Let me pick your brains...
03-15-2007, 12:15 PM
#1
Gets Weekends Off
Thread Starter
Joined APC: Oct 2006
Position: pǝʇɹǝʌuı
Posts: 228
Let me pick your brains...
Hey all,
I'm doing a project and need some information. Here are the guidelines for the project.
B-744 w/ 15,000 lbs of fuel
How far can you fly and land safely. FAR's do NOT have to apply.
Basically the task is to take off, fly as long as possible and land. All engines must be running when you have fuel. IE you can't fly on two engines only but you can run out of gas and glide to your airport.
I'm going over performance and aero stuff to figure out what speed to climb at, how high to climb, and what speed to descend at.
So, what are your thoughts on how I should plan to go about doing this? What can I get away with turning off, what don't I need to fly, etc etc.
Thanks in advance!
I'm doing a project and need some information. Here are the guidelines for the project.
B-744 w/ 15,000 lbs of fuel
How far can you fly and land safely. FAR's do NOT have to apply.
Basically the task is to take off, fly as long as possible and land. All engines must be running when you have fuel. IE you can't fly on two engines only but you can run out of gas and glide to your airport.
I'm going over performance and aero stuff to figure out what speed to climb at, how high to climb, and what speed to descend at.
So, what are your thoughts on how I should plan to go about doing this? What can I get away with turning off, what don't I need to fly, etc etc.
Thanks in advance!
03-16-2007, 04:22 PM
#2
Flying Farmer
Joined APC: Jul 2006
Position: Turbo-props' and John Deere's
Posts: 3,160
Would it even make it from the gate to the runway with that much fuel
Seriously though, I wouldn't be suprised if that much fuel will get you to the top of your climb, or you wouldn't be there for long. Not sure what that big boy burns an hour, but figure even more for the climb phase, and the fact that in turbines, going to altitude will always increase your range regardless, instead of leveling off at a lower altitude, I'd imagine that'll be the proper technique.
Seriously though, I wouldn't be suprised if that much fuel will get you to the top of your climb, or you wouldn't be there for long. Not sure what that big boy burns an hour, but figure even more for the climb phase, and the fact that in turbines, going to altitude will always increase your range regardless, instead of leveling off at a lower altitude, I'd imagine that'll be the proper technique.
03-17-2007, 05:03 AM
#4
Gets Weekends Off
Thread Starter
Joined APC: Oct 2006
Position: pǝʇɹǝʌuı
Posts: 228
it has nothing to do with the environment...i'm just talking about the airplane besides the 15,000 lbs of gas. literally i need to run at the highest peak of efficiency...and then coast into a landing. so the question remains...what all can i get away with turning off? and how should i get to destination x?
03-17-2007, 08:50 AM
#5
Flying Farmer
Joined APC: Jul 2006
Position: Turbo-props' and John Deere's
Posts: 3,160
No offense man, but this question is Wayyyyyyyyyy Left field, hence the lack of response. You need the following info, and maybe you should just ask for this
Climb-Rate(@ lowest weight)
Climb-Fuel-Burn-Rate
Climb-Speed
Glide-Ratio
Glide-Speed
Climb-Rate(@ lowest weight)
Climb-Fuel-Burn-Rate
Climb-Speed
Glide-Ratio
Glide-Speed
03-17-2007, 10:52 AM
#6
Gets Weekends Off
Joined APC: Dec 2005
Position: FO, looking left
Posts: 1,057
I have a background in aerospace engineering. What I will tell you below will give you the exact answer and a certain A+ on the project. However, I do realize it will be tough to understand. So I am going to try and step through this as simply as possible. If it is too tough to understand, I can give you another model that is not exact, but certainly reasonable based upon assumptions.
For simplicity, lets assume standard temperature/pressure, and no wind. In this condition, here is the mission profile I would recommend. First, climb to the top of the trophosphere (called the trophopause). That is at approximately 36,000ft. You would cruise there until you run out of fuel, and then glide at best L/D until you reach the ground.
