1
00:00:00,000 --> 00:00:00,680


2
00:00:00,680 --> 00:00:05,710
Let's say we have a pipe again--
this is the opening--

3
00:00:05,710 --> 00:00:07,380
and we have fluid going
through it.

4
00:00:07,380 --> 00:00:14,860
The fluid is going with a
velocity of v1, the pressure

5
00:00:14,860 --> 00:00:19,560
entering the pipe is P1, and
then the area of this opening

6
00:00:19,560 --> 00:00:22,770
of the pipe is A1.

7
00:00:22,770 --> 00:00:27,030
It could even go up, and
the other end is

8
00:00:27,030 --> 00:00:28,280
actually even smaller.

9
00:00:28,280 --> 00:00:31,790


10
00:00:31,790 --> 00:00:36,940
The fluid-- the liquid-- is
exiting the pipe with velocity

11
00:00:36,940 --> 00:00:42,950
v2, the pressure that it
exerts as it goes out.

12
00:00:42,950 --> 00:00:44,880
If there was a membrane on the
outside, how much pressure

13
00:00:44,880 --> 00:00:47,980
would it exert on it as it
pushes it out on the adjacent

14
00:00:47,980 --> 00:00:53,580
water is P2, and the area of
the smaller opening-- it

15
00:00:53,580 --> 00:00:56,300
doesn't have to be
smaller-- is A2.

16
00:00:56,300 --> 00:01:05,710
Let's say that this opening is
at a height, on average, of

17
00:01:05,710 --> 00:01:14,790
h1, and the water exiting this
opening is on average at a

18
00:01:14,790 --> 00:01:15,960
height of h2.

19
00:01:15,960 --> 00:01:18,720
We won't worry too much about
the differential between the

20
00:01:18,720 --> 00:01:21,260
top of the pipe and the bottom
of the pipe-- we'll assume

21
00:01:21,260 --> 00:01:23,610
that these h's are much
bigger relative to

22
00:01:23,610 --> 00:01:24,860
the size of the pipe.

23
00:01:24,860 --> 00:01:28,020


24
00:01:28,020 --> 00:01:30,150
With that set up-- and remember,
there's fluid going

25
00:01:30,150 --> 00:01:38,070
through this thing-- let's go
back to what keeps showing up,

26
00:01:38,070 --> 00:01:40,260
which is the law of conservation
of energy, which

27
00:01:40,260 --> 00:01:43,830
is in any closed system, the
amount of energy that you put

28
00:01:43,830 --> 00:01:47,290
into something is equal to
the amount of energy

29
00:01:47,290 --> 00:01:48,605
that you get out.

30
00:01:48,605 --> 00:01:57,030
So energy in is equal
to energy out.

31
00:01:57,030 --> 00:02:00,390
What's the energy that you put
into a system, or that the

32
00:02:00,390 --> 00:02:02,970
system starts off with
at this end?

33
00:02:02,970 --> 00:02:11,210
It's the work that you input
plus the potential energy at

34
00:02:11,210 --> 00:02:17,180
that point of the system, plus
the kinetic energy at that

35
00:02:17,180 --> 00:02:18,430
point of the system.

36
00:02:18,430 --> 00:02:20,970


37
00:02:20,970 --> 00:02:22,840
Then we know from the
conservation of energy that

38
00:02:22,840 --> 00:02:30,780
that has to equal the output
work plus the output potential

39
00:02:30,780 --> 00:02:35,110
energy plus the output
kinetic energy.

40
00:02:35,110 --> 00:02:37,110
A lot of times in the past,
we've just said that the

41
00:02:37,110 --> 00:02:40,710
potential energy input plus the
kinetic energy input is

42
00:02:40,710 --> 00:02:43,680
equal to the potential energy
output plus the kinetic energy

43
00:02:43,680 --> 00:02:48,260
output, but the initial energy
in the system can

44
00:02:48,260 --> 00:02:49,580
also be done by work.

45
00:02:49,580 --> 00:02:52,640
So we just added work to this
equation that says that the

46
00:02:52,640 --> 00:02:55,650
energy in is equal to
the energy out.

47
00:02:55,650 --> 00:02:57,900
With that information, let's
see if we can do anything

48
00:02:57,900 --> 00:03:00,800
interesting with this pipe
that I've drawn.

