Cold weather climb performance: fact or fiction?
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Put at its simplest, flat-rating means that you artificially put a cap on the power the engines produce, usually because the engine was not specifically designed for that type of aircraft.
Britten-Norman did this, for example, on the Allison 250 that powers the Turbine Islander. If they allowed the pilot to use all the power available in the engine, the airframe would rip itself apart.
Britten-Norman did this, for example, on the Allison 250 that powers the Turbine Islander. If they allowed the pilot to use all the power available in the engine, the airframe would rip itself apart.
I think that's a rather more general definition than Hawk37 intended when he wrote:
He was suggesting that the thrust produced was independent of temperature over the range of interest. That seems consistent with the general principle that the thrust is artificially capped, thus the capped thrust can be produced over a range of temperatures rather than having a maximum at only the most favourable. "Flat" conjures up a graph of power or thrust being independent of something.
So I'm at a loss as to understand how the 737-400s engines can be flat-rated to a particular value, and yet produce more static thrust on a cold day.
I'll bite my tongue over the produces more lift comment.
assuming the engines are flat rated throughout the temperature we’re considering
So I'm at a loss as to understand how the 737-400s engines can be flat-rated to a particular value, and yet produce more static thrust on a cold day.
I'll bite my tongue over the produces more lift comment.
I still think Hawk has added 2 & 2 to equal 5. I just can't quite put my finger on where/why. 
Even allowing for a constant thrust as temp. rises the airframe will require more power to fly. That means that a lesser proportion of the fixed engine output will be available for climb performance.
What am I missing?
Even allowing for a constant thrust as temp. rises the airframe will require more power to fly. That means that a lesser proportion of the fixed engine output will be available for climb performance.
What am I missing?
What am I missing?
Which is why it's useful, for systems where the constant thrust approximation is closer than a constant power approximation, it makes sense to divide everything through by TAS. So for props we deal with power required, power available and rate of climb, while for jets we deal with thrust required, thrust available and angle of climb.
Of course constant thrust is an approximation, but I think Hawk37 is spot on in his last post in calculating the magnitude of the imperfection.
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This is one of those really infuriating problems......each time I think that I have solved it, I discover a fault in my argument. But I think that we can solve it if we first restate it to take out the controversy regarding flat rating.
Let's imagine that we have an aircraft fitted with instruments that give direct indications of CAS, TAS, thrust, power available, power required and climb angle.
We start a climb at Vmd using less than the maximum thrust available. The power available gauge will give us an indication that is equal to thrust x TAS. The power required gauge will give us an indication that is equal to drag x TAS. As we pass through a pressure altitude of our choice we note the readings on our gauges.
We then descend and move to an area of higher ambient temperature. The higher ambient temperature will cause the air denity to be lower. This in turn will reduce the thrust at any given engine RPM and increase the TAS at any given CAS.
We now repeat the climb using the same value of CAS and thrust. To get this same thrust we will need to open the throttles a little bit to compensate for the density-induced loss of thrust. As we pass through the selected pressure altitude we note the readings on our various gauges.
If we have adjusted the throttles to the correct position, the thrust will be the same as in our original climb. Because we are using the same CAS we will also have the same drag as in our original climb. Because drag and thrust are unchanged the excess thrsut will also be unchanged. Angle of climb is proportional to (drag - thrust) divided by weight. So if the weight of the aircraft has not changed (we did an in-flight refuel before repeating the test), our angle of climb will be the same as in the original test.
But the reduced air density will have increased the TAS at our selected CAS. We will therefore be climbing up the same slope at a greater TAS, so our ROC will have increased. This is Hawk's original conjecture and it is true. The problem of course is why is it true?
To understand this we need to look at what determines our ROC. ROC is equal to (power available minus power required) divided by weight. So if our weight has not changed and our ROC has increased, then our excess power (power available minus power required) must have increased.
So where did we get this extra excess power from? We got it by opening the throttles to have the same thrust at the higher TAS.
Power available is equal to thrust x TAS and power required is equal to drag x TAS. Let us suppose that the higher temperature in our second test was sufficieint to increase the TAS by 5%. This 5% increase in TAS will affect both the power available (thrust x TAS) and in the power required (drag x TAS). This means that power available and power required will both have increased by 5%. It might be tempting to think that the difference between the two (the excess power) is unchanged, but this has also increasd by 5%.
This can be demonstrated by trying an example. Let's suppose the original power available was 2000 horsepower and the original power required was 1000 horsepower. The excess power was 1000 horse power. If we increase power available and power required by 5% we get 2100 available and 1050 required. The excess has increased to 1050, which is a 5% increase from the original 1000.
So in our second climb experiment the 5% increase in TAS causes a 5% increase in excess power, whcih causes a 5% increase in ROC.
We need to be a little bit careful however in deciding what all of this means for aircraft performnce. The only reason we were able to get the highr ROC is because we were able to open the throttles a little bit more to maintain constant thrust. The fact remains that aircraft performance is generally better at low ambient temperatures.
All of this is of course pure conjecture!!!!!
Let's imagine that we have an aircraft fitted with instruments that give direct indications of CAS, TAS, thrust, power available, power required and climb angle.
