluid inside so Pascal
00:27:49
principle states that the pressure on
00:27:51
the left side this left side must be
00:27:53
equal to the pressure on the right side
00:27:56
so when of course for example if you
00:28:00
have this hydraulic press okay if we
00:28:03
apply force here on the left side okay
00:28:06
for this cross sectional area in sub 1
00:28:09
we are actually producing the pressure
00:28:12
because this force is acting
00:28:14
perpendicular to the cross sectional
00:28:16
area in someone and according to Pascal
00:28:18
if we create pressure here the pressure
00:28:20
we created it must be also equal to the
00:28:23
pressure that is produced here so often
00:28:27
if we're going to apply force here fluid
00:28:30
will be the space so that there will be
00:28:32
a force upward F sub 2 that is acting
00:28:35
went of much bigger cross sectional area
00:28:38
of this hydraulic press a sub 2 so what
00:28:41
the pressure remains the same according
00:28:44
to Pascal's
00:28:45
alright so what are we I'm trying to say
00:28:48
is that because of the
00:28:51
pressure on the left side and on the
00:28:53
right side we have this following
00:28:54
formula okay
00:28:56
so f1 is equals to a 1 that is pressure
00:28:59
1 that is equal to f2 over a 2 because
00:29:03
the pressures are equal so because the
00:29:05
increase in pressure is the same on the
00:29:07
both sides is more force on f1 to do
00:29:10
this a much greater force f2 so let me
00:29:14
just go back to our image okay so let us
00:29:18
assume that the smaller area we call
00:29:21
that a 1 and the smaller force to go the
00:29:23
f1 and the bigger area we produce or the
00:29:26
bigger force we produce is f2 and the
00:29:29
bigger area side of the hydraulic press
00:29:31
is what we call the a - so let's just
00:29:34
try to ok so for us to really see the
00:29:41
beauty of Pascal is full so we have
00:29:44
let's say in the p1 is equal to p2
00:29:47
according to Pascal's principle and if
00:29:49
we're going to elaborate this F 1 over a
00:29:53
1 that is equal to F 2 over a tune
00:29:58
suppose we want to find the f1 this f1
00:30:02
is the force that we are going to apply
00:30:04
on the left side of a to carry or to
00:30:07
lift half a car with a higher area a sub
00:30:11
2 so let's go let's let's say we we need
00:30:15
to the isolate F 1 so we have a1 / a2 x
00:30:21
f2 by and let's say class that the area
00:30:26
of one or the smaller disc on or on the
00:30:29
left side of the hydraulic press is
00:30:32
equal let's say - what do we call this
00:30:41
let's say 100
00:30:46
meter squared and well of course a to
00:30:51
let's say one meters square me okay so
00:31:01
let me just make this clear so the what
00:31:06
are we trying to get here is f2 instead
00:31:08
of F 1 I'm so sorry
00:31:10
so we have to get F 2 because F 2 is
00:31:13
where we have to leave the card that is
00:31:15
the right side where we have to leave
00:31:17
the heart so when we get F 2 F 2 or the
00:31:21
first
00:31:21
Anto should be what a2 over a1 x the
00:31:27
first one let's say that a 2 or the area
00:31:30
to which the car is being to be lifted
00:31:34
up let's say that the area of that is
00:31:36
that say 100 square meter
00:31:38
alright let's just exaggerate this right
00:31:41
so and the area of 1/2 which we are
00:31:44
going to apply the force of a this area
00:31:48
is the smaller area on the left side of
00:31:50
our hydraulic jack let's say that is 1
00:31:52
square meter okay
00:31:55
and what what if what are we going to do
00:31:58
is to apply 1 Newton of force on the
00:32:00
left side of the hydraulic press so if
00:32:04
we're going to compute for the first two
00:32:06
we have a to the area is 100 meter
00:32:09
square over 1 meter squared then
00:32:12
multiplied that by 1 Newton
00:32:14
all right let's say 1 unit on what will
00:32:21
happen is that upon evaluating the f2
00:32:24
what will happen is that this would
00:32:26
cancel and we are left with 100 times 1
00:32:30
