1
00:00:00,080 --> 00:00:11,380
In
2
00:00:11,380 --> 00:00:19,250
the last class we had seen how efficiencies
affect the performance of gas turbine engines
3
00:00:19,250 --> 00:00:27,730
that is turbo jet engine in all these classes
in all our analysis and now there is an assumption
4
00:00:27,730 --> 00:00:37,750
that we made that CP is constant ? is the
same right and if we take F into account also
5
00:00:37,750 --> 00:00:44,730
purely ratio we have assumed to be very much
less than 1 and therefore we have neglected
6
00:00:44,730 --> 00:00:54,780
it now we have done all this and our analysis
does not seem to include if there is bleeding
7
00:00:54,780 --> 00:00:59,680
from some of the compressor stage after the
low pressure low pressure compressor stage
8
00:00:59,680 --> 00:01:04,030
if there is bleeding how do we account for
it okay.
9
00:01:04,030 --> 00:01:11,750
ICP’s are different if ? the ratio of specific
heats is different how do we account for that
10
00:01:11,750 --> 00:01:18,670
now what we do is we will do a step by step
procedure that is we will go from the inlet
11
00:01:18,670 --> 00:01:25,030
to the outlet and we put down what do we know
and what we need to get and try and see if
12
00:01:25,030 --> 00:01:31,250
we can get it even with all these complexities
that are there in an actual engine now actual
13
00:01:31,250 --> 00:01:37,063
engine is not as simplistic as we assume that
efficiencies are all one and all right. So
14
00:01:37,063 --> 00:01:43,970
we will see how we can still analyze an actual
engine with this.
15
00:01:43,970 --> 00:02:28,420
Firstly the TS diagram would be
16
00:02:28,420 --> 00:02:32,910
this is for a cycle with all efficiencies
being one now if you have an actual cycle
17
00:02:32,910 --> 00:03:17,569
this would be 2’ 3’ the .ted line shows
the actual cycle right now what we are going
18
00:03:17,569 --> 00:03:25,090
to do in this analysis is if we know conditions
at the inlet can we get whatever parameters
19
00:03:25,090 --> 00:03:50,270
we were looking for that is if we know P0
T0 M0 this is the inlet conditions and then
20
00:03:50,270 --> 00:03:58,910
let us say we know the compressor pressure
ratio pC then the other parameter that we
21
00:03:58,910 --> 00:04:10,739
know is the turbine Inlet temperature then
the exhaust area a7 and all the efficiencies
22
00:04:10,739 --> 00:04:17,669
I will call it as ? I if we know compressor
efficiency intake efficiency then efficiency
23
00:04:17,669 --> 00:04:26,599
of the burner nozzle turbine efficiency is
all this if I know it I wil put this in ? I.
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00:04:26,599 --> 00:04:40,449
Okay now if we know this can we calculate
now from this
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00:04:40,449 --> 00:04:47,449
what is the parameter that we have interested
in f that is the thrust of the engine then
26
00:04:47,449 --> 00:05:10,259
the ISP of the engine then the other non-dimensional
quantities that we looked at
27
00:05:10,259 --> 00:05:22,469
so we will try and see how to get to this
knowing this one okay so as a first step we
28
00:05:22,469 --> 00:05:28,310
also need to know in addition to this we need
to know what else we need to know thermodynamic
29
00:05:28,310 --> 00:05:55,339
properties right.
Like thermodynamic data that we require our
30
00:05:55,339 --> 00:06:34,580
CP of turbine then CP of compressor then CP
of burner then heat of combustion of the fuel
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00:06:34,580 --> 00:06:52,490
that is Q right then we need to know ? right
if we know this and if we know this can we
32
00:06:52,490 --> 00:07:01,949
calculate thrust ISP non-dimensional thrust
and non dimensional ISCB this is under the
33
00:07:01,949 --> 00:07:06,749
condition wherein you could have bleed you
could have a patience ease being different
34
00:07:06,749 --> 00:07:23,039
from one and also all these cps b0 = 1 and
? being also different so as a first step.
