1
00:00:00,080 --> 00:00:11,500
In
2
00:00:11,500 --> 00:00:17,920
the last class we had looked at the cycle
analysis for a ramjet and we had sat out the
3
00:00:17,920 --> 00:00:26,140
cycle analysis for a turbojet, let us continues
the cycle analysis for the turbojet.
4
00:00:26,140 --> 00:00:41,760
If you look at the Ts diagram firstly let
us assume that all processes are 100% efficient
5
00:00:41,760 --> 00:00:53,240
that is we will assume an ideal cycle initially
and we also started something, wherein the
6
00:00:53,240 --> 00:01:18,280
flow is optimally expanded in the nozzle.
So these were the two things that we are going
7
00:01:18,280 --> 00:01:25,600
to look at if we take an ideal cycle then
there is compression in the intake itself,
8
00:01:25,600 --> 00:01:43,210
so I will call 0 to 2 and then through the
compressor, then you have heat addition then
9
00:01:43,210 --> 00:02:06,130
you have expansion through the turbine then
you have expansion again through the nozzle
10
00:02:06,130 --> 00:02:14,030
this is the Ts diagram we had okay.
Now we had derived certain things in the last
11
00:02:14,030 --> 00:02:37,400
class we had derived that t b is nothing but
?b / ? C ?0 and
12
00:02:37,400 --> 00:03:12,880
we have got this expression for F/ m. a b0
this is the expression that we had okay. Now
13
00:03:12,880 --> 00:03:19,660
let us try and find out how to get these ratios
just like the previous time when we had done
14
00:03:19,660 --> 00:03:25,420
this cascading in ramjets let us do the cascading
and find out how we get these temperatures.
15
00:03:25,420 --> 00:04:01,170
Now to get T 7/ T 0 can you cascade sub T
7 / TT 7 x TT 7 / Tt6 / TT 5 / T 4 T4 / Tt3
16
00:04:01,170 --> 00:04:23,740
/ okay again the
17
00:04:23,740 --> 00:04:31,550
subscript T here indicates stagnation conditions
and if you take a look at this all these cancels
18
00:04:31,550 --> 00:04:45,110
out and you get T E 7 / T0 okay. So we also
know that the last term is nothing but ? 0
19
00:04:45,110 --> 00:04:56,669
from our previous class, so we have this is
nothing but a ratio of stagnation to static,
20
00:04:56,669 --> 00:05:08,059
so we can express this in terms of Mach number
as 1 + ? - 1/ 2 m7 2 okay.
21
00:05:08,059 --> 00:05:17,050
Now what is TT 7 / TT6 this is flow through
the nozzle again if the efficiencies are 1
22
00:05:17,050 --> 00:05:25,939
the stagnation temperature ratio will be the
same, so this is 1 and TT6 / TT5 this is flow
23
00:05:25,939 --> 00:05:31,650
through the afterburner anyway we are not
considering in this analysis this analysis
24
00:05:31,650 --> 00:05:45,689
is without the afterburner. So this will again
be 1 and TT5 / TT4 s flow through the turbine
25
00:05:45,689 --> 00:05:50,060
now flow through the turbine because it is
again an isentropic process this ratio will
26
00:05:50,060 --> 00:06:10,939
also be what will that ratio be? TT5 / TT4.
One you have forgotten something we derived
27
00:06:10,939 --> 00:06:25,320
in the last class that this is t T and TT4
/ TT3 is process through the combustor, so
28
00:06:25,320 --> 00:06:47,719
I will call it t B x Tt3 / TT to is flow through
the compressor again this is t C x again you
29
00:06:47,719 --> 00:07:03,969
get the diffuser this ratio because the flow
is isentropic is 1 and x ? 0 okay. So we know
30
00:07:03,969 --> 00:07:25,830
that t b is this if we substitute it there
we will get T7 / T 0 =t T x t B is T B / ? C
31
00:07:25,830 --> 00:07:45,270
? 0 x t C ß 0 here is a ratio that we have
this t C ? 0 and this t C ? 0 cancels off
32
00:07:45,270 --> 00:08:00,069
and we get ? B t T divided / 1 + ? - 1 / 2
m 72 okay. Now similarly we will do for the
33
00:08:00,069 --> 00:08:15,259
pressures.