Newton so that we are able to produce
00:32:33
100 Newton only by applying 1 Newton so
00:32:38
no wonder why this hydraulic jump would
00:32:41
actually perform the lifting of the
00:32:44
horse because small pores that we are
00:32:48
going to apply on the left side of the
00:32:50
hydraulic press would create a greater
00:32:54
force on the right
00:32:56
because of the Equality impression okay
00:33:00
so what I'm trying to say is this is our
00:33:02
first one one new phone available opted
00:33:06
by a small areas of what that is the our
00:33:08
one square meter then the force that
00:33:11
will be produced on the other side or
00:33:13
the right side of the hydraulic press
00:33:15
would be 100 movin imagine from one year
00:33:19
told to 100 Newton with with perfect
00:33:21
design or be the good design of the
00:33:23
diameter or the area of this hydraulic
00:33:27
presses way we can actually live up cars
00:33:31
in a very easy way okay so of course the
00:33:36
same amount of fluid dibs on the left
00:33:39
and enters on the right side this is
00:33:40
actually before you a one factor
00:33:42
displacement one is equal to a 1 need no
00:33:45
volume of the fluid is left or is
00:33:49
actually the whole this looks a force of
00:33:59
magnitude 1 well of course we have seen
00:34:01
this a while ago
00:34:02
oops so well of course the work done on
00:34:04
the left side of the hydraulic press is
00:34:07
also equal to the work done on the right
00:34:09
side of the hydraulic press so let's try
00:34:13
to have some sample problems okay so we
00:34:16
have a part that weighs 10 kilo Newton
00:34:18
is space on the 1 meter ranges based on
00:34:21
a hydraulic press how much force should
00:34:24
be exerted in the 5 cm Vegas piston to
00:34:27
lift the bar let me just see if I have a
00:34:30
face here okay so I haven't updated here
00:34:34
so we have to answer this accordant
00:34:40
weighs 10 kilo Newton is placed on the 1
00:34:43
meter radius well actually in in solving
00:34:47
hydraulic presses problems so we must
00:34:51
consider the high the viscosity or even
00:34:56
is equals to people and we must always
00:34:58
do 4 is the smaller radius all right if
00:35:03
we find the smaller radius which will in
00:35:06
this case 1 meter
00:35:08
and five centimeters the smaller radius
00:35:10
is 5 centimeter so five centimeters
00:35:14
should be our r1
00:35:16
that should be the left side okay so
00:35:20
what I'm trying to say is that we have
00:35:22
our 1 is given by 5 centimeters the
00:35:27
problem is asking is how much force
00:35:30
should be exerted in the 5 centimeters
00:35:32
so if that is the R 1 this should be the
00:35:35
first one to which the force is being
00:35:37
applied
00:35:38
okay in order to lift the car on the
00:35:41
other side so that's the unknown DF 1
00:35:43
and we have here 10 kilonewtons
00:35:46
by the way that could you kilonewtons is
00:35:48
also force that is the F 2 10 kilo
00:35:52
Newtons and our R 2 is also D 1 meter
00:35:57
okay
00:35:58
so applying Pascal's principle we have P
00:36:00
1 is equals to P 2 that is F 1 over a 1
00:36:04
is equals to F 2 over a 2 and we are
00:36:08
asked to find for the F 1 so f 1 will
00:36:12
now be equal to a1 over a2 x d F 2 so we
00:36:20
have now the area we have to be really
00:36:22
careful in this we are given that I do
00:36:25
snap the area but to get the area we are
00:36:30
asked to find the cross section area so
00:36:33
the cross-sectional area is actually a
00:36:36
circle okay so we have pi we have to
00:36:40
convert this 5cm into meters so that we
00:36:44
have 0.05 meters squared over pi
00:36:50
multiplied by area of 2 that is 1 meter
00:36:54
squared we have nothing we have no
00:36:57
problem with that and that is basically
00:36:59
multiplied by the weight of the card
00:37:02
okay so what are we going to do here is
00:37:05
simply as we can see PI's will cancel ok
00:37:09
and multiply that by 10 kilonewton so we
00:37:14
have the answered 25-year furnace so
