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hawk37
3rd Dec 2003, 05:13
There seems to be the visual perception that for a jet, the colder the temperature, the greater the climb angle and rate after takeoff. Even if one requires the aircraft to be in the same configuration, and thrust (assuming the engines are flat rated throughout the temperature weíre considering), this seems to be the perception. Having pondered this for awhile, Iíve come up with the following, and am asking anyone to help with comments, correct, and generally additional info:

Consider a jet climbing just after take off, clean, at the min drag speed. For noise abatement climbs, I thought this would be close to reality. This calibrated speed should be independent of the OAT.

As for the climb angle, Iím surmising that it must be the same regardless of the temperature. ??

However, on a warmer day, the TAS will be higher, which means the rate of climb is higher on a warm day than a cold day. This seems opposite to the visual illusion.

Any fallacies?

Hawk37

Runaway Gun
3rd Dec 2003, 05:45
Mr Hawk,

Remember the ole formula: Lift=CL 1/2 Rho Vsquared S ??

Well on a colder day, the air is more dense. Therefore the Density part of the formula (Rho) is greater.

Whilst this does achieve more engine thrust, even if this is flat rated, you do get more air flowing over them wings.

Therefore the climb performance does increase markedly, unless the cold weather means that the pilots drank more through the winter to pass away those cold lonely nights, and put on a few extra pounds, which negated all the above.. :p

Hot weather will decrease the climb performance (almost to Nil climb angle) when you are out in the desert on a very hot day.

Strangely enough, the desert heat didn't stop me drinking the nights away either...

alatriste
3rd Dec 2003, 05:46
Climbing on a hot day at certain IAS or CAS, associated TAS will be higher and therefore DRAG. If drag is increased, then Power needed will be increase and so, excess of power will be reduced.( even assuming available thrust remain constant cause of flat rated engine operating below reference temperature).
If excess thrust is reduced, then both rate and angle of climb will be reduced when operating on a hot day.
So it is a fact that aeroplanes climb better on cold days.
REGARDS

Crossunder
3rd Dec 2003, 14:12
alatriste: drag does not increase that much just because TAS increases (unless you're at high Mno). But, as you say, a higher TAS will require more power, thus decreasing performance on a hot day.

hawk37
3rd Dec 2003, 20:04
Further to my original post, an acquaintance has shown me Falcon 50 climb performance that shows at both
V2 and V final segment, the climb gradient DOES NOT vary with temperature, within the flat rating temperatures.
Boeings/airbusses different?
SOOOOO, since V final is practically the calibrated airspeed for minimum drag speed, one can
conclude that the warmer it is, the better the climb rate at that speed. (for flat rated operations).


Runaway
thanks. While I agree with most of your points, Iím not sure they explain the situation I
described, that is an aircraft at min drag speed, within a temperature range that the engines are flat rated.

Reference to your formula, while rho does change with temperature, so does TAS, since the V in your
formula is true airspeed. Warmer day is less rho, but since calibrated airspeed for min drag stays the
same, the tas goes up. This explains why lift then is the same, since weíre talking about same climb angle
and weight.

Iím not sure about "more air flowing over the wings" as you put it. And engine thrust stays the same for
flat rating operation.


alatriste,
However, in this example, the aircraft is climbing at min drag, or put another way at a calibrated airspeed
that allows the most excess thrust. The excess thrust is the same regardless of the temperature, for flat rated operation.
Since the climb gradient is the same, warm or cold, the warmer temperature gives a higher TAS.
Accordingly, the aircraft will climb at a higher rate of climb on a warmer day. Hence the paradox that I think I can explain

Crossunder
Power is not the issue here. Weíre talking about the aircraft flying at a speed for maximum excess thrust,
since the aircraft is flying at minimum drag, and hence getting the best GRADIENT.

Drag does not increase if you fly at minimum drag speed, except at high machs.
If excess power was the consideration, then we would be looking speeds for RATE OF CLIMB.

I can only conclude then that jets should climb at a higher rate on a warm
day at V fs, than on a cold day, when flat rated. Further dissentions?

Hawk37

Runaway Gun
3rd Dec 2003, 20:40
OK Hawk, you've caught me out. My theory is a little rusty, but my personal experience is that colder days will give better climb performance than hot days. Simple as that.

hawk37
3rd Dec 2003, 22:52
Runaway, thats just my point. Sounds "simple as that", yet the math, the performance manuals, the
aerodynamics etc. tell us the gradient is temperature independent, and that the climb rate
will be greater on a warm day. All perception on the other hand, is that climb rate, even angle is
better on a cold day. I alway add the caveat that this is for flat rated thrust, climbing at V
minimum drag, which for most jets is very close to V final segment. No aerodynamicists with some points?
Hawk37

Black Baron
3rd Dec 2003, 22:56
The effect of increasing temperature is the same as increasing altitude.

The aircraft will have travelled a lesser distance on a cold day, in the same time span, yet has produced the same thrust, so it must be climbing at a steeper angle.

