Introduction to Airbreathing Propulsion

Prof. Ashoke De
Department of Aerospace Engineering
Indian Institute of Technology – Kanpur

Lecture – 15
Introduction to Gas Turbine Engines (Contd.,)

Okay, so let us looking at this different performance parameter and we are talking about overall
efficiency, thermal efficiency, propulsion efficiency and we have looked at how they vary and
another important factor that we have looked at is the take-off thrust because take-off thrust is
one of the important design parameter for any aircraft engine. As I already mentioned this is
the maximum thrust that one engine can produce.

And this required to be produced at the engine when it is on the static condition or rather adjust
on the run at the ground level while it is going to take-off with the full payload including the
fuel load complete fuel load and everything. So that is the maximum thrust that one engine can
produce and how that can be varied that we have already seen. Now we are moving ahead with
the other factors like range and endurance and all this.
(Refer Slide Time: 01:19)

So, what we will discuss is the range of aircraft. So, that is we are going to look at it. So the
range when we talk about range it means it says how far an aircraft can travel with a given
supply of fuel. Given supply of fuel so this is also an important performance parameter and
range actually the range depends on L / D ratio which lift to drag ratio and thrust generated by
the engine.
Now, if we ignore let us say if we ignore the climb too and descent from cruise altitude that
means we are ignoring the climb and descent from the cruise altitude and we can only consider
the then consider the aircraft to be at level flight. So, when you talk about the level flight the
engines thrust and the vehicle drag must be equal and lift on the aircraft should be equal to its
weight.

That means if we look at the free body let us say this is the aircraft then this is thrust, this is
drag, this is weight and this is lift so they should be equal. So
𝐷 𝑚𝑔
𝑇=𝐷=𝐿∗ =
𝐿 (𝐿)
𝐷
because we have L is W is mg. Now here m is the instantaneous mass, g is the acceleration due
to gravity and D is drag force which is L / D or rather lift to drag ratio.

Then we have thrust power that would be

𝑚𝑔𝑢
𝑇ℎ𝑟𝑢𝑠𝑡 𝑝𝑜𝑤𝑒𝑟 = 𝑇𝑢 =
𝐿
(𝐷 )

and what we have seen that overall efficiency is

𝑇𝑢
η0 =
𝑚̇𝑓 𝑄𝑅
So if we equate this term what we can write
𝑚𝑔𝑢
𝑚̇𝑓 𝑄𝑅 η0 =
𝐿
(𝐷 )

So that gives us
𝑚𝑔𝑢
𝑚̇𝑓 =
𝐿
𝑄𝑅 η0 (𝐷)

(Refer Slide Time: 06:17)

Now 𝑚̇𝑓 is also is part of overall aircraft mass and this is what 𝑚̇𝑓 is also getting consumed so
dm / dt. Since fuel is consumed then what we can write that
𝑑𝑚 𝑑𝑠 𝑢𝑑𝑚
− ∙ =−
𝑑𝑠 𝑑𝑡 𝑑𝑠
where S one can assume that the this is the distance along the flight path. So now we put this
back in this 𝑚̇𝑓 expression so what we get
𝑢𝑑𝑚 𝑚𝑔𝑢
− =
𝑑𝑠 𝐿
𝑄𝑅 η0 (𝐷)

Now, if we integrate
𝑚2 (𝑓𝑖𝑛𝑎𝑙 𝑚𝑎𝑠𝑠) 𝑆
𝑑𝑚 𝑔
∫ =∫ − 𝑑𝑠
𝑚1 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑎𝑠𝑠) 𝑚
𝐿
0 𝑄𝑅 η0 (𝐷)

So this becomes
𝑚2 𝑔𝑆
𝑙𝑛 =−
𝑚1 𝐿
𝑄𝑅 η0 (𝐷)

So we will get
𝑄𝑅 η0 𝐿 𝑚1
𝑆= ( )𝑙𝑛
𝑔 𝐷 𝑚2
So this is known as Breguet’s range formula. So what we see here that S is proportional to eta
0 that means if eta 0 is higher so we have higher range.
(Refer Slide Time: 09:51)

Now, if we consider f is small and pressure term is neglected then what we can write
𝑚̇𝑎 [𝑢𝑒 − 𝑢]
η0 =
𝑚̇𝑓 𝑄𝑅

which is
[𝑢𝑒 − 𝑢]
η0 =
𝑓𝑄𝑅

So now we take derivative to the eta 0 with respect to u for a given ue. So what we can write
this is
𝑑η0 [𝑢𝑒 − 2𝑢]
= =0
𝑑𝑢 𝑓𝑄𝑅

that gives u is ue / 2 so we have already seen that eta 0 is max when u = ue / 2 and that is the
same time also range is max because range is also proportional to the eta 0.