00:37:20
it's very
00:37:22
amazing why because we can lift the car
00:37:25
on the right side of the hydraulic press
00:37:28
by just simply applying 25 Newtons on
00:37:32
the left side of the hydraulic press how
00:37:35
cool is that right so even if in even
00:37:38
you you can slowly lift her so the best
00:37:43
example is that is that the mechanics we
00:37:45
can see here we can see them trying to
00:37:48
lift the car by water for the hydraulic
00:37:50
press anybody continuously applying
00:37:53
force so that they are able to lift the
00:37:56
heart that's one basic application of
00:37:58
the hydraulic press number two for
00:38:00
problem number two in a fairly used in a
00:38:03
service station compressed air exerts a
00:38:05
force on a small teaspoon that has a
00:38:07
radius of 5 cm this pressure is
00:38:10
transmitted by a liquid to a piston that
00:38:12
has a radius of 15 cm so we have to
00:38:14
range assess again and 5 cm is the R 1
00:38:18
because it's smaller are twisting 15 cm
00:38:21
so what are we going to do is to write
00:38:22
the event R 1 is 5 cm R 2 is 15 cm okay
00:38:29
so what force must the compressed air
00:38:32
exert to lift a car weighing 13,300
00:38:37
Newtons about 13,000 300 Newton's where
00:38:40
the force is on the right side and that
00:38:44
should be the weight of the bar that
00:38:46
should be F 2 all right
00:38:50
30,000 300 Newton what air pressure
00:38:52
produces this 4 so we have two questions
00:38:54
here what force should be exerted on the
00:38:58
left side of the hydraulic press in
00:39:00
order for this 13,300 Newton off part of
00:39:03
it if that up that is again F 1 and we
00:39:07
have derive again the F 1 by the formula
00:39:10
F 1 is equal to a 1 over a 2 multiplied
00:39:14
by F 2 so we have now a 1
00:39:18
alright the area of one cross sectional
00:39:20
area we have to convert this again into
00:39:22
meters so I r1 0.05 meters squared over
00:39:30
area of 2 pi over 0 point
00:39:34
in meters squared and that should be
00:39:37
multiplied by the force to thirteen
00:39:40
thousand three hundred Newtons that is
00:39:43
the weight of the car 0.05 squared 0.15
00:39:48
square multiplied by 13 thousand three
00:39:53
hundred and that should be equal to one
00:39:57
four seven seven point seven seven eight
00:40:04
that's that's Arabic up to three
00:40:07
decimals right so that is the amount of
00:40:10
force that is needed in order to leave
00:40:14
the are in a bar lift in service station
00:40:16
so number two the question is actually
00:40:19
what air pressures produces this force
00:40:21
so simply the problem is asking for us
00:40:24
what is the pressure well since this is
00:40:26
hydraulic press the pressure in the left
00:40:30
side is equal to the pressure to the
00:40:31
right side so that we can calculate any
00:40:34
of the pressure let's say pressure what
00:40:36
is equals to F 1 over a 1 and our F 1 is
00:40:40
basically what we have computed here one
00:40:43
four seven seven point seven seven eight
00:40:47
Newton's divided by the area one area
00:40:51
one is by 0.05 meter squared and that
00:40:58
should result one answer or let's see
00:41:04
one eighty a so 180 a point sixteen
00:41:11
times ten bits of tree or that is low
00:41:14
phosphorus all right
00:41:17
that is 188 point 16 kilo Pascal's and
00:41:21
for pressure to we have f2 over a - all
00:41:28
right so what do you expect if we're
00:41:30
going to compute P - since this is a
00:41:33
hydraulic press the pressure on the left
00:41:35
side is equal to the pressure on the
00:41:37
right side so this two pressure must be
00:41:39
yes you are correct that must be equal
00:41:42
so we have thirteen thousand three
00:41:44
hundred Newton or f2 over eight
00:41:47
then glass is 0.15 meters width okay so
00:41:55
13300 is 5 times 0.15 squared that is
00:42:00
also equal to 1/8 8.16 kilo Pascal's
00:42:08
okay so that because that works on the
00:42:12
face of all of the Pascal's law or
00:42:16