I think it is easier to work backwards in this problem. So the first thing I will walk you through is the power off glide:
First you would slow to best L/D speed. Then the aircraft would be put at the maximum range glide angle. This can be solved using.
Tan(theta) = 1/(Lift/Drag)
You can either research the maximum L/D ratio for the 747, or you can solve for it exactly. If you want to solve for max L/D exactly, I quoted it below.
Once you solve for the best angle theta, you would use:
Distance = Altitude / Tan(theta)
This will give you a glide distance if feet along the ground. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
Next is the cruise portion:
This equation we will use is called the "Brequet Range Formula."
You need to research:
1. Wing area of the 747 (ft^2). I found 5500ft^2
2. Empty weight of the 747. I found 393,000lbs.
3. The thrust specific fuel consumption (C_t).
If you cannnot find this value for C_t (it is often tough to find), you should research fuel burn per hour and thrust of the engines. The Pratt and Whitney engines have 63,300 lbf of thrust each (times 4 engines).
To solve for thrust specific fuel consumption, use the equation:
TSFC = (fuel burn per hour of all 4 engines)/ (lbs of thrust of all 4 engines).
The density at 36,000ft is .00071.
Weight initial = Empty weight of 747 + 15,000lbs of gas - fuel burned up during climb (standby, we will solve for this below, then come back up here)
Weight final = Empty weight of 747
We can now plug these into the equation.
Range = 2* sqrt(2 / (density*wingarea) ) * (1/TSFC) * (Cl^.5 / Cd) * ( (Weight initial)^.5 - (Weight final)^.5 )
This will give you a range in feet. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
The really tricky part is the climb. I would recommend doing one of two things. The easiest and best is to ask someone what the fuel burn of 747 is up to 36,000ft. The harder way is to solve for it exactly. I posted how to do it below, however it is VERY tricky. If you can't figure this out, just use a guess for fuel burned in climb, and how far the airplane traveled during climb.
CLIMB:
You need to look up best rate of climb of a 747 (also called Vy). You will need to convert it to ft/s. 1mph = 1.466 ft/s, and 1kt = 1.27ft/s. Every calculation below for velocity should be for Vy in ft/s.
Rate of climb= (Power available - Power required) / Weight
Power required = Thrust required * velocity
Thrust required = Weight / (Lift/Drag)
You can either look up best L/D for the 747, or you can calculate it (this is shown in the quoted box above. It is the same max L/D ratio).
Power available = Thrust available * velocity
The Pratt and Whitney engines have 63,300 lbf of thrust available each (times 4 engines).
Using the rate of climb formula above, you can solve for rate of climb of 747 at sea level. This answer will be in ft/s.
To solve for distance traveled during the climb, we need to first solve for angle of climb. Use the equation:
Sin(theta) = (Rate of climb) / (velocity)
Solve for theta
Distance traveled during climb = velocity*Cos(theta)
This is in feet. Divide by 5280 to get to miles, or divide by 6000 to get nautical miles.
Lastly, we need to solve for fuel burn during climb.
Fuel burn per second = (TSFC * thrust available)/3600
To find out how many seconds it takes to climb to 36,000ft, use:
Time to climb = 36,000/ (rate of climb)
Fuel burn during climb = (fuel burn per second)*(time to climb)
Now, go up above to the cruise portion and subtract fuel burn during climb to get weight initial of climb.
You now have distance during climb, distance during cruise, and distance during glide. Add them up and you have your answer.
Error in the model
In this ideal model, you would use a computer program that would compute rate of climb, L/D, thrust available, and thrust required at all altitudes between 0ft and 36,000ft. However, that is way too time consuming for me to mention on here. This error takes rate of climb at sea level, and assumes it is constant to altitude. Bad, I know, but the most accurate you can get without using a computer. The cruise portion and power off glide portion are almost exact. The problem is with the climb portion. If you could figure out both:
1. How much fuel a 747 burns climbing to 36,000ft
2. What distance it travels during that time
you could ignore the whole climb section and just use those variables. However, the cruise portion and gliding portion are pretty realistic.
Hope this helps. Let me know if you need some clarification.