In the desert jets take forever to groan off the runway and then have very poor climb performance, in Antartica where its well below zero, their CAS is greater than their TAS, so they have excellent climb performance.
:ok:

hawk37
3rd Dec 2003, 23:25
Baron,you say
"The lower the temperature the lower the TAS required for a given CAS, TAS requires thrust. Less thrust required for straight and level means a surplus of thrust."

Not sure what you mean, did you leave a word out?
We're talking minimum drag, so any change in TAS will mean more thrust is required, whether it's an increase or decrease in tas.

There will be no case of "less thrust required", since you already at minimum drag.

I thought my argument was very logical. Is there any sentence you disagree with? Just trying to explain this paradox, that's all. If I'm wrong, belieive me, I'll be the first to agree.

thanks, Hawk37

Black Baron
3rd Dec 2003, 23:49
To achieve the same CAS on a hot day as opposed to a cold day, you will have a greater ground speed.
Greater groundspeed equals longer take off run, and a greater distance to achieve the same altitude, which means a reduced angle of climb.
I edited my previous post to remove nonsense which refferred to props and horsepower available. For the same CAS drag must be the same, so thrust required must be the same, excess thrust the same, so rate of climb the same.



:ok:

bookworm
4th Dec 2003, 00:40
excess thrust the same, so rate of climb the same

Nope, fell at the last unfortunately! Excess thrust at higher speed generates more power, and therefore a higher rate of climb -- just enough, in fact, to compensate for the extra distance travelled and make the gradient of climb the same.

If you take it from the basic equations for equilibrium,

T - D = L*tan(angle_of_climb)

So the angle of climb ends up depending on (T - D)/L

D/L is (near enough) the same at the same CAS, so the angle of climb ends up depending on thrust available.

I know very little about turbine engines, but is the assumption that the thrust is flat rated throught the temperature ranges considered actually true? For example I've got a chart here for a P&W JT4 which is flat rated to 60 degF, after which thrust falls. Perhaps modern turbofans have a wider range.

hawk37
4th Dec 2003, 01:48
Baron, thanks for the reply
ref your quote
"Greater groundspeed equals longer take off run, and a greater distance to
achieve the same altitude, which means a reduced angle of climb "

Iíve meant climb only, no wind, at V min drag, so Iíve not included take off, or ground. Seems we
disagree on the climb angle. My understanding was that it is independent of temperature, for
constant thrust ie flat rated engine. Does anyone have a Boeing or Airbus Final segment chart
to verify this. As I said previously, the Dassault Falcon 50 chart shows this, but it could be wrong.

Your quote
"For the same CAS drag must be the same, so thrust required must be the same, excess thrust the same, so rate of climb the same

I agree with you first two phrases, the third should mean climb gradient the same, no? After
all, itís excess horsepower that determines climb rate, and excess thrust that determines the climb
gradient. And bookworm seems to agree climb gradient is the same

I still keep getting that the aircraft climbs at a better rate, at min drag with flat rated engines
??

Bookworm, good to hear from you. You must be referrring to barons quote above
My understanding is that flat rated means just that. As temperature would rise, the engine will
speed up in order to maintain the thrust. Until the predetermined temperature, usually ISA plus
something, say isa plus 15, equals 30 deg C. This is at sea level, adjusts for Pressure altitude. I
believe most modern turbofans are somewhat flat rated to isa + 15 deg or more.

Iíve been attempting to show that a flat rated aircraft at min drag speed will climb at a better
rate when the temperature is warm, versus cold. If you say the gradient is the same, which I
thought, then it seems the next logical step is that the rate of climb is higher when warm.

And so, it is just an illusion that the colder it is, the better the rate and gradient. Paradox solved

Keith.Williams.
4th Dec 2003, 03:02
The fact that flat rating gives constant STATIC thrust over a range of ambient temperatures, does not mean that it gives the same thrust at each possible value of CAS. This is because the thrust is related to TAS, which at any given CAS varies with ambient temperature.

For the purposes of POF we often make the simplifying assumption that jet thrust is constant with changing TAS. This is not actually true but for most purposes is "close enough for government work".

Thrust is proportional to the mass flow of air passing through the engines multiplied by the acceleratrion that this air is given by the engines. This acceleration is equal to the exhaust speed minus the TAS of the aircraft. As an aircraft accelerates down the runway the difference between exhaust speed and TAS decreases, causing thrust to decrease. The magnitude of this decrease depends upon the ratio of exhaust speed to TAS. So if we assume that exhaust speed is constant at all ambient temperatures, then anything which increases the TAS at any point in our take-off will reduce the thrust.

As stated previously the sine of the climb angle is equal to (thrust minus drag) divided by aircraft weight. The drag at Vmd is not affected by ambient temperature, but the TAS and hence thrust are affected. As ambient temperature increases, the increasing TAS at Vmd causes thrust to decrease. This decreases the maximum climb angle that can be achieved.

ROC is equal to excess power divided by weight, where power available is thrust x TAS and power required is Drag x TAS. As ambient temperature increases, the increased TAS and reduced thrust at Vmd cause power available to decrease while power required increases. The overall effect is that ROC also decreases as ambient temperature increases.

