1
00:00:00,000 --> 00:00:01,090


2
00:00:01,090 --> 00:00:04,940
Where we left off, we had this
canister, because it had a

3
00:00:04,940 --> 00:00:07,790
closed top and it had a vacuum
above the fluid.

4
00:00:07,790 --> 00:00:15,380
The fluid on top had an area
of A1, and I poked a little

5
00:00:15,380 --> 00:00:18,490
hole with a super-small
area A2.

6
00:00:18,490 --> 00:00:20,800
I said that the area of
A2 is so small, it's

7
00:00:20,800 --> 00:00:22,980
1/1,000 of area 1.

8
00:00:22,980 --> 00:00:25,330
Then we used the continuity
equation.

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00:00:25,330 --> 00:00:29,040
We said the velocity, the rate
at which the surface is moving

10
00:00:29,040 --> 00:00:34,510
up here, V1, times area 1, the
whole surface area of the

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00:00:34,510 --> 00:00:37,800
liquid, has to be equal to the
output velocity, which we're

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00:00:37,800 --> 00:00:39,840
trying to figure out as a
function of everything else

13
00:00:39,840 --> 00:00:41,580
times this output area.

14
00:00:41,580 --> 00:00:45,090


15
00:00:45,090 --> 00:00:46,990
I made a mistake.

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00:00:46,990 --> 00:00:48,820
I don't know if I did this in
the last video, or I did this

17
00:00:48,820 --> 00:00:50,570
is in the mistake video.

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00:00:50,570 --> 00:00:56,340
So we know that the initial top
velocity times this top

19
00:00:56,340 --> 00:01:01,520
area is equal to the output
velocity times-- instead of

20
00:01:01,520 --> 00:01:10,770
writing area 2, we could write
area 1 over 1,000.

21
00:01:10,770 --> 00:01:13,580
You can get rid of the area 1 on
both sides, and then you're

22
00:01:13,580 --> 00:01:21,475
saying that the velocity up here
is equal to the rate at

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00:01:21,475 --> 00:01:24,640
which the top of the surface
moves down and is equal to

24
00:01:24,640 --> 00:01:29,580
1/1,000 of the velocity of the
liquid spurting out of this

25
00:01:29,580 --> 00:01:32,460
little hole.

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00:01:32,460 --> 00:01:38,160
With that, we actually have the
three variables for the

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00:01:38,160 --> 00:01:40,540
left-hand side of Bernoulli's
equation.

28
00:01:40,540 --> 00:01:42,920
What are the variables on
the left-hand side?

29
00:01:42,920 --> 00:01:49,490
What is the pressure at this
point where we have a hole?

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00:01:49,490 --> 00:01:50,630
This is an important thin.

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00:01:50,630 --> 00:01:52,770
When we talk about Bernoulli's--
let me rewrite

32
00:01:52,770 --> 00:01:53,550
Bernoulli's equation.

33
00:01:53,550 --> 00:02:13,010
It's P1 plus rho gh1 plus rho V1
squared over 2 is equal to

34
00:02:13,010 --> 00:02:23,650
P2 plus rho gh2 plus rho
V2 squared over 2.

35
00:02:23,650 --> 00:02:26,750
We figured out all of these
terms. Now let's figure out

36
00:02:26,750 --> 00:02:29,660
the things that we have
to input here.

37
00:02:29,660 --> 00:02:32,080
What is the pressure
at point two?

38
00:02:32,080 --> 00:02:32,970
This is the important thing.

39
00:02:32,970 --> 00:02:35,170
You might want to say, and this
was my initial reaction,

40
00:02:35,170 --> 00:02:37,800
too, and that why I made a
mistake, is that what 's the

41
00:02:37,800 --> 00:02:40,250
pressure at this depth
in the fluid?

42
00:02:40,250 --> 00:02:42,690
That's not what Bernoulli's
equation is telling us.

43
00:02:42,690 --> 00:02:45,330
Bernoulli's equation is telling
us actually what is

44
00:02:45,330 --> 00:02:47,743
the external pressure
at that hole.

45
00:02:47,743 --> 00:02:52,670


46
00:02:52,670 --> 00:02:55,530
When we did the derivation, we
were saying how much work--

47
00:02:55,530 --> 00:02:57,440
this was kind of the work
term, although we played

48
00:02:57,440 --> 00:02:59,300
around with it a little bit.

