1
00:00:00,110 --> 00:00:19,070
Good afternoon in the last class we have seen
how ISP and non-dimensional thrust depend
2
00:00:19,070 --> 00:00:27,520
on efficiency as well as ? B and ?0 okay in
this class we will go further and find out
3
00:00:27,520 --> 00:00:37,720
what happens can we get a ramjet which can
start below Mach number one is that possible
4
00:00:37,720 --> 00:00:43,590
or a subsonic in other words subsonic ramjets
are they possible and then we will also look
5
00:00:43,590 --> 00:00:51,820
at whether ramjets are self-starting we know
for sure ramjets are not self-starting whether
6
00:00:51,820 --> 00:00:57,190
our equations will bring it out is what we
are going to see right and we will also look
7
00:00:57,190 --> 00:01:02,680
at what happens to thrust per unit mass flow
rate and other things as we go along in this
8
00:01:02,680 --> 00:01:05,059
class.
9
00:01:05,059 --> 00:01:23,909
Now if you remember we had derived expression
and said that 1 + ? – 1/ 2 m 42 = e x okay
10
00:01:23,909 --> 00:01:32,930
this was the expression that we derived in
the last class now if you put ß = 1 that
11
00:01:32,930 --> 00:01:46,799
is ideal case if we look at an ideal case
what happens here we get this relationship
12
00:01:46,799 --> 00:02:01,520
that m4 =M0. So does it mean that at the exit
Mach number is the same as the inlet Mach
13
00:02:01,520 --> 00:02:09,700
number then it would not produce any thrust
is that the case what do you think will happen
14
00:02:09,700 --> 00:02:17,150
we are looking at an ideal case and this case
exit and Inlet Mach numbers are the same,
15
00:02:17,150 --> 00:02:29,031
so do you think it will produce thrust or
not produce thrust, yeah you are right if
16
00:02:29,031 --> 00:02:35,980
you look at Mach number Mach numbers might
be the same what we are interested in this
17
00:02:35,980 --> 00:02:42,590
exit velocity and inlet velocity okay.
So Mach number being the same does not mean
18
00:02:42,590 --> 00:03:04,760
anything because if you know that at the exit
of the nozzle temperatures can be 500 to 100Kelvin
19
00:03:04,760 --> 00:03:22,670
greater than Inlet, so if you have the flow
through the nozzle where and the temperatures
20
00:03:22,670 --> 00:03:27,239
are higher and the speed of sound at that
location will be higher and therefore you
21
00:03:27,239 --> 00:03:36,950
will be able to still produce first if efficiencies
are one. Now we derived expression for non
22
00:03:36,950 --> 00:04:20,949
dimensional thrust as okay this was a non-dimensional
thrust equation now in this if we put e = 1
23
00:04:20,949 --> 00:04:45,439
this simplifies to okay.
24
00:04:45,439 --> 00:04:55,949
So for efficiencies being or one unity ideal
system you get this expression for non-dimensional
25
00:04:55,949 --> 00:05:05,189
thrust and similarly one can get an expression
for ISP now if you look at this expression
26
00:05:05,189 --> 00:05:17,520
for non-dimensional thrust what you see here
is if?B increases what happens to the non-dimensional
27
00:05:17,520 --> 00:05:24,550
thrust ? B remember is under the designers
control that is you can choose what is the
28
00:05:24,550 --> 00:05:31,530
exit temperature at the exit of the combustor
by burning suitable amounts of fuel there
29
00:05:31,530 --> 00:05:37,880
is an upper limit that is if you are using
if you are using kerosene air there is an
30
00:05:37,880 --> 00:05:42,229
upper limit of two thousand three hundred
Kelvin which is the stoichio metric combustion
31
00:05:42,229 --> 00:05:49,999
temperature but other than that you can vary
the air fuel ratio and get different temperatures
32
00:05:49,999 --> 00:06:00,009
this number is under your control right.