To understand why some aircraft are affected less than others we need to consider the type of engine used. The degree to which thrust decreases with increasing TAS depends upon the ratio of the exhaust speed to the TAS. In high by-pass engines with their low average exhaust speed each knot of TAS causes a significant drop in thrust. But in a low by-pass or no by-pass engine the higher average exhast gas speed reduces the effect of each knot of airspeed.



Edited by Keith Williams to correct a number of tooping erriots.

bookworm
4th Dec 2003, 03:35
Nice explanation Keith, thank you.

To understand why some aircraft are affected less than others we need to consider the type of engine used. The degree to which thrust decreases with increasing TAS depends upon the ratio of the exhaust speed to the TAS. In high by-pass engines with their low average exhaust speed each knot of TAS causes a significant drop in thrust. But in a low by-pass or no by-pass engine the higher average exhast gas speed reduces the effect of each knot of airspeed.

I guess there's a continuum from almost-constant-thrust turbojet through low-bypass turbofans, high-bypass turbofans to almost-constant-power turboprops.

hawk37
4th Dec 2003, 04:13
So, Keith, if I follow, can you confirm then that you saying that climb GRADIENT at min drag, flat rated, will decrease (slightly) as temp increases
And if so, then can you make a determination on the climb RATE as temp increases, same conditions?

Do the perfomance charts show this climb gradient decrease as temp increases? Boeing/airbus, or both?

Do typical hi bypass engines have enough of a thrust decrease to show this decrease in climb gradient?

many thanks for your time. Your last post must have taken a considerable amount of time. Well done?

Black Baron
4th Dec 2003, 07:47
From the tables for a Boeing 767-300 with GE CF-80 turbofans
CLB 2 (reduced thrust climb power)
weight 160 tonnes
climbing for Fl 370
ISA +10 and cooler 24 mins, 151 nm
ISA + 15 27 mins, 177 nm
ISA + 20 32 mins, 218 nm
at ISA +10 and cooler, it can make FL 390, at ISA +15 it can only make FL 370.
As previously stated, temperature reduces climb performance.
:ok:

HotDog
4th Dec 2003, 09:45
Figures for 747-200 RB524-D4 engines, brake rel wt 380 tonnes, climb to FL 300
ISA +10 & below - 28mins
ISA + 15 - 34mins
ISA + 20 - 44mins

During climb out of Dubai or Abu Dhabi in the summer, it is not uncommon to hit a temperature inversion at around 1500ft where the ROC drops to zero.

Tinstaafl
4th Dec 2003, 10:51
Even if the engines are flat rated and produce exactly the same thrust at the different temperatures the a/c WILL have a faster TAS for a given IAS as the temperature rises.

A faster TAS requires more power ie more of the engine(s) output is being used ergo less is available to provide climb rate. The lower RoC & the faster TAS results in a reduced AoC.

hawk37
4th Dec 2003, 22:17
Baron, weíve been referring to climb at V minimum drag, approximately final segment climb speed,
and with flat rated engines. Thatís why I had put the phrase "same conditions". Certainly a typical
climb to FL 370 is temperature dependent.
Keiths post seems to nail it pretty well. Having read
it a further 7 or 8 times, it now makes even more sense.
The assumption that thrust is "flat rated" does not apply to changes in TAS. Hence while climbing
at V min drag, which is a calibrated air speed, temperature increase causes a measureable loss in
thrust available, for other than pure jets.
Secondly, if I understand him correctly, there is an increase in Power required (note that Thrust required
will stay the same) due to the increase tas as temperature increases.
Third, and Iím not 100% sure of this, (KEITH can you help here) there is an effect on Power
available, as temperature increases, due to the increase tas. (P=T x V). thrust may slightly
decrease due to tas increase as temp rises, but velocity in the formula also has an INCREASE. I
think Keith is saying that the total effect on Power available though is that it decreases.

Iíd say the jury is out. Even at V min drag, expect a decrease in the climb gradient

for BARON and HOTDOG,

Do either of you two have a V final segment climb GRADIENT chart, and are you capable of providing
any figures on how much the gradient changes based on temperature?

For simplicity, it looks like Dassault publishes theirís as a straight line, ie no change in climb
gradient with temp, when considering the flat rated portion of engine operation

Does anyone has a link showing Power available, required, and Thrust available, VS velocity,
for different TEMPERATURES?

Many thanks for your efforts

bookworm
5th Dec 2003, 03:38
The figures Baron and HotDog quote demonstrate that the engine is not capable of supplying that flat-rated thrust (static or otherwise) over the entire temperature and altitude range.

What would be more useful would be figures indicating the climb gradient at sea level over a range of temperatures.

Keith.Williams.
5th Dec 2003, 03:39
The only set of charts to which I have regular access are those in the CAP 698 which is published by the CAA for use in JAR ATPL exams. These are reputedly based on a version of the 737, but I know not which.