Ryan
For simplicity, lets assume standard temperature/pressure, and no wind. In this condition, here is the mission profile I would recommend. First, climb to the top of the trophosphere (called the trophopause). That is at approximately 36,000ft. You would cruise there until you run out of fuel, and then glide at best L/D until you reach the ground.
I think it is easier to work backwards in this problem. So the first thing I will walk you through is the power off glide:
First you would slow to best L/D speed. Then the aircraft would be put at the maximum range glide angle. This can be solved using.
Tan(theta) = 1/(Lift/Drag)
You can either research the maximum L/D ratio for the 747, or you can solve for it exactly. If you want to solve for max L/D exactly, I quoted it below.
Once you solve for the best angle theta, you would use:
Distance = Altitude / Tan(theta)
This will give you a glide distance if feet along the ground. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
If you truly wanted to calculate Lift/Drag ratio at Vy, you would need to look up the aspect ratio (AR), oswald efficiency (e), and Cd,o.
Coefficient of lift = Weight / (dynamic pressure*wing area)
Dynamic pressure = .5*density*velocity^2 [again, velocity is your Vy]
Density at sea level is .0023769
Coefficient of drag = Cd,o + (Coefficient of Lift)^2 / (3.14*e*AR)
Lift/Drag = Coefficient of lift/Coefficient of drag
You could do this, but I would recommend just looking up best L/D ratio.
Coefficient of lift = Weight / (dynamic pressure*wing area)
Dynamic pressure = .5*density*velocity^2 [again, velocity is your Vy]
Density at sea level is .0023769
Coefficient of drag = Cd,o + (Coefficient of Lift)^2 / (3.14*e*AR)
Lift/Drag = Coefficient of lift/Coefficient of drag
You could do this, but I would recommend just looking up best L/D ratio.
This equation we will use is called the "Brequet Range Formula."
You need to research:
1. Wing area of the 747 (ft^2). I found 5500ft^2
2. Empty weight of the 747. I found 393,000lbs.
3. The thrust specific fuel consumption (C_t).
If you cannnot find this value for C_t (it is often tough to find), you should research fuel burn per hour and thrust of the engines. The Pratt and Whitney engines have 63,300 lbf of thrust each (times 4 engines).
To solve for thrust specific fuel consumption, use the equation:
TSFC = (fuel burn per hour of all 4 engines)/ (lbs of thrust of all 4 engines).
The density at 36,000ft is .00071.
Weight initial = Empty weight of 747 + 15,000lbs of gas - fuel burned up during climb (standby, we will solve for this below, then come back up here)
Weight final = Empty weight of 747
We can now plug these into the equation.
Range = 2* sqrt(2 / (density*wingarea) ) * (1/TSFC) * (Cl^.5 / Cd) * ( (Weight initial)^.5 - (Weight final)^.5 )
This will give you a range in feet. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
The really tricky part is the climb. I would recommend doing one of two things. The easiest and best is to ask someone what the fuel burn of 747 is up to 36,000ft. The harder way is to solve for it exactly. I posted how to do it below, however it is VERY tricky. If you can't figure this out, just use a guess for fuel burned in climb, and how far the airplane traveled during climb.
CLIMB:
You need to look up best rate of climb of a 747 (also called Vy). You will need to convert it to ft/s. 1mph = 1.466 ft/s, and 1kt = 1.27ft/s. Every calculation below for velocity should be for Vy in ft/s.
Rate of climb= (Power available - Power required) / Weight
Power required = Thrust required * velocity
Thrust required = Weight / (Lift/Drag)
You can either look up best L/D for the 747, or you can calculate it (this is shown in the quoted box above. It is the same max L/D ratio).
Power available = Thrust available * velocity
The Pratt and Whitney engines have 63,300 lbf of thrust available each (times 4 engines).
Using the rate of climb formula above, you can solve for rate of climb of 747 at sea level. This answer will be in ft/s.
To solve for distance traveled during the climb, we need to first solve for angle of climb. Use the equation:
Sin(theta) = (Rate of climb) / (velocity)
Solve for theta
Distance traveled during climb = velocity*Cos(theta)
This is in feet. Divide by 5280 to get to miles, or divide by 6000 to get nautical miles.