35
00:07:23,039 --> 00:07:36,719
Step 1 I know IT0 I know I N0 what can I calculate
from these two
36
00:07:36,719 --> 00:07:44,499
stagnation temperature right I can calculate
T T 0 and from that I can calculate it or
37
00:07:44,499 --> 00:08:01,289
not okay so I can write ?0 if you remember
was 1 +? - 1 / 2 m02 I know M0 so I can get
38
00:08:01,289 --> 00:08:20,009
?0 okay right and I can also from this get
TT 0 which is nothing but T0 into ? 0 and
39
00:08:20,009 --> 00:08:34,400
similarly I know p0 so and I also know ?0
so I can get PT0 the stagnation pressure at
40
00:08:34,400 --> 00:08:46,800
Inlet as p0 into ? 0 to the power of ? / ? - 1
okay.
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00:08:46,800 --> 00:09:01,201
So we were able to calculate the stagnation
conditions here right now the next step what
42
00:09:01,201 --> 00:09:09,320
happens between 0 and 2 there is the intake
here stagnation temperature remains the same
43
00:09:09,320 --> 00:09:15,860
stagnation pressure changes because of diffuser
efficiencies or efficiency of the intake right
44
00:09:15,860 --> 00:09:23,800
so if I know all the efficiencies right that
is diffuser efficiency burner efficiency compressor
45
00:09:23,800 --> 00:09:29,810
efficiency turbine efficiency and then nozzle
efficiency I can calculate these parameter
46
00:09:29,810 --> 00:09:38,440
so let us do that I need PT 2 that is pressure
at this point.
47
00:09:38,440 --> 00:09:57,360
So I will get that by ? diffuser into PT0
okay and
48
00:09:57,360 --> 00:10:20,330
I can get t C = p C ? - 1 / ? so I know PT2
I also know TT 2 is nothing but TT0 so I know
49
00:10:20,330 --> 00:10:34,340
TT 0 so I know fresh stagnation quantities
at 2 so I can calculate quantities at 3 right
50
00:10:34,340 --> 00:10:51,800
I know conditions at 2 now I can calculate
at 3 / PT 3 is nothing but pC into PT2 right
51
00:10:51,800 --> 00:11:03,750
I know PT2 I can calculate PT3 and I want
this temperature right for an actual cycle
52
00:11:03,750 --> 00:11:30,610
so I can calculate PT 3’ as equal to okay.
So
53
00:11:30,610 --> 00:11:41,480
I know TT 2 I know tc know ? c so I can get
this TT 3’ so I know now the conditions
54
00:11:41,480 --> 00:12:10,310
at the inlet of the combustor.
So next I need to know PT4 what is PT 4 there
55
00:12:10,310 --> 00:12:16,730
is a pressure drop as you go from the inlet
of the combustor to the exit because of a
56
00:12:16,730 --> 00:12:22,730
rally process wherein you are adding heat
therefore pressure will drop okay so I know
57
00:12:22,730 --> 00:12:37,950
the efficiency of the burner ? B into PT 3
will give me PT 4 and what is TT4, TT4 is
58
00:12:37,950 --> 00:13:09,540
nothing but the turbine Inlet temperature
that is a given condition so now we know conditions
59
00:13:09,540 --> 00:13:18,110
at this point that is temperatures and pressures
we know efficiency of the turbine okay.
60
00:13:18,110 --> 00:13:22,760
And we also know the compressor pressure ratio
so using this we can calculate what is the
61
00:13:22,760 --> 00:14:31,290
condition at p okay what we have used here
is that the CPs need not be the same okay
62
00:14:31,290 --> 00:14:39,930
CPs compressor CP of turbine that is the burnt
gases need not be the same so we were counted
63
00:14:39,930 --> 00:14:46,810
for it and we have also taken a condition
where an F need not be very small or if we
64
00:14:46,810 --> 00:15:02,540
do not want to neglect that part or be happy
with that error we can take into account here
65
00:15:02,540 --> 00:15:30,050
knowing TT p’ I can calculate TT5 as TT4
– efficiency of the turbine and from this
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00:15:30,050 --> 00:15:58,580
I can calculate the PT5 PT5 is nothing but
67
00:15:58,580 --> 00:16:07,170
okay this quantity is IT sorry this quantity
is ?t.
