1
00:00:11,309 --> 00:00:22,690
In the last few classes we had looked at non
dimensional thrust and ISP for cases where
2
00:00:22,690 --> 00:00:29,820
in the efficiencies was all unity right, let
us now look at the case where efficiencies
3
00:00:29,820 --> 00:01:12,729
are not equal to unity and let us see how
the equations change there.
4
00:01:12,729 --> 00:01:28,819
Now if you look at the TS diagram for a turbojet
with efficiencies what you will
5
00:01:28,819 --> 00:01:51,679
let us consider only the simple cycle no after
burner this is a cycle with the 100% efficiency,
6
00:01:51,679 --> 00:03:09,590
now if you have non unity efficiencies, now
if you have non unity efficiencies you will
7
00:03:09,590 --> 00:03:26,570
get a cycle like this
the dashed ones are for non unity efficiencies,
8
00:03:26,570 --> 00:03:33,800
now when we did the analysis for a ramjet
we saw that we were able to handle efficiencies
9
00:03:33,800 --> 00:03:53,400
of the non moving parts namely intake then
burner then nozzle right.
10
00:03:53,400 --> 00:04:05,780
Here in addition to these we have the compressor
and the turbine okay, so in addition we have
11
00:04:05,780 --> 00:04:25,360
good thing about this is we are going to look
at it in two different ways we are going to
12
00:04:25,360 --> 00:04:32,180
handle efficiencies with relate related with
the non moving parts in the same way as we
13
00:04:32,180 --> 00:04:39,740
did in the ramjet analysis that is we are
going to look at intake burner and nozzle
14
00:04:39,740 --> 00:04:47,210
same way as in done yet here we are going
to look at it a little differently just to
15
00:04:47,210 --> 00:05:00,420
refresh your memory the intake efficiencies
can be varying from 0.62 0.9 and is a strong
16
00:05:00,420 --> 00:05:18,280
function of the Mach number then we have compressor
efficiency which can be of the order of 0.2187
17
00:05:18,280 --> 00:05:49,730
then the burner is 0.932 0.95 then turban
efficiencies and finally nozzle efficiencies
18
00:05:49,730 --> 00:06:00,479
of the order of
19
00:06:00,479 --> 00:06:10,910
okay notice.
One thing that both these two are lower than
20
00:06:10,910 --> 00:06:18,919
these two right there is a difference between
the two of them that is both these have an
21
00:06:18,919 --> 00:06:24,770
adverse pressure gradient to cope with which
is why they will have a typically diffusers
22
00:06:24,770 --> 00:06:32,990
and compressors will have a lower efficiency
compared to our turbines and nozzles okay,
23
00:06:32,990 --> 00:06:50,379
now let us do our cycle analysis cycle analysis
the expression for
24
00:06:50,379 --> 00:07:04,800
thrust per unit mass flow rate right.
Or non dimensional thrust what does this expression
25
00:07:04,800 --> 00:07:48,280
now this is the same as earlier that is t7
- xt0 into right this is for
26
00:07:48,280 --> 00:08:00,740
optimally expanded flow in the nozzle, so
therefore d7 = P0 that term vanishes out the
27
00:08:00,740 --> 00:08:10,370
pressure thrust term vanishes out we have
to find expressions for these two ratios to
28
00:08:10,370 --> 00:08:25,360
find expression for T7 / T0 before we got
here let us first look at what is different
29
00:08:25,360 --> 00:08:38,800
from the earlier duty cycle, now if you remember
in the previous case when we analyze ramjet
30
00:08:38,800 --> 00:08:45,380
efficiencies we dealt with slightly differently.
We said for the non moving parts if you look
31
00:08:45,380 --> 00:08:52,890
at intake burner and nozzle we included the
efficiency terms in the pressure with cascading
32
00:08:52,890 --> 00:09:00,250
the reason for that being that, if you assume
that it goes to the same stagnation temperature,
33
00:09:00,250 --> 00:09:11,230
let us say we are looking at intake on a TS
diagram if you are looking at only an intake
34
00:09:11,230 --> 00:09:26,820
process this is for a 100% efficient cycle,
now the assumption that we made was an honest
35
00:09:26,820 --> 00:09:36,010
entropic process would go something like this
to do dash where in the temperatures are the
36
00:09:36,010 --> 00:09:39,050
same.