Lastly, we need to solve for fuel burn during climb.
Fuel burn per second = (TSFC * thrust available)/3600
To find out how many seconds it takes to climb to 36,000ft, use:
Time to climb = 36,000/ (rate of climb)
Fuel burn during climb = (fuel burn per second)*(time to climb)
Now, go up above to the cruise portion and subtract fuel burn during climb to get weight initial of climb.
You now have distance during climb, distance during cruise, and distance during glide. Add them up and you have your answer.
Error in the model
In this ideal model, you would use a computer program that would compute rate of climb, L/D, thrust available, and thrust required at all altitudes between 0ft and 36,000ft. However, that is way too time consuming for me to mention on here. This error takes rate of climb at sea level, and assumes it is constant to altitude. Bad, I know, but the most accurate you can get without using a computer. The cruise portion and power off glide portion are almost exact. The problem is with the climb portion. If you could figure out both:
1. How much fuel a 747 burns climbing to 36,000ft
2. What distance it travels during that time
you could ignore the whole climb section and just use those variables. However, the cruise portion and gliding portion are pretty realistic.
Hope this helps. Let me know if you need some clarification.
Ryan
03-17-2007, 11:07 AM
#7
Gets Weekends Off
Joined APC: Oct 2006
Position: G2 gear slammer
Posts: 308
Originally Posted by ryane946
I have a background in aerospace engineering. What I will tell you below will give you the exact answer and a certain A+ on the project. However, I do realize it will be tough to understand. So I am going to try and step through this as simply as possible. If it is too tough to understand, I can give you another model that is not exact, but certainly reasonable based upon assumptions.
For simplicity, lets assume standard temperature/pressure, and no wind. In this condition, here is the mission profile I would recommend. First, climb to the top of the trophosphere (called the trophopause). That is at approximately 36,000ft. You would cruise there until you run out of fuel, and then glide at best L/D until you reach the ground.
I think it is easier to work backwards in this problem. So the first thing I will walk you through is the power off glide:
First you would slow to best L/D speed. Then the aircraft would be put at the maximum range glide angle. This can be solved using.
Tan(theta) = 1/(Lift/Drag)
You can either research the maximum L/D ratio for the 747, or you can solve for it exactly. If you want to solve for max L/D exactly, I quoted it below.
Once you solve for the best angle theta, you would use:
Distance = Altitude / Tan(theta)
This will give you a glide distance if feet along the ground. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
Next is the cruise portion:
This equation we will use is called the "Brequet Range Formula."
You need to research:
1. Wing area of the 747 (ft^2). I found 5500ft^2
2. Empty weight of the 747. I found 393,000lbs.
3. The thrust specific fuel consumption (C_t).
If you cannnot find this value for C_t (it is often tough to find), you should research fuel burn per hour and thrust of the engines. The Pratt and Whitney engines have 63,300 lbf of thrust each (times 4 engines).
To solve for thrust specific fuel consumption, use the equation:
TSFC = (fuel burn per hour of all 4 engines)/ (lbs of thrust of all 4 engines).
The density at 36,000ft is .00071.
Weight initial = Empty weight of 747 + 15,000lbs of gas - fuel burned up during climb (standby, we will solve for this below, then come back up here)
Weight final = Empty weight of 747
We can now plug these into the equation.
Range = 2* sqrt(2 / (density*wingarea) ) * (1/TSFC) * (Cl^.5 / Cd) * ( (Weight initial)^.5 - (Weight final)^.5 )
This will give you a range in feet. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
The really tricky part is the climb. I would recommend doing one of two things. The easiest and best is to ask someone what the fuel burn of 747 is up to 36,000ft. The harder way is to solve for it exactly. I posted how to do it below, however it is VERY tricky. If you can't figure this out, just use a guess for fuel burned in climb, and how far the airplane traveled during climb.
CLIMB:
You need to look up best rate of climb of a 747 (also called Vy). You will need to convert it to ft/s. 1mph = 1.466 ft/s, and 1kt = 1.27ft/s. Every calculation below for velocity should be for Vy in ft/s.