49
00:02:59,300 --> 00:03:01,950
But if we look at the water
that's spurting out of the

50
00:03:01,950 --> 00:03:10,670
hole, it's not doing any work,
because it's not actually

51
00:03:10,670 --> 00:03:13,720
exerting force against anything
so it's not actually

52
00:03:13,720 --> 00:03:14,940
doing work.

53
00:03:14,940 --> 00:03:17,770
When we think about the
pressure, the output pressure,

54
00:03:17,770 --> 00:03:20,180
it's not the pressure at that
depth of the fluid.

55
00:03:20,180 --> 00:03:24,480
You should think of it as the
external pressure at the hole.

56
00:03:24,480 --> 00:03:28,340
In this case, there is no
external pressure at the hole.

57
00:03:28,340 --> 00:03:32,610
Let's say that if we closed
the hole, then

58
00:03:32,610 --> 00:03:34,120
at that point, sure.

59
00:03:34,120 --> 00:03:36,870
The pressure would be the
pressure that's being exerted

60
00:03:36,870 --> 00:03:39,750
by the outside of the canister
to contain the water, in which

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00:03:39,750 --> 00:03:41,200
case, we would end up
with no velocity.

62
00:03:41,200 --> 00:03:42,570
The water wouldn't
spurt anywhere.

63
00:03:42,570 --> 00:03:44,270
But now we're seeing the
external pressure is zero.

64
00:03:44,270 --> 00:03:47,480
That's what the hole essentially
creates.

65
00:03:47,480 --> 00:03:51,990
We're going to say that P2 is
zero, so this pressure was

66
00:03:51,990 --> 00:03:54,850
zero, because we're
in a vacuum.

67
00:03:54,850 --> 00:03:58,440
P2 is also zero, so both
of these are zero.

68
00:03:58,440 --> 00:04:01,430
Remember, that's the
external pressure.

69
00:04:01,430 --> 00:04:05,180
P1 is the external pressure to
the input to the pipe, and you

70
00:04:05,180 --> 00:04:06,670
can view this as a pipe.

71
00:04:06,670 --> 00:04:13,900


72
00:04:13,900 --> 00:04:23,600
I could redraw it as a pipe that
looks like it has a big

73
00:04:23,600 --> 00:04:25,770
hole on the top, and it goes
down to some level to a

74
00:04:25,770 --> 00:04:28,510
super-small hole like that.

75
00:04:28,510 --> 00:04:30,750
This would be a vacuum, and the
fluid is just going in and

76
00:04:30,750 --> 00:04:34,260
it's spurting out of this end.

77
00:04:34,260 --> 00:04:37,090
Anyway, the pressure going into
the pipe is zero, and we

78
00:04:37,090 --> 00:04:39,030
said since we put a hole, the
pressure coming out of the

79
00:04:39,030 --> 00:04:43,030
pipe is zero, so we're
doing no work.

80
00:04:43,030 --> 00:04:46,790
What is this term?

81
00:04:46,790 --> 00:04:50,580
This was the potential energy
term, and we said that h1 is

82
00:04:50,580 --> 00:04:51,980
equal to h.

83
00:04:51,980 --> 00:04:56,220
We're saying that this is zero
height, so now this simplifies

84
00:04:56,220 --> 00:05:04,620
to rho times gravity times h
plus rho times V1 squared.

85
00:05:04,620 --> 00:05:12,370
V1, we said, is equal to this,
so this is rho over 2 times V2

86
00:05:12,370 --> 00:05:16,170
over 1,000 squared.

87
00:05:16,170 --> 00:05:19,490
I just substituted V2
over 1,000 for V1.

88
00:05:19,490 --> 00:05:22,570
That equals the pressure at
the hole, the external

89
00:05:22,570 --> 00:05:30,390
pressure at the hole, which
is zero, plus h2.

90
00:05:30,390 --> 00:05:33,720
This is h2 right here, which
we said is zero.

91
00:05:33,720 --> 00:05:36,870
We determined that the hole was
poked at height zero, so

92
00:05:36,870 --> 00:05:38,940
this is also zero.

93
00:05:38,940 --> 00:05:42,370
That equals this kinetic
energy-like term.