The engines used in these charts are flat rated up to ISA +15 degrees C. The climb performance charts show a steady decline in climb performance at any given altitude up to the flat rating temperature limit. This is followed by an abrupt increase in performance degradation at higher temperatures. The overall implication of this being that flat rating reduces the detrimental effects of increasing temperatures, but does not entirely eliminate these effects.

If we ignore flat rating to simplify the arguments we will see that increasing temperature has two effects. These are a decrease in power available and an increase in power required. The overall result of increasing ambient temperature is therefore a reduction in excess power and hence a reduction in best possible ROC.

The principal cause of the reduction in power availabe is the reduction in thrust. In a non-flat rated engine the thrust reduction is caused by a reduction in mass flow due to reduced air density. In a flat rated engine it is caused by the reduced acceleration of the air, which is in turn caused by the increased TAS to CAS ratio.

The principal cause of the increase in power required is the increase in TAS at any given CAS. Power required is equal to drag x TAS. If we assume that drag at any given CAS remains constant, then the power required is proportional to any temperature-induced change in TAS.

The drag equation includes TAS squared, so multiplying drag by TAS to get power required, gives something that is proportional to TAS cubed. So any increase in TAS at any given CAS will cause an even bigger increase in power required.

To visualise the effects of temperature we need to sketch power required and power available curves. Power required looks like a drag curve which has been rotated in an anti-clockwise direction..A bit like a NIKE tick.

For a jet aircraft if we simplify the situation by assuming that thrust is constant at all values of TAS, the power available is a straight line starting at the origin (0,0) and moving up towards the top right hand corner of the chart. At low altitude these two lines will cross at two points. At these crossing points the power available is equal to the power required. At all higher and lower speeds the power available is less than the power required. So these two points are the minimum and maximum speeds for which sufficient power is available.

At all speeds between the two crossing points the power available is greater than the power required. This excess power can be used to provide a rate of climb. The excess power and hence ROC are proportional to the vertical distance between the two power curves. ROC will be greatest at the speed at which the vertical distance between the two lines is greatest.

If we draw a tangent from the origin to touch the under surface of the power available curve we can use this to predict the effects of changes in altitude or temperature. All subsequent power available curves are similar to the first and all curves touch the same tangent. Increasing altitude or increasing temperature cause the power required curve to slide up the tangent towards the right hand end of the chart.

The power available curve always starts at the origin because at this point it is equal to thrust multiplied by zero TAS. Its angle is dependent upon altitude and temperature. Increasing altitude and increasing temperarture both cause this curve to rotate clockwise about the origin, thereby reducing power available at any given TAS.

The overall effect of these changes is that increasing altitude or increasing temperature will increase power required, reduce power available and reduce ROC.

The use of flat rated engines reduces the rate of reduction in power available but has no effect on the rate of increase in the power required.

mbga9pgf
5th Dec 2003, 05:01
Turbofan/Turbojet thrust increases with reduced Compressor inlet temperature, as more work can be done with greater heat input into the compressed airflow. the limiting factor of course being Tmax experienced by the turbine blades (Turbine inlet Temp), which is set by the material they are made of. More thrust=Greater climb performance.

Black Baron
5th Dec 2003, 06:25
Thankyou Dorset , excellent post.
Not sure about the climb gradient questions hawk, surely the figures for distance to top of climb represent the gradient?
:ok:

bookworm
5th Dec 2003, 15:49
One point I don't follow Keith:

The drag equation includes TAS squared, so multiplying drag by TAS to get power required, gives something that is proportional to TAS cubed. So any increase in TAS at any given CAS will cause an even bigger increase in power required.

Surely the drag depends on the dynamic pressure and therefore the CAS squared?

HotDog
5th Dec 2003, 17:22
Hawk,

"Do either of you two have a V final segment climb GRADIENT chart, and are you capable of providing
any figures on how much the gradient changes based on temperature? "

No, I do not have a V final segment climb gradient chart. All I can tell you from personal experience of around 20,000 hours of operating a variety of jet aircraft with RR, GE and P&W engines, mostly in the tropics, is that climb performance deteriorates with a rise in temperature.

hawk37
5th Dec 2003, 21:14
Wow, Keith, quite the post. You took a tremendous amount of time to write that, an amount of time that will only be surpassed by the number of times I re-read it. I appreciate your efforts.

What still bothers me is that I can make an argument, based on THRUST, and not power, that is still academically unclear to me.

Consider thrust changes with tas and temp as being small enough to not be a factor, perhaps as with early non bypass jet engines on a cool day.

Regardless of temperature, climbing at V min drag gives a constant amount of excess thrust. This constant amount of excess thrust gives a constant climb angle. This constant climb angle gives a higher rate of climb on cool days than very cold days.

Iím still assuming this ACADEMIC argument is 100% logical Iím also assuming that it is not what happens in the REAL world for ONE simple reason. As you pointed out, attaining constant thrust with changes in tas and temperature do not happen.

I'm also assuming that if engine manufacturers came up with engine controls that could maintain a constant thrust throughout some defined envelope, that my argument of higher rate on a warmer day, at V min drag would be correct

thanks

Edited for formatting only

Tinstaafl
5th Dec 2003, 22:41
Even if the excess thrust relationship is unchanged AND the climb rate is unchanged**, the a/c will be at a faster TAS as temp. rises. Faster TAS vs same RoC = shallower angle.