49
00:03:00,800 --> 00:03:04,960
So what's the work that's being
put into this system?

50
00:03:04,960 --> 00:03:10,200
Work is force times distance,
so let's just focus on this.

51
00:03:10,200 --> 00:03:18,410
It's the force in times the
distance in, and so over a

52
00:03:18,410 --> 00:03:22,050
period of time, t, what
has been done?

53
00:03:22,050 --> 00:03:24,740
We learned in the last video
that over a period of time, t,

54
00:03:24,740 --> 00:03:29,510
the fluid here might have
moved this far.

55
00:03:29,510 --> 00:03:31,300
What is this distance?

56
00:03:31,300 --> 00:03:35,740
This distance is the input
velocity times whatever amount

57
00:03:35,740 --> 00:03:40,880
of time we're dealing with, so
T-- so that's the distance.

58
00:03:40,880 --> 00:03:43,500
What's the force?

59
00:03:43,500 --> 00:03:47,170
The force is just pressure times
area, and we can figure

60
00:03:47,170 --> 00:03:52,630
that out by just dividing
force by, area and then

61
00:03:52,630 --> 00:03:57,890
multiply by area, so we get
input force divided by area

62
00:03:57,890 --> 00:04:00,820
input, times area input.

63
00:04:00,820 --> 00:04:03,870
It's divided and multiplied by
the same number-- that's

64
00:04:03,870 --> 00:04:05,060
pressure, that's area.

65
00:04:05,060 --> 00:04:10,010
It's equal to the input distance
over that amount of

66
00:04:10,010 --> 00:04:17,779
time, and that's velocity times
time, so the work input

67
00:04:17,779 --> 00:04:33,240
is equal to the input pressure
times the input area times

68
00:04:33,240 --> 00:04:37,136
input velocity times time.

69
00:04:37,136 --> 00:04:39,730


70
00:04:39,730 --> 00:04:46,390
What is this area times velocity
times time, times

71
00:04:46,390 --> 00:04:48,050
this distance?

72
00:04:48,050 --> 00:04:51,670
That's the volume of fluid
that flowed in over that

73
00:04:51,670 --> 00:04:53,940
amount of time.

74
00:04:53,940 --> 00:04:58,400
So that equals the volume of
fluid over that period of

75
00:04:58,400 --> 00:05:04,660
time, so we could call that
volume in, or volume i--

76
00:05:04,660 --> 00:05:07,200
that's the input volume.

77
00:05:07,200 --> 00:05:17,110
We know that density is just
mass per volume, or that

78
00:05:17,110 --> 00:05:21,060
volume times density is equal
to mass, or we know that

79
00:05:21,060 --> 00:05:25,410
volume is equal to mass
divided by density.

80
00:05:25,410 --> 00:05:28,050
The work that I'm putting into
the system-- I know I'm doing

81
00:05:28,050 --> 00:05:33,110
a lot of crazy things, but it'll
make sense so far-- is

82
00:05:33,110 --> 00:05:36,880
equal to the input pressure
times the amount of volume of

83
00:05:36,880 --> 00:05:39,280
fluid that moved over
that period of time.

84
00:05:39,280 --> 00:05:43,020
That volume of fluid is equal to
the mass of the fluid that

85
00:05:43,020 --> 00:05:46,590
went in at that period of time,
and we'll call that the

86
00:05:46,590 --> 00:05:54,190
input mass, divided
by the density.

87
00:05:54,190 --> 00:05:57,740
Hopefully, that makes a
little bit of sense.

88
00:05:57,740 --> 00:06:01,140
As we know, the input volume
is going to be equal to the

89
00:06:01,140 --> 00:06:04,070
output volume, so the input
mass-- because the density

90
00:06:04,070 --> 00:06:06,250
doesn't change-- is equal to the
output mass, so we don't

91
00:06:06,250 --> 00:06:08,800
have to write an input and
output for the mass.

92
00:06:08,800 --> 00:06:11,290
The mass is going to be
constant; in any given amount

93
00:06:11,290 --> 00:06:13,220
of time, the mass that enters
the system will be equivalent

94
00:06:13,220 --> 00:06:16,460
to the mass that exits
the system.

95
00:06:16,460 --> 00:06:18,800
There we go: we have an
expression, an interesting

96
00:06:18,800 --> 00:06:21,810
expression, for the work being
put into the system.