The stagnation temperatures of 2 and 2 dash
37
00:09:39,050 --> 00:09:50,490
are the same only that these pressures are
different right these are constant pressure
38
00:09:50,490 --> 00:09:58,880
lines these are different and therefore we
looked at handling this terms with only efficiencies
39
00:09:58,880 --> 00:10:11,230
coming up in the pressure terms fine the flow
is at most stagnation it is not standing,
40
00:10:11,230 --> 00:10:24,520
now soon then at the end of a diffuser intake
do not 2.2 we are assuming that it is a stagnation
41
00:10:24,520 --> 00:10:31,100
temperature okay but we know st north is the
content of energy in the flow yes okay if
42
00:10:31,100 --> 00:10:37,139
some losses are there.
So energy should not be equal to the ideal
43
00:10:37,139 --> 00:10:48,339
energy like HT not of the ideal Idol is not
equal to H Dino dash true so only say in real
44
00:10:48,339 --> 00:10:57,839
systems what we look at is when we are looking
at an intake okay if the process were isentropic
45
00:10:57,839 --> 00:11:05,180
you would have a certain pressure recovery
okay at the end of the isentropic compression
46
00:11:05,180 --> 00:11:09,970
you would have a certain pressure recovery,
if the process is real what is the pressure
47
00:11:09,970 --> 00:11:13,709
recovery or what is the pressure at the end
of such a process is what you are looking
48
00:11:13,709 --> 00:11:19,810
at so from that perspective.
If you look at this here you are trying to
49
00:11:19,810 --> 00:11:27,300
capture what is the pressure recovery this
efficiency would then indicate whether it
50
00:11:27,300 --> 00:11:37,510
is 100% efficient 90% 60% efficient the pressure
terms contain the efficiency part right and
51
00:11:37,510 --> 00:11:45,870
here, if you look at this case we have only
turbine and compressor to take care of right,
52
00:11:45,870 --> 00:11:52,110
now in the turbine and compressor what we
do is we know that the power of the turbine
53
00:11:52,110 --> 00:11:57,279
the work or the power produced by the turbine
must be equal to the power consumed by the
54
00:11:57,279 --> 00:12:02,470
compressor right because of the balance between
the two.
55
00:12:02,470 --> 00:12:09,490
So if you remember we then say t C and tT
are two different things we try to combine
56
00:12:09,490 --> 00:12:15,260
them through one equation, so if we take efficiencies
of turbine and compressor there we are going
57
00:12:15,260 --> 00:12:24,529
to complete the cycle, so even in this analysis
we will carry out a cascading wherein we will
58
00:12:24,529 --> 00:12:29,420
deal with efficiencies in terms of pressure
for the non moving parts and when we come
59
00:12:29,420 --> 00:12:35,880
to the compressor and turbine we look at what
happens between the power of the compressor
60
00:12:35,880 --> 00:12:41,370
and the turbine with efficiencies okay.
Now if you look at the compressor turbine
61
00:12:41,370 --> 00:12:49,610
and power balance or before we go that our
efficiency is defined for the compressor and
62
00:12:49,610 --> 00:13:19,779
turbine.
?C is defined as a T3 - T2 T3 – T2 okay
63
00:13:19,779 --> 00:13:43,620
and similarly the turbine efficiency is defined
as Tt 4 - Tt 5 / T 4 – 5 – okay this is
64
00:13:43,620 --> 00:13:56,850
obvious from this figure where this is Tt
5 dash this is Tt drag 3 dash okay, so let
65
00:13:56,850 --> 00:14:05,899
us now try and do the cycle analysis and try
to get these two ratios, so from here if you
66
00:14:05,899 --> 00:14:12,760
are going to do a similar analysis to what
we did in Ramsey okay would whatever we had
67
00:14:12,760 --> 00:14:23,269
derived in the previous classes 47 /T0 be
any different see compressive turbine power
68
00:14:23,269 --> 00:14:29,290
balance will consider efficiency is there
but otherwise in the temperatures while cascading
69
00:14:29,290 --> 00:14:40,101
we do not look at efficiency true.
