Perf Problems part 1
TightYorksherMan
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Perf Problems part 1
Here is a few performance questions i am having problems with.
Many thanks,
PERF = Two identical turbojet aeroplanes (whose SFC is assumed to be the constant) are in a holding pattern at the same altitude. The mass of the first one is 95,000kg and it hourly fuel consumption is equal to 3100kg/h. Since the mass of the second one is 105,000kg, its hourly fuel consumption is:
a) 3787kg/hr
b) 3426kg/hr
c) 3259kg/hr <- answer
d) 3602kg/hr
PERF = Using climb performance chart for SE aeroplane (fig 2.3 SEP1), determine the ground distance to reach a height of 2000ft above reference zero in the following conditions:
Given:
OAT at takeoff = 25C
Airport pressure alt = 1000ft
A/C mass = 3600lbs
Speed = 100KIAS
Wind component = 15kts headwind
a) 18832ft
b) 18347ft <- answer
c) 21505ft
d) 24637ft
I have used the graph and come out with a climb gradient of 9.5% and ROC of 1125ft/min
Having problems from now on!
Many thanks,
Jinkster
Many thanks,
PERF = Two identical turbojet aeroplanes (whose SFC is assumed to be the constant) are in a holding pattern at the same altitude. The mass of the first one is 95,000kg and it hourly fuel consumption is equal to 3100kg/h. Since the mass of the second one is 105,000kg, its hourly fuel consumption is:
a) 3787kg/hr
b) 3426kg/hr
c) 3259kg/hr <- answer
d) 3602kg/hr
PERF = Using climb performance chart for SE aeroplane (fig 2.3 SEP1), determine the ground distance to reach a height of 2000ft above reference zero in the following conditions:
Given:
OAT at takeoff = 25C
Airport pressure alt = 1000ft
A/C mass = 3600lbs
Speed = 100KIAS
Wind component = 15kts headwind
a) 18832ft
b) 18347ft <- answer
c) 21505ft
d) 24637ft
I have used the graph and come out with a climb gradient of 9.5% and ROC of 1125ft/min
Having problems from now on!
Many thanks,
Jinkster
Join Date: Apr 2003
Location: UK
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Perf Prob....
Jinkster,
this is a classic one, and no matter what ANYONE tells you, it is a simple ratio question.....
95,000 / 3100 = 30.64
105,000 / 30.64 = 3426Kg/hr
Now i know that this not tha answer that u have selected there but it is the correct one....this type of question always comes up, and has just been re worded....which feedback was it from?...I am sure that i am going to get alot 'comments' back about this post, that it is not correct and what not, but trust me on this, i have it on very good authority....its just a ratio...
keeno
this is a classic one, and no matter what ANYONE tells you, it is a simple ratio question.....
95,000 / 3100 = 30.64
105,000 / 30.64 = 3426Kg/hr
Now i know that this not tha answer that u have selected there but it is the correct one....this type of question always comes up, and has just been re worded....which feedback was it from?...I am sure that i am going to get alot 'comments' back about this post, that it is not correct and what not, but trust me on this, i have it on very good authority....its just a ratio...
keeno
Join Date: Aug 2001
Location: UK, In the middle at the bottom and sometimes in LHR!
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Jinkster,
A mate of mine did the PERF brush-up at Oxford, there is a similar question in there feedback and he shwed me how to work it out:
Two identical turbojets are at the same altitude and same speed and have the same specific fuel consumption. Plane 1 weighs 130000kg and has a fuel flow of 4300kg/hr. Plane 2 weighs 115000kg. what is the fuel flow?
a) 3804kg/hr
b) 4044kg/hr
c) 3364kg/hr <= correct answer
d) 3530kg/hr
(115000)2 / (130000)2 =0.7825 X 4300kg/hr = 3364kg/hr.
If you don't square the two weights you will get answer (a)
So with your question if you square the two weights you get answer (a),and if you don't you get answer (b).
Hope this will be of some help.