We start a climb at Vmd using less than the maximum thrust available. The power available gauge will give us an indication that is equal to thrust x TAS. The power required gauge will give us an indication that is equal to drag x TAS. As we pass through a pressure altitude of our choice we note the readings on our gauges.
We then descend and move to an area of higher ambient temperature. The higher ambient temperature will cause the air denity to be lower. This in turn will reduce the thrust at any given engine RPM and increase the TAS at any given CAS.
We now repeat the climb using the same value of CAS and thrust. To get this same thrust we will need to open the throttles a little bit to compensate for the density-induced loss of thrust. As we pass through the selected pressure altitude we note the readings on our various gauges.
If we have adjusted the throttles to the correct position, the thrust will be the same as in our original climb. Because we are using the same CAS we will also have the same drag as in our original climb. Because drag and thrust are unchanged the excess thrsut will also be unchanged. Angle of climb is proportional to (drag - thrust) divided by weight. So if the weight of the aircraft has not changed (we did an in-flight refuel before repeating the test), our angle of climb will be the same as in the original test.
But the reduced air density will have increased the TAS at our selected CAS. We will therefore be climbing up the same slope at a greater TAS, so our ROC will have increased. This is Hawk's original conjecture and it is true. The problem of course is why is it true?
To understand this we need to look at what determines our ROC. ROC is equal to (power available minus power required) divided by weight. So if our weight has not changed and our ROC has increased, then our excess power (power available minus power required) must have increased.
So where did we get this extra excess power from? We got it by opening the throttles to have the same thrust at the higher TAS.
Power available is equal to thrust x TAS and power required is equal to drag x TAS. Let us suppose that the higher temperature in our second test was sufficieint to increase the TAS by 5%. This 5% increase in TAS will affect both the power available (thrust x TAS) and in the power required (drag x TAS). This means that power available and power required will both have increased by 5%. It might be tempting to think that the difference between the two (the excess power) is unchanged, but this has also increasd by 5%.
This can be demonstrated by trying an example. Let's suppose the original power available was 2000 horsepower and the original power required was 1000 horsepower. The excess power was 1000 horse power. If we increase power available and power required by 5% we get 2100 available and 1050 required. The excess has increased to 1050, which is a 5% increase from the original 1000.
So in our second climb experiment the 5% increase in TAS causes a 5% increase in excess power, whcih causes a 5% increase in ROC.
We need to be a little bit careful however in deciding what all of this means for aircraft performnce. The only reason we were able to get the highr ROC is because we were able to open the throttles a little bit more to maintain constant thrust. The fact remains that aircraft performance is generally better at low ambient temperatures.
All of this is of course pure conjecture!!!!!
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KEITH, Glad to hear that someone agrees my original conjecture is true. (Pardon to BOOKWORM, he may have already hinted so too).
I’ve been away, or would have posted earlier.
While I have to agree with you that "generally" aircraft performance is better at low temps, it seems this conjecture supports the opposite for operations within engine flat rating and at V md.
If I follow you Keith, then using your example of a 5% increase in TAS due to a higher temp (a reasonable situation), and maintaining V md CAS, seems to result in a 5 % increase in ROC, assuming the SAME thrust output of the engine.
** 5 % increase in RATE of CLIMB when the OAT is warmer (flat rated, Vmd). **
And as BOOKWORM has repeatedly pointed out, an engine can be expected to put out the SAME static thrust over its flat rated temperature range.
Which leaves only the decrease in thrust output as TAS increases to further affect the conjecture.
So…..I’d like to take your computations a bit further…
The graph I have for a 731 shows approximately a 1.2% decrease in TA over a 10 KTAS speed increase at a typical V md. See previous post. If we assume that typically an aircraft requires 33% of its max thrust at V md, then a 1.2% decrease in TA becomes a 1.8% decrease in excess thrust. This 1.8% decrease in excess thrust will decrease Keith’s 5% increase in ROC.
But by how much?
Well, I believe that sin(gradient) = excess thrust/weight
Thus sin(gradient) decreases by 1.8%
At the low angles we’re considering, sin(x) is proportional to x, so climb gradient itself will decrease 1.8%.
Thus the ROC will decrease 1.8%
Note that the ROC decrease is the same as the gradient decrease since TAS is the same, at the 5% increase Keith proposed earlier.
So, going further with your conjecture, Keith (5% ROC increase), I further calculate that the decrease in thrust as TAS increases will lower this 5% ROC by 1.8%, to get 3.2%.
The conjecture of better climb rate at higher OAT’s (within engine flat rating) at V md seems substantiated!! A 3.2% climb rate increase due to temperature increase
Now I’ve flown jets enough to shake my head and say it just aint so, but mathematically, what’s wrong with this argument?
As TINSTAAFL put it, "Hawk has added 2+2 to equal 5. I just can’t put my finger on it"
I’ve been away, or would have posted earlier.
While I have to agree with you that "generally" aircraft performance is better at low temps, it seems this conjecture supports the opposite for operations within engine flat rating and at V md.
