1
00:00:00,080 --> 00:00:12,280
In
2
00:00:12,280 --> 00:00:18,380
the last class we had seen how to get the
expression for non-dimensional thrust for
3
00:00:18,380 --> 00:00:24,410
a turbojet when we had the condition that
the flow is optimally expanded through the
4
00:00:24,410 --> 00:00:33,520
muscle optimally expanded through the nozzle
is a condition that is most times not satisfied
5
00:00:33,520 --> 00:00:40,640
the other condition that is the flow is choked
as it leaves the nozzle is the most probable
6
00:00:40,640 --> 00:00:49,830
one in case of turbo jets okay turbo jets
as I said in the previous classes other than
7
00:00:49,830 --> 00:00:59,790
what was used on Concorde all other turbojet
engines are primarily a noisy nozzle.
8
00:00:59,790 --> 00:01:08,150
They do not use a convergent divergent nozzle
this is because the exit pressure after the
9
00:01:08,150 --> 00:01:15,040
turbine is very low so you do not have a scope
for using a convergent divergent nozzle okay
10
00:01:15,040 --> 00:01:23,720
so let us look at the case where the flow
is choked.
11
00:01:23,720 --> 00:01:52,880
Now at the exit of the convergent nozzle
12
00:01:52,880 --> 00:02:00,409
what does this mean if we accept of the converging
nozzle the flow is – what does it imply
13
00:02:00,409 --> 00:02:26,610
to our thrust equation if you remember our
thrust equation was F = m. A into right this
14
00:02:26,610 --> 00:02:33,459
is our thrust equation if the flow at the
exit of the nozzle is choked then what it
15
00:02:33,459 --> 00:02:42,110
means is this part is not going to zero in
all the previous expressions that we derived
16
00:02:42,110 --> 00:02:48,300
we had assumed that the at the exit of the
nozzle the exit pressure is equal to the ambient
17
00:02:48,300 --> 00:02:51,270
pressure that is the condition for optimally
expanded flow.
18
00:02:51,270 --> 00:03:00,710
Now we are saying that the flow is only choked
and p7 need not be equal to P0 so here p7
19
00:03:00,710 --> 00:03:17,940
? P0 so therefore if we produce a pressure
thrust also typically
20
00:03:17,940 --> 00:03:35,180
pressure thrust is around 20 to 25% of the
overall thrust so there is not a small portion
21
00:03:35,180 --> 00:03:42,629
so in this class let us look at how to do
this analysis when we have this condition
22
00:03:42,629 --> 00:03:53,370
before we go that what do you mean by flow
in the nozzle is choked when do we say that
23
00:03:53,370 --> 00:04:17,879
when do we say that the flow through the nozzle
is choked P7 =1.2 P0 = 1 this condition chopped
24
00:04:17,879 --> 00:04:28,000
and then what do we mean when we say that
the flow is to go through the nozzle is going
25
00:04:28,000 --> 00:04:34,349
yes.
So if you have a nozzle a converging nozzle
26
00:04:34,349 --> 00:04:41,490
at the exit of it if the Mach number is one
then we say the flow is choked the reason
27
00:04:41,490 --> 00:04:49,279
for that is the reason why we say the flow
is choked is what happens when you have Mach
28
00:04:49,279 --> 00:04:58,740
number one here whatever happens in this side
okay you cannot transmit any disturbance or
29
00:04:58,740 --> 00:05:04,419
any information upstream because the flow
has already reached the speed of some disturbances
30
00:05:04,419 --> 00:05:10,289
propagate upstream at the speed of sound right,
so if the flow is already reached speed of
31
00:05:10,289 --> 00:05:18,430
sound at the exit they can no longer the flow
cannot feel anything in the downstream conditions.
32
00:05:18,430 --> 00:05:26,309
So if M =1 is reached then the flow becomes
independent of downstream conditions and is
33
00:05:26,309 --> 00:05:37,289
only a function of upstream conditions right
so which is why some people when they plot
34
00:05:37,289 --> 00:05:42,360
the mass flow rate with back pressure they
say it reaches a maximum that is a little
35
00:05:42,360 --> 00:05:49,770
confusing you just need to say that it becomes
independent of the back pressure okay typically
36
00:05:49,770 --> 00:05:54,779
if you take any nozzle when the flow is not
choked the mass flow rate is determined by
37
00:05:54,779 --> 00:06:04,490
both the pressures only upon the nozzle being
choked then it becomes independent of the
38
00:06:04,490 --> 00:06:08,589
downstream condition and becomes only a function
of the upstream condition okay.