68
00:16:07,170 --> 00:16:16,380
This whole thing is PIT so PT4 into pT is
what gives me PT5 now I having known this
69
00:16:16,380 --> 00:16:39,550
PT6 in case we are using the after burner
okay even otherwise also there is an efficiency
70
00:16:39,550 --> 00:16:45,810
there is a pressure drop across the jet pipe
so we can take that into account here into
71
00:16:45,810 --> 00:17:04,050
PT5 gives me PT6 and PT 6 = TT 5 right if
there is no afterburner if we switch on the
72
00:17:04,050 --> 00:17:09,299
afterburner then we need to take that into
account here that will be specified in addition
73
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to TIT that will be specified to us that will
be known to us.
74
00:17:13,350 --> 00:17:25,419
So if it is given then we have to take into
account that part also and in the next step
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00:17:25,419 --> 00:17:53,549
we calculate what happens through the nozzle
76
00:17:53,549 --> 00:18:02,380
okay so what we have done is starting from
here and knowing efficiencies and CP we have
77
00:18:02,380 --> 00:18:10,320
been able to calculate all these parameters
right now we need to calculate thrust right
78
00:18:10,320 --> 00:18:17,809
there are two conditions that can exist one
is depending on the nozzle pressure ratio
79
00:18:17,809 --> 00:18:27,039
we could either use a converging diverging
nozzle wherein we will assume P7 = P0 or even
80
00:18:27,039 --> 00:18:31,789
in a converging nozzle a special case P7 = P0
or nozzle is choked.
81
00:18:31,789 --> 00:19:28,419
Let us first take the case where P7 =P0 this
can be for both CD as well as convergent nozzles
82
00:19:28,419 --> 00:20:08,059
so if P7 = to P0 then I can calculate Mach
number n7 is equal to
83
00:20:08,059 --> 00:20:20,480
okay so I know Mach number at the exit what
can I find out I need also a7 right if I know
84
00:20:20,480 --> 00:20:34,960
a7 then I can calculate velocities right so
a7 is nothing but under v?RT 7 but I do not
85
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know T7 as yet so let us get T7 as what do
we know TT7 / T7 = T7 is equal to okay.
86
00:20:56,080 --> 00:21:10,130
So I
know m7 I can get I know TT7 so I can get
87
00:21:10,130 --> 00:21:16,120
T7 and since I know T7 I can substitute it
here I get as even now knowing Mach number
88
00:21:16,120 --> 00:21:32,870
and a7 I can calculate the velocity at the
exit that is V7 also need to calculate the
89
00:21:32,870 --> 00:21:45,320
mass flow-rate okay so mass flow rate we can
mass flow rate leaving the I can do two cases
90
00:21:45,320 --> 00:22:11,739
one is m. a = m. a into 1 + F = ?7 right if
I do not want to neglect F I can write this
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00:22:11,739 --> 00:22:26,909
expression I know this to be P7 / RT7 into
V7 T7 seven
92
00:22:26,909 --> 00:22:35,420
now here I know all the quantities I can get
m . a into 1 +F right. I still do not know
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00:22:35,420 --> 00:22:54,460
what is f we will get to that shortly so how
do we get F if you remember when we did ISP
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00:22:54,460 --> 00:23:09,919
calculations what did we do energy balance
across the burner.
95
00:23:09,919 --> 00:24:25,859
So we will do that right this is the expression
that we get if we do energy balance across
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00:24:25,859 --> 00:24:33,669
the combustor we know this quantity m. a into
1 + F what we have just now calculated this
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00:24:33,669 --> 00:24:39,129
will be given to us these two things are what
you have seen earlier TT3 - is what we calculated
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00:24:39,129 --> 00:24:55,259
TT4is given to us so we can calculate m. F
right now F is nothing but M 0f / m. a so
99
00:24:55,259 --> 00:25:06,419
I know m. f from this I know m0a + 1 / 1 +
F and using these two I can calculate m . a
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00:25:06,419 --> 00:25:21,249
right that clear so from here we get m. a.