P 7 / P 0 now he has room all efficiencies
34
00:08:15,259 --> 00:08:44,730
to be 1so the first one this would be if you
cascade it will be P 7 / PT 7 x PT 7 / PT
35
00:08:44,730 --> 00:09:12,430
6 PT 6 / PT5 okay this is the cascading and
if we plug in the values then this is the
36
00:09:12,430 --> 00:09:29,750
ratio of static to stagnation condition, so
you get 1 + ?- 1 / 2 m72 ?/ ?- 1 then this
37
00:09:29,750 --> 00:09:40,020
is flow through nozzle, so you get for all
efficiencies being one you get one then you
38
00:09:40,020 --> 00:09:47,530
have flow through afterburner this is again
one what does this flow through turbine you
39
00:09:47,530 --> 00:10:01,760
get PIT okay, pressure ratio across turbine
and what happens in an ideal cycle to the
40
00:10:01,760 --> 00:10:10,030
pressure ratio across the combustor.
For an ideal cycle the pressure ratio across
41
00:10:10,030 --> 00:10:16,760
the combustor it will be is isentropic isobaric
process so the pressure is the same, so this
42
00:10:16,760 --> 00:10:32,990
ratio would be 1 + PT3 / PT2 this is again
p C ratio pressure ratio across the compressor
43
00:10:32,990 --> 00:10:42,120
and PT2 / PT0 is flow through diffuser for
a ideal process this is 1 and lastly this
44
00:10:42,120 --> 00:11:04,110
is nothing but ? 0 ?/ ?- 1 okay. Now I can
change this to ?t and we will get a new ratio,
45
00:11:04,110 --> 00:11:35,450
so I will do that I will get tT t C ? 0 divided
/ 1 + okay this is the ratio that we get.
46
00:11:35,450 --> 00:11:41,190
And if you remember we have taken the case
where the flow is optimally expanded through
47
00:11:41,190 --> 00:11:53,490
the nozzle so what is P 7/ P 0, so we can
rewrite this as this ratio as 1 + ?- 1 / 2
48
00:11:53,490 --> 00:12:11,470
m 72 must be =t T C ? 0 all right I can take
out the powers without any problem and therefore
49
00:12:11,470 --> 00:12:20,250
I can write it like this. Now remember in
this expression T7 / T 0 in the denominator
50
00:12:20,250 --> 00:12:29,320
I have 1 + ?- 1/ 2 m 7 2 which is what I have
got that, so if I plug in this value I will
51
00:12:29,320 --> 00:12:58,740
get T 7 / T 0 =? B t T divided / t e t C the
t 0, so I will get
52
00:12:58,740 --> 00:13:04,570
this is the temperature ratio.
Now to find the Mach number ratio we will
53
00:13:04,570 --> 00:13:28,780
have to take this expression and derive the
ratio for Mach numbers.
54
00:13:28,780 --> 00:14:15,710
So we know that 1 + ?- 1 /2 m 7 2 =t P t C
? 0 and we also know 1 + ?- 1 / 2 m 02 = ? 0
55
00:14:15,710 --> 00:14:38,080
so from these two expressions I can write
M7 =t C t D ? 0 - 1 okay and similarly M 0
56
00:14:38,080 --> 00:15:16,620
we had done this earlier = okay and the ratio
M 7/ M 0 will then become okay. So this is
57
00:15:16,620 --> 00:15:22,240
the ratio of Mach numbers and that is the
ratio of temperatures that we were looking
58
00:15:22,240 --> 00:15:32,500
for we have got those two and if we plug them
x the expression for thrust F / M.A 0 okay
59
00:15:32,500 --> 00:16:25,029
that is what we were looking for.
So F/ M . a 0 =what we got was M 0 x p7 / T
60
00:16:25,029 --> 00:17:05,500
0 x m7 / M 0 - 1 and if we substitute the
ratios 47/ T 0 and m7 / M 0 we will get okay,
61
00:17:05,500 --> 00:17:35,759
this is the expression for non-dimensional
thrusts that we were looking for, now is this
62
00:17:35,759 --> 00:17:52,110
expression complete is there something that
is still missing, say here you have got t
63
00:17:52,110 --> 00:18:00,770
C & t T right but we know that compressor
and turbine and Power Balance must be there,
64
00:18:00,770 --> 00:18:05,999
so they cannot be independent of each other
they must be related to each other and therefore
65
00:18:05,999 --> 00:18:09,580
you can write one expression for connecting
these right.
66
00:18:09,580 --> 00:18:34,720
So there is a relationship for connecting
t C n t T that is through the compressor
67
00:18:34,720 --> 00:18:47,580
turbine power balance, now if we do the compressor
turbine power balance we will get this ratio
68
00:18:47,580 --> 00:19:09,100
connecting ? C to t T. Now we know that flow
through compressor is M. a CP TT3 - TT 2 this
69
00:19:09,100 --> 00:19:33,629
must = M.a x 1 + F E TT4 - TT 5, now there
are 2 things that we are going to make an
70
00:19:33,629 --> 00:19:40,350
assumption on if you really take a look at
this is primarily air getting compressed in
71
00:19:40,350 --> 00:19:48,700
the compressor whereas for the flow through
the turbine you have product gases.