39
00:06:08,589 --> 00:06:19,990
So
now let us look at the case on our hand wherein
40
00:06:19,990 --> 00:06:27,430
you have to do the analysis for the flow when
the reconversion nozzle is choked and this
41
00:06:27,430 --> 00:06:33,289
is our thrust equation and we cannot neglect
the pressure thrust part okay.
42
00:06:33,289 --> 00:06:45,560
So we can again use or we can say that F is
still very much less than one because most
43
00:06:45,560 --> 00:06:53,529
times the main gas turbine engine operates
with somewhere around 0.02 or 0.04 okay so
44
00:06:53,529 --> 00:07:02,449
purely ratio of 0.04 so f is very much less
than 1 so this goes to 0.
45
00:07:02,449 --> 00:07:26,930
So you can rewrite our expression as f = ma.v0
into V7/ V0 -1 + I will take out P0 here so
46
00:07:26,930 --> 00:07:45,399
I will get P0A7okay now the this portion we
can simplify as we had done earlier but in
47
00:07:45,399 --> 00:07:58,520
addition we have another component here so
what is it that we do V0 = e0 M 0 right so
48
00:07:58,520 --> 00:08:14,919
if you put that and M 0 is f is equal to m
. a a0 M 0 and V7 / V0 I can express it as
49
00:08:14,919 --> 00:08:54,740
a7 okay now using a familiar stuff that is
we know that we assume that ? and r do not
50
00:08:54,740 --> 00:09:02,180
change across the gas turbine engine so you
get a7 by a not as an expression in terms
51
00:09:02,180 --> 00:09:29,130
of temperatures so let us do that I can rewrite
this as f/ m.a v0 = M 0 into under v T7 / T
52
00:09:29,130 --> 00:10:03,430
0 into M7/ M 0 - 1 + B0 A7/ m . a a0 okay.
Now what do we do further we would do is the
53
00:10:03,430 --> 00:10:14,690
condition that we know we know that m7 = 1
okay at the exit of the nozzle the flow is
54
00:10:14,690 --> 00:10:22,470
took that as Mach number is 1 okay so we can
substitute that here and rewrite our expression
55
00:10:22,470 --> 00:11:18,160
as.
So n7 goes to one here I can bring in this
56
00:11:18,160 --> 00:11:26,490
M0 so I will get only under v T7 / T0 - this
will become M0 so this is the expression that
57
00:11:26,490 --> 00:11:39,030
we have now we need to find ratio of T 7 / T0
and also P 7 / P0 in the earlier case P7 / P0
58
00:11:39,030 --> 00:11:44,540
we used to take it as equal to 1 and find
using that find the ratio of Mach numbers
59
00:11:44,540 --> 00:11:51,990
now that is not equal to 1 so you need to
find two ratios of temperatures and pressures
60
00:11:51,990 --> 00:12:01,370
now coming to T7 /T0 we have done this exercise
earlier.
61
00:12:01,370 --> 00:12:06,870
Now that just that the nozzle exit is choked
do you think this ratio is going to change
62
00:12:06,870 --> 00:12:28,680
so we need to do a first generation of this
ratio you think it is going to change what
63
00:12:28,680 --> 00:12:40,660
will be conditions only thing let us see how
it is going to change TT7 / PT7 into TT 7
64
00:12:40,660 --> 00:13:20,350
/ TT 6 into TT 6 / TT5 TT 5 / TT 4 okay now
what is this is the only place probably it
65
00:13:20,350 --> 00:13:27,710
can change all the rest of the terms are similar
to the previous conditions right only thing
66
00:13:27,710 --> 00:13:36,910
that will change is probably here right here
we know this is the ratio of we can express
67
00:13:36,910 --> 00:13:44,790
this is ratio of static to stagnation.