If I follow you Keith, then using your example of a 5% increase in TAS due to a higher temp (a reasonable situation), and maintaining V md CAS, seems to result in a 5 % increase in ROC, assuming the SAME thrust output of the engine.
** 5 % increase in RATE of CLIMB when the OAT is warmer (flat rated, Vmd). **
And as BOOKWORM has repeatedly pointed out, an engine can be expected to put out the SAME static thrust over its flat rated temperature range.
Which leaves only the decrease in thrust output as TAS increases to further affect the conjecture.
So…..I’d like to take your computations a bit further…
The graph I have for a 731 shows approximately a 1.2% decrease in TA over a 10 KTAS speed increase at a typical V md. See previous post. If we assume that typically an aircraft requires 33% of its max thrust at V md, then a 1.2% decrease in TA becomes a 1.8% decrease in excess thrust. This 1.8% decrease in excess thrust will decrease Keith’s 5% increase in ROC.
But by how much?
Well, I believe that sin(gradient) = excess thrust/weight
Thus sin(gradient) decreases by 1.8%
At the low angles we’re considering, sin(x) is proportional to x, so climb gradient itself will decrease 1.8%.
Thus the ROC will decrease 1.8%
Note that the ROC decrease is the same as the gradient decrease since TAS is the same, at the 5% increase Keith proposed earlier.
So, going further with your conjecture, Keith (5% ROC increase), I further calculate that the decrease in thrust as TAS increases will lower this 5% ROC by 1.8%, to get 3.2%.
The conjecture of better climb rate at higher OAT’s (within engine flat rating) at V md seems substantiated!! A 3.2% climb rate increase due to temperature increase
Now I’ve flown jets enough to shake my head and say it just aint so, but mathematically, what’s wrong with this argument?
As TINSTAAFL put it, "Hawk has added 2+2 to equal 5. I just can’t put my finger on it"
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But in "real life", the acceleration to a higher TAS would be done by pitching down, because you'd use a constant CAS/IAS climb? I've never heard of anyone who uses a "constant thrust climb", only IAS or Mno? Isn't that why your theories don't fit your in-flight observations?
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Crossunder, this constant thrust" climb, as you refer to it as, is actually a best angle climb.
Not sure if it’s considered in the airline world that much, but corporates flying out of some
difficult fields will use it to attain the climb gradient necessary to ensure obstacles are
avoided. Additionally, I use it as an initial climb speed when initial climb altitude is only a
few thousand feet (less than 4000 or so), and expecting a turn of 90 degrees or more. This is
when it certainly appears that the colder it is, the higher the climb rate and gradient, although
this thread seems to have substantiated just the opposite.
The "pitch down" as you refer to it, now no longer is following the best gradient climb, and so is
outside the parameters of the conjecture of this thread.
So no, I don’t think that’s why this theory doesn’t fit in flight observations.
Not sure if it’s considered in the airline world that much, but corporates flying out of some
difficult fields will use it to attain the climb gradient necessary to ensure obstacles are
avoided. Additionally, I use it as an initial climb speed when initial climb altitude is only a
few thousand feet (less than 4000 or so), and expecting a turn of 90 degrees or more. This is
when it certainly appears that the colder it is, the higher the climb rate and gradient, although
this thread seems to have substantiated just the opposite.
The "pitch down" as you refer to it, now no longer is following the best gradient climb, and so is
outside the parameters of the conjecture of this thread.
So no, I don’t think that’s why this theory doesn’t fit in flight observations.
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Hello again Hawk,
I think that the problem might lie in the way that you are interpreting your 731 thrust graph.
If the graph is just for changes in thrust against changes in TAS, without including any atmospheric changes, then the results will be invalid for your increasing temperature scenario.
Let's start by assuming that your graph is based on constant air density. In this case any increase in TAS will result in an equal increase in CAS and dynamic pressure. This in turn will result in an increase in mass airflow through the engines. This means that the thrust reduction caused by the reducing acceleration of the airflow, will be partly offset by an increase in thrust due to increased mass flow.
If we assume that your 10 Kt increase in TAS represents a 5% increase, then we will also get a 5% increase in mass flow which will tend to increase thrust by 5%. But your graph shows an overall thrust decrease of 1.2%, which is the combined effects of reduced acceleration and increased mass flow. So the thrust reduction caused purely by the reduced acceleration must be something like 6.2%.
Now let's consider the situation where the increased TAS is caused by a reduction in density due to a temperature increase. If as you suggest, we keep CAS is constant, then dynamic pressure will aso be conctant. This means that the reduction in density is exactly balanced by the increase in TAS squared (in our dynamic pressure = 1/2RhoV squared).
This constant dynamic pressure means that there will be no increase in mass airflow through the engines. So for our 5% increase in TAS, the only change in thrust will be the 6.2% caused by reduced acceleration of the airflow. This will clearly cause a reduction in climb performance, which is what we actually see.
I could of course be wrong!!!!!!
I think that the problem might lie in the way that you are interpreting your 731 thrust graph.
If the graph is just for changes in thrust against changes in TAS, without including any atmospheric changes, then the results will be invalid for your increasing temperature scenario.