principle so hydrostatic pressure so we
00:42:22
are going to derive the hydrostatic
00:42:24
pressure for this video and we're going
00:42:29
to actually verify if the change in
00:42:33
height as we go deeper into the fluid
00:42:36
actually affects deep pressure all right
00:42:39
so I would like you to consider and to
00:42:43
listen very carefully
00:42:44
all right in this discussion
00:42:48
for example class we have this figure
00:42:52
all right
00:42:53
so for example I have an open pan and
00:42:56
this open pump is absolutely one filled
00:43:01
with water and I get a small amount of
00:43:05
let's say a QB size of water here all
00:43:09
right that water is sitting still doing
00:43:12
nothing and if I'm going to magnify this
00:43:15
cube that I get a portion of small
00:43:18
portion of water if you portion of water
00:43:21
we're going to zoom it in this is the
00:43:24
figure hey I hope you understand what
00:43:27
I'm talking about and if we're going to
00:43:29
look at this figure class if you're
00:43:31
going to look at this figure okay what
00:43:35
will happen class is that according to
00:43:38
our discussion we have pressure is
00:43:40
equals to F over a and if we're going to
00:43:43
manipulate the equation we have the
00:43:44
force is supposed to do the pressure PA
00:43:47
so at the very bottom of the tube of the
00:43:52
small tube water when we have a force f1
00:43:56
acting on the very bottom faith that is
00:44:00
called the f1 here
00:44:01
and at the very top of the the cube a
00:44:05
small element water water element that
00:44:08
we got here is we have also for school
00:44:11
okay and of course this F is actually a
00:44:15
height Y let's label that as height Y
00:44:18
the very bottom of the view height Y and
00:44:21
the top where the F 2 is being
00:44:25
experienced let's call that y plus Delta
00:44:27
Y so that the height is become now what
00:44:31
Delta Y this height becomes the Delta Y
00:44:35
okay so what are we going to do class is
00:44:38
well of course since force is acting
00:44:42
perpendicular to that water or cube of
00:44:47
water smooth element of water well we
00:44:50
must have also a force acting on the
00:44:52
sides correct but what happens is that
00:44:56
if we have a force acting here on the
00:44:58
side there is also a force acting here
00:45:00
on the other side so that it is
00:45:02
perpendicular to the surface of the Pew
00:45:04
but these two forces cancel each other
00:45:07
so we need not to include that sir but
00:45:10
why why do we have to consider the force
00:45:12
is acting on the Morison and wearing the
00:45:15
y-axis those things those forces will
00:45:18
never cancel because this water element
00:45:22
this element that we got on this open am
00:45:27
is absolutely has been with that is what
00:45:30
we got the mg so that if we're going to
00:45:33
get the summation then we're going to
00:45:36
get the summation of forces along Y that
00:45:38
should be equal to 0 because this fluid
00:45:40
element is static not moving away so
00:45:43
that if we're going to get the summation
00:45:45
of forces along Y positive going upward
00:45:49
we have F 1 negative going downward they
00:45:52
have F 2 that is minus mg so that these
00:45:55
two F 1 F 2 will never concern because
00:45:58
of the presence of the way and that
00:46:00
should be equal to zero okay so if we're
00:46:06
going to this is a cube by the way this
00:46:09
is of view so the surface or the basis
00:46:12
of the few are all squares and they are
00:46:14
all in
00:46:15
okay so if we're going to manipulate
00:46:17
this force F sub 1 and F sub 2
00:46:20
toward birth this away so this F 1 is
00:46:24
also equal to PA and so we have to
00:46:27
include the a that is the area of the
00:46:30
faces of the view that we called up a
00:46:32
okay and this F 1 or the pressure is
00:46:36
absolutely use the F 1 F 1 is at the
00:46:39
bottom force at the bottom of the
00:46:43
element so we have what include thus add
00:46:47