So we can express it in terms of Mach numbers
68
00:13:44,790 --> 00:13:58,010
m72 then this is 1 this is again 1 after burner
first one is nozzle this after burner this
69
00:13:58,010 --> 00:14:14,430
is tT flow through turbine then t B flow through
combustor or burner then t C flow through
70
00:14:14,430 --> 00:14:28,550
compressor and this is flow through intake
which is again 1 and this is ?0 okay now here
71
00:14:28,550 --> 00:14:57,870
we know that m7 = 1 so we can get an expression
for T7 / T 0 as t T okay now we also know
72
00:14:57,870 --> 00:15:29,060
what is t B if you look into your notes the
previous class we derived this.
73
00:15:29,060 --> 00:15:58,170
What is tT t/ ?0 okay and if you plug that
in you will get T7 / T0 = tT instead of t
74
00:15:58,170 --> 00:16:27,610
B I have ? b by t C we cannot into tC?0 okay
so this cancels off and I am left with ?B
75
00:16:27,610 --> 00:16:39,990
okay so this is the expression for T7/ T0
for the condition when the nozzle is choked
76
00:16:39,990 --> 00:16:44,710
now what happens to the compressor turbine
power balance does it change because of this
77
00:16:44,710 --> 00:16:53,710
or whatever were derived remains the same
see what is happening is something downstream
78
00:16:53,710 --> 00:16:59,190
of the turbine right.
So whatever we are derived is upstream of
79
00:16:59,190 --> 00:17:04,671
this so this should not change so what we
had derived for the compressor turbine power
80
00:17:04,671 --> 00:17:37,539
balance does not change
81
00:17:37,539 --> 00:18:06,259
so we derived that from compressor turbine
power balance he had derived that tT = 1 – ?0/?
82
00:18:06,259 --> 00:18:21,259
B into tC - 1 this
remains the same okay now we have to find
83
00:18:21,259 --> 00:18:53,399
an expression for tT here so let us do that
I can rewrite this as tT is equal to ?B - and
84
00:18:53,399 --> 00:19:27,900
using this expression into this equation I
will get T7 / T0 = 2 into ?B so this and this
85
00:19:27,900 --> 00:19:53,710
cancels off and finally left with 2 into ?B
– tC?0 okay.
86
00:19:53,710 --> 00:20:02,660
So we have been able to get the temperature
ratio here that we were looking for right
87
00:20:02,660 --> 00:20:09,460
now what are the other things that we need
to derive here we have got this part temperature
88
00:20:09,460 --> 00:20:16,119
ratio done we still have to find the pressure
ratio and we have to derive an expression
89
00:20:16,119 --> 00:20:57,429
for this quantity here okay so let us do that
now.
90
00:20:57,429 --> 00:21:22,220
So again cascading pressures you are looking
for an expression for P7 /P 0 we can write
91
00:21:22,220 --> 00:22:03,750
it as P7 / PT7 into PT7 /PT6 PT6 / PT5 okay
now when we do this in the previous case we
92
00:22:03,750 --> 00:22:08,470
had assumed this I mean we had assumed the
flow to be optimally expanded so this became
93
00:22:08,470 --> 00:22:19,700
1 now it is a not equal to one so we will
get the first term is ratio of static to stagnation
94
00:22:19,700 --> 00:22:35,210
conditions so I can express it in terms of
Mach numbers okay the next one indicates flow
95
00:22:35,210 --> 00:22:44,379
through nozzle we have assumed all efficiencies
we are doing this analysis assuming all efficiencies
96
00:22:44,379 --> 00:22:53,460
to be 1 so this is 1 again the next is flow
through the after burner we have not switched
97
00:22:53,460 --> 00:23:02,880
it on so this is again 1 and what is PT5 / PT4
this is flow through turbine.
98
00:23:02,880 --> 00:23:15,590
So this is this is pressure ratio so this
is PI T into PT4 / PT3 is flow through combustor
99
00:23:15,590 --> 00:23:21,980
the pressure is the same because we are assuming
an ideal cycle here so this is 1 and this
100
00:23:21,980 --> 00:23:31,860
is flow through compressor so this is p C
and this ratio is for flow through diffuser
101
00:23:31,860 --> 00:23:39,519
or intake this is 1 because we are assuming
efficiencies to be unity and the last term
102
00:23:39,519 --> 00:23:55,380
is T20 to the power of ? / ? - okay.