Now I know I m0 a m. a into1 + F all these
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00:25:21,249 --> 00:25:50,210
quantities so my thrust equation is f = m.
a into 1 + f v7 - v0 + a7 a7 - v 0 this goes
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00:25:50,210 --> 00:26:08,440
to 0 because we assumed p7 = V 0 so here we
know this we know this we know this now what
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we do not know yet is V0 now I know M 0 right
and I also know T0 so I can calculate V0 firstly
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V 0 is nothing but a0 M0 that is equal to
okay so we know both T 0 and M0 so I can calculate
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00:26:41,249 --> 00:26:50,260
V0 so I will know everything here right in
this equation all these quantities are now
106
00:26:50,260 --> 00:27:00,820
known and therefore I can get thrust and once
I get thrust I can also get ISP very easily
107
00:27:00,820 --> 00:27:17,249
ISP is nothing but right thrust per unit mass
flow rate of fuel I know mass flow rate of
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fuel.
I have calculated here I know thrust so I
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can get is P now what are the other quantities
the non-dimensional quantities I know I know
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it also so I can get is P / e0 I also can
get m . F / m . a a0 so all these quantities
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00:27:39,639 --> 00:27:48,200
we can get by doing this analysis going step
by step one can also look at there are cases
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00:27:48,200 --> 00:27:56,440
there are certain engines where in a portion
of the compressor compressed bled to cool
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00:27:56,440 --> 00:28:03,700
either the turbine.
Turbine blades or you use it for air conditioning
114
00:28:03,700 --> 00:28:11,529
the aircraft certain quantity of the compressed
air is taken out especially after the low
115
00:28:11,529 --> 00:28:21,029
pressure stage so you can then suitably write
the equation for power balance across the
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00:28:21,029 --> 00:28:25,489
turbine and compressor taking that into account
because there will be a reduced mass flow
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00:28:25,489 --> 00:28:32,250
rate through the second stage of the compressors
and also through the turbine.
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00:28:32,250 --> 00:28:38,210
So going by the step-by-step procedure you
can calculate that and account for it also
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00:28:38,210 --> 00:28:43,720
right so in this way we can do even for a
realistic engine all these calculations it
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00:28:43,720 --> 00:28:49,580
is not as if that what we did was something
that is not applicable to any realistic engine
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00:28:49,580 --> 00:28:58,320
this is applicable to any kind of realistic
engine with bleed and other things also okay
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00:28:58,320 --> 00:29:05,600
now we have looked at turbine it we have looked
at turbofan sorry the turbojet and turbojet
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00:29:05,600 --> 00:29:16,519
we have considered all the cases that is afterburner
on then water methanol injection conversion
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00:29:16,519 --> 00:29:23,369
sorry one minute I need to still we have only
considered one case P7 = P0 .
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00:29:23,369 --> 00:29:31,679
We need to consider the other case wherein
we have choked condition so let us do that
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00:29:31,679 --> 00:30:38,080
first case when nozzle is choked we know that
m7 =1. So if M7 = 1then I know that T7 = 2
127
00:30:38,080 --> 00:31:18,320
x TT 7 / ? +1 right, so I also can similarly
get P 7s now I know t7 using this I can get
128
00:31:18,320 --> 00:31:49,769
a7 = and I know m 7 = 1 so I can get V7 = m7
x a7 and similarly m. E x 1 + F is equal to
129
00:31:49,769 --> 00:32:28,320
? 7 / RT 7 a 7 v 7 and a similar expression
for m. F Q = m. a x1 + f CP combustor
130
00:32:28,320 --> 00:32:44,489
so I know from here I can get F which is nothing
but m . F / m. a I know m. a I know m. F I
131
00:32:44,489 --> 00:33:03,649
know F I know v7what else do we need V 0 we
know has nothing but M0 e 0 that is M 0 x
132
00:33:03,649 --> 00:33:44,899
so knowing all this I can get again f 0 okay.
I know p7 I am given P 0 m. a is known v7
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00:33:44,899 --> 00:33:53,859
is known V 0 is known a seven P 7 and P 0
so I can get thrust from here I can get is
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00:33:53,859 --> 00:34:07,230
P as f / m. F and the other non-dimensional
quantities that we wanted, so we have looked
135
00:34:07,230 --> 00:34:17,159
at the cases where in for ramjet as well as
turbojet we have looked at cases where efficiencies
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00:34:17,159 --> 00:34:23,740
are equal to 1not equal to 1 and for a turbojet
we have looked at case wherein with after
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00:34:23,740 --> 00:34:30,820
burner without afterburner choke nozzle optimally
expanded flow then we have also looked at
138
00:34:30,820 --> 00:34:37,340
case with water methanol injection all this
we have looked at.