the pressure up the height Y because it
00:46:52
is at the very bottom that in our height
00:46:55
there we designate as wife - again the
00:46:59
area is the same for all tools okay then
00:47:02
we have F to the prayer the force at the
00:47:06
very foot of the very fact that is what
00:47:08
we call the y plus delta y avait that is
00:47:11
the height so we have to have P the
00:47:15
pressure at P y plus Delta Y at the very
00:47:19
top of the liquid element minus mg so we
00:47:25
have to convert em as Delta Y times a
00:47:29
times D density because as we are going
00:47:32
to recall Rho is equals to mass over
00:47:35
volume and if you're going to manipulate
00:47:37
this or mass is equal to volume times
00:47:40
density and in this case to get the
00:47:42
volume of this cube simply we have to
00:47:46
multiply the area away the area any of
00:47:50
the base times height and the height we
00:47:52
have considered us Delta Y and of course
00:47:56
we have the included density of that
00:47:59
material right so P so this term a times
00:48:03
Delta Y is the volume and this Rho is
00:48:06
the density so that we are going to
00:48:08
replace that by our M with a delta Y ro
00:48:14
multiplied by force the G is equal to
00:48:18
zero okay now we can cancel what the
00:48:22
area's here a way to look this screen so
00:48:25
we have py minus py plus there
00:48:29
why minus Delta Y Rho G is equal to zero
00:48:34
and if you're going to rearrange this
00:48:37
equation a ramming sneeze equation I'm
00:48:41
going to rate reality negative Delta Y
00:48:43
Rho G okay that is equal to what
00:48:47
positive py plus Delta Y minus py and if
00:48:52
we're going to divide this both sides by
00:48:55
or multiply this by 1 over Delta Y what
00:49:00
will happen is we'll have negative Rho G
00:49:02
here on the left side so we have py plus
00:49:06
Delta Y minus py over the Hawaii alright
00:49:11
so and again if we are going to take the
00:49:15
limits of both sides for example let me
00:49:18
just rewrite this alright we're going to
00:49:21
derive the to get the limits of both
00:49:24
sides so the limit of Delta Y as Delta y
00:49:29
equal to 0 the limit of negative Rho G
00:49:31
and we have to apply him here the limit
00:49:34
of py plus Delta Y minus P Y divided by
00:49:38
Delta Y a step away approaches zero let
00:49:41
us recall that that this is the
00:49:43
definition of basically the derivative
00:49:45
okay so and if we offer a value in the
00:49:49
limit of this since that's how I know
00:49:51
that how are you then this should be
00:49:53
constant so what will happen is that we
00:49:55
have now negative Rho G is equal to what
00:50:00
the derivative of pressure with respect
00:50:03
to Delta Y right
00:50:08
so that what will happen here glass is
00:50:11
that let's continue let me just erase
00:50:13
this so we have now the Rho G negative
00:50:17
Rho G is equal to DP dy so let me just
00:50:21
write negative Rho G suppose we change
00:50:25
in pressure with respect to Y we're
00:50:27
going to cross multiply this negative
00:50:29
Rho G dy is equal to DP and if we're
00:50:34
going to integrate this to all right
00:50:38
integrating this from p1 to people
00:50:42
integrating this phone height 1 - hi -
00:50:45
what will happen we have a definite
00:50:48
integral so it is easy to evaluate the
00:50:52
definite integral integral we have the
00:50:55
integral of dy is y and that is
00:50:58
evaluating limits from y1 to y2 and the
00:51:02
integral of DS is simply P evaluating
00:51:05
limits from p1 to p2 so what will happen
00:51:08
is that the negative Rho G is now equal
00:51:12
to Y sub 2 minus y sub 1 and equal to P
00:51:16
sub 2 minus P sub 1 and if I'm going to
00:51:20
multiply this all by negative 1 what do
00:51:24
I have to get is let's say Rho G times
00:51:28
y2 minus y1 is equal to p1 minus p2 now
00:51:35
this is the hydrostatic or the change in
00:51:38
pressure whenever we change our height
00:51:41
all right so this Y plane y1 can be
00:51:45