So I can write P7/ P 0 = this is 1 so I will
103
00:23:55,380 --> 00:24:25,299
get 2 / ? + 1 into tT ? / ? - 1 into t C?
by
104
00:24:25,299 --> 00:24:38,499
okay
and we do know that t T and tC are related
105
00:24:38,499 --> 00:24:47,810
through compressor turbine Power Balance and
if we were to take that into account we can
106
00:24:47,810 --> 00:25:20,080
show that the ratio Ps7 / P0 would be.
Taking into account
107
00:25:20,080 --> 00:25:31,609
compressor turbine power balance
108
00:25:31,609 --> 00:25:37,950
we saw that the compressor turbine power balance
was unaffected and we could get this expression
109
00:25:37,950 --> 00:26:33,460
for tT now using this in that expression there
we can get okay right okay I have only substituted
110
00:26:33,460 --> 00:26:45,080
for tT there tC? 0 / Y and so this is the
expression that we have now we have been able
111
00:26:45,080 --> 00:26:53,109
to get expression for two quantities the last
quantity that we need an expression for is
112
00:26:53,109 --> 00:27:09,669
we need an expression for
okay so how do we go about doing this how
113
00:27:09,669 --> 00:27:27,429
do we get this expression yes they just like
to get an expression for F you said he said
114
00:27:27,429 --> 00:27:32,179
compressor turbine power balance.
And then to get an expression for tT in terms
115
00:27:32,179 --> 00:27:39,519
of tC sorry in the last case energy balance
across the combustor what was what we used
116
00:27:39,519 --> 00:27:50,100
to get an expression for F to get an expression
for tT we said compressor turbine power balance
117
00:27:50,100 --> 00:27:56,960
and we got the expression similarly is there
something that we can do here to get this
118
00:27:56,960 --> 00:28:08,280
expression
yeah what we need to do is we need to look
119
00:28:08,280 --> 00:28:52,419
at mass flow rate through the nozzle we will
see how we can use that to get this expression.
120
00:28:52,419 --> 00:28:57,799
Now mass flow rate through the nozzle what
is the expression that we know.
121
00:28:57,799 --> 00:29:09,879
We know that m.a into 1 + F that is mass of
F + mass of fuel burnt this must be equal
122
00:29:09,879 --> 00:29:27,559
to 07 v7 into a7 right now again we can make
this approximation that F is very much less
123
00:29:27,559 --> 00:29:41,730
than 1 so therefore this goes to 0 so I get
m. a = ??7 v7 a self now what is ??7 in terms
124
00:29:41,730 --> 00:29:57,410
of pressure. e is equal to using the equation
of state I can write P7 / RT7 for ?? 7 into
125
00:29:57,410 --> 00:30:12,289
V7 I can rewrite it as what do not number
into a7 right.
126
00:30:12,289 --> 00:30:28,950
So into A7 do I get what we are looking for
is P0 into a7 / m . a a0 okay so what do we
127
00:30:28,950 --> 00:30:46,950
need to do here this we know is this quantity
is what is this because this is flow is
128
00:30:46,950 --> 00:31:08,700
so m7 = 1 so I am left with what is a7, a7
is nothing but ? RT7okay so if I substitute
129
00:31:08,700 --> 00:31:39,059
for a7 here rewrite m .a I will get P 7 under
v ? divided by this would not be there RT7
130
00:31:39,059 --> 00:31:50,509
into a 7 okay now what do I need to do I need
to multiply by if I multiply by a0 on both
131
00:31:50,509 --> 00:32:31,559
sides I get one part that is m . e a 0 okay.