And one it is very obvious that if you increase
33
00:06:00,009 --> 00:06:11,029
? B you get a larger f / m. a0 now what this
value F by m.a0 indicates is that you will
34
00:06:11,029 --> 00:06:23,699
see as this number is higher the size of the
ramjet becomes smaller, so
35
00:06:23,699 --> 00:06:40,360
larger value of
36
00:06:40,360 --> 00:06:52,249
means smaller size of
which is desirable because your drag will
37
00:06:52,249 --> 00:07:01,550
then be smaller okay. So you need a larger
value of this and if?B increases you will
38
00:07:01,550 --> 00:07:18,999
find that as a consequence f / m.a a 0 also
increases okay, now when we were discussing
39
00:07:18,999 --> 00:07:25,129
ramjets earlier in the introductory classes
I had made a point that transits are not self-starting
40
00:07:25,129 --> 00:07:30,990
that is they need something to take them to
a slightly higher Mach number and then they
41
00:07:30,990 --> 00:07:36,979
will start from their own let us see if we
can bring that out from this equation.
42
00:07:36,979 --> 00:07:50,080
If we put M0 = 0 here what happens M0 is the
Mach number at the inlet if we put that to
43
00:07:50,080 --> 00:07:55,929
zero that is if you are holding the vehicle
stationary what happens to the thrust this
44
00:07:55,929 --> 00:08:20,739
goes to zero, so
45
00:08:20,739 --> 00:08:38,149
not self-starting or in other words static
thrust is zero, so if you put the ramjet on
46
00:08:38,149 --> 00:08:48,089
a thread stand and you do not do you do not
give the flow that is you do not induce the
47
00:08:48,089 --> 00:08:53,290
flow yourself then it would not be able to
produce any thrust even if you switch on the
48
00:08:53,290 --> 00:09:04,259
fuel flow rate okay.
We know now that it does not produce any static
49
00:09:04,259 --> 00:09:12,139
thrust is there any value of Mach number at
which it will start producing some thrust
50
00:09:12,139 --> 00:09:17,440
that is our next exercise we know that static
thrust is zero here let us now find out what
51
00:09:17,440 --> 00:09:41,500
happens at what Mach number can we get some
meaningful thrust okay, now we know that efficiencies
52
00:09:41,500 --> 00:09:46,750
if you remember efficiencies will always be
less than one and when we multiply all of
53
00:09:46,750 --> 00:09:59,110
them the product range between product that
is e range between 0.84and 0.97 right so efficiency
54
00:09:59,110 --> 00:10:07,260
will always be less than one, so if you incorporate
it and try and see what happens at what Mach
55
00:10:07,260 --> 00:10:13,810
number will we get some thrust okay.
To derive that we need to look at this expression
56
00:10:13,810 --> 00:10:54,420
here okay this is the expression and we know
that if it has to produce positive thrust
57
00:10:54,420 --> 00:11:03,890
this number should be greater than zero right,
so if it has to be greater than zero then
58
00:11:03,890 --> 00:11:10,750
obviously the product of the first two numbers
must be greater than one or if you look at
59
00:11:10,750 --> 00:11:17,860
the terms in the bracket this should be greater
than one. So we will put that dump I will
60
00:11:17,860 --> 00:11:37,480
also include the 1 + F term here they remember
we had neglected this term for our convenience
61
00:11:37,480 --> 00:12:10,990
okay. So for
62
00:12:10,990 --> 00:12:37,839
positive thrusts 1 + f should be greater than
one okay.
63
00:12:37,839 --> 00:13:04,410
Now we can square both sides and cross multiply
and we will get this condition that should
64
00:13:04,410 --> 00:13:18,850
be greater than ? B okay, even though F is
smaller here this quantity can be slightly
65
00:13:18,850 --> 00:13:28,060
larger therefore I have taken that into account
in this equation okay now this is a less cumbersome
66
00:13:28,060 --> 00:13:36,260
equation to solve and bring out what should
be the value of Mach number at which it will
67
00:13:36,260 --> 00:13:41,839
give positive thrust so the easier way is
you plot a graph and then try and find out
68
00:13:41,839 --> 00:13:49,759
where it leads you okay and that is what is
done in this plot here.
69
00:13:49,759 --> 00:13:55,019
What is plotted as non-dimensional thrust
variation with flight Mach number that is
70
00:13:55,019 --> 00:14:04,769
as you increase M 0what happens to f by m.a
a0 you notice that there are four lines here
71
00:14:04,769 --> 00:14:14,199
each corresponding to a different value of
efficiency this is for e = 1, so you have
72
00:14:14,199 --> 00:14:20,230
its starting from zero itself okay whereas
all the other lines are clubbed somewhere
73
00:14:20,230 --> 00:14:26,889
close to one where efficiencies are around
0.8 six okay.