72
00:19:48,700 --> 00:19:55,980
If you remember earlier when we said ? and
r are the same for both air and exhaust gases
73
00:19:55,980 --> 00:20:02,299
right if ? and our concert is the same for
both air and exhaust gases then CP what should
74
00:20:02,299 --> 00:20:20,920
happen to CP should also be the same for both
of them, so assuming CP to be constant and
75
00:20:20,920 --> 00:20:27,220
also we will make another assumption that
we have been doing all along that the fuel
76
00:20:27,220 --> 00:20:38,159
a ratio F is very much less than 1 when for
turbojet, because typically this will be of
77
00:20:38,159 --> 00:20:44,590
the order of 0.02 2.04.
So even if you neglect it you want to make
78
00:20:44,590 --> 00:20:57,740
a 2 or 4 % error okay, so with these two assumptions
this will simplify to TT 3 - TT 2 must be
79
00:20:57,740 --> 00:21:43,350
=TT 4 - TT5. So what we do from here is we
know / cascading what the ratios are so we
80
00:21:43,350 --> 00:22:30,919
just have two more divided both sides / T
0 let us divide both sides / T not okay.
81
00:22:30,919 --> 00:22:45,809
Now what does TT 3 / T 0 again if we cascade
I will get TT 3 / TT 2 x TT 2 / TT 0 x T T
82
00:22:45,809 --> 00:23:10,230
0 / T 0 what is TT 3 / TT 2 this is flow through
compressor, so this will be t T so t C and
83
00:23:10,230 --> 00:23:29,210
this ratio is 1 across the diffuser. So and
this is T 0 right TT 0 / T 0 is T 0 and this
84
00:23:29,210 --> 00:23:50,239
would be T T 2 / TT 0 x TT 0 / T 0 this again
is T T 2 / T T 0 is 1 and this would be ? 0
85
00:23:50,239 --> 00:24:15,419
right and what is T4 / T 0 this is ? B this
is from our definition and TT5 / TT0 I can
86
00:24:15,419 --> 00:24:39,660
write it as T T v / TT 4 x TT 4 / T 0 what
is this 85 / TT 4 t T this is t T x ? V. So
87
00:24:39,660 --> 00:25:03,590
I end up getting if I take out ?0 as common
? 0 x t C - 1must be = ? B x1 - t T, so therefore
88
00:25:03,590 --> 00:25:49,739
I can write the expression for t TS.
I can write t T as =okay, so this is the expression
89
00:25:49,739 --> 00:25:59,179
for t T now if I plug in this expression for
t T in this expression for F /M dot a you
90
00:25:59,179 --> 00:26:07,289
note that is the non-dimensional thrust before
we do that what we see here is we need this
91
00:26:07,289 --> 00:26:15,760
ratio all the in t C terms are only in this
ratio so let us look at this first term in
92
00:26:15,760 --> 00:26:39,160
the under the square root sign and let us
try to simplify that. So t C ? B ? 0 right
93
00:26:39,160 --> 00:27:13,970
/ t C ? 0 - I can write the numerator combining
these terms these are the terms containing
94
00:27:13,970 --> 00:27:20,129
t C and t T.
So if I combine these terms and rewrite it
95
00:27:20,129 --> 00:27:32,369
I will get it like this okay which is I can
simplify this further and write it as t C
96
00:27:32,369 --> 00:27:52,899
? 0 cancels off here, I will get ? B– ? 0
and now if I substitute for t T in this expression
97
00:27:52,899 --> 00:28:33,799
okay I will get this
as = ? B - ? 0 t C, so that is now if I plug
98
00:28:33,799 --> 00:28:41,149
back this expression x that and rewrite my
expression I will get the non-dimensional
99
00:28:41,149 --> 00:29:35,950
thrust as F / M.a a 0 must be =this is the
final expression for a non-dimensional thrust
100
00:29:35,950 --> 00:29:54,769
totally.
See how T and t C are connected now ? b is
101
00:29:54,769 --> 00:29:59,850
a controlling factor is okay with you because
it is a turbine Inlet temperature, so that
102
00:29:59,850 --> 00:30:06,529
is fine with you what you are looking for
is why is it that t t has gone out it out
103
00:30:06,529 --> 00:30:12,989
had to go out because the compressor and turbine
there is a power balance between them and
104
00:30:12,989 --> 00:30:19,799
we have assumed the processes to be 100% efficient.