97
00:06:21,810 --> 00:06:25,010
What is the potential energy
of the system on

98
00:06:25,010 --> 00:06:26,350
the left-hand side?

99
00:06:26,350 --> 00:06:36,050
The potential energy of the
system is going to be equal to

100
00:06:36,050 --> 00:06:44,180
that same mass of fluid that I
talked about times gravity

101
00:06:44,180 --> 00:06:50,500
times this input height-- the
initial height-- times h1.

102
00:06:50,500 --> 00:06:57,170
The initial kinetic energy of
the fluid equals the mass of

103
00:06:57,170 --> 00:07:00,170
the fluid-- this mass right
here, of that same cylinder

104
00:07:00,170 --> 00:07:04,020
volume that I keep pointing to--
times the velocity of the

105
00:07:04,020 --> 00:07:06,300
fluid squared.

106
00:07:06,300 --> 00:07:08,620
We remember this from kinetic
energy divided by 2.

107
00:07:08,620 --> 00:07:11,410
So what's the total energy at
this point in the system over

108
00:07:11,410 --> 00:07:12,280
this period of time?

109
00:07:12,280 --> 00:07:15,600
How much energy has gone
into the system?

110
00:07:15,600 --> 00:07:19,080
It's going to be the work
done, which is the input

111
00:07:19,080 --> 00:07:23,120
pressure-- I'm running out
of space, so let me

112
00:07:23,120 --> 00:07:24,330
erase all of this.

113
00:07:24,330 --> 00:07:27,265
I'll probably have to run out of
time, too, but that's OK--

114
00:07:27,265 --> 00:07:28,515
it's better than
being confused.

115
00:07:28,515 --> 00:07:53,620


116
00:07:53,620 --> 00:07:55,270
Back to what we were doing.

117
00:07:55,270 --> 00:07:59,950
So, the total energy going into
the system is the work

118
00:07:59,950 --> 00:08:01,760
being done into the system,
and I rewrote it in this

119
00:08:01,760 --> 00:08:06,920
format, which is the input
pressure-- we'll call that

120
00:08:06,920 --> 00:08:15,000
P1-- times the mass divided by
the density of the liquid,

121
00:08:15,000 --> 00:08:16,130
whatever it is.

122
00:08:16,130 --> 00:08:21,920
This is work in plus-- and
what's the potential energy?

123
00:08:21,920 --> 00:08:25,720
I wrote it right here-- that's
just mgh, where m is the mass

124
00:08:25,720 --> 00:08:30,187
of this volume of fluid, h is
its average height, and you

125
00:08:30,187 --> 00:08:34,539
could almost think of how high
the center of mass above the

126
00:08:34,539 --> 00:08:36,850
surface of the planet.

127
00:08:36,850 --> 00:08:40,090
Since we have a g here, we
assume we're on Earth, so this

128
00:08:40,090 --> 00:08:45,440
is h1, because the height
actually changes, so this is

129
00:08:45,440 --> 00:08:53,270
potential energy input plus
the kinetic energy mv1

130
00:08:53,270 --> 00:08:56,130
squared over 2.

131
00:08:56,130 --> 00:08:59,920
That is the kinetic
energy input.

132
00:08:59,920 --> 00:09:02,930


133
00:09:02,930 --> 00:09:08,250
We know that this has
to equal the energy

134
00:09:08,250 --> 00:09:09,500
coming out of the system.

135
00:09:09,500 --> 00:09:33,110


136
00:09:33,110 --> 00:09:34,700
This is going to be equal
to the same thing

137
00:09:34,700 --> 00:09:38,070
on the output side.

138
00:09:38,070 --> 00:09:41,040
This is going to be equal to
the work out, so that'll be

139
00:09:41,040 --> 00:09:46,530
the output pressure times the
mass divided by the density

140
00:09:46,530 --> 00:09:52,470
plus the output potential
energy, which will just be mg

141
00:09:52,470 --> 00:10:00,440
h2, plus the outbound kinetic
energy, which will be mv2

142
00:10:00,440 --> 00:10:02,740
squared divided by 2.

143
00:10:02,740 --> 00:10:04,790
I just realized I'm
out of time.

144
00:10:04,790 --> 00:10:06,040
I will continue this
in the next video.

145
00:10:06,040 --> 00:00:00,000
See you soon.