So let us do that
132
00:32:31,559 --> 00:32:48,869
I get m .a a0 is equal to again a 0 is nothing
but right so I will use that on the right
133
00:32:48,869 --> 00:33:20,070
hand side so I get P7 into ? under v RT0 okay
so R and R cancels off here so I am left with
134
00:33:20,070 --> 00:33:47,249
P7 a7 into ? into okay
135
00:33:47,249 --> 00:34:18,080
right now what do we need to do again so if
I want m . a / a0 / P7 P0 by a7 right so I
136
00:34:18,080 --> 00:34:43,530
can do this I can write this as m . a0 / P0
a7 = P7 / P0 into ? into under v T7 T0 / T7
137
00:34:43,530 --> 00:34:49,659
okay.
So we will been able to reduce m.aa 0 into
138
00:34:49,659 --> 00:34:56,840
two known quantities right we already know
the expression for P0 / P7 / P0 and we also
139
00:34:56,840 --> 00:35:02,860
know expression for 7 /T 0 so we have been
able to reduce it to this form now let us
140
00:35:02,860 --> 00:35:24,090
go back and substitute it and see what we
can get
141
00:35:24,090 --> 00:35:33,490
if you remember our expression for
our expression for.
142
00:35:33,490 --> 00:35:58,100
F/m . a a not great watt was under v T7 / T0
- M 0 this part was taken care of and sorry
143
00:35:58,100 --> 00:36:22,750
P0 a7 by m . A a0 into what was the P7 / P0
-1so we have got expression for this as well
144
00:36:22,750 --> 00:36:41,540
as this so when we substitute we get this
part remains as is P7 / E0 - M 0 + P 7 / P7
145
00:36:41,540 --> 00:37:18,250
P0 a7 / m . a a0 is 1 /? into okay.
Now we can simplify here and rewrite this
146
00:37:18,250 --> 00:37:58,960
expression as okay now we know expression
for T 0T 7 /T 0 and we also have no expression
147
00:37:58,960 --> 00:38:05,680
for P7 / P0 so we will substitute that and
see what is the final form that we can get
148
00:38:05,680 --> 00:40:21,790
it.
Into 1 - is a really big expression okay this
149
00:40:21,790 --> 00:40:31,730
is the final expression that we get now if
we were to substitute here what we did earlier
150
00:40:31,730 --> 00:40:48,640
that is does this produce static thrust or
not. So let us do that for static thrust M0
151
00:40:48,640 --> 00:42:32,590
= 0 and ?0 =1 so if you substitute that we
will get yeah there is the expression that
152
00:42:32,590 --> 00:42:38,920
one can get and further simplification is
possible on this I will leave it as an exercise
153
00:42:38,920 --> 00:42:52,310
to you okay you can take out ? + 1 and further
simplified and this part also you can take
154
00:42:52,310 --> 00:42:58,530
out so we have been able to derive this expression
for non-dimensional thrust.
155
00:42:58,530 --> 00:43:08,420
Now this is a fairly complicated expression
compared to that we derived for optimally
156
00:43:08,420 --> 00:43:13,910
expanded flow through the nozzle and much
more complicated than the expression for tan
157
00:43:13,910 --> 00:43:26,170
jet okay. Now the next part that we need to
address is what is ISP okay so ISP part what
158
00:43:26,170 --> 00:44:04,370
we will do is ISP we know the expression for
ISP b/ a0 as.
159
00:44:04,370 --> 00:44:12,350
Right and we have spent a considerable amount
of effort trying to derive this expression
160
00:44:12,350 --> 00:44:20,310
we already got this expression now what we
need is 1 / 8 now in the previous case when
161
00:44:20,310 --> 00:44:24,050
the flow was optimally expanded through the
nozzle we had derived this expression for
162
00:44:24,050 --> 00:44:29,540
1 / F now what we need to look at is what
is the difference between the expression that
163
00:44:29,540 --> 00:44:38,690
we derived there and will there be a difference
here now again if you look at the flow process
164
00:44:38,690 --> 00:44:42,430
what is happening in the nozzle is much more
downstream than what is happening through
165
00:44:42,430 --> 00:44:47,600
the combustor.