So the compressor power must be = the turbine
105
00:30:19,799 --> 00:30:26,269
power, so when you do that you can eliminate
t t and that is what we have done here.
106
00:30:26,269 --> 00:30:36,019
So it will be only dependent on t C ? 0 and
? B now we have done all these calculations
107
00:30:36,019 --> 00:30:44,580
right what we need to do is cross check whether
what we have got is correct, how do we do
108
00:30:44,580 --> 00:31:01,649
that? How do we cross check what you got is
corrector not and still does not Ellison what
109
00:31:01,649 --> 00:31:10,669
do we know you will always assume previous
class is true right or previous class is correct.
110
00:31:10,669 --> 00:31:41,249
So we know the results for ramjet right what
is the result for ramjet
111
00:31:41,249 --> 00:32:01,429
the result for Ramjet.
That was F/ M.a A 0 =M 0 ? B mine / ? 0 - 1
112
00:32:01,429 --> 00:32:19,919
? =1 this is for okay this was the result
that we had for R and it, now what is the
113
00:32:19,919 --> 00:32:35,610
difference between ramjet and a turbojet?
That you have a compressor right compressor
114
00:32:35,610 --> 00:32:45,470
and therefore a turbine, so what happens if
you put t C to be =1 compressor pressure ratio
115
00:32:45,470 --> 00:32:51,619
or PI C is 1 or t C is one compressor turbine
D re temperature ratio is one or compressor
116
00:32:51,619 --> 00:32:56,059
pressure ratio is 1 that is what is a ramjet
right.
117
00:32:56,059 --> 00:33:06,970
So if you put t C =one here what happens ? 0
? 0 cancels off and you get t C this is T
118
00:33:06,970 --> 00:33:33,330
B / ? right, so let us do that the budget
becomes a ramjet 1 now C = 1 so when I substitute
119
00:33:33,330 --> 00:34:02,710
t C =1 in the expression for ramjet or turbojet,
you get ? B - ? 0 x 1 + ? 0 - ? B / ? 0 t
120
00:34:02,710 --> 00:34:22,030
C is again 1. Now it is obvious that you can
cancel out this and you get you can take out
121
00:34:22,030 --> 00:34:51,190
? B common so you get M0 and if you take out
? B as common what you get is 1 - 1 / ? 0
122
00:34:51,190 --> 00:35:22,320
okay, this is anyway the same as you can take
one / ? 0 out you will again get.
123
00:35:22,320 --> 00:35:54,180
So I get which is the same as rammed it so
in that sense it is consistent okay what we
124
00:35:54,180 --> 00:36:01,590
have derived is consistent with when we put
t Z =1 we get the ramjet result okay. Now
125
00:36:01,590 --> 00:36:10,020
there is another thing that we need to look
at what is that? We know that turbo Jets have
126
00:36:10,020 --> 00:36:19,400
a static thrust right the budgets do produce
a static thrust, now we need to find out whether
127
00:36:19,400 --> 00:36:23,030
the expressions that we have derived do show
that.
128
00:36:23,030 --> 00:36:31,200
Now in this expression here what happens what
is the condition for static thrust, if you
129
00:36:31,200 --> 00:36:45,610
put M 0 = 0 then you get 0 new suddenly we
have done all this extravagant calculations
130
00:36:45,610 --> 00:36:51,210
and suddenly found out that we are on the
wrong side, our equations do not bring out
131
00:36:51,210 --> 00:36:58,250
that fact that it still can produce you know
static thrust looks like it produces zero
132
00:36:58,250 --> 00:37:13,290
thrust or is there a catch to it let us see.
We know that
133
00:37:13,290 --> 00:37:29,020
what is ?0 one + ?- 1 / 2 m 0 2 okay.
So if you can you have M 0 and ? 0 - 1 so
134
00:37:29,020 --> 00:37:56,530
M 0 / v? 0 -1 = what you get here vright,
so we can use that here and rewrite the expression
135
00:37:56,530 --> 00:38:36,790
let us do that, so I get
136
00:38:36,790 --> 00:38:46,580
okay is this correct right. We have substituted
for this in the earlier equation and now using
137
00:38:46,580 --> 00:38:56,130
this if you put here M 0 = 0 this goes to
0, so the expression for static thrust would
138
00:38:56,130 --> 00:39:31,350
be
139
00:39:31,350 --> 00:40:31,600
static trust for turbojet.