74
00:14:26,889 --> 00:14:35,970
So if you have a non unity efficiency that
is for real systems you notice that it is
75
00:14:35,970 --> 00:14:46,880
very difficult to have a subsonic transit
okay whatever be the ? B that you can go up
76
00:14:46,880 --> 00:14:54,949
to write ? B is the ratio of combustion exit
temperature combustor exit temperature to
77
00:14:54,949 --> 00:15:01,879
the inlet temperature even if you add more
heat even if you keep on adding more heat
78
00:15:01,879 --> 00:15:08,579
it does not matter we brain varied ? B from
6 to 10 almost by a factor of 2even if you
79
00:15:08,579 --> 00:15:15,329
add more heat it does not matter you still
cannot have something like a subsonic ramjet
80
00:15:15,329 --> 00:15:22,470
this is purely determine by efficiencies for
non-unity efficiencies you cannot have subsonic
81
00:15:22,470 --> 00:15:31,440
ramjets.
And what you will notice here is that all
82
00:15:31,440 --> 00:15:42,680
these systems have their thrust- or the non-dimensional
thrust is maximum as somewhere between two
83
00:15:42,680 --> 00:16:15,889
to three okay, so
it is Maxima for F by a Mach number range
84
00:16:15,889 --> 00:16:29,610
of two to three right but we need to take
this result a little more carefully what we
85
00:16:29,610 --> 00:16:38,329
have looked at in this analysis is this is
only the thrust produced by the ramjet we
86
00:16:38,329 --> 00:16:45,470
have not accounted for the net thrust that
is thrust minus the drag right there is a
87
00:16:45,470 --> 00:16:51,970
drag component also that determines whether
the vehicle will move forward or not and that
88
00:16:51,970 --> 00:17:04,299
will make it go even further right if you
look at this graph in this graph at this Mach
89
00:17:04,299 --> 00:17:09,970
number f by m. a0 is this much there will
be drag right.
90
00:17:09,970 --> 00:17:15,709
So that drag will bring down this value and
probably you will start producing net thrust
91
00:17:15,709 --> 00:17:22,250
somewhere at a slightly higher Mach number
right, so typically you will not find any
92
00:17:22,250 --> 00:17:53,409
sub sonic ram dux that is the underlying message
from this graph okay.
93
00:17:53,409 --> 00:18:12,179
So then we consider
net thrust that is thrust minus the drag portion
94
00:18:12,179 --> 00:18:45,779
then the starting Mach number
will be higher and you cannot have
95
00:18:45,779 --> 00:19:06,600
subsonic
ramjets or efficiency is less than what now
96
00:19:06,600 --> 00:19:19,240
in this graph we see that F / m.a a 0 varies
in this fashion and if I were to look at ISP
97
00:19:19,240 --> 00:20:12,710
variation remember is P / a 0 is nothing but
M 0 into okay now ISP what would be a preferable
98
00:20:12,710 --> 00:20:20,590
thing having a larger value of ISP or a smaller
value of ISP you remember SFC we need a smaller
99
00:20:20,590 --> 00:20:29,110
value and ISP being one over SFC we need a
larger value of ISP if you notice here you
100
00:20:29,110 --> 00:20:36,179
have ? b appearing here as well as here right
in two places in this expression and here
101
00:20:36,179 --> 00:20:40,190
it is under root sign and here it is not under
root sign okay.
102
00:20:40,190 --> 00:20:49,149
So what do you expect to happen if ? B increases
what happens to ISP is P should decrease okay
103
00:20:49,149 --> 00:20:53,870
and what are what should happen with efficiencies
if efficiencies are less than one what should
104
00:20:53,870 --> 00:21:01,740
happen to ISPs it should also again decrease
and the same thing is plotted here in this
105
00:21:01,740 --> 00:21:06,030
graph.
106
00:21:06,030 --> 00:21:16,110
What you see here is ISP on the y-axis versus
Mach number the flight Mach number for an
107
00:21:16,110 --> 00:21:27,899
efficiency of 0.86 with different ? B's as
? B is increased from 6 to 10 you see that
108
00:21:27,899 --> 00:21:35,549
the ISPs are lower okay, so which means SF
C's will be higher or fuel consumption will
109
00:21:35,549 --> 00:21:44,050
be more which is not a desirable feature so
you would like to operate such that ? B is
110
00:21:44,050 --> 00:21:53,279
lower so that your fuel consumption is less
but then if you go back to the previous curve
111
00:21:53,279 --> 00:22:00,919
you will find that if ? B is less your f /m.
a a 0 will also decrease or the thrust will
112
00:22:00,919 --> 00:22:07,660
also decrease.
