A340-600 weights
Prof. Airport Engineer
Thread Starter
Join Date: Oct 2000
Location: Australia (mostly)
Posts: 726
Likes: 0
Received 0 Likes
on
0 Posts
A340-600 weights
I'm doing a design check for a new taxiway and want to examine it for the A340-600. If anyone can fill in the missing weights, I'd appreciate it: I've done the usual search of PPRUNE, and found nothing, and most of the reference manuals don't have the data (yet):
Operating empty weight = 177,000 kg
Maximum zero fuel weight = ?
Maximum landing weight = ?
Maximum takeoff weight = 365,000 kg
Also can anyone give me an indication of the A340-600 fuel consumption per hour in cruise please – total all engines ?
Operating empty weight = 177,000 kg
Maximum zero fuel weight = ?
Maximum landing weight = ?
Maximum takeoff weight = 365,000 kg
Also can anyone give me an indication of the A340-600 fuel consumption per hour in cruise please – total all engines ?
Join Date: Dec 2001
Location: what U.S. calls ´old Europe´
Posts: 941
Likes: 0
Received 0 Likes
on
0 Posts
A340-600 : MTOW 368t, MZFW 259t, MLW 245t
A340-500 : MTOW 372t, MZFW 243t, MLW 230t.
A new High Gross Weight variant is comming soon
A340-600HGW : MTOW 380t, MZFW 265t, MLW 251t
A340-500HGW : MTOW 380t, MZFW 246t, MLW 232t
hope that helps.
A340-500 : MTOW 372t, MZFW 243t, MLW 230t.
A new High Gross Weight variant is comming soon
A340-600HGW : MTOW 380t, MZFW 265t, MLW 251t
A340-500HGW : MTOW 380t, MZFW 246t, MLW 232t
hope that helps.
Join Date: Jul 2004
Location: Sydney NSW
Posts: 513
Likes: 0
Received 0 Likes
on
0 Posts
A340-600 fuel consumption
Weight FL290 FL310 FL330 FL350 FL370 FL390
tonnes Fuel consumption in metric tonnes per hour M0.83
355 10.6
345 10.41
336 10.12 10.04
327 9.94 9.78
318 9.72 9.56
309 9.52 9.28 9.22
300 9.28 9.09 8.96
290 9.13 8.87 8.71
280 8.91 8.66 8.51 8.39
272 8.73 8.43 8.25 8.17
264 8.53 8.27 8.07 7.93
255 8.34 8.06 7.83 7.68 7.58
245 8.19 7.86 7.62 7.47 7.32
236 8.04 7.66 7.44 7.24 7.1
227 7.87 7.51 7.26 7.05 6.88 6.79
218 7.71 7.36 7.07 6.83 6.66 6.54
Assumes 4704 sq ft wing area, aspect ratio 9.2, Oswald's "e" 0.86, typical sfc 0.587lb/lb-hr and 7% compressibility drag at M0.83, Cdo being 0.014 (777 being 0.013 and 747 being 0.015).
Sfc is not constant and neither is compressibility but it is good back of fag packet stuff at intvu if I did the XL spreadsheet aright. About what I'd expect, 10.5 tonnes per hour ToC and 6.5 tonnes/hr ToD on a long haul.
My pleasure... The Enicalyth
tonnes Fuel consumption in metric tonnes per hour M0.83
355 10.6
345 10.41
336 10.12 10.04
327 9.94 9.78
318 9.72 9.56
309 9.52 9.28 9.22
300 9.28 9.09 8.96
290 9.13 8.87 8.71
280 8.91 8.66 8.51 8.39
272 8.73 8.43 8.25 8.17
264 8.53 8.27 8.07 7.93
255 8.34 8.06 7.83 7.68 7.58
245 8.19 7.86 7.62 7.47 7.32
236 8.04 7.66 7.44 7.24 7.1
227 7.87 7.51 7.26 7.05 6.88 6.79
218 7.71 7.36 7.07 6.83 6.66 6.54
Assumes 4704 sq ft wing area, aspect ratio 9.2, Oswald's "e" 0.86, typical sfc 0.587lb/lb-hr and 7% compressibility drag at M0.83, Cdo being 0.014 (777 being 0.013 and 747 being 0.015).