But they appear as tau C and pouty right if
70
00:14:40,101 --> 00:14:46,410
you look at the expression there like appear
as tC and tT when you include that power balance
71
00:14:46,410 --> 00:14:52,660
Gaussian tC tt cannot be independent terms,
so when you include them in the power balance
72
00:14:52,660 --> 00:15:06,320
that is when you get the real expression for
both of connecting both of them okay, so let
73
00:15:06,320 --> 00:15:56,060
us do the cascading.
So firstly cascading temperatures okay we
74
00:15:56,060 --> 00:16:11,660
want an expression 47 / d not similar to the
previous cases T7 / Tt7 into Tt7 by Tt686
75
00:16:11,660 --> 00:17:00,910
by Tt 5 Tt4 / Tt3 not okay now as in the previous
case what is this is a ratio of stagnation
76
00:17:00,910 --> 00:17:10,530
to static, so I can express it in terms of
Mach number
77
00:17:10,530 --> 00:17:21,580
this is flow through jet pipe this is one
now this is flow through nozzle one is flow
78
00:17:21,580 --> 00:17:36,170
through jet pipe again one this is tt sorry
tt and this is flow through the main combustor
79
00:17:36,170 --> 00:17:47,130
one and this is, now see this is one flow
through in teak and this is ? not right.
80
00:17:47,130 --> 00:18:02,140
So it is similar to what we had earlier derived
that is sorry this has to be Tt4 / tB and
81
00:18:02,140 --> 00:18:20,070
then if we put it in terms of ?B the expression
that we get this we know that tB and ? B are
82
00:18:20,070 --> 00:18:57,670
related that is so from here I get tB would
be equal to ?B by t C ?0 and if i substitute
83
00:18:57,670 --> 00:19:47,210
it finally I will get T7 / T0 is equal to
tD tc listen this cancels out I get okay,
84
00:19:47,210 --> 00:20:07,980
now I need an expression for mach number ratio
which I will get by cascading pressures.
85
00:20:07,980 --> 00:21:12,430
Okay, so we know that P7 /P0 = 1 is equal
to P7 / PT7 / PT6 into PT6 / PT5/ PT4 okay
86
00:21:12,430 --> 00:21:21,070
now when we cascade pressures, if you remember
what we did with ramjet will get efficiencies
87
00:21:21,070 --> 00:21:42,530
here so this is 1 is equal to 1 by 1 + ?- 1
/ 2 m72 into PT7 by PT6 this is flow through
88
00:21:42,530 --> 00:21:54,010
nozzle, so this has an efficiency that is
the nozzle efficiency okay, now a flow through
89
00:21:54,010 --> 00:22:02,750
the jet pipe also has an efficiency let me
call it ?JP, if it is on it will be different
90
00:22:02,750 --> 00:22:09,340
value if it is off be a slightly different
value.
91
00:22:09,340 --> 00:22:22,320
Then what is this PT5 / PT4 for this is pT
into ? burner this is through the combustor
92
00:22:22,320 --> 00:22:33,580
then this is through the compressor, so this
is pT see this is intake into ?0 to the power
93
00:22:33,580 --> 00:22:44,810
of right, now just like when we did ramjet
analysis we will club all these efficiencies
94
00:22:44,810 --> 00:23:11,130
into one and we will define ? ? if I define
it this way I can express all the terms here
95
00:23:11,130 --> 00:23:18,560
as a power of ? by ? - 1 okay is as the intelligent
way of putting it here, so that we will get
96
00:23:18,560 --> 00:23:29,330
the same powers
97
00:23:29,330 --> 00:23:34,480
so pT I know is nothing but tT to the power
of ? by ? - 1.