replaced as I owe - height 1 because we
00:51:49
are dealing with the high okay so we
00:51:53
have proven that as we go on deeper the
00:51:56
there is a pressure change because of
00:51:59
the change in high alright so that is
00:52:03
our formula so I am I have derived it
00:52:06
here okay so now we have gotten this
00:52:10
formula right so there is a change in
00:52:14
pressure whenever there is a change in
00:52:16
high okay
00:52:21
sotto continue our discussion we're
00:52:23
going to talk about the who enforces and
00:52:25
neck turkey leaders principle on the
00:52:27
latter part of our argued to with
00:52:29
mechanics so have you ever tried to push
00:52:31
a boat down underwater but it's very
00:52:33
hard to do because there is absolutely
00:52:36
what people deeply reinforce it is
00:52:38
extremely difficult to do to hold down
00:52:41
push them in both underwater because of
00:52:44
the large upward force exerted by the
00:52:47
water on the board and that's what of
00:52:49
the buoyant force the buoyant force
00:52:51
actually the force that the water or the
00:52:55
liquid exerted on the object and
00:52:58
immersed object that is what totally
00:53:00
buoyant force and we of course once
00:53:03
again is the outward force exerted by a
00:53:05
fluid on any immersed object so for
00:53:08
example we have a cylinder here and we
00:53:10
have a witness in integral shape here
00:53:13
that is being immersed on this water
00:53:16
move it so we have two forces acting on
00:53:19
its top and on its bottom and the we are
00:53:22
forced by formula is actually the
00:53:24
magnitude or the resultant of the forces
00:53:28
exerted on the top and the bottom of the
00:53:32
pages of by the liquid so in other words
00:53:34
FB is equal to f1 or f2 minus f1 the
00:53:39
result plan of the forces acting on the
00:53:42
top and on the bottom of the object okay
00:53:46
and we have the Archimedes principle
00:53:47
Archimedes principle only states that
00:53:49
the magnitude of the buoyant force on an
00:53:53
object by the liquid is always equal to
00:53:56
the weight of the fluid displaced by the
00:53:58
object so we have here the formula
00:54:00
according to work a B this principle we
00:54:02
have the boolean force is equals to the
00:54:04
magnitude of the buoyant force equal to
00:54:08
the weight of the fluid displaced okay
00:54:10
so let's take note of this this is an
00:54:13
important concept or principle okay so
00:54:18
as I'm said available in the video what
00:54:21
if we have the determining factor
00:54:24
whether the object will float or sink
00:54:27
when it is submerged in point liquid is
00:54:30
that its density so here is the Vepr
00:54:32
meaning
00:54:33
if the density of the fluid is greater
00:54:35
than the density of the object then the
00:54:37
object will probe alright the Gardeners
00:54:39
of their size the gardens of the river
00:54:41
wait as long as the density of the fluid
00:54:44
is greater than the density of the
00:54:45
object the object will flow on the other
00:54:47
hand if the density of the object is
00:54:50
greater than the density of the fluid to
00:54:53
which it is immersed the object will see
00:54:55
so our grandfather would actually talk
00:55:00
something about us which is much massive
00:55:04
okay is it the one peso going or the
00:55:07
ship if you answer the ship why does the
00:55:12
one peso Polly sing and the ship nothing
00:55:16
it is much massive than the one pencil
00:55:19
body okay because the one peso point is
00:55:22
absolutely has a greater density than
00:55:26
the density of the fluid or the water so
00:55:29
that the one peso going on with sink and
00:55:33
the the ship will actually by convention
00:55:37
or without any large amount of waves
00:55:41
well technically that ship is made of
00:55:44
wood and the the density of wood is
00:55:48
greater than the density of the fluid
00:55:50
water or the sea water then we can
00:55:52
conclude that that ship would actually
00:55:54