Let's start by assuming that your graph is based on constant air density. In this case any increase in TAS will result in an equal increase in CAS and dynamic pressure. This in turn will result in an increase in mass airflow through the engines. This means that the thrust reduction caused by the reducing acceleration of the airflow, will be partly offset by an increase in thrust due to increased mass flow.
If we assume that your 10 Kt increase in TAS represents a 5% increase, then we will also get a 5% increase in mass flow which will tend to increase thrust by 5%. But your graph shows an overall thrust decrease of 1.2%, which is the combined effects of reduced acceleration and increased mass flow. So the thrust reduction caused purely by the reduced acceleration must be something like 6.2%.
Now let's consider the situation where the increased TAS is caused by a reduction in density due to a temperature increase. If as you suggest, we keep CAS is constant, then dynamic pressure will aso be conctant. This means that the reduction in density is exactly balanced by the increase in TAS squared (in our dynamic pressure = 1/2RhoV squared).
This constant dynamic pressure means that there will be no increase in mass airflow through the engines. So for our 5% increase in TAS, the only change in thrust will be the 6.2% caused by reduced acceleration of the airflow. This will clearly cause a reduction in climb performance, which is what we actually see.
I could of course be wrong!!!!!!
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Keith,
As in my earlier post, the graph is of thrust vs mach, at 59 deg F (below the 76 deg F flat rate)
and sea level (US standard atmosphere). So it is as you surmised.
It’s taken awhile for me to respond because I’ve taken some time to try and understand the thrust
equation for a jet engine, without success. I’m having trouble convincing myself that a 5%
increase in mass flow results from a 5% increase in TAS. A simple situation is to consider an
aircraft just after the start of the take off roll, say at 5 KTAS. There will be a certain mass
flow through the engine (= area x rho x TAS ?). At 10 kts, we’ve had a 100% increase in TAS (from
5 to 10 kts). Does this mean we’ve had a 100% increase in mass flow? Can it be that this would
only be correct over small TAS percent increases?
Other than perhaps this, I think you’ve been able
to explain to me the cold weather climb paradox.
Your quote
"I could of course be wrong"
No, I don’t think so
As in my earlier post, the graph is of thrust vs mach, at 59 deg F (below the 76 deg F flat rate)
and sea level (US standard atmosphere). So it is as you surmised.
It’s taken awhile for me to respond because I’ve taken some time to try and understand the thrust
equation for a jet engine, without success. I’m having trouble convincing myself that a 5%
increase in mass flow results from a 5% increase in TAS. A simple situation is to consider an
aircraft just after the start of the take off roll, say at 5 KTAS. There will be a certain mass
flow through the engine (= area x rho x TAS ?). At 10 kts, we’ve had a 100% increase in TAS (from
5 to 10 kts). Does this mean we’ve had a 100% increase in mass flow? Can it be that this would
only be correct over small TAS percent increases?
Other than perhaps this, I think you’ve been able
to explain to me the cold weather climb paradox.
Your quote
"I could of course be wrong"
No, I don’t think so
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I think that as with most things. it is a bit more complicated than that!
You are correct in saying that mass flow is equal to compressor inlet area x density x inlet air velocity. But inlet air velocity is not always the same as the TAS of the aircraft and the density inside the intake is not constant even if we keep the pressure altitude and temperature constant. This is because at very low TAS values the air is being sucked into the intake and at very high TAS values it is being rammed in.
My Rolls Royce jet engine book gives a typical air velocity of 500 ft/sec (about 300 Knots) at the front face of the compressor of a pure jet engine at max RPM. This will obviously be rather lower for a modern high bypass machine so let's assume a velocity of say 200 Knots.
Assuming we can set take-off power before releasing the brakes at the start of the take-off roll the TAS will be zero but the air velocity at the front face of the fan will be 200 Knots. This is achieved by the fan creating a very low pressure in the intake. This low pressure draws in the air from outside, accelerating it from zero to 200 Knots. But this low pressure reduces the density of the air, so although we have the required 200 Knot air velocity, the mass flow has been reduced by the reduced air density.
As we accelerate down the runway the increasing TAS reduces the amount by which the engine needs to accelerate the air to get the required 200 knots at the face of the fan. So the air density increases slightly, which in turn causes mass flow to increase. Although this tends to increase the thrust, this effect is outweighed by the reducing overall acceleration of air passing through the engine (Jet velocity - TAS). The overall effect of all of this is that thrust decreases as the aircraft accelerates down the runway.
Now getting back to your example of accelerating from 5 Knots to 10 Knots. Although this represents a 100% increase in TAS, it does not give a 100% increase in mass flow or thrust. All it really does is to reduce the pressure drop that the engine is creating in the inlet in order to draw in the air. There will of course be an increase in mass flow but it will be less than 100%.
But if we look at the overall trend, increasing TAS tends to increase thrust by increasing mass flow. But increasing TAS also tends to reduce thrust by increasing intake drag (once the air starts being rammed into the intake) and reducing overall acceleration. The thrust increasing tendency due to ram effect is proportional to TAS squared, while the decreasing effect due to intake drag and reduced acceleration is proportional to TAS. So at some value of TAS, the thrust increasing tendency due to ram effect balances the thrust decreasing tendency due to intake drag and reduced acceleration. At this speed the thrust stops dropping. If we continue to increase TAS the thrust starts to increase.