So you should design such that this is a large
113
00:22:07,660 --> 00:22:15,169
quantity so that your size becomes okay and
then the other thing will automatically get
114
00:22:15,169 --> 00:22:28,220
fixed okay now we looked at how the non-dimensional
thrust varies with ? B let us look at what
115
00:22:28,220 --> 00:22:35,929
happens with M 0 you have already seen that
in graph let's try and bring out if we can
116
00:22:35,929 --> 00:22:41,529
do some analysis and find out what happens
with Mach number Mach number if you notice
117
00:22:41,529 --> 00:22:48,740
here it appears this is the expression that
we are looking for it expressed it you find
118
00:22:48,740 --> 00:22:54,020
Mach number in M 0 and ? 0 it appears in both
places.
119
00:22:54,020 --> 00:23:00,510
So it is not obvious as ? B is we noted that
with the increase in ? B it is this becomes
120
00:23:00,510 --> 00:23:07,039
smaller which is what we found NIC curves
but Mach number is not such a straightforward
121
00:23:07,039 --> 00:24:02,630
relationship let us try and find out what
happens there, that is what we are trying
122
00:24:02,630 --> 00:24:08,580
to find out us is there an optimal Mach number
at which if we fly it seem that it should
123
00:24:08,580 --> 00:24:25,220
lie somewhere between two and three can we
bring that out through some analysis here.
124
00:24:25,220 --> 00:24:36,200
So what do we do we take a derivative with
respect to M 0, so firstly we will assume
125
00:24:36,200 --> 00:24:44,590
e to be 1 so that we do not end up with a
large or a cumbersome equation so if you look
126
00:24:44,590 --> 00:25:06,539
at e = 1 the expression for F / M 0 a 0 would
be M0 x v? B by ? 0 -1 right so now if we
127
00:25:06,539 --> 00:25:40,720
take a derivative of this
with respect to M0 et me also include 1 +
128
00:25:40,720 --> 00:26:00,769
s now if we take a derivative we will get
1 x the rest of the things + F v? B / ? 0
129
00:26:00,769 --> 00:26:29,349
– 1 – ½ m0 x 1 + F ? B ? B does not vary
with M 0 so ? B does not get change here so
130
00:26:29,349 --> 00:26:45,970
? 0 to the power of 3 / 2 x ? – 1 m0 okay.
And for a Maxima this must be equal to 0 so
131
00:26:45,970 --> 00:27:18,529
if we equate this to 0 then both these two
must be equal we will get 1 + f = 0 v ? B
132
00:27:18,529 --> 00:28:18,760
/? 0 – ½ x ? -1 m02 okay, what is this
quantity here what is the definition of ? 0
133
00:28:18,760 --> 00:28:46,740
? 0 is so what you have here is ? 0 - 1 so
if you put that you will get 1 +F ? V – ? 0
134
00:28:46,740 --> 00:29:04,960
x and do the simplification that you get this
? 0 -1 is divided by this must be equal to
135
00:29:04,960 --> 00:29:33,250
1 or ? B by ? 0 this becomes ? 0 - ? 0 + 1
so ? 0 ? 0 cancels out you get 1/ ? 0is equal
136
00:29:33,250 --> 00:30:08,000
to 1 ? 0 to the power of 3 / 2 must be equal
to okay this is the expression that we get
137
00:30:08,000 --> 00:30:20,500
for maximum value of f / m. aa0 occurs when
? 0 is in this is the value of ? 0 again.
138
00:30:20,500 --> 00:30:25,990
Now you plug in some typical values we will
find out what we had discovered earlier that
139
00:30:25,990 --> 00:30:57,910
it ranges somewhere between 2.5 to 3.
140
00:30:57,910 --> 00:31:10,850
So if we take et3 to be 2300Kelvin which is
what you will get as the adiabatic flame temperature
141
00:31:10,850 --> 00:31:17,191
if you use kerosene air system at special
metric condition and I will take T not to
142
00:31:17,191 --> 00:31:26,740
be 230 Kelvin that is as you go higher in
altitude temperature drops, so at some particular
143
00:31:26,740 --> 00:31:46,990
ask you will get this temperature and therefore
my ? B will be N and if I use seven okay you
144
00:31:46,990 --> 00:32:04,500
will find that ? 0 you will get as somewhere
between 2.252 2.15 this depends on what value
145
00:32:04,500 --> 00:32:13,960
of F you consider and correspondingly this
will also change a little okay and therefore
146
00:32:13,960 --> 00:32:26,250
M 0 to be somewhere between 2.42 0.5 okay.