Now let us consider a turbofan engine which is essentially with double propellant stream. So,
here we need to replace because turbofan has one cold stream and hot stream so we need to
replace 𝑢𝑒 by the thrust averaged quantity like ̅̅̅
𝑢𝑒 . So, where ̅̅̅
𝑢𝑒 is defined as
𝑚̇𝑎ℎ (1 + 𝑓)𝑢𝑒ℎ + 𝑚̇𝑎𝑐 𝑢𝑎𝑐
𝑢𝑒 =
̅̅̅
𝑚̇𝑎ℎ (1 + 𝑓) + 𝑚̇𝑎𝑐
so using this average ue and this fuel air ratio.
(Refer Slide Time: 13:51)
One write that
𝑚̇𝑎𝑐 [𝑢𝑒 − 𝑢]
η0 = (1 + ) 𝑢
𝑚̇𝑎ℎ 𝑓𝑄𝑅

Again, for a given ue η0 max would be ue bar / 2. So we can take the simple derivative of that
and find out that this would be ue / 2. So eta 0 also depends on flight Mach number. So one
can plot η0 like let us say 1, 2, 3 like this, this is flight Mach number 0.2, 0.4, 0.6. So eta 0 this
varies like this.

So η0 increases as flight Mach number increases and at the same time what we can look at is
the L / D variation how that goes with flight Mach number 1, 2, 3 this is flight Mach number
10, 20, 30 so this is how L / D with the flight Mach number and other things Mach number this
is 5 and this is η0 into L / D. So the range should increase with Mach number how L / D also
depends on Mach number that means range should increase with flight Mach number.

However, L / D also depends on M. Now L / D for supersonic is much less than subsonic and
L / D is maximum at high subsonic. For the product of η0 L / D this is also maximum at high
subsonic.
(Refer Slide Time: 17:27)
Now, for a given fuel consumption rate the maximum range is obtained at high subsonic flight.
For a given fuel consumption rate maximum range is obtained at high subsonic flight. Now the
other things like commercial aircraft fly at high subsonic speed for two reason one is
economical and second is that efficient operations. So this is where that all our civilian
commercial aircraft that fly most of the time at the high subsonic.

So, typically they operate around Mach 0.7 to 0.8 in that range so this is high subsonic range.
So what happens that higher Mach number shock forms in front of aircraft nose. Now once that
happens if there is a shock formation so we have already seen that across a shock the
temperature increases. So when the temperature would increase obviously the thermal load will
increase.

So thermal load on structure increases with M. So to avoid this the payload also so now these
two thermal load are to be resistant the structure need to be heavier to withstand all this. This
is one of the immediate consequence that one can have as soon as this is obviously having said
that it is quite obvious that supersonic flights are much less economical for example when
Concorde can take 100 passenger.

And use more fuel than a Boeing 777 which can carry roughly around 500 passengers so they
are not economical either. Again looking back to that so looking back to that range equation
what we get or obtain that
𝑄𝑅 η0 𝐿 𝑚1
𝑆= ( )𝑙𝑛
𝑔 𝐷 𝑚2
 𝑇𝑢
η0 =
𝑚̇𝑓 𝑄𝑅
𝑇 𝑢 𝐿 𝑚1
𝑆= ( )𝑙𝑛
𝑚̇𝑓 𝑔 𝐷 𝑚2

(Refer Slide Time: 22:23)

𝑇 𝑇
So which tells that 𝑚̇ is an important parameter. So is an important parameter and this is
𝑓 𝑚̇𝑓

the specific thrust that means the thrust generated per unit mass of fuel consumed. Now the
𝑚̇𝑓
other the reverse of it which is this is called the thrust specific fuel consumption that is
𝑇

TSFC that means fuel consumed per unit thrust. So just to produce one unit of thrust how much
fuel is consumed that is called the TSFC.

And this is one of the parameter which is often provided for providing an engine or
configuration or something I mean if you look at or search in internet how this engine specs
are provided any engine manufacturer they provide this kind of information the TSFC, PRC,
efficiencies and then how much thrust it produces and such like that. Just to look at this things
again these are generic terms just to extend these things for turbojet.

Just to show you how things can be written for TSFC

 𝑚̇𝑓 𝑚̇𝑓
𝑇𝑆𝐹𝐶 = =
𝑇 [𝑚̇𝑎 (1 + 𝑓)𝑢𝑒 − 𝑢]
So this is for a turbojet how one can estimate the TSFC value. So, similarly for turbofan or
turbo propeller let us say for turbofan or turboprop you can just replace this 𝑢𝑒 by 𝑢𝑒 average.

So one thing is clear here that TSFC depend strongly on 𝑢𝑒 that means what is the exit velocity.
Now let us say for any turbine engine producing shaft power. So for turbine engine producing
shaft power what happens? So we define another quantity called the BSFC which is
𝑚̇𝑓
𝐵𝑆𝐹𝐶 =
𝑃𝑠
that is called this is fuel consumption per unit of shaft power. To consider the thrust produced
by hot gases also.

So this BSFC stand for brake specific fuel consumption this is brake specific fuel consumption.
Now equivalent brake specific fuel consumption so this is we write BSFC is
𝑚̇𝑓
𝐵𝑆𝐹𝐶 =
𝑃𝑒𝑠

which is equivalent shaft power. So that would be

𝑚̇𝑓
𝐵𝑆𝐹𝐶 =
𝑃𝑠 + 𝑇𝑢
so this is equivalent brake specific fuel consumption. Now what we can see here that Te is the
thrust produced by the so Te is the thrust produced by the turbine engine exhaust.

So, typically the engine brake specific fuel consumption would be some range like typical range
of they should be 0.27 to 0.36 roughly kg per kilowatt hour so that is the typical value of BSFC
which is there. So you can see how TSFC and all these things they can be calculated and also
the BSFC and range. So, we will stop here and continue this discussion in this next lecture.