At some point along the TAS line a 5% increase in TAS gives a 5% increase in mass flow. But this will be true only over a very narrow TAS band. At lower speeds the mass flow will increase is less than the TAS increase and at higher speeds the mass flow will increase is more than the TAS increase.
If our Vmd climb is done at about 200 to 250 Kts, this is (probably) about the right speed for a direct 1:1 link between TAS and mass flow through a high bypass turbofan. Al figures quoted in this post are of course very much approximations.
You are correct in saying that mass flow is equal to compressor inlet area x density x inlet air velocity. But inlet air velocity is not always the same as the TAS of the aircraft and the density inside the intake is not constant even if we keep the pressure altitude and temperature constant. This is because at very low TAS values the air is being sucked into the intake and at very high TAS values it is being rammed in.
My Rolls Royce jet engine book gives a typical air velocity of 500 ft/sec (about 300 Knots) at the front face of the compressor of a pure jet engine at max RPM. This will obviously be rather lower for a modern high bypass machine so let's assume a velocity of say 200 Knots.
Assuming we can set take-off power before releasing the brakes at the start of the take-off roll the TAS will be zero but the air velocity at the front face of the fan will be 200 Knots. This is achieved by the fan creating a very low pressure in the intake. This low pressure draws in the air from outside, accelerating it from zero to 200 Knots. But this low pressure reduces the density of the air, so although we have the required 200 Knot air velocity, the mass flow has been reduced by the reduced air density.
As we accelerate down the runway the increasing TAS reduces the amount by which the engine needs to accelerate the air to get the required 200 knots at the face of the fan. So the air density increases slightly, which in turn causes mass flow to increase. Although this tends to increase the thrust, this effect is outweighed by the reducing overall acceleration of air passing through the engine (Jet velocity - TAS). The overall effect of all of this is that thrust decreases as the aircraft accelerates down the runway.
Now getting back to your example of accelerating from 5 Knots to 10 Knots. Although this represents a 100% increase in TAS, it does not give a 100% increase in mass flow or thrust. All it really does is to reduce the pressure drop that the engine is creating in the inlet in order to draw in the air. There will of course be an increase in mass flow but it will be less than 100%.
But if we look at the overall trend, increasing TAS tends to increase thrust by increasing mass flow. But increasing TAS also tends to reduce thrust by increasing intake drag (once the air starts being rammed into the intake) and reducing overall acceleration. The thrust increasing tendency due to ram effect is proportional to TAS squared, while the decreasing effect due to intake drag and reduced acceleration is proportional to TAS. So at some value of TAS, the thrust increasing tendency due to ram effect balances the thrust decreasing tendency due to intake drag and reduced acceleration. At this speed the thrust stops dropping. If we continue to increase TAS the thrust starts to increase.
At some point along the TAS line a 5% increase in TAS gives a 5% increase in mass flow. But this will be true only over a very narrow TAS band. At lower speeds the mass flow will increase is less than the TAS increase and at higher speeds the mass flow will increase is more than the TAS increase.
If our Vmd climb is done at about 200 to 250 Kts, this is (probably) about the right speed for a direct 1:1 link between TAS and mass flow through a high bypass turbofan. Al figures quoted in this post are of course very much approximations.
Last edited by Keith.Williams.; 27th Dec 2003 at 06:28.
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Keith, thanks for your insights. I kept drafting replies, but my problem is I don’t understand what I read from various internet sources (that’s probably my problem right there). I tried looking up your rolls royce book on amazon, but couldn’t find it.
Can we assume a nearly constant velocity at the front face of a compressor of a pure jet, from TAS zero to high subsonic?
If so, with the aircraft at high subsonic, the air is slowed and compressed a great deal just prior to the compressor. This must give a higher mass flow. But then, if the acceleration of the air is "jet velocity minus TAS" as you put it, then at high subsonic, there will be very little acceleration, going to zero as tas reaches jet velocity. (?)
But then I’ve read the aircraft TAS is essentially the same as the jet velocity entering the engine
I’ve also read of a contribution to thrust of the pressure difference between inlet and exhaust, times the exhaust area. But this can’t be very large I presume.
So how does the concorde cruise M2.0 with subsonic (or sonic at the most, I thought) exhaust speed?
Any thoughts much appreciated
Can we assume a nearly constant velocity at the front face of a compressor of a pure jet, from TAS zero to high subsonic?
If so, with the aircraft at high subsonic, the air is slowed and compressed a great deal just prior to the compressor. This must give a higher mass flow. But then, if the acceleration of the air is "jet velocity minus TAS" as you put it, then at high subsonic, there will be very little acceleration, going to zero as tas reaches jet velocity. (?)
But then I’ve read the aircraft TAS is essentially the same as the jet velocity entering the engine
I’ve also read of a contribution to thrust of the pressure difference between inlet and exhaust, times the exhaust area. But this can’t be very large I presume.