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00:34:37,340 --> 00:34:45,210
Now let us look at the next two classes of
engines that is turbofan and turboprop, turboprop
140
00:34:45,210 --> 00:34:55,370
is very simple we do not need to do any analysis
except that all the power that is developed
141
00:34:55,370 --> 00:35:00,690
in the turbine must be equal to the power
of the compressor as well as the power of
142
00:35:00,690 --> 00:35:08,930
the fan propeller right. So if we do that
you will get the turboprop engine that is
143
00:35:08,930 --> 00:35:16,110
not a big problem but a turbofan engine needs
some effort so let us look at the turbofan
144
00:35:16,110 --> 00:37:21,000
engine first.
145
00:37:21,000 --> 00:38:20,720
So this is a turbojet engine and to this we
add the turbofan part both of them all of
146
00:38:20,720 --> 00:38:34,120
them are mounted on the same shaft right now
our typical conditions where this was zero
147
00:38:34,120 --> 00:38:48,160
after the intake we had to write this was
three, four, five, six, in this is seven whatever
148
00:38:48,160 --> 00:39:04,060
we had done for the turbojet remains as this
now for the turbofan
149
00:39:04,060 --> 00:39:09,200
you have the fan in front also and there is
a certain amount of air that is bypassing
150
00:39:09,200 --> 00:39:16,800
and is going through only the fan portion
okay, so if you look at condition after the
151
00:39:16,800 --> 00:39:32,820
fan will call it as eight okay and we will
call the exit condition here as nine at the
152
00:39:32,820 --> 00:39:39,910
exit of the finest nine the bypass exit as
nine okay.
153
00:39:39,910 --> 00:39:48,880
So you have 0 to 2, 2 to 8, 8 to 9 that is
the additional part in the turbofan engine
154
00:39:48,880 --> 00:39:52,070
the rest of it to the turbojet engine right.
155
00:39:52,070 --> 00:39:59,720
So how does the thrust expression look like
for a turbofan engine F is equal to m. a x
156
00:39:59,720 --> 00:40:31,800
1 + F V 7 -V0 + a 7 e 7 - P0 all this is turbojet
art
157
00:40:31,800 --> 00:41:00,830
now in addition to this you will also have
a x m. v9 - V0+ a 9 p 9 - V0 this is because
158
00:41:00,830 --> 00:41:15,570
of the fan or the bypassed stand a is the
159
00:41:15,570 --> 00:41:24,200
okay now we will make an assumption that the
flow is optimally expanded through both the
160
00:41:24,200 --> 00:41:29,070
nozzles there are two nozzles now one is this
nozzle and the other one is this nozzle.
161
00:41:29,070 --> 00:41:47,020
So we will make an assumption where we say
flow is optimally expanded, so therefore we
162
00:41:47,020 --> 00:42:06,110
get p7 = P 9 = P0 okay, so this part remains
the same as what we did for turbojet analysis
163
00:42:06,110 --> 00:42:14,600
we now have to only look at this part because
of our assumption here this goes to 0 this
164
00:42:14,600 --> 00:42:21,250
goes to 0, so I have to only take care of
this term all the rest of the analysis that
165
00:42:21,250 --> 00:42:32,050
we have done for this for a turbo jet with
optimally expanded flow is valid.
166
00:42:32,050 --> 00:42:41,830
So in order to do this what do we do what
are the things that we need V 9 / V0 is what
167
00:42:41,830 --> 00:43:04,280
we will get so we need T 9 / T0 and V 9 / V0
that is from here we will get m9 / M 0, so
168
00:43:04,280 --> 00:43:10,550
knowing these two we can also get this portion
non-dimensional is this portion and get this
169
00:43:10,550 --> 00:43:25,130
value okay so
what is T 9 / T0 you have teen T 9 / TT 9
170
00:43:25,130 --> 00:44:10,580
TT 9 / T eight okay right now I know this
quantity is ? not fine this is ratio of static
171
00:44:10,580 --> 00:44:23,990
to stagnation conditions, so I will put this
as 1 + 1/ ? - 1 M 92 next is flow through
172
00:44:23,990 --> 00:44:30,500
this duct here okay this is similar to the
jet pipe.