one would actually prove okay so the
00:55:57
only determining factor whether the
00:56:00
object will sink or will float is d
00:56:03
density okay so the ability to sink or
00:56:07
float is completely independent on the
00:56:09
dimension of the object or the shape of
00:56:11
the object the only determining factor
00:56:14
is the density so we have here for the
00:56:16
first events the sinking stone we can
00:56:19
conclude that the density of this stone
00:56:21
is greater than the density of the water
00:56:23
to which it is being immersed and we
00:56:25
have the flow thing we move regardless
00:56:28
of their size regard regardless of their
00:56:31
shape okay so it's only determining
00:56:35
factor whether it will float or not is d
00:56:38
that's it so we have now deeper newless
00:56:41
equation when fluids are moving the
00:56:45
we more complicated than when they are
00:56:47
starting so suppose we have a fluid
00:56:50
flowing and if I this is our pipe
00:56:52
alright if the fluid is starting then
00:56:54
forget about v1 and v2 then and forget
00:56:58
about this hydrostatic pressure all
00:57:00
right we will no longer use that because
00:57:02
this formula is only for fluids that are
00:57:06
studies so how about if they move for
00:57:09
example we have a pipe here okay and
00:57:12
this pipe is filled with fluid so I'm so
00:57:16
sorry so fluid so that when the fluid
00:57:20
flows here it is flowing to a
00:57:22
cross-sectional area one and it has a
00:57:24
velocity V sub one
00:57:26
so when this fluid flows here and
00:57:29
eventually flows you there will be a
00:57:31
change in pressure because actually the
00:57:34
cross-sectional area here is not the
00:57:36
same as the cross-sectional area here
00:57:37
and of course there will be a change in
00:57:40
velocity why because there is also a
00:57:44
change in the opening of the soap or the
00:57:46
area of the tube so the we have the
00:57:48
velocity one here and we have the
00:57:50
velocity v2 here so of course if we make
00:57:53
a height of which we use this as our
00:57:56
reference to y1 the white one and we
00:57:59
have y2 here so there is also the
00:58:01
presence of what what potential energy
00:58:05
because of the height the velocity
00:58:07
kinetic energy the presence of kinetic
00:58:10
energy because of the velocity and of
00:58:12
course the presence of pressure
00:58:14
so in Bernoulli's equation rugby players
00:58:16
we have the potential energy because
00:58:18
there is a height y 1 and y 2 and we
00:58:21
have velocity 1 and velocity - which
00:58:23
reminds us of kinetic energy and we have
00:58:26
force the pressure due to difference in
00:58:29
area and the force of the fluid acting
00:58:31
on that area okay so I will not be
00:58:34
really deriving you the Bernoulli's
00:58:36
equation because it is a hideous
00:58:38
there I felt wrong okay but rather I
00:58:41
would like to take note I would like you
00:58:43
to take note of this formula which is
00:58:45
the painless Bernoulli's equation they
00:58:47
are p1 plus 1/2 the density of that
00:58:51
fluid V sub 1 squared away the velocity
00:58:53
1 of that first velocity of the fluid
00:58:56
plus Rho which again
00:58:58
the density G is the gravitational force
00:59:00
and height one is the height of the tube
00:59:03
with respect to a certain reference that
00:59:05
is equal to p2 plus one half Rho V sub 2
00:59:07
squared plus Rho H so this is the famous
00:59:11
Bernoulli's equation F and of course
00:59:13
before we solve the the usually the
00:59:16
problem is solving for the velocity at
00:59:18
the other end of the tube or the pipe or
00:59:21
insulting the pressure so in other words
00:59:23
before we can solve the pressure we have
00:59:28
to take note of what republican-led
00:59:30
equation and the continuity equation is
00:59:32
is simply a 1 V 1 where a 1 is the
00:59:36
cross-sectional area for 1 and a 2 is
00:59:38
the cross-sectional area for 2 V 1 is
00:59:41
the velocity 1 and the velocity 2 during