98
00:23:34,480 --> 00:24:01,450
I will put all of them so I will get 1 + m7
c now C ? not if you look at this expression
99
00:24:01,450 --> 00:24:12,200
we wanted in the denominator that value, so
we have got this so now I can write T7 by
100
00:24:12,200 --> 00:24:48,110
t0 as equal took which I can cancel the t
terms write it as ?0 now we are interested
101
00:24:48,110 --> 00:24:57,840
in Mach number ratios and the other thing
that I know about us 1 + ? - 1 / 2 m02 is
102
00:24:57,840 --> 00:25:17,870
equal to t? so using these two expressions
I get the mach number ratio as n 7 by m0 = ? tau
103
00:25:17,870 --> 00:25:37,560
d tC ?0- 1 now we know both the ratios temperature
as well as Mach number we substitute them
104
00:25:37,560 --> 00:26:06,180
in the equation and find out how it looks
like.
105
00:26:06,180 --> 00:26:19,820
So when we put them in this non-dimensional
thrust equation m dot.a not
106
00:26:19,820 --> 00:26:58,450
we get m0 ? be by ? okay, if we put efficiencies
is equal to one here we recover back the turbojet
107
00:26:58,450 --> 00:27:05,150
equation right you will put this is equal
to 1 which is what is the turbojet equation
108
00:27:05,150 --> 00:27:26,660
right now we need to find an expression connecting
tC n tT right, so this comes from t from
109
00:27:26,660 --> 00:27:49,350
compressor turbine power balance
we get m .CP Tt3 dash - CT to this is the
110
00:27:49,350 --> 00:28:05,380
actual power consumed by the compressor this
must be equal to m.a CP Tt 4 - Tt p dash this
111
00:28:05,380 --> 00:28:15,140
is the actual power delivered by the turbine
okay.
112
00:28:15,140 --> 00:28:27,660
So here we have assumed CP to be constant
yes you can take a mechanical efficiency which
113
00:28:27,660 --> 00:28:35,050
will turn out to be that you will have to
take one by mechanical efficiency here you
114
00:28:35,050 --> 00:28:46,160
can include that you can assume it to be one,
if you want to include it you can put I there
115
00:28:46,160 --> 00:28:57,740
so CP is equal to constant and the other assumption
is that I am sorry to write here 1 + F is
116
00:28:57,740 --> 00:29:02,550
because there is a larger mass flow rate and
the other assumption that we will make is
117
00:29:02,550 --> 00:29:15,840
F is very much less than 1 okay.
Now using this will get we also know how to
118
00:29:15,840 --> 00:29:25,150
connect efficiency of the compressor to this
right do you remember what was the efficiency
119
00:29:25,150 --> 00:29:51,240
Tt3 by Tt 2 /okay so I get I want this term
I want this term sorry so I will get it as
120
00:29:51,240 --> 00:30:17,120
Tt3- T2 / ? C and similarly a wine efficiency
was Tt4 - Tt p dash divided by, so if I take
121
00:30:17,120 --> 00:30:23,650
these efficiency terms into account these
equations will be able to rewrite as follows
122
00:30:23,650 --> 00:31:18,920
okay right.
I think this mechanical efficiency needs to
123
00:31:18,920 --> 00:31:26,950
be included here and not here because if mechanical
efficiencies are non unity which is less than
124
00:31:26,950 --> 00:31:33,780
1 then the compressor requires more than what
the turbine can give so you have to include
125
00:31:33,780 --> 00:31:49,650
it here so et be if you cross multiply it
will be a fraction compressor will only get
126
00:31:49,650 --> 00:31:58,250
a fraction of what the turbine develops okay
fine so where does mechanical efficiency come
127
00:31:58,250 --> 00:32:14,640
in here in this equation right.