Rate of climb= (Power available - Power required) / Weight
Power required = Thrust required * velocity
Thrust required = Weight / (Lift/Drag)
You can either look up best L/D for the 747, or you can calculate it (this is shown in the quoted box above. It is the same max L/D ratio).
Power available = Thrust available * velocity
The Pratt and Whitney engines have 63,300 lbf of thrust available each (times 4 engines).
Using the rate of climb formula above, you can solve for rate of climb of 747 at sea level. This answer will be in ft/s.
To solve for distance traveled during the climb, we need to first solve for angle of climb. Use the equation:
Sin(theta) = (Rate of climb) / (velocity)
Solve for theta
Distance traveled during climb = velocity*Cos(theta)
This is in feet. Divide by 5280 to get to miles, or divide by 6000 to get nautical miles.
Lastly, we need to solve for fuel burn during climb.
Fuel burn per second = (TSFC * thrust available)/3600
To find out how many seconds it takes to climb to 36,000ft, use:
Time to climb = 36,000/ (rate of climb)
Fuel burn during climb = (fuel burn per second)*(time to climb)
Now, go up above to the cruise portion and subtract fuel burn during climb to get weight initial of climb.
You now have distance during climb, distance during cruise, and distance during glide. Add them up and you have your answer.
Error in the model
In this ideal model, you would use a computer program that would compute rate of climb, L/D, thrust available, and thrust required at all altitudes between 0ft and 36,000ft. However, that is way too time consuming for me to mention on here. This error takes rate of climb at sea level, and assumes it is constant to altitude. Bad, I know, but the most accurate you can get without using a computer. The cruise portion and power off glide portion are almost exact. The problem is with the climb portion. If you could figure out both:
1. How much fuel a 747 burns climbing to 36,000ft
2. What distance it travels during that time
you could ignore the whole climb section and just use those variables. However, the cruise portion and gliding portion are pretty realistic.
Hope this helps. Let me know if you need some clarification.
Ryan
For simplicity, lets assume standard temperature/pressure, and no wind. In this condition, here is the mission profile I would recommend. First, climb to the top of the trophosphere (called the trophopause). That is at approximately 36,000ft. You would cruise there until you run out of fuel, and then glide at best L/D until you reach the ground.
I think it is easier to work backwards in this problem. So the first thing I will walk you through is the power off glide:
First you would slow to best L/D speed. Then the aircraft would be put at the maximum range glide angle. This can be solved using.
Tan(theta) = 1/(Lift/Drag)
You can either research the maximum L/D ratio for the 747, or you can solve for it exactly. If you want to solve for max L/D exactly, I quoted it below.
Once you solve for the best angle theta, you would use:
Distance = Altitude / Tan(theta)
This will give you a glide distance if feet along the ground. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
Next is the cruise portion:
This equation we will use is called the "Brequet Range Formula."
You need to research:
1. Wing area of the 747 (ft^2). I found 5500ft^2
2. Empty weight of the 747. I found 393,000lbs.
3. The thrust specific fuel consumption (C_t).
If you cannnot find this value for C_t (it is often tough to find), you should research fuel burn per hour and thrust of the engines. The Pratt and Whitney engines have 63,300 lbf of thrust each (times 4 engines).
To solve for thrust specific fuel consumption, use the equation:
TSFC = (fuel burn per hour of all 4 engines)/ (lbs of thrust of all 4 engines).
The density at 36,000ft is .00071.
Weight initial = Empty weight of 747 + 15,000lbs of gas - fuel burned up during climb (standby, we will solve for this below, then come back up here)
Weight final = Empty weight of 747
We can now plug these into the equation.
Range = 2* sqrt(2 / (density*wingarea) ) * (1/TSFC) * (Cl^.5 / Cd) * ( (Weight initial)^.5 - (Weight final)^.5 )
This will give you a range in feet. Divide by 5280 to get miles, or divide by 6000 to get nautical miles.