**Of course the RoC will be reduced, making the gradient worse, but for the sake of argument...


Hawk, I get the impression you're trying to use a Thrust &/or Power &/or Drag graph for a particular circumstance (WAT) and possibly with unsuitable axis (IAS/CAS?) then applying it in slightly different conditions eg different TAS? Just a thought...

bookworm
6th Dec 2003, 00:38
Even if the excess thrust relationship is unchanged AND the climb rate is unchanged**, the a/c will be at a faster TAS as temp. rises. Faster TAS vs same RoC = shallower angle.

But the climb rate wouldn't be unchanged if the excess thrust is unchanged. It would be increased. The rate of climb depends on excess power, which is excess thrust times TAS. With TAS as a factor in both power and rate of climb, when dealing with approximately constant thrust it makes sense to divide through by TAS and deal with thrust and angle of climb.

hawk37
6th Dec 2003, 01:42
Tinstaffl, I didnít say the climb rate is unchanged. In my scenario, which is academic, the excess thrust is unchanged, which implies the climb gradient is unchanged, which implies the ROC is increased for a temp increase. And actually, I wasnít using any graph, just a 3 step hypothesis that I thought is clear to follow, but difficult to explain. See my previous post.
Oh, I see bookworm posted a reply so yours also.
Here's the 3 steps again

1. Regardless of temperature, climbing at V min drag gives a constant amount of excess thrust.
2. This constant amount of excess thrust gives a constant climb angle.
3. This constant climb angle gives a higher rate of climb on cool days than very cold days.

Caveat: dealing with a case of near CONSTANT thrust output from the engine, eg non bypass jets.
(See Keith's earlier posts about why this is not entirely possible).I use "cool" and "very cold"
to avoid the flat rated questions.

Bookworm, your question about Keith's post was

"Surely the drag depends on the dynamic pressure and therefore the CAS squared?"

Not sure what you mean. I get dynamic pressure is .5 * V ^ 2 * p

and therefor to tas squared. Drag = thrust required, and P = T x V, so Power is proportional to tas cubed.

this help?

Keith.Williams.
6th Dec 2003, 03:21
Hawk,

I think that the fault in your argument lies in the fact that the thrust does not remain unchanged as you climb. Thrust is proportional to mass airflow multiplied by the acceleration the engine gives to that air. The acceleration is equal to exhaust velocity minus TAS. As we climb the air density and hence mass airflow at any given engine RPM reduce, causing thrust to reduce.

In addition to this, the increasing TAS to CAS ratio causes the acceleration we give the air to reduce. This again reduces thrust. So even if we keep the same CAS and hence the same drag, excess thrust and climb gradient will decrease. This process continues until we hit the absolute ceiling at which maximum thrust equals drag, and excess thrust and best climb gradient are zero.

We can visualise this effect by drawing a drag and thrust available curve on a single chart. The drag curve will be the typical bucket shape. If we simplify the situation by assuming constant thrust at all airspeeds, thrust will be represented by a horizontal straight line. (In reality it would be a bucket a bit like the drag curve, but much shallower).

At low altitude the thrust line will be some distance above the bottom of the drag curve and will cross it at two points. These are the minimum and maximum speeds for which sufficient thrust is available. Excess thrust is the vertical distance between the two lines and this will be greatest at Vmd. Excess thrust is proportional to best climb gradient so this will occur at Vmd.

If we have marked our speed range in CAS we will find that the drag curve does not change with changing altitude. But as explained above, the thrust available will decrease as altitude increases. This can be represented by repeatedly drawing further thrust lines, each lower than its predecessor. When the thrust line just touches the bottom of the drag curve we are at our absolute ceiling.

It is tempting to think that we can treat excess power and excess thrust as if they were unrelated. This is not the case. As we climb, the power required increases and the power available decreases. This causes excess power and best ROC to fall to zero at the absolute ceiling. So at the absolute ceiling we have no excess thrust, no excess power, and the best climb gradient and best ROC are zero. Worse than this, there is only one speed at which we have enough thrust and power to fly. For a jet this is speed Vmd.

If we repeat the process described above, but this time with power available and power required, we can see how excess power and ROC change. As altitude increases the power required curve keeps being repeated ever higher. At the same time the power available line rotates clockwise about the origin, becoming shallower with each increase in altitude. When the power available line just touches the power required curve we are at the absolute ceiling. We have just enough power to fly straight and level at a single airspeed. If we superimpose the power and drag curves on the same chart we will see that the single speed at which we have enough thrust to fly is also the single speed at which we have enough power. For a jet this speed is Vmd. At this altitude it should really be called Vonly, but for some strange reason it is not.


Bookworm,

Regarding your comment "Surely the drag depends on the dynamic pressure and therefore the CAS squared?"