Sfc is not constant and neither is compressibility but it is good back of fag packet stuff at intvu if I did the XL spreadsheet aright. About what I'd expect, 10.5 tonnes per hour ToC and 6.5 tonnes/hr ToD on a long haul.
My pleasure... The Enicalyth
A340-600 : MTOW 368t, MZFW 259t, MLW 245t
A340-500 : MTOW 372t, MZFW 243t, MLW 230t.
A new High Gross Weight variant is comming soon
A340-600HGW : MTOW 380t, MZFW 265t, MLW 251t
A340-500HGW : MTOW 380t, MZFW 246t, MLW 232t
A340-500 : MTOW 372t, MZFW 243t, MLW 230t.
A new High Gross Weight variant is comming soon
A340-600HGW : MTOW 380t, MZFW 265t, MLW 251t
A340-500HGW : MTOW 380t, MZFW 246t, MLW 232t
Prof. Airport Engineer
Thread Starter
Join Date: Oct 2000
Location: Australia (mostly)
Posts: 726
Likes: 0
Received 0 Likes
on
0 Posts
Thanks for the info everyone on the A340 weights - that is a big help. It's rapid exit taxiways, and geometrically the aircraft can't use them for takeoff (i.e. fully laden), so the landing weight is the design weight (and it makes quite a difference).
I've taken supercarb's comment and used:
A340-600 : MTOW 368t, MLW 259t, MZFW 245t
A340-500 : MTOW 372t, MLW 243t, MZFW 230t.
Being conservative of course, I also checked it can take occasional aircraft at MTOW without
I've taken supercarb's comment and used:
A340-600 : MTOW 368t, MLW 259t, MZFW 245t
A340-500 : MTOW 372t, MLW 243t, MZFW 230t.
Being conservative of course, I also checked it can take occasional aircraft at MTOW without
Join Date: May 2003
Posts: 409
Likes: 0
Received 0 Likes
on
0 Posts
Enicalyth,
Reference your calculations, I think I understand how you did them, but may I ask how the 7% compressibility was factored in? Is Cd just multiplied by 1.07 with CL left unchanged?
Are values of Cdo and e readily available for other aircraft, and is this fascinating data available via special flight test papers/aerodynamic studies?
I'd love to know more!! Anything available to email me?
Hawk
Reference your calculations, I think I understand how you did them, but may I ask how the 7% compressibility was factored in? Is Cd just multiplied by 1.07 with CL left unchanged?
Are values of Cdo and e readily available for other aircraft, and is this fascinating data available via special flight test papers/aerodynamic studies?
I'd love to know more!! Anything available to email me?
Hawk
Join Date: Jul 2004
Location: Sydney NSW
Posts: 513
Likes: 0
Received 0 Likes
on
0 Posts
hello hawk - enicalyth here
I guessed that at about FL350 the A340 would have 7% compressibity i.e. cd tot = cdo + cdi +cdcc = 1.07(cdo + cdi). cdtot is the sum of all the wetted areas plus a fiddle factor for protuberances times a finishing coefficient, usually 0.0036. This area is then divided by the wing area to give cdo. I usually add the trim drag coefft into cdo, for example Boeing quote 0.0144 for their 744 cdo but I take 0.015 which is spookily accurate when it come to fuel consumption. cdi the induced drag is the square of the lift coefft divided by pi x aspect ratio x Oswald's "e". For the DC10 "e" was 0.83, the 744 "e" is 0.86 and the 777 "e" is 0.89 about as high as you can go. cdo for a DC9 is 0.02, a 737 0.0176, a 777 0.013. In cruise when lift coeffts go above 0.55 the aircraft tends to have little buffet margin left so the highest first cruise altitude is usually the density altitude where cl = 0.55 for the chosen speed. likewise when weight falls off in cruise you will step up when the new cl is 0.55. eventually you cannot step up any more and cl rolls off. when it gets down to about 0.4 compressibilty has dropped from about 7-7.5% to 5-6%, speed and weight and altitude dependent. 737NGs, 777s, A330/340s can run at L/D of 19-ish when cl is 0.5-0.55; 744s 17.5. specific fuel consumption is 0.65lb/lb-hr for 737NG at FL350 cl 0.5-0.55, 0.59 for a 744 and 0.57 for a 777. the distance you travel in cruise is (v knots/sfc) x L/D x natural logarithm(start weight/finish weight). So your A340-500 at 480 kts FL350 will travel 20.5 nm per 1000lb from the equation distance = (480/.587)*19.55*0.001283 assuming I am right in guessing compressibility at 7%. I made cl 0.522 and cdtot 0.0267 which is the 19.55 term in above equation. It feels right to me but I can only speak with authority on the 777 and the 744.