So the expression for 1 / F that we derived
166
00:44:47,600 --> 00:45:11,430
earlier is the same so I can use 1 / F that
we had derived earlier as Q/ CPT 0 into okay
167
00:45:11,430 --> 00:45:21,130
this expression remains the same so I know
that 1/ F is known F / M0 a is known so if
168
00:45:21,130 --> 00:45:29,710
you substitute this expression there sorry
this expression you will get is P okay right
169
00:45:29,710 --> 00:45:38,240
now having derived these expressions let us
look at what is it that we can understand
170
00:45:38,240 --> 00:45:43,920
from these expressions remember we did the
same exercise we found out what is the Mach
171
00:45:43,920 --> 00:45:51,580
number at which the non-dimensional thrust
would be a maxima depending on what values
172
00:45:51,580 --> 00:45:59,710
of ?B and other things for the turbojet let
us do the similar exercise for turbojet okay.
173
00:45:59,710 --> 00:46:08,960
Now in a turbojet we will again look at the
condition where the flow through the nozzle
174
00:46:08,960 --> 00:46:15,040
is optimally expanded simply because it is
something that is easier to do in the classroom
175
00:46:15,040 --> 00:46:25,250
doing something on this is a little more complicated
so I will look at what we can find out using
176
00:46:25,250 --> 00:46:37,660
the expression that we derived for optimally
expanded flow.
177
00:46:37,660 --> 00:46:56,890
So if you look at the expression for optimally
expanded flow through nozzle f / m . a a0
178
00:46:56,890 --> 00:47:34,730
was M 0 into ? B - t C? 0 +? 0 – okay. Right
this was the expression that we had if you
179
00:47:34,730 --> 00:47:49,370
see here as ? ba increases what should happen
yes ?B increases what should happen to F/
180
00:47:49,370 --> 00:47:56,930
m. a a0 this also increases because you can
pull out the ? B terms and you see that you
181
00:47:56,930 --> 00:48:05,910
get 1- 1 / t C ?0 so as ?B increases it is
fairly obvious from this expression that as
182
00:48:05,910 --> 00:48:18,450
a consequence f / m . e a0 also increases
right which means what that if you have a
183
00:48:18,450 --> 00:48:24,070
large enough ?B then the size of the engine
is going to be smaller and smaller for the
184
00:48:24,070 --> 00:48:30,010
same thrust.
Okay fine or if you keep the dimension same
185
00:48:30,010 --> 00:48:37,740
right then your thrust is going to increase
if you increase ?B okay this is preferred
186
00:48:37,740 --> 00:48:48,250
because you will have lesser drag on the engine
okay then this is fairly clear then what we
187
00:48:48,250 --> 00:48:53,760
need to look at is what happens to what is
there an optimal value for the compression
188
00:48:53,760 --> 00:49:03,120
ratio if I fix ? B and ?0 is there an optimal
value for compression ratio okay if I am flying
189
00:49:03,120 --> 00:49:08,670
at a cruise Mach number let us say if I am
flying at a cruise Mach number then my ?0
190
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gets fixed remember when in this previous
class one of the previous class we had derived
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an expression wherein we looked at what is
the range and how the overall efficiency gets
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affected by it right.
In that we had said most of the flight takes
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place in the cruise range right so if you
substitute ?0 for cruise and you know the
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00:49:35,020 --> 00:49:43,530
? B value then you can find an optimal expression
for t C let us do that in the next few minutes
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so we are looking for optimal value of t C
given ? B and ? 0 so how do we go about it
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again take a derivative with respect to t
C so you get- ? B tC? not this goes to the
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power of – ½ into the derivative of the
terms containing tC that is the first term
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is 0.
Then you have a - ? 0 then again you have
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00:51:12,650 --> 00:51:33,490
+ 0 + because this becomes - so you have ?b/
tC2? -not okay right derivative with one numb
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goes to 0 so you get this expression now this
goes to the denominator because you have multiplying
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I mean you have a power of- ½ so what you
need to look at is only this part of the expression
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00:51:47,970 --> 00:51:56,270
because for this to be for f by m. a0 to be
maxima this should go to 0 this can only go
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00:51:56,270 --> 00:52:24,410
to 0 if the numerator goes to 0 okay so this
is the numerator in this so for Maxima ?0
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00:52:24,410 --> 00:53:05,161
must be equal to 0.
Which means that t C must be equal to okay
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right sorry yeah so you get this condition
so we will stop here and continue in the next
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class thank you.