So how does the concorde cruise M2.0 with subsonic (or sonic at the most, I thought) exhaust speed?
Any thoughts much appreciated
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The copy of the Rolls Royce book "The Jet Engine" that I have here at home is a 1973 version. I examined it for an ISBN number but it (my old copy) does not have one. You can certainly get it from POOLEYS or TRANSAIR and probably PILOTWAREHOUSE through their websites. It does not go into the theory to any really great depth (no atom splitting or gene splicing equations), but it gives a good overall description of how jet engines work.
The air velocity at the face of the fan will vary with RPM. But should ideally be constant at any given RPM, regardless of TAS. To achieve this the engine sucks in the air when the aircraft TAS is very low. So before brake release in the take-off run, although the aircraft TAS is zero, the velocity at the face of the fan may be about 200 or 300 knots. But the pressure and air density in the intake are very low.
As aircraft TAS increases, there is less of a need to suck in the air, so pressure in the intake gradually rises towards ambient. At higher aircraft speeds the air is forced into the intake at a speed greater than that required by the fan. So the air inlet is designed to decelerate the air, converting as much of the excess velocity (kinetic energy) as possible into static pressure. But unless the divergence of the intake duct can be varied in flight, the airspeed at the fan face will vary to some extent with TAS.
For low speed flight (up to typical airliner cruise speeds) the intakes are of the slightly divergent PITOT type. When the TAS is greater than the required compressor inlet speed the divergent intake slows the air thereby increasing its static pressure and density.
For higher speeds (transonic and supersonic) a PITOT intake would create too much drag and be very inefficient. This is because the individual shock waves that form around the lip combine to form a strong shockwave right across the throat of the intake. The air passing through this shockwave would be decelerated very abruptly. Although this would increase its static pressure and density, it would also cause a large increase in temperature. This increased inlet temperature would reduce the thermal efficiency of the engine. In addition to this, excess air would spill around the lips of the intake, causing lots of turbulence and drag.
To overcome these problems, supersonic aircraft use either multi-shock spike-type intakes (as in the MIG21 and SR71), or variable area moving ramp intakes (as in CONCORDE, and most modern supersonic combat aircraft). These intakes use a series of oblique shockwaves to gradually decelerate and compress the air. Also, by moving the internal ramp, these intakes change their frontal area and convergence/diverenge ratios to match the incoming airspeed. The more gradual deceleration produced by these methods causes much lower increases in temperature. The overall effect is a very large increase in pressure before the air enters the compressor.
The exhaust systems of high speed aircraft are also designed to achieve very high exhaust gas velocities. In a simple convergent duct type propelling nozzle, the maximum possible velocity at the throat of the nozzle is the local speed of sound. This is 661 knots at 15 degrees C, but is equal to 39.84 x the square root of the absolute temperature.
At ISA msl this gives 38.94 x the square root of (15 + 273) = 661 Knots. But the temperature in a jet pipe is much higher than ambient. If we assume a 400 degree C jet pipe temperature, this will give a local speed of sound of 1010 Knots. This is why aircraft with simple convergent propelling nozzles can achieve supersonic speeds. The TAS of the aircraft may be more than 661 Knots, but the exhaust velocity is much greater, so the engine is still producing thrust.
But if we just keep opening the throttle we will eventually reach a point where the velocity at the throat of the nozzle is equal to the local speed of sound. If we now open the throttle even more, the jet pipe pressure will continue to increase, but the exhaust gas velocity at the throat of the nozzle will remain constant. Instead of going faster, the air simply leaves the nozzle with some excess static pressure. This pressure is then wasted as the air expands outside of the aircraft. This situation is called a CHOKED NOZZLE condition.
It is in this choked nozzle condition that we get a small amount of pressure thrust due to the pressure difference across the nozzle. Pressure thrust is equal to the nozzle area multiplied by the difference between the ambient pressure and that in the jetpipe. But this is a very inefficient method of creating thrust, so it is best avoided if possible.
If we want to go really fast we need to get even greater exhaust velocities. This can be done in two ways. Firstly if we use reheat by burning extra fuel in the jetpipe we will increase the exhaust gas temperature. This will increase the local speed of sound at the nozzle thereby enabling our simple convergent nozzle to achieve higher exhaust velocities. But the exhaust gas velocity at the throat of the nozzle is still only equal to the (now rather higher) local speed of sound.
To get truly supersonic exhaust velocities we need to use a more complex propelling nozzle in the form of a convergent duct followed by a divergent duct. The convergent duct accelerates the gas to local speed of sound. This sonic gas then flows into the divergent duct, which continues to accelerate it to supersonic velocities. The explantion of this effect is not really very complicated, but works best with diagrams, so I won't try it here. But to describe the resilts we need simply reverse the effects of ducts on low speed flow. At subsonic speeds a convergent duct increases velocity, decreases temperature and decreases static pressure. At supesonic speeds these effects are produced by a divergent duct.
If you look at any modern supersonic aircraft, the propulsion system will usually include variable area intakes, reheat and variable area convergent-divergent propelling nozzles.