We need to be a bit careful here. If we change the CAS at any given altitude and temperature it is true that drag is (in part at least) proportional to CAS (or more correctly EAS) squared. But it is more complicated than that, because total drag is made up of induced drag and profile drag. Induced is proportrional to one over EAS squared and profile is proportional to EAS squared. Worse still the coefficient of induced is proportional to one over the fourth power of EAS and the coefficient of profile is more or less constant at most angles of attack. The overall relationship is therefore a bit complicated.

But the situation we were considering above was one of increasing temperature. In this case if we keep the altitude constant, we see TAS increases while CAS remains constant.

Looking at the drag equation Drag = Cd 1/2 Rho V squared S, the 1/2RhoVsquared is the dynamic pressure and the V squared is TAS squared. The ASI captures dynamic pressure and gives us a CAS indication proportional to it. So if we fly at constant CAS we are flying at constant dynamic pressure. But if temperature increases, the reducing air density Rho must be matched by a corresponding increase in TAS squared, in order to keep the dynamic pressure and CAS constant. This is the cause of the increasing ratio of TAS to CAS as temperature increases.

So as temperature increases at any given CAS, the drag stays constant but the TAS increases. Power required is equal to Drag times TAS. As temperature increases at any given CAS, the drag remains constant but is multpilied by an increased TAS. Because TAS appears as TAS squared in the drag, when this is multiplied by TAS to give power required we get something that is proportional to TAS cubed. The same thing happens as altitude increases.

hawk37
6th Dec 2003, 04:47
Keith, reference your quote "I think that the fault in your argument lies in the fact
that the thrust does not remain unchanged as you climb"

Iím talking instantaneous climb gradient, instantaneous climb rate. I know that normally
thrust decreases as you climb. Iím not trying to suggest this scenario can occur at any other speed
other than at, or very close to V md. And certainly, I understand best ROC is at max excess power etc.

My scenario has been at ONE altitude, say 1000 ft msl., and only at V md. And, most importantly,
with CONSTANT THRUST AS TAS CHANGES WITH TEMPERATURE.

Iím still donít see why my argument isnít sound.

IE
1. Regardless of temperature, climbing at V min drag gives a constant amount of excess thrust.
2. This constant amount of excess thrust gives a constant climb angle.
3. This constant climb angle gives a higher rate of climb on cool days than very cold days

What I SUSPECT is the problem, is that the caveat that Iíve put in, that thrust be considered
constant with a change in temperature (note, Iím not also adding altitude, since Iím looking at
instantaneous gradient/rate) IS TOO FAR OFF from the real world, even in pure jets, to have the
effect Iím expecting on gradient, and ROC.

Black Baron
6th Dec 2003, 15:20
You are correct that if thrust is constant, and TAS increases through temperature then angle remains constant, and rate increases.

If you were talking rocket engined aircraft this would be usual.

To the gas turbine engine a rise in temperature causes a rise in power required to produce the same thrust;
T=mv, but kinetic energy is 1/2mv2.
A drop in air density with temperature increase, requires an increase in V to maintain thrust. But that is squared into power required, whilst the drop in mass is halved.

Then the TAS is increasing so the prop/fan/jet speed must again be increased, again requiring more power yet.

An increase in temperature causes an increase in power required, a reduction in power available, requires higher compressor speeds, brings you closer to your EGT limits.

So as far as a gas turbine goes you are comparing a reduced power climb on a cool day to a max power climb on a warm day.


:ok:

bookworm
6th Dec 2003, 16:50
Keith

I think this TAS cubed thing is just a lack of clarity about what we're holding constant and what we're considering as a variable. We're probably saying the same thing in different ways.

Neglecting compressibility, and using a slight variant on your terminology:

dynamic pressure = 1/2Rho(TAS)squared = 1/2Rho_standard(CAS)squared

(where Rho_standard is a reference constant, density at ISA SL)

or TAS = d * CAS where d is the square root of the density ratio.

So

Drag = Cd * dynamic pressure * area

and

Power required = Drag * TAS

So you can write that as:

Cd * 1/2Rho_standard(CAS)squared * TAS * area

in other words proportional to (retaining density as a variable)

CAS squared * TAS

You can, if you want, write this as

Power required is proportional to CAS cubed * d

or you can write it as

Power required is proportional to TAS cubed / d squared

But for a constant CAS climb, the key is that the d comes in once not three times.

I think the complex shape of the drag curve is a bit of red herring, isn't it? Since lift depends on speed through the dynamic pressure too, the a point at constant CAS on the curves represents a particular drag-to-lift ratio, and therefore, for an aircraft of a particular weight a particular drag-to-weight ratio, regardless of density ratio. Turning that into a power required involves multiplying by the TAS or d only once.

hawk

I think Keith answered this the first time round, but let me share my understanding of his explanation by putting it another way. The thrust available depends explicitly on both temperature and TAS. Since TAS also depends on temperature for a constant CAS, there is also an indirect dependence on temperature.

If thrust is flat-rated, it means that the explicit dependence on temperature is removed. The static thrust (TAS = CAS = 0) will not depend on temperature. But for non-zero CAS, there is still that indirect temperature dependence. Thus the thrust available at climb CAS does vary with temperature even for a flat-rated engine.