So we are interested in t C and t T so to
128
00:32:14,640 --> 00:33:02,170
get that we divide both sides by T0 pt4/ T
0-Tt 5 /T0what is TT three by T 0P 3 by dt
129
00:33:02,170 --> 00:33:31,860
2 into tt 2 / t 0 right what is this DT tube
this is one for flow through diffuser or intake
130
00:33:31,860 --> 00:33:54,390
this is what is this now see and this is ? not
so this term becomes ?0 and similarly on this
131
00:33:54,390 --> 00:34:17,770
side what is Tt 4 / T 0 this is ? B and this
term would be Tt 5 / Tt 4 into Tt4 by T 0
132
00:34:17,770 --> 00:34:41,550
okay so Tt 4 /T0 is ? B this is so now T into
? B.
133
00:34:41,550 --> 00:35:41,250
So we get eater not by ? B into ? see you
tatty the mechanical right or t T =-1 ? 0
134
00:35:41,250 --> 00:35:57,550
/ ? B okay now if you look at this expression
what happens if efficiencies are not equal
135
00:35:57,550 --> 00:36:16,340
to1 then this term increases right and you
will get a lesser pouty fine that means that
136
00:36:16,340 --> 00:36:24,310
you will have lower and lower pressure at
the end of the turbine right that correct
137
00:36:24,310 --> 00:36:34,980
sorry if you have unity you will have three
turn out by ? B into t C- 1.
138
00:36:34,980 --> 00:36:45,120
But if you have no n unity then what happens
this term is large and therefore 1 minus this
139
00:36:45,120 --> 00:36:50,600
will be small which is again what I said earlier
that is correct so you will get pouty to be
140
00:36:50,600 --> 00:37:00,200
smooth right and then t T is small the pressure
ratio the pressure at the end of turbine is
141
00:37:00,200 --> 00:37:08,010
low and sometimes if the efficiency is a very
poor one might actually end up having a turbine
142
00:37:08,010 --> 00:37:12,480
at the end of which there is no power left
for X or there is nothing left for expansion
143
00:37:12,480 --> 00:37:16,710
through the nozzle which means there will
be no thrust that is produced.
144
00:37:16,710 --> 00:37:35,870
So bone needs to be careful about efficiencies
here so that we do not end up in such a situation
145
00:37:35,870 --> 00:37:47,880
now we have been able to derive expressions
for non dimensional thrust for all situations
146
00:37:47,880 --> 00:37:56,960
that is efficiency is not equal to 1 efficiency
is equal to 1and optimally expanded flow as
147
00:37:56,960 --> 00:38:04,060
well as nozzle being choked and we have also
looked at afterburner without afterburner
148
00:38:04,060 --> 00:38:11,030
and what is the other mode of water methanol
injection also we have looked at.
149
00:38:11,030 --> 00:38:18,720
And now here in this case is P by a not that
does not change the expression does not change
150
00:38:18,720 --> 00:38:28,380
but F /m dot a knot will be different because
you now have efficiency terms coming the expression
151
00:38:28,380 --> 00:38:46,040
for is p by a naught would be
152
00:38:46,040 --> 00:38:56,000
I sorry I think we need to derive this see
there is efficiency term involved and therefore
153
00:38:56,000 --> 00:39:05,060
the TT 4 x because of which we need to relook
at what is the is p expression.
154
00:39:05,060 --> 00:39:15,180
So if we look at is P by a 0 we had in the
previous classes looked at this expression
155
00:39:15,180 --> 00:39:50,900
this is 1 / F xF / m dot e a not okay now
how do we get 1 / F we get 1 / F by looking
156
00:39:50,900 --> 00:40:13,180
at the energy balance across the burner so
from
157
00:40:13,180 --> 00:40:39,000
across the burner I get m dot F into Q = m
dot into 1 +F remember in the earlier expression
158
00:40:39,000 --> 00:40:46,980
that we wrote when the efficiency is where
one we were looking at dt3 only right.