NB I forgot to say what weight didn't I ?...780000lb!!! When the weight falls to 480000lb expect 28.2nm/1000
Best regards to hawk from the enicalyth
NB I forgot to say what weight didn't I ?...780000lb!!! When the weight falls to 480000lb expect 28.2nm/1000
Best regards to hawk from the enicalyth
Last edited by enicalyth; 17th Jul 2004 at 09:46.
Join Date: May 2003
Posts: 409
Likes: 0
Received 0 Likes
on
0 Posts
Hi Dan.
To save Eniclyth any more work, I’ll try. But I hate trying to explain things I don’t really know a lot about though!!
“I guessed that at about FL350 the A340 would have 7% compressibity i.e. cd tot = cdo + cdi +cdcc = 1.07(cdo + cdi).”
The coefficient of drag for the aircraft (CD) is the sum of the parasitic drag coefficient (CDO), induced drag coefficient (CDI) plus the compressibility drag coefficient correction (CDCC). I’ve used capitals to designate that we’re referring to a complete aircraft, not to just an airfoil section.
“cdtot is the sum of all the wetted areas plus a fiddle factor for protuberances times a finishing coefficient, usually 0.0036. This area is then divided by the wing area to give cdo. I usually add the trim drag coefft into cdo, for example Boeing quote 0.0144 for their 744 cdo but I take 0.015 which is spookily accurate when it come to fuel consumption.”
The above paragraph is unfamiliar to me.
**Eniclyth, can you expand a bit please?***
I’ve always taken the total CD as the sum of the three coefficients of drag, using a representative area as simply the wing area, ie Total Drag D = .5 * rho * V squared * wing area * CD
“cdi the induced drag is the square of the lift coefft divided by pi x aspect ratio x Oswald's "e". For the DC10 "e" was 0.83, the 744 "e" is 0.86 and the 777 "e" is 0.89 about as high as you can go.”
Perhaps I’d add that the aspect ratio and e factor are both in the denominator, ie CDI=(CL squared)/(pi * e * AR)
The math that produced the above was dependant on a perfectly elliptical lift distribution along the span, ie zero at the tips, increasing elliptically to the root. The spitfire is often cited as an example, but I believe it only had an elliptical wing planform, and not an elliptical lift distribution, although the planform would have helped it attain a near elliptical distribution. Some factors that reduce this perfect elliptical distribution are changing airfoil shape across the span (chord and camber), twist, and of course straight leading and trailing edges and tips (simpler to manufacturer). And so…. The Oswald efficiency factor e was included to allow for these real life affects. Typically ranges as per E’s post.