So you get okay so this is a nonzero quantity,
140
00:40:31,600 --> 00:40:38,180
so therefore we can safely say that whatever
we have derived is correct and it produces
141
00:40:38,180 --> 00:40:46,450
nonzero static thrust okay, even when M 0
=0 what happens when M 0 = 0what happens to
142
00:40:46,450 --> 00:40:58,910
? 0 ? 0 sorry when M 0 = 0 it means that ? 0
should be 1, so you substitute it here again
143
00:40:58,910 --> 00:41:32,140
you can rewrite this expression ? 0. So this
cannot go to zero and therefore we have a
144
00:41:32,140 --> 00:41:50,870
positive static thrust okay.
The static thrust itself is some air moving
145
00:41:50,870 --> 00:41:59,870
in the compressor itself some air force sucked
in the compressor said quote some velocity
146
00:41:59,870 --> 00:42:11,960
yes is it M0 really 0 okay what does they
takeoff velocity of an aircraft okay let us
147
00:42:11,960 --> 00:42:39,700
do that.
See so you have takeoff
148
00:42:39,700 --> 00:42:54,140
velocity around 250 km/h what is the speed
of sound at that condition m/s so you convert
149
00:42:54,140 --> 00:43:24,520
it x kilometers per second kilometers per
hour one. So I will assume it to be fine around
150
00:43:24,520 --> 00:43:37,300
1200 km/h, so you calculate the Mach number
based on this what will be. So what does this
151
00:43:37,300 --> 00:43:43,860
sacrosanct about 0.3Mach number which we say
is the regime where we differentiate that
152
00:43:43,860 --> 00:43:48,440
the flow is incompressible and then the flow
becomes compressible what is this limit of
153
00:43:48,440 --> 00:43:56,580
0.3duration?
So if you look at this number here Mach number
154
00:43:56,580 --> 00:44:09,790
is 1 to 1 so if we do 1+ ?- 1 / 2 this is
M 0 2 is nothing but ? 0 how different will
155
00:44:09,790 --> 00:44:22,150
it be from 1 this will be anyway 1 so even
while the aircraft is taking off with very
156
00:44:22,150 --> 00:44:32,400
high velocities of around 240 250 km/h the
Mach number is around 0.2 to 0.3 which means
157
00:44:32,400 --> 00:44:42,240
that it is still ? 0 is still around 1 okay.
So therefore this is consistent what we have
158
00:44:42,240 --> 00:44:52,800
done here is consistent right okay. Now let
us look at what happens to the other quantity
159
00:44:52,800 --> 00:45:32,930
of interest to us that is B okay.
In the previous class when we were looking
160
00:45:32,930 --> 00:45:46,270
at ramjet we had already done this exercise
and we noted that ISP / E 0 which is the speed
161
00:45:46,270 --> 00:46:05,010
of sound we can write this as 1 / F x F / M.e
0 right and we said that we put in a lot of
162
00:46:05,010 --> 00:46:12,220
effort to find out this expression, so it
is meaningful to use this to get the ISP.
163
00:46:12,220 --> 00:46:21,580
So we know this part and we need to evaluate
what this quantity right is so let us do that
164
00:46:21,580 --> 00:46:33,040
or we can evaluate this / again looking at
the power balance in the combustor from energy
165
00:46:33,040 --> 00:46:44,320
balance
166
00:46:44,320 --> 00:47:10,450
across combustor.
I can write M. F Q = M. x 1 + F right and
167
00:47:10,450 --> 00:47:16,920
what we will again do is we are trying to
find an expression for F we will say that
168
00:47:16,920 --> 00:47:25,790
F here in comparison to 1small, so therefore
we will neglect that we can do that even while
169
00:47:25,790 --> 00:47:37,010
trying to derive this expression for F. So
f is nothing but M. F / me.
170
00:47:37,010 --> 00:47:52,410
So using this
171
00:47:52,410 --> 00:48:15,180
so I can rewrite my expression as 1 / F must
be =Q / CP I will take out T 0 as common then
172
00:48:15,180 --> 00:48:39,720
I will be left with TT4 / T not okay, what
is TT4/ T 0 this is ? B
173
00:48:39,720 --> 00:49:05,360
what is this T 3 / TT 2 x T 2 / T T 0 T 0
this is ? 0 this is 1 this is t C, so you
174
00:49:05,360 --> 00:49:29,340
get 1 / F = Q / C PT 0 x ? B - t C T done
okay. So this is the expression into now we
175
00:49:29,340 --> 00:50:39,740
have got 1 / F there so I can write the complete
expression for is P / E 0 as =? 0 t C + ? 0,
176
00:50:39,740 --> 00:50:46,240
so this is the expression that we have for
Isp by you know we will stop here and will
177
00:50:46,240 --> 00:50:48,460
continue in the next class thank you.