173
00:44:30,500 --> 00:44:38,300
So if we assume all efficiencies to be 1 which
is what we will do so here we are assuming
174
00:44:38,300 --> 00:44:56,431
also ? = 1 so TT 9 / T T 8 would be 1 and
TT 8 / TT 2 is process across this fan I will
175
00:44:56,431 --> 00:45:21,880
define a new ratio that is I will define t
F is equal to TT 8 / DT 2 and p F = PT 8 / PT2,
176
00:45:21,880 --> 00:45:38,440
so this would be tau F would be equal to PI
F to the power of ? - 1 / ? okay coming back
177
00:45:38,440 --> 00:45:57,520
here I will get this is tau F into this is
T T 2 by T not it is 1 so I get tau F ? 0
178
00:45:57,520 --> 00:46:24,470
by 1 + ? - 1 by 2 m 92 okay when we cascade
pressures will get this ratio also so let
179
00:46:24,470 --> 00:46:48,780
us do that.
180
00:46:48,780 --> 00:47:16,040
This will be p9 / PT 9 - PT 9 / PT 8PT 8 / PT
2 PT 2 / PT0 P0 / P0 okay this quantity is
181
00:47:16,040 --> 00:47:28,900
nothing but ?0 to the power of ? / ? - 1 this
is equal to 1 and what is this is ratio of
182
00:47:28,900 --> 00:47:43,800
static to stagnation
what about this PT 9 by PT 8this is 1 right
183
00:47:43,800 --> 00:48:01,810
so this is 1 and PT 8 by PT 2 this is PI F
and again PT 2 by PT 0 is one because we have
184
00:48:01,810 --> 00:48:09,330
assumed all efficiencies to be 1 into ? 0
to the power of ? / ? - 1 so if I change this
185
00:48:09,330 --> 00:48:21,650
to tau F I will get all of them in powers
of ? by ? - 1 so I will get 1 + ? - 1 by 2
186
00:48:21,650 --> 00:48:41,250
m 9 2 is equal to t F ? 0.
So I can come back here
187
00:48:41,250 --> 00:48:58,370
and I know that this is again tau F ? R 0
/ t F ? 0 so this is 1 which means that if
188
00:48:58,370 --> 00:49:05,270
all efficiencies are 1 t 9 will be t 8 rights.
189
00:49:05,270 --> 00:49:13,410
Because here in the intake you are compressing
it then you have a fan to increase the pressure
190
00:49:13,410 --> 00:49:17,650
and again you are expanding the flow if there
are no efficiencies involved there is no heat
191
00:49:17,650 --> 00:49:37,070
addition here so T9 must be equal to T 0 okay
and I also know that 1 + ? - 1 by 2m02 = ? 0
192
00:49:37,070 --> 00:49:48,580
so using these two I can get the ratio of
Mach numbers which will be m9 by now F ? 0
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00:49:48,580 --> 00:50:09,890
- 1 divided by ? 0 - 1 so finally I can write
the expression for thrust.
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00:50:09,890 --> 00:50:40,620
We non-dimensional is trust you do not need
this F is less than one very much less than
195
00:50:40,620 --> 00:51:36,630
one, so this part is the turbojet part that
we had already seen tau C t T ? 0 – 1 okay
196
00:51:36,630 --> 00:51:42,630
this is the expression that we get now we
need to do the compressor turbine power balance
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00:51:42,630 --> 00:51:58,580
because tau F t C and t T are all related
okay, so let us do that turbine compressor
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00:51:58,580 --> 00:52:57,350
and fan + a m. a PT 8 - TT - if we deduce
from this similar to what we had done if you
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00:52:57,350 --> 00:53:04,740
remember we had found an expression for tau
T in terms of tau C by cancelling out m . a
200
00:53:04,740 --> 00:53:12,210
with assuming that F is very much less than
1 and CP being the same.
201
00:53:12,210 --> 00:53:32,350
We can reduce this to t T = 1 - ? 0 / 0B t
C - 1 this was if only there was compressor
202
00:53:32,350 --> 00:53:53,520
turbine power balance in addition you have
the other term – a okay, so if we plug back
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00:53:53,520 --> 00:54:00,290
this into this expression here we will get
the overall expression for thrust non-dimensional
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00:54:00,290 --> 00:54:03,850
thrust okay, we will stop here and continue
in the next place, thank you.