“cdo for a DC9 is 0.02, a 737 0.0176, a 777 0.013. In cruise when lift coeffts go above 0.55 the aircraft tends to have little buffet margin left so the highest first cruise altitude is usually the density altitude where cl = 0.55 for the chosen speed. likewise when weight falls off in cruise you will step up when the new cl is 0.55. eventually you cannot step up any more and cl rolls off. when it gets down to about 0.4 compressibilty has dropped from about 7-7.5% to 5-6%, speed and weight and altitude dependent. 737NGs, 777s, A330/340s can run at L/D of 19-ish when cl is 0.5-0.55; 744s 17.5. “
I think the above is straight forward
“specific fuel consumption is 0.65lb/lb-hr for 737NG at FL350 cl 0.5-0.55, 0.59 for a 744 and 0.57 for a 777. the distance you travel in cruise is (v knots/sfc) x L/D x natural logarithm(start weight/finish weight). So your A340-500 at 480 kts FL350 will travel 20.5 nm per 1000lb from the equation distance = (480/.587)*19.55*0.001283 assuming I am right in guessing compressibility at 7%. I made cl 0.522 and cdtot 0.0267 which is the 19.55 term in above equation. It feels right to me but I can only speak with authority on the 777 and the 744.
NB I forgot to say what weight didn't I ?...780000lb!!! When the weight falls to 480000lb expect 28.2nm/1000”
Above is where the math starts. **I think** Enicalyth has figured Drag from .5 rho S CD Vsquared. In cruise, this equals thrust required. Then knowing lbs of fuel burned per lb of thrust generated, per hour, he figures out fuel consumption. I didn't think one needed Breguet’s range formula for a jet aircraft but E's use on natural logs perhaps indicates so. I’ve yet to find a quiet few hours to do my math to come up with E’s numbers.
If you’ve got a calculator in the cockpit, you’ve really no need for flight manual data.
Make any sense Dan?
Enicalyth, sorry you got no reply on your 737 data request. Of course, I'd of helped, but I figured C172 data wasn't what you needed.
Hawk
To save Eniclyth any more work, I’ll try. But I hate trying to explain things I don’t really know a lot about though!!
“I guessed that at about FL350 the A340 would have 7% compressibity i.e. cd tot = cdo + cdi +cdcc = 1.07(cdo + cdi).”
The coefficient of drag for the aircraft (CD) is the sum of the parasitic drag coefficient (CDO), induced drag coefficient (CDI) plus the compressibility drag coefficient correction (CDCC). I’ve used capitals to designate that we’re referring to a complete aircraft, not to just an airfoil section.
“cdtot is the sum of all the wetted areas plus a fiddle factor for protuberances times a finishing coefficient, usually 0.0036. This area is then divided by the wing area to give cdo. I usually add the trim drag coefft into cdo, for example Boeing quote 0.0144 for their 744 cdo but I take 0.015 which is spookily accurate when it come to fuel consumption.”
The above paragraph is unfamiliar to me.
**Eniclyth, can you expand a bit please?***
I’ve always taken the total CD as the sum of the three coefficients of drag, using a representative area as simply the wing area, ie Total Drag D = .5 * rho * V squared * wing area * CD
“cdi the induced drag is the square of the lift coefft divided by pi x aspect ratio x Oswald's "e". For the DC10 "e" was 0.83, the 744 "e" is 0.86 and the 777 "e" is 0.89 about as high as you can go.”
Perhaps I’d add that the aspect ratio and e factor are both in the denominator, ie CDI=(CL squared)/(pi * e * AR)
The math that produced the above was dependant on a perfectly elliptical lift distribution along the span, ie zero at the tips, increasing elliptically to the root. The spitfire is often cited as an example, but I believe it only had an elliptical wing planform, and not an elliptical lift distribution, although the planform would have helped it attain a near elliptical distribution. Some factors that reduce this perfect elliptical distribution are changing airfoil shape across the span (chord and camber), twist, and of course straight leading and trailing edges and tips (simpler to manufacturer). And so…. The Oswald efficiency factor e was included to allow for these real life affects. Typically ranges as per E’s post.