Keith.Williams.
6th Dec 2003, 19:17
Bookworm,

If we hold pressure altitude and temperature constant then the relationship between CAS and TAS is also constant. Under these circumstances it is true to say that power required is proportional to CAS cubed just as much as it is proportional to TAS cubed.

But if we wish to consider real world situations where pressure altitude and temperature change, the ratio of CAS to TAS changes. Under these circumstances it is more useful to say that power required is proportional to TAS cubed.

To illustrate this let's use msl and 40000 ft in the ISA. At msl the CAS equals the TAS. At 40000 ft the CAS is about half of the TAS. If we climb from msl to 40000 ft at any given CAS the CAS cubed will be unchanged but the TAS cubed will be 8 times its original value.

The power required will in fact only have increased to twice it original value because the density will be about 1/4 of its msl value. Under these circumstances it would be reasonable to say that power required is proportional to Cd x density x TAS cubed. But it would not be at all accurate to say that power required is proportional to Cd x density x CAS cubed.

In summary if we keep pressure altitude and temperature constant, then power required is proportional to CAS cubed in exactly the same way that it is proportional to TAS cubed. But if we want to allow for changes in altitude and density we must use TAS cubed.


Hawk,

I think that I can see what you are saying, but again you are starting with a false assumption. Thrust will not remain constant as temperature increases. It will decrease, causing excess thrust and climb performance to decrease. I suspect that you are using the fact that very high exhaust speeds do not suffer much of an acceleration reduction for each knot of TAS increase. But they still suffer this to some extent. Although we might imagine that the old pure jets would benefit from this effect, they did not do so. This is partly because they did not enjoy the benefits of flat rated engines and partly because increasing temperature reduces thrust in all aircraft propulsion systems (jet or prop).

bookworm
6th Dec 2003, 19:59
The power required will in fact only have increased to twice it original value because the density will be about 1/4 of its msl value. Under these circumstances it would be reasonable to say that power required is proportional to Cd x density x TAS cubed. But it would not be at all accurate to say that power required is proportional to Cd x density x CAS cubed.

Looks like we agree on the equations, Keith. We'll just have to agree to differ as to how we describe those in words.

hawk37
6th Dec 2003, 21:10
Keith, thanks, and your quote "I think that I can see what you are saying, but again you are
starting with a false assumption. Thrust will not remain constant as temperature increases."

I was considering flat rating here. Based on all comments though, I'll assume my scenario would be
true if thrust was somehow constant with TAS. But it isn't, and the effect is significant
enough that it's effect on my academic scenario is that the reverse is true. I'm done. thanks
Bookworm too and all

Black Baron
7th Dec 2003, 09:52
Boeing 767 -300 cf6-80
brw 150 tonnes
climb to FL 370

ISA +10> clb / 18 minutes over 115 miles

ISA +10> clb 2/ (reduced climb) 21 minutes over 132 miles

ISA +15 clb / (fulll climb ) 20 minutes over 130 miles

ISA +15 clb 2 / 24 minutes over 151 miles

ISA +20 clb / 23 minutes over 155 miles

ISA +20 clb 2 / 27 minutes over 179 miles

clb2 represents a 20% thrust deduction to 10,000 feet decreasing linearly to Max climb thrust at FL 300

I think it is then fair to say that a 5'C rise in temperaure equals a 20% or greater loss of thrust available, and visa versa.

:ok:

hawk37
8th Dec 2003, 09:47
Baron, I don't think this figures are any where near V min drag. Also, way out of the flat rating
range. And 20% loss in thrust available with a 5 deg C rise sound WAAAAAAY too much. Was that a published number?

For Keith, I have found a thrust vs mach graph for the Garret 731 at sea level, 59 deg F. Its a 2.79
bypass fan. The graph has thrust in lbs on the y axis, and mach on the x axis. The curve starts at
3700 lbs at zero mach, and slopes down to the right. Other values are 2800 lbs at mach .3, and 2200 lbs at mach .6.
Concerning thrust decrease with tas near V min drag, say mach .3, about 198 ktas, if I draw a
tangent, the slope of the line works out to be 3.4 lbs of thrust per knot.

So with a 10 ktas increase near V min drag, the decrease in thrust from the engine is 34 lbs.
Based on the 2800 lbs it is producing, this is a 1.2 percent decrease in thrust available. Regarding the excess thrust available, this percent would be considerably higher, of course.
Now it you had a model to calculate drag, the decrease in excess thrust due to tas increase could give you the decrease in climb gradient. Iíll leave that up to Baron to calculate

timzsta
11th Dec 2003, 07:33
Hawk37 - when Black Baron talks about a 20% reduction in thrust he is correct. What he has done in his "reduced climb" is derated his engines through the FMC. This allows us to lengthen engine life, reduce wear and tear etc etc. I dont have the figures for the 767 but I do for the aircraft I am familar with, 737-400, from doing my ATPL exams.