00:59:43
the fluid is moving alright so let's try
00:59:48
to solve a problem about the gasoline
00:59:50
problem the gasoline entering pipe
00:59:52
entering the pipe below at 3 meters per
00:59:55
second here I add 140 kilo Pascal's
00:59:59
compute the speed and pressure at which
01:00:01
the liquid meets the Bible all right
01:00:03
compute the pressure and the velocity of
01:00:06
the fluid at this point the the density
01:00:10
of doesn't mean 680 kilogram per cubic
01:00:12
meter
01:00:14
why not mistaken I am happy paste it
01:00:17
will write again our formula is you have
01:00:23
taken note of that or continuty equation
01:00:25
we have a1 v1 is equal to a2 v2 we once
01:00:30
used first the continuity equation to
01:00:33
find the velocity of the fluid at that
01:00:35
point at the second point to which it
01:00:37
lives the pipe so in order for us to get
01:00:40
v2 v3 simply equal to a1 / a2 times V V
01:00:48
sub 1
01:00:49
okay so a1 is we can see here from the
01:00:52
event
01:00:53
the radius is 4 cm we have to convert
01:00:55
that into meters and the cross-sectional
01:00:58
area of the tube or the pipe is after
01:01:01
the game yes that's right encircled by
01:01:05
0.04 meters squared by 4 cm 0.04 me
01:01:11
/ pi the area of the other one is 0.07
01:01:17
meters squared that's the Regis I'm so
01:01:20
sorry the area is pi times R squared PI
01:01:24
R squared so we have here the formula
01:01:28
and multiplied by the velocity 1
01:01:31
velocity 1 is also d13 meters per second
01:01:37
so if we're going to compute that PI's
01:01:40
will cancel so we have zero point zero
01:01:44
four squared divided by zero point zero
01:01:46
seven squared times 3 and that is 0.98
01:01:50
meters per second so this is now our
01:01:55
velocity pool all right how do we find
01:01:58
the pressure oh we're going to use the
01:02:01
Bernoulli's equation where in fresh or 1
01:02:04
plus 1/2 Rho P sub 1 squared plus Rho G
01:02:10
H 1 is equal to P sub 2 plus 1/2 Rho V
01:02:15
sub 2 squared plus Rho G H sub 2 ok and
01:02:21
we are up 30 we are after to find D in
01:02:26
sub 2 so we have to isolate y sub 2 okay
01:02:30
so we have P sub 1 all right plus 1/2
01:02:35
Rho if we're going to factor it out
01:02:38
transpose this on the left side of a and
01:02:42
factor out 1/2 Rho of a 1/2 Rho that
01:02:45
should be V 1 V 1 squared minus minus V
01:02:50
sub 2 squared plus Rho G away again
01:02:55
we're going to transpose this on the
01:02:58
left side all right so that we can
01:03:00
factor Rho G that is i1 minus i2 so the
01:03:06
pressure 1 is 140 kilo Pascal's plus 1/2
01:03:11
the ROE is 680 that's the density of the
01:03:16
gasoline V sub 1 is 3 squared 3 meter
01:03:20
squared
01:03:21
meter per second squared I would not
01:03:24
like the unit soon so that we could
01:03:27
consume detailed space so three squared
01:03:30
minus zero point nine a a squared plus
01:03:34
zero again 680 kilograms per cubic meter
01:03:38
times the G that is 9.81 m/s^2
01:03:43
multiplied by height one - hi - and
01:03:45
according to our figure height one is
01:03:48
six meters
01:03:49
all right the reference is the height is
01:03:52
zero so six minus zero or simply six so
01:03:57
we're going to compute the pressure - of
01:03:59
that okay let's try it on page 148 up
01:04:03
asked us plus one-half 680 times three
01:04:07
squared minus zero point nine eight
01:04:09
squared plus six eight T times 9.81
01:04:13
times 6 and that is 182 or 183 oops
01:04:23
so our answer should be one one hundred
01:04:29
eighty two point seventy six kilo
01:04:32
pascals one hundred eighty two point
01:04:37
seventy six kilo Pascal's so that's all
01:04:42
okay so thank you so much for listening
01:04:45
here are our references university
01:04:47
physics by survey and modern physics in
01:04:51
person physics by young in Friedman and
01:04:53
pieces look like great terrific Envato
01:04:55
that is a philippine order so thank you
01:04:57
so much for listening and god bless you