147
00:32:26,250 --> 00:32:34,450
So you will notice that most systems most
real systems although all these analysis we
148
00:32:34,450 --> 00:32:40,510
have carried out with e to be one even if
you put in the efficiencies we have seen in
149
00:32:40,510 --> 00:32:49,080
the graphs that Mach number range for high
value of F / m. a a 0 ranges between two to
150
00:32:49,080 --> 00:32:55,000
three which is where you will find most operating
systems okay most systems would want to go
151
00:32:55,000 --> 00:33:00,270
to that condition and then fly in that Mach
number range okay.
152
00:33:00,270 --> 00:33:10,549
So that you have a larger value of F/ m. a
a 0 okay this finishes our discussions on
153
00:33:10,549 --> 00:33:23,740
ramjets now let us move on to the next engine
that is the turbojet okay to do rate has a
154
00:33:23,740 --> 00:33:30,830
slightly involved system it has a compressor
and a turbine, so we need to account for that
155
00:33:30,830 --> 00:33:40,029
here okay now when we look at turbojets there
are two things that we can look at one is
156
00:33:40,029 --> 00:33:45,700
when the nozzle is choked and the other is
when the nozzle or when the flow through the
157
00:33:45,700 --> 00:33:50,000
nozzle is optimally expanded.
Firstly we will consider the case wherein
158
00:33:50,000 --> 00:33:55,500
it is the flow through the nozzle is optimally
expanded and we will try to find out what
159
00:33:55,500 --> 00:34:36,640
derive what equations will get.
160
00:34:36,640 --> 00:35:19,840
The turbojets at about that similar to a ramjet
we will have an intake
161
00:35:19,840 --> 00:35:36,180
and then we will have a compressor
162
00:35:36,180 --> 00:36:06,990
which is connected to a wine and you have
a combustion chamber in between and then you
163
00:36:06,990 --> 00:36:25,510
have a jet pipe or an afterburner and then
convergent nozzle. So let me again call these
164
00:36:25,510 --> 00:36:45,570
stations as one two three four five six and
seven this is the zero to two is the intake
165
00:36:45,570 --> 00:37:05,800
two to three is compressor three to four is
combustor
166
00:37:05,800 --> 00:37:31,300
this is the turbine and this is the afterburner
and then the converging nozzle as I said earlier
167
00:37:31,300 --> 00:37:39,900
we do not use typically CD nozzle the only
aircraft or a rail system that has a turbojet
168
00:37:39,900 --> 00:37:51,310
which uses a convergent divergent nozzle is
the what does the engine Olympus engine write
169
00:37:51,310 --> 00:38:01,640
on what is that a Concorde aircraft okay.
So other than that typically only a converging
170
00:38:01,640 --> 00:38:07,980
nozzle is used because the pressure at the
exit of the turbine is very low and it would
171
00:38:07,980 --> 00:38:13,490
not be useful if you have a convergent divergent
nozzle and therefore only a convergent nozzle
172
00:38:13,490 --> 00:38:34,910
is used, now here we will take up the case
where P 7 = P 0 okay, that is Emily expanded
173
00:38:34,910 --> 00:38:45,540
flow
174
00:38:45,540 --> 00:38:51,180
notice here that you have compressor extra
and the turbine extra. So we need to define
175
00:38:51,180 --> 00:39:09,780
certain other parameter that is we will define
compressor pressure ratio.
176
00:39:09,780 --> 00:39:26,030
I will call it p C, p C is nothing but PT
3 / PT 2 and E again here indicates stagnation
177
00:39:26,030 --> 00:39:49,640
conditions from this we can define t C that
is nothing but t t3 / TT 2 which is p C ? – 1/
178
00:39:49,640 --> 00:40:26,230
? okay, and similarly I can define a turbine
pressure ratio, so from this I can define
179
00:40:26,230 --> 00:40:47,120
t T which is nothing but TT5 / DT4 okay. So
now with this and we know that from a previous
180
00:40:47,120 --> 00:41:02,950
calculation ?0 is nothing but TT 0 / T 0 that
is 1 + ? – 1 n02 right.