94
00:05:42,370 --> 00:05:44,150
It's not exactly
kinetic energy.

95
00:05:44,150 --> 00:05:51,710
It's rho times V squared
divided by 2.

96
00:05:51,710 --> 00:05:54,080
One thing that we can
immediately see is that we

97
00:05:54,080 --> 00:05:55,860
have all these rhos on both
sides of the equation, so we

98
00:05:55,860 --> 00:06:02,560
can divide both sides by rho and
get rid of all of those.

99
00:06:02,560 --> 00:06:05,050
Then we can multiply both sides
of the equation by 2,

100
00:06:05,050 --> 00:06:17,500
and we get 2gh plus V2 squared
over-- what's 1,000 squared--

101
00:06:17,500 --> 00:06:18,750
over 1 million.

102
00:06:18,750 --> 00:06:22,590


103
00:06:22,590 --> 00:06:28,840
That is equal to V2 squared.

104
00:06:28,840 --> 00:06:32,050
We could do the exact thing.

105
00:06:32,050 --> 00:06:36,700
We could subtract 1 over 1
million V2 squared from both

106
00:06:36,700 --> 00:06:41,510
sides, and we would get 0.999999
V2 squared, but let's

107
00:06:41,510 --> 00:06:43,970
just say for the sake of
simplicity, or let's say, if

108
00:06:43,970 --> 00:06:47,430
this wasn't 1,000, but 1
million, and that this surface

109
00:06:47,430 --> 00:06:50,780
was much bigger, we see that
this term becomes very, very,

110
00:06:50,780 --> 00:06:52,090
very small.

111
00:06:52,090 --> 00:06:54,980
If that hole is one millionth
of the surface area, then it

112
00:06:54,980 --> 00:07:00,250
becomes really insignificant,
so we can ignore this term

113
00:07:00,250 --> 00:07:02,200
because it just makes things
complicated, and we're

114
00:07:02,200 --> 00:07:04,290
assuming this is a really,
really large number, and that

115
00:07:04,290 --> 00:07:10,130
this hole is much smaller than
the surface area of the fluid.

116
00:07:10,130 --> 00:07:11,910
This is like poking a
hole in Hoover Dam.

117
00:07:11,910 --> 00:07:14,290
Hoover Dam is backing up this
huge lake, and you poke a hole

118
00:07:14,290 --> 00:07:16,940
in it, so that hole is going to
be this very small fraction

119
00:07:16,940 --> 00:07:18,160
of the surface area
of the fluid.

120
00:07:18,160 --> 00:07:21,490
You can only make this
assumption when that output

121
00:07:21,490 --> 00:07:24,090
hole is much smaller than
the input hole.

122
00:07:24,090 --> 00:07:26,770
With that said, what is
the output velocity?

123
00:07:26,770 --> 00:07:30,130
The velocity-- you just take
the square root of both

124
00:07:30,130 --> 00:07:36,390
sides-- is the square
root of 2gh.

125
00:07:36,390 --> 00:07:40,610
That is the output velocity.

126
00:07:40,610 --> 00:07:47,330
What is the amount of liquid
that flows out per second?

127
00:07:47,330 --> 00:07:48,470
We figured that out already.

128
00:07:48,470 --> 00:07:54,300
It's a column of fluid that
comes out, so per second, the

129
00:07:54,300 --> 00:07:56,930
length of the column of fluid
will be the velocity times

130
00:07:56,930 --> 00:08:01,600
time, and then the cross-section
of that column

131
00:08:01,600 --> 00:08:04,010
is equal to the output area.

132
00:08:04,010 --> 00:08:08,720
If I wanted to know the flow
coming out, the flow coming

133
00:08:08,720 --> 00:08:13,740
out or the flux coming out would
be equal to the hole's

134
00:08:13,740 --> 00:08:17,670
area times the hole's
output velocity.

135
00:08:17,670 --> 00:08:24,420
That would equal the area times
the square root of 2gh.

136
00:08:24,420 --> 00:08:28,430
We could use that actually solve
problems in the future

137
00:08:28,430 --> 00:08:30,850
if we had actual numbers.

138
00:08:30,850 --> 00:08:32,539
I only have a minute
and a half left.

139
00:08:32,539 --> 00:08:34,620
I'll see you in the
next video.

140
00:08:34,620 --> 00:00:00,000