The air velocity at the face of the fan will vary with RPM. But should ideally be constant at any given RPM, regardless of TAS. To achieve this the engine sucks in the air when the aircraft TAS is very low. So before brake release in the take-off run, although the aircraft TAS is zero, the velocity at the face of the fan may be about 200 or 300 knots. But the pressure and air density in the intake are very low.
As aircraft TAS increases, there is less of a need to suck in the air, so pressure in the intake gradually rises towards ambient. At higher aircraft speeds the air is forced into the intake at a speed greater than that required by the fan. So the air inlet is designed to decelerate the air, converting as much of the excess velocity (kinetic energy) as possible into static pressure. But unless the divergence of the intake duct can be varied in flight, the airspeed at the fan face will vary to some extent with TAS.
For low speed flight (up to typical airliner cruise speeds) the intakes are of the slightly divergent PITOT type. When the TAS is greater than the required compressor inlet speed the divergent intake slows the air thereby increasing its static pressure and density.
For higher speeds (transonic and supersonic) a PITOT intake would create too much drag and be very inefficient. This is because the individual shock waves that form around the lip combine to form a strong shockwave right across the throat of the intake. The air passing through this shockwave would be decelerated very abruptly. Although this would increase its static pressure and density, it would also cause a large increase in temperature. This increased inlet temperature would reduce the thermal efficiency of the engine. In addition to this, excess air would spill around the lips of the intake, causing lots of turbulence and drag.
To overcome these problems, supersonic aircraft use either multi-shock spike-type intakes (as in the MIG21 and SR71), or variable area moving ramp intakes (as in CONCORDE, and most modern supersonic combat aircraft). These intakes use a series of oblique shockwaves to gradually decelerate and compress the air. Also, by moving the internal ramp, these intakes change their frontal area and convergence/diverenge ratios to match the incoming airspeed. The more gradual deceleration produced by these methods causes much lower increases in temperature. The overall effect is a very large increase in pressure before the air enters the compressor.
The exhaust systems of high speed aircraft are also designed to achieve very high exhaust gas velocities. In a simple convergent duct type propelling nozzle, the maximum possible velocity at the throat of the nozzle is the local speed of sound. This is 661 knots at 15 degrees C, but is equal to 39.84 x the square root of the absolute temperature.
At ISA msl this gives 38.94 x the square root of (15 + 273) = 661 Knots. But the temperature in a jet pipe is much higher than ambient. If we assume a 400 degree C jet pipe temperature, this will give a local speed of sound of 1010 Knots. This is why aircraft with simple convergent propelling nozzles can achieve supersonic speeds. The TAS of the aircraft may be more than 661 Knots, but the exhaust velocity is much greater, so the engine is still producing thrust.
But if we just keep opening the throttle we will eventually reach a point where the velocity at the throat of the nozzle is equal to the local speed of sound. If we now open the throttle even more, the jet pipe pressure will continue to increase, but the exhaust gas velocity at the throat of the nozzle will remain constant. Instead of going faster, the air simply leaves the nozzle with some excess static pressure. This pressure is then wasted as the air expands outside of the aircraft. This situation is called a CHOKED NOZZLE condition.
It is in this choked nozzle condition that we get a small amount of pressure thrust due to the pressure difference across the nozzle. Pressure thrust is equal to the nozzle area multiplied by the difference between the ambient pressure and that in the jetpipe. But this is a very inefficient method of creating thrust, so it is best avoided if possible.
If we want to go really fast we need to get even greater exhaust velocities. This can be done in two ways. Firstly if we use reheat by burning extra fuel in the jetpipe we will increase the exhaust gas temperature. This will increase the local speed of sound at the nozzle thereby enabling our simple convergent nozzle to achieve higher exhaust velocities. But the exhaust gas velocity at the throat of the nozzle is still only equal to the (now rather higher) local speed of sound.
To get truly supersonic exhaust velocities we need to use a more complex propelling nozzle in the form of a convergent duct followed by a divergent duct. The convergent duct accelerates the gas to local speed of sound. This sonic gas then flows into the divergent duct, which continues to accelerate it to supersonic velocities. The explantion of this effect is not really very complicated, but works best with diagrams, so I won't try it here. But to describe the resilts we need simply reverse the effects of ducts on low speed flow. At subsonic speeds a convergent duct increases velocity, decreases temperature and decreases static pressure. At supesonic speeds these effects are produced by a divergent duct.
If you look at any modern supersonic aircraft, the propulsion system will usually include variable area intakes, reheat and variable area convergent-divergent propelling nozzles.
Last edited by john_tullamarine; 2nd Jan 2004 at 07:42.
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thanks for your reply Keith. Your quote:
"The thrust increasing tendency due to ram effect is proportional to TAS squared, while the decreasing effect due to intake drag and reduced acceleration is proportional to TAS"
I’ve been spending some time considering your quote, and am wondering if it is becauseof the following reason. We know Mass flow = p*V*A. Now at all speeds, V is essentially constant, say 200 kts. Thus p is less than static for low TAS. At higher speeds, mass flow is affected by above ambient p, which we know is caused by ram. Can we say ram pressure = dynamic pressure = .5 * p ambient * V squared? And this is the reason you state mass flow is proportional to TAS squared?