181
00:41:02,950 --> 00:41:20,920
And based on this I can get M0 = v 2/ ? – 1
? 0 – 1 the other parameter that we need
182
00:41:20,920 --> 00:41:43,740
to define is ? B in this case I will define
? B as t t4 /T 0 okay so ? B is nothing but
183
00:41:43,740 --> 00:42:15,930
t t4 / T 0 which I can rewrite as tt 4 /tt
3 x dt3 / TT 2 x T T 2 / T T 0 T t 0 / t0
184
00:42:15,930 --> 00:42:29,670
what is tt 4 / TT 3, this I will call it as
t b similar to what we had done earlier I
185
00:42:29,670 --> 00:42:47,090
will call this as t b and tt3 / TT 2 again
we have hid here it is p C and no t C sorry
186
00:42:47,090 --> 00:42:58,811
and what is TT 0 T T 2/ T T 0 if we assume
isentropic processes this will be 1 this is
187
00:42:58,811 --> 00:43:16,770
flow through this is flew through intake and
just like in the previous case if the process
188
00:43:16,770 --> 00:43:35,250
is isentropic we can get this to be one and
lastly little this is nothing but ? 0 okay.
189
00:43:35,250 --> 00:44:02,310
So I get my expression for ? B or tau B I
can write it as okay, now we want to derive
190
00:44:02,310 --> 00:44:14,890
expressions for thrust so the thrust equation
is f / f = m . a x in this case the exit velocity
191
00:44:14,890 --> 00:44:47,440
is at v7 okay and intake is at 0 so you get
v7 into okay we are assume P 7 = P 0 and therefore
192
00:44:47,440 --> 00:44:58,850
this goes to 0 and will also make the other
assumption that F is very much less than 1
193
00:44:58,850 --> 00:45:05,270
this is more true in a run in a turbojet compared
to a ramjet because in the main combustor
194
00:45:05,270 --> 00:45:12,920
it will be much lower than 0.067 because you
are worried about the turbine Inlet temperature
195
00:45:12,920 --> 00:45:17,400
so you want it to be much lower than the stoichio
metric condition.
196
00:45:17,400 --> 00:45:35,290
So f is less than 1 so I can neglect this
part and I get F = m. a v7 - V 0 and we will
197
00:45:35,290 --> 00:45:53,810
do the similar procedure as we did last time
so I can write this as m. aa0 m0 x m7 / m0
198
00:45:53,810 --> 00:46:12,270
x v t7 / t0 – 1 again we have used the relationship
for speed of sound here okay and we have again
199
00:46:12,270 --> 00:46:19,180
made the assumption that ? and r are the same
for exhaust gases as well as incoming air,
200
00:46:19,180 --> 00:46:51,810
so assuming ? and r to be same for gases and
air okay. So we get this relationship now
201
00:46:51,810 --> 00:47:12,530
again we need to look at these two ratios
or we can write the non-dimensional thrusters.
202
00:47:12,530 --> 00:47:54,690
We can write the expression for
non-dimensional thrusters m0 x m7 / m0 okay
203
00:47:54,690 --> 00:48:03,950
so we are again left with the task of trying
to find these two ratios again we will follow
204
00:48:03,950 --> 00:48:14,160
a similar procedure of cascading temperatures
and cascading pressures right, now what I
205
00:48:14,160 --> 00:48:21,420
want you to think about is all the while I
have been telling you that you are looking
206
00:48:21,420 --> 00:48:29,530
at flow through intake and I said we are looking
at isentropic conditions in the previous case
207
00:48:29,530 --> 00:48:39,530
that we took up ramjet we looked at all the
temperature ratios and efficiency is never
208
00:48:39,530 --> 00:48:49,390
figured in that place right why does the efficiency
only figure when we are looking at pressures
209
00:48:49,390 --> 00:48:57,040
and why does not it figure here.
Now in the previous case also we looked at
210
00:48:57,040 --> 00:49:05,240
a realistic system right, so we cannot look
at realistic systems and say the flow through
211
00:49:05,240 --> 00:49:11,910
the intake is isentropic. So what really is
the crux there why are we taking this is one
212
00:49:11,910 --> 00:49:20,170
and why do we take the other thing wherein
when we look at pressures we are looking at
213
00:49:20,170 --> 00:49:26,430
efficiencies here okay we will discuss this
in the next class okay, thank you very much.