“cdo for a DC9 is 0.02, a 737 0.0176, a 777 0.013. In cruise when lift coeffts go above 0.55 the aircraft tends to have little buffet margin left so the highest first cruise altitude is usually the density altitude where cl = 0.55 for the chosen speed. likewise when weight falls off in cruise you will step up when the new cl is 0.55. eventually you cannot step up any more and cl rolls off. when it gets down to about 0.4 compressibilty has dropped from about 7-7.5% to 5-6%, speed and weight and altitude dependent. 737NGs, 777s, A330/340s can run at L/D of 19-ish when cl is 0.5-0.55; 744s 17.5. “
I think the above is straight forward
“specific fuel consumption is 0.65lb/lb-hr for 737NG at FL350 cl 0.5-0.55, 0.59 for a 744 and 0.57 for a 777. the distance you travel in cruise is (v knots/sfc) x L/D x natural logarithm(start weight/finish weight). So your A340-500 at 480 kts FL350 will travel 20.5 nm per 1000lb from the equation distance = (480/.587)*19.55*0.001283 assuming I am right in guessing compressibility at 7%. I made cl 0.522 and cdtot 0.0267 which is the 19.55 term in above equation. It feels right to me but I can only speak with authority on the 777 and the 744.
NB I forgot to say what weight didn't I ?...780000lb!!! When the weight falls to 480000lb expect 28.2nm/1000”
Above is where the math starts. **I think** Enicalyth has figured Drag from .5 rho S CD Vsquared. In cruise, this equals thrust required. Then knowing lbs of fuel burned per lb of thrust generated, per hour, he figures out fuel consumption. I didn't think one needed Breguet’s range formula for a jet aircraft but E's use on natural logs perhaps indicates so. I’ve yet to find a quiet few hours to do my math to come up with E’s numbers.
If you’ve got a calculator in the cockpit, you’ve really no need for flight manual data.
Make any sense Dan?
Enicalyth, sorry you got no reply on your 737 data request. Of course, I'd of helped, but I figured C172 data wasn't what you needed.
Hawk
Join Date: May 2003
Posts: 409
Likes: 0
Received 0 Likes
on
0 Posts
Oh, and Dan
I forgot to mention, I don't think you can use that CL = .55 value given by Eniclyth for any calculations. You should calculate it for accuracy. That's what weight is for. ie
W=L=.5 * Vsquared*rho*S*CL
Hawk
I forgot to mention, I don't think you can use that CL = .55 value given by Eniclyth for any calculations. You should calculate it for accuracy. That's what weight is for. ie
W=L=.5 * Vsquared*rho*S*CL
Hawk
Join Date: Jul 2004
Location: Sydney NSW
Posts: 513
Likes: 0
Received 0 Likes
on
0 Posts
its a drag, hawk
Yeah, total drag. Calculate the surface area of the whole aircraft and say it comes to 21000sq ft. Now paint it so that it has a surface roughness coefficient of say 0.0036. Then its flat plate area is meant to be 0.0036 x 21000 = 76 sq ft. Your inspectors look at the paint finish through a microscope and say “Yes that matches the definition of a 0.0036 finish!” Now measure the drag coefficient cdo of the whole plane for real and it comes in at mebbe 15% more!! That is the fiddle factor because aerials stick out, windows and doors aren’t flush and there are gaps here and there. You can now apply the fiddle factor and say that the effective wetted area is 1.15 x 21000 = 24150 sq ft and the effective flat plate area is 1.15 x 76= 87 sq ft. Let’s say your aircraft is a 744 for which Boeing tell you the wing reference area is 5825 sq ft. What is the real drag coefficient cdo? It is 87/5825 = 0.015! That's a bit rushed but okay for the back of a fag packet. And now a last word on the 777 where aspect ratio is 8.4, “e” is 0.89 wing area is 4605sq ft, parasite drag is 0.013 and compressibility drag is say 7%. You are at FL330 in still air on a standard day clocking 488kts and weigh 550000lbs. I make my lift coefft 0.442 therefore my induced coefft is 0.0083 and my compressible coefft is 0.0015. So L/D is 0.442/(0.013+0.0083+0.0015) =19.4 and my drag is 550000/19.4 = 28350lb. Fuel flow is sfc x drag = 0.57 x 28350 outta the service shop = 16160lb/hr or 7.33 tonnes per hour. Is it? Yes. I’m a happy bunny. Isn’t it? I find out why. Currently I estimate my sfc at 0.587 and the paint job is nearer 0.0042 than 0.0036 in finish coefft. Not useless knowledge, just the tools to give the Enicalyth the edge and the courage to say this baby needs a service and a scrub up.