159
00:40:46,980 --> 00:41:03,380
Because if you look at what happens due to
non isentropic process
160
00:41:03,380 --> 00:41:13,580
this temperature would be higher than or different
than the earlier temperature so you have a
161
00:41:13,580 --> 00:41:37,560
situation where n is different so therefore
that will come here
162
00:41:37,560 --> 00:41:42,700
so we make the assumption that we are done
earlier that is f is very much less than 1
163
00:41:42,700 --> 00:42:12,880
so I can take out this quantity and I get
my expression for 1 / f =dt4 /T 0 okay now
164
00:42:12,880 --> 00:42:19,290
we know that compressor efficiency is ? = t
t3 - t2 okay.
165
00:42:19,290 --> 00:42:33,210
So using this three dash material will be
one by one by as this society thank you this
166
00:42:33,210 --> 00:43:04,810
goes to the denominator so it will be
167
00:43:04,810 --> 00:43:15,640
thank you we know efficiency is denoted in
this fashion so I can connect from Katie three
168
00:43:15,640 --> 00:43:43,240
dash to TT three here
169
00:43:43,240 --> 00:44:01,560
so 1 / f = q by cpt not before I do that I
will just do a small manipulation with this
170
00:44:01,560 --> 00:44:15,270
expression PT 3by TT 2 -1 into TP two I can
write this expression rewrite this as this.
171
00:44:15,270 --> 00:44:38,310
And TT three dash by TT 2-1this in this cancels
out I am interested in this expression so
172
00:44:38,310 --> 00:45:27,620
I will get TT three dash by TT 2 = TT three
by TT TT three dash by TT two would be 1 minus
173
00:45:27,620 --> 00:46:03,590
what is this TT three by TT to this is see
a so very 1 plus this goes here you get this
174
00:46:03,590 --> 00:46:14,490
now we have been able to get the expression
for TT three dash by TT 2 we can use it here
175
00:46:14,490 --> 00:46:45,110
and we can write TT three dash by D0 is equal
to
176
00:46:45,110 --> 00:46:49,440
this expression is what we have derived just
now what is this, this is nothing.
177
00:46:49,440 --> 00:47:34,710
But ? not so I get gt3 by T 0= t C -1 / ETAC
into ?0 so finally my 1 / F term comes out
178
00:47:34,710 --> 00:48:19,320
to be theta B - ? 0 1 + 0 C -1okay this is
the expression that we get so what happens
179
00:48:19,320 --> 00:49:02,730
to if you plug this in by a not expression
you get is p / e0 = q by cpt not
180
00:49:02,730 --> 00:49:09,530
okay this is the expression that we get for
is p by a not now if you see in this expression
181
00:49:09,530 --> 00:49:29,070
what happens if ? C = 1 this part increases
and therefore ? B -this would be small right.
182
00:49:29,070 --> 00:49:38,810
So can I now say it is advantageous to have
a non efficient compression process if you
183
00:49:38,810 --> 00:49:44,860
look at this expression it is a it is in a
sense it disguises some things you have not
184
00:49:44,860 --> 00:49:51,350
written the full expression for f by m dot
AE not where inefficiency terms are also there
185
00:49:51,350 --> 00:49:58,770
right if you look at this it appears as if
this part will be reduced will be increased
186
00:49:58,770 --> 00:50:09,190
and therefore the entire term will be smaller.
And as a consequence is P will be higher but
187
00:50:09,190 --> 00:50:18,960
if ? C becomes large means is a is a low value
not large it is a low value then what happens
188
00:50:18,960 --> 00:50:25,110
is the pressure ratio across the compressor
or the pressure at the end of the computer
189
00:50:25,110 --> 00:50:32,920
turbine becomes smaller and smaller they will
not have any power left for expansion through
190
00:50:32,920 --> 00:50:41,530
the nozzle right so although it looks very
deceiving here it is not the actual case you
191
00:50:41,530 --> 00:50:47,440
need to plug in the values for ETA C and other
things into this expression and then see how
192
00:50:47,440 --> 00:50:51,400
it looks like we look at it in the next class
thank you.