The 737-400 comes usually with engines that are flat rated at 23,500lbs static thrust. However we can de-rate the engines to 22k or 20k through the N1 page on the FMC. We do this when we can still safely achieve our performance requirements with the use of less thrust, in order to reduce wear and tear on engine, extend engine life and reduce noise. At London Stansted (where I dispatch) over the last few days daytime temp has not got much over 5 deg celcius. So its pretty much ISA - 10 deg which is great for performance (the fog however isnt great for flying!!!). With a long runway and light pax load we dont need 23.5k, so you can de-rate to 22k. If we went for 20k you can see that we have reduced thrust by approx 15%.

On a cold day the air is more dense so the engine produces more thrust. The wing produces more lift because the air is more dense, so the climb gradient is higher.

The other thing you need to consider is that performance charts for GRADIENT of climb are only really applicable at low altitude when you want to clear an obstacle in the take off flight patch - ie as at London City, all those big buildings nearby.

With a jet airliner what you are really interested in is getting to cruise altitude in the shortest TIME as that make the aircraft more economical. Hence you aim to fly a jet at best rate of climb speed for all but the initial part of the climb.

Remeber max angle of climb (GRADIENT) - max altitude gain over shortest distance over the ground. (Vx)
Max rate of climb - largest altitude gain for shortest TIME in the air (Vy).
On the 737 FMC you can select from the climb page max angle or max rate climb.

I think the problem, Hawk37 , is that you are trying to make a hypothetical "text book/labatory" situation fit the real world. Interesting discussion though and useful performance refresher as it was some months ago I sat (and passed) the exam.

bookworm
12th Dec 2003, 06:29
The 737-400 comes usually with engines that are flat rated at 23,500lbs static thrust. However we can de-rate the engines to 22k or 20k through the N1 page on the FMC.
....
On a cold day the air is more dense so the engine produces more thrust.


So, what does flat rated mean then?

Captain Stable
12th Dec 2003, 17:34
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.

bookworm
12th Dec 2003, 19:19
I think that's a rather more general definition than Hawk37 intended when he wrote:

assuming the engines are flat rated throughout the temperature weíre considering

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. :)

Tinstaafl
12th Dec 2003, 22:04
I still think Hawk has added 2 & 2 to equal 5. I just can't quite put my finger on where/why. :confused:

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?

bookworm
12th Dec 2003, 23:48
What am I missing?

That the "engine output" is not fixed. If the thrust is constant, then just as the power required (drag * TAS) increases with density altitude at constant CAS, so does the power available (thrust * TAS).

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.

Tinstaafl
13th Dec 2003, 03:03
Ta Bookworm. Obvious once it's pointed out, isn't it? :O

Keith.Williams.
13th Dec 2003, 22:28
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!!!!!

hawk37
19th Dec 2003, 04:57
KEITH, Glad to hear that someone agrees my original conjecture is true. (Pardon to BOOKWORM, he may have already hinted so too).
Iíve 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.
SoÖ..Iíd like to take your computations a bit furtherÖ

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 Keithís 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 weíre 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 OATís (within engine flat rating) at V md seems substantiated!! A 3.2% climb rate increase due to temperature increase

Now Iíve flown jets enough to shake my head and say it just aint so, but mathematically, whatís wrong with this argument?

As TINSTAAFL put it, "Hawk has added 2+2 to equal 5. I just canít put my finger on it"

Crossunder
19th Dec 2003, 20:47
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?

hawk37
20th Dec 2003, 01:30
Crossunder, this constant thrust" climb, as you refer to it as, is actually a best angle climb.
Not sure if itís 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 donít think thatís why this theory doesnít fit in flight observations.

Keith.Williams.
20th Dec 2003, 03:02
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!!!!!!

hawk37
25th Dec 2003, 04:45
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.
Itís taken awhile for me to respond because Iíve taken some time to try and understand the thrust
equation for a jet engine, without success. Iím 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, weíve had a 100% increase in TAS (from
5 to 10 kts). Does this mean weíve 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 youíve been able
to explain to me the cold weather climb paradox.
Your quote
"I could of course be wrong"
No, I donít think so

Keith.Williams.
27th Dec 2003, 03:00
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.

hawk37
31st Dec 2003, 21:44
Keith, thanks for your insights. I kept drafting replies, but my problem is I donít understand what I read from various internet sources (thatís probably my problem right there). I tried looking up your rolls royce book on amazon, but couldnít 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 Iíve read the aircraft TAS is essentially the same as the jet velocity entering the engine

Iíve also read of a contribution to thrust of the pressure difference between inlet and exhaust, times the exhaust area. But this canít 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

Keith.Williams.
1st Jan 2004, 03:45
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.

hawk37
5th Jan 2004, 02:23
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"

Iíve 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?
Iím 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).

Keith.Williams.
6th Jan 2004, 03:00
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.

qnc3guy
10th Jan 2004, 14:38
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!

LEM
20th Mar 2004, 10:06
Although unable to carefully read ALL this thread to the end :bored: , 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....
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
-10įC day, your N1 will be 89.6.
On a +30įC 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


:ouch: !