I’m not sure why your quote states the decreasing effect on acceleration comes out proportional to TAS though. I come up with V exhaust minus V freestream.
Fascinating explanation on supersonic cruise by use of propelling nozzles. Know I see why the concorde course cruise at M2.0 (and without afterburner).
"The thrust increasing tendency due to ram effect is proportional to TAS squared, while the decreasing effect due to intake drag and reduced acceleration is proportional to TAS"
I’ve been spending some time considering your quote, and am wondering if it is becauseof the following reason. We know Mass flow = p*V*A. Now at all speeds, V is essentially constant, say 200 kts. Thus p is less than static for low TAS. At higher speeds, mass flow is affected by above ambient p, which we know is caused by ram. Can we say ram pressure = dynamic pressure = .5 * p ambient * V squared? And this is the reason you state mass flow is proportional to TAS squared?
I’m not sure why your quote states the decreasing effect on acceleration comes out proportional to TAS though. I come up with V exhaust minus V freestream.
Fascinating explanation on supersonic cruise by use of propelling nozzles. Know I see why the concorde course cruise at M2.0 (and without afterburner).
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You are correct in saying that ram effect is proportional to TAS squared. This is because it is the dynamic pressure that is increasing static pressure in the intake. this in turn is what is increasing the air density in the intake and hence mass airflow through the engine.
You are also correct in saying that thrust is proportional to exhaust gas velocity minus freestream TAS. But this means that (if we ignore the benefits of ram effect), for each knot increase in freestream TAS we get the same reduction in thrust. So the thrust reduction is proportional to TAS.
We need to be a bit careful about which V we use in considering air massflow and the acceleration given to it. The mass flow is equal to air density at the compressor face x comressor frontal area x air VELOCITY AT THE COMPRESSOR FACE. But the acceleration given to the air is equal to exhaust gas velocity minus FREESTREAM TAS. As discussed earlier, velocity at the compressor face and freestream TAS are ferequently different things.
You are also correct in saying that thrust is proportional to exhaust gas velocity minus freestream TAS. But this means that (if we ignore the benefits of ram effect), for each knot increase in freestream TAS we get the same reduction in thrust. So the thrust reduction is proportional to TAS.
We need to be a bit careful about which V we use in considering air massflow and the acceleration given to it. The mass flow is equal to air density at the compressor face x comressor frontal area x air VELOCITY AT THE COMPRESSOR FACE. But the acceleration given to the air is equal to exhaust gas velocity minus FREESTREAM TAS. As discussed earlier, velocity at the compressor face and freestream TAS are ferequently different things.
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Gentlemen;
May I say that I have read this thread through a couple of times..... very very slowly in some parts... and I have found your lines of reasoning both interesting and educational as well.
Just wanted to say thankyou. It has been a great review!
May I say that I have read this thread through a couple of times..... very very slowly in some parts... and I have found your lines of reasoning both interesting and educational as well.
Just wanted to say thankyou. It has been a great review!
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Although unable to carefully read ALL this thread to the end 
, my impressions are the following:
1) Someone seems to ignore that with increased temperature, to keep the thrust constant up to the flat rated limit, engine RPM increase.
Higher temperature=higher inlet TAS, but also=higher output TAS.
On a 733, 22K flat rated engine, if you take off at sea level on a
-10°C day, your N1 will be 89.6.
On a +30°C N1 will be 96.1.
The air is less dense, OK, but the engine will accelerate it at a greater speed, to produce the same thrust until the flat rated limit.
2) Everybody knows, by direct experience, that the climb rate is better on cold days.
But here we are not talking about the enroute climb phase, but about the very initial takeoff phase.
The "desert experience" might be misleading someone: in the desert you are mostly often way above the flat rated limit.
Flat rated engines produce the same thrust up to ISA+...(15, usually).
Simple as that.
If we maintain the same V2 after liftoff, the warmer air will give higher TAS, thus higher climb rate.
Don't confuse this "picture" with very high desert temperatures and with enroute climb performances.
LEM
!
, my impressions are the following:1) Someone seems to ignore that with increased temperature, to keep the thrust constant up to the flat rated limit, engine RPM increase.
...So if we assume that exhaust speed is constant at all ambient temperatures....
On a 733, 22K flat rated engine, if you take off at sea level on a
-10°C day, your N1 will be 89.6.
On a +30°C N1 will be 96.1.
The air is less dense, OK, but the engine will accelerate it at a greater speed, to produce the same thrust until the flat rated limit.
2) Everybody knows, by direct experience, that the climb rate is better on cold days.
But here we are not talking about the enroute climb phase, but about the very initial takeoff phase.
The "desert experience" might be misleading someone: in the desert you are mostly often way above the flat rated limit.
Flat rated engines produce the same thrust up to ISA+...(15, usually).
Simple as that.
If we maintain the same V2 after liftoff, the warmer air will give higher TAS, thus higher climb rate.
Don't confuse this "picture" with very high desert temperatures and with enroute climb performances.
LEM
!