The really tricky part is the climb. I would recommend doing one of two things. The easiest and best is to ask someone what the fuel burn of 747 is up to 36,000ft. The harder way is to solve for it exactly. I posted how to do it below, however it is VERY tricky. If you can't figure this out, just use a guess for fuel burned in climb, and how far the airplane traveled during climb.
CLIMB:
You need to look up best rate of climb of a 747 (also called Vy). You will need to convert it to ft/s. 1mph = 1.466 ft/s, and 1kt = 1.27ft/s. Every calculation below for velocity should be for Vy in ft/s.
Rate of climb= (Power available - Power required) / Weight
Power required = Thrust required * velocity
Thrust required = Weight / (Lift/Drag)
You can either look up best L/D for the 747, or you can calculate it (this is shown in the quoted box above. It is the same max L/D ratio).
Power available = Thrust available * velocity
The Pratt and Whitney engines have 63,300 lbf of thrust available each (times 4 engines).
Using the rate of climb formula above, you can solve for rate of climb of 747 at sea level. This answer will be in ft/s.
To solve for distance traveled during the climb, we need to first solve for angle of climb. Use the equation:
Sin(theta) = (Rate of climb) / (velocity)
Solve for theta
Distance traveled during climb = velocity*Cos(theta)
This is in feet. Divide by 5280 to get to miles, or divide by 6000 to get nautical miles.
Lastly, we need to solve for fuel burn during climb.
Fuel burn per second = (TSFC * thrust available)/3600
To find out how many seconds it takes to climb to 36,000ft, use:
Time to climb = 36,000/ (rate of climb)
Fuel burn during climb = (fuel burn per second)*(time to climb)
Now, go up above to the cruise portion and subtract fuel burn during climb to get weight initial of climb.
You now have distance during climb, distance during cruise, and distance during glide. Add them up and you have your answer.
Error in the model
In this ideal model, you would use a computer program that would compute rate of climb, L/D, thrust available, and thrust required at all altitudes between 0ft and 36,000ft. However, that is way too time consuming for me to mention on here. This error takes rate of climb at sea level, and assumes it is constant to altitude. Bad, I know, but the most accurate you can get without using a computer. The cruise portion and power off glide portion are almost exact. The problem is with the climb portion. If you could figure out both:
1. How much fuel a 747 burns climbing to 36,000ft
2. What distance it travels during that time
you could ignore the whole climb section and just use those variables. However, the cruise portion and gliding portion are pretty realistic.
Hope this helps. Let me know if you need some clarification.
Ryan
You realize we are pilots right?
03-17-2007, 11:20 AM
#8
Gets Weekends Off
Joined APC: Dec 2005
Position: FO, looking left
Posts: 1,057
Originally Posted by bluebravo
You realize we are pilots right?
Plug and chug.
03-17-2007, 03:59 PM
#9
Gets Weekends Off
Joined APC: Jul 2005
Posts: 2,242
Originally Posted by oldveedubs
Hey all,
I'm doing a project and need some information. Here are the guidelines for the project.
B-744 w/ 15,000 lbs of fuel
How far can you fly and land safely. FAR's do NOT have to apply.
Basically the task is to take off, fly as long as possible and land. All engines must be running when you have fuel. IE you can't fly on two engines only but you can run out of gas and glide to your airport.
I'm going over performance and aero stuff to figure out what speed to climb at, how high to climb, and what speed to descend at.
So, what are your thoughts on how I should plan to go about doing this? What can I get away with turning off, what don't I need to fly, etc etc.
Thanks in advance!
I'm doing a project and need some information. Here are the guidelines for the project.
B-744 w/ 15,000 lbs of fuel
How far can you fly and land safely. FAR's do NOT have to apply.
Basically the task is to take off, fly as long as possible and land. All engines must be running when you have fuel. IE you can't fly on two engines only but you can run out of gas and glide to your airport.
I'm going over performance and aero stuff to figure out what speed to climb at, how high to climb, and what speed to descend at.
So, what are your thoughts on how I should plan to go about doing this? What can I get away with turning off, what don't I need to fly, etc etc.
Thanks in advance!
-LAFF
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