Join Date: May 2003
Posts: 409
Likes: 0
Received 0 Likes
on
0 Posts
Enicalyth, thanks for the reply, and apologies for being so slow myself.
I “did the math” for the 340-500 and came up with CL of .525 (very close to your value of .522), 19.45 for L/D (again, very close to your 19.55 value) and 20.4 nm per 1000 lb at 780,000 lb (vice your 20.5).
Reference your “last word on the 777”, after 2 pages of calculations, I managed to generate your figures (only needed one piece of info you didn’t include; density at 33,000 ft). Not sure if I could do it in the cockpit though.
What still puzzles me is your explanation (in 2 different posts) of the use of total wetted area in the determination cdtot. First, if we consider D=.5 rho V squared S Cdtot where S is wing area, then this seems to determine Cdtot. Yet your quote
“Cdtot is the sum of all wetted areas plus fiddle factor time finishing coefficient...”
seems to be re defining cdtot.
Second, to
“then divide by wing area to get cdo”
again seems to be a paradox since cdo is already defined by cdtot=cdo + cdi + cdcc.
Can you help please?
Hawk
I “did the math” for the 340-500 and came up with CL of .525 (very close to your value of .522), 19.45 for L/D (again, very close to your 19.55 value) and 20.4 nm per 1000 lb at 780,000 lb (vice your 20.5).
Reference your “last word on the 777”, after 2 pages of calculations, I managed to generate your figures (only needed one piece of info you didn’t include; density at 33,000 ft). Not sure if I could do it in the cockpit though.
What still puzzles me is your explanation (in 2 different posts) of the use of total wetted area in the determination cdtot. First, if we consider D=.5 rho V squared S Cdtot where S is wing area, then this seems to determine Cdtot. Yet your quote
“Cdtot is the sum of all wetted areas plus fiddle factor time finishing coefficient...”
seems to be re defining cdtot.
Second, to
“then divide by wing area to get cdo”
again seems to be a paradox since cdo is already defined by cdtot=cdo + cdi + cdcc.
Can you help please?
Hawk
Join Date: Dec 2003
Location: Dubai
Posts: 212
Likes: 0
Received 0 Likes
on
0 Posts
What enicalyth means is that:
If the guys from the deicing ask you “Capt there is a bit of hoar frost all over the plane; do you want us to deice just the wing, stab and rudder ?”
You better tell them to do the whole plane or you may run out of fuel in the middle over the atlantic.
If the guys from the deicing ask you “Capt there is a bit of hoar frost all over the plane; do you want us to deice just the wing, stab and rudder ?”
You better tell them to do the whole plane or you may run out of fuel in the middle over the atlantic.
Join Date: Sep 2002
Location: Europe
Posts: 182
Likes: 0
Received 0 Likes
on
0 Posts
> Also can anyone give me an indication of the A340-600 fuel consumption per hour in cruise please – total all engines ?
I am afraid nobody can do that... Have a look at the (poor) fuel flow xmitter accuracy (ATA 73 TFU)...
LDG_GEAR _MONITOR: A340-600s can handle approx. 159 tonnes.
Take the A340-600
away from me, and give me an A340-300/A330-200 
!
J.V.
I am afraid nobody can do that... Have a look at the (poor) fuel flow xmitter accuracy (ATA 73 TFU)...
LDG_GEAR _MONITOR: A340-600s can handle approx. 159 tonnes.
Take the A340-600
away from me, and give me an A340-300/A330-200 !J.V.