M
A mate of mine did the PERF brush-up at Oxford, there is a similar question in there feedback and he shwed me how to work it out:
Two identical turbojets are at the same altitude and same speed and have the same specific fuel consumption. Plane 1 weighs 130000kg and has a fuel flow of 4300kg/hr. Plane 2 weighs 115000kg. what is the fuel flow?
a) 3804kg/hr
b) 4044kg/hr
c) 3364kg/hr <= correct answer
d) 3530kg/hr
(115000)2 / (130000)2 =0.7825 X 4300kg/hr = 3364kg/hr.
If you don't square the two weights you will get answer (a)
So with your question if you square the two weights you get answer (a),and if you don't you get answer (b).
Hope this will be of some help.
M
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Jinkster,
Your first question is addressed in different ways by different schools.
The OATS method is to note that Cdi is proportional to Cl squared.
If speed stays constant then any % increase in weight will cause that% squared increase in induced drag. Fuel flow in a jet is proportional to thrust so the fuel flow will change in proportion to the change in drag.
So in this question we can use (105000/95000)squared = 1.2216. If we assume that all of the drag is increased by this amount then the new fuel flow = 3100 x 1.2216 = 3786.98 (option a).
But all of the above ignores the fact that total drag also includes profile drag, which does not vary with wieght.
The question states that the aircraft are in a holding pattern so we may assume that they are at something close to Vmd (max endurance speed). At this speed induced drag is half of the total drag. So we can say that half of the drag increases by a factor of 1.2216, while the other half stays unchanged.
In effect this will increase total drag by 0.2216/2 = 0.1108. So the total drag (and fuel flow) at the higher weight will be 1.1108 x that at the lower weight. so at 105000 Kg the aircarft will burn 3443 Kg/hr, which is closest to option b.
The alternative (simpler) solution is to consider the total drag and to note that for any small % change in weight at Vmd, there will be approximately that same % change in total darg and jet fuel flow. This gives the simple equation:
FF1/W1 = FF2/W2 for conditions 1 and 2
Rearranging this gives FF2 = FF1 (W2/W1)
Inserting the given numbers gives FF2 = 3100 (105000/95000) = 3426 (option b)
I have discussed this type of question with the examiners who have advised me to "consider all of the drag". So I believe that option b is what they are looking for.
For your second question you need to look at the method and equations on page 12of the CAP 698. I cannot get any of the options exactly, but option b is the closest
Your first question is addressed in different ways by different schools.
The OATS method is to note that Cdi is proportional to Cl squared.
If speed stays constant then any % increase in weight will cause that% squared increase in induced drag. Fuel flow in a jet is proportional to thrust so the fuel flow will change in proportion to the change in drag.
So in this question we can use (105000/95000)squared = 1.2216. If we assume that all of the drag is increased by this amount then the new fuel flow = 3100 x 1.2216 = 3786.98 (option a).
But all of the above ignores the fact that total drag also includes profile drag, which does not vary with wieght.
The question states that the aircraft are in a holding pattern so we may assume that they are at something close to Vmd (max endurance speed). At this speed induced drag is half of the total drag. So we can say that half of the drag increases by a factor of 1.2216, while the other half stays unchanged.
In effect this will increase total drag by 0.2216/2 = 0.1108. So the total drag (and fuel flow) at the higher weight will be 1.1108 x that at the lower weight. so at 105000 Kg the aircarft will burn 3443 Kg/hr, which is closest to option b.
The alternative (simpler) solution is to consider the total drag and to note that for any small % change in weight at Vmd, there will be approximately that same % change in total darg and jet fuel flow. This gives the simple equation:
FF1/W1 = FF2/W2 for conditions 1 and 2
Rearranging this gives FF2 = FF1 (W2/W1)
Inserting the given numbers gives FF2 = 3100 (105000/95000) = 3426 (option b)
I have discussed this type of question with the examiners who have advised me to "consider all of the drag". So I believe that option b is what they are looking for.
For your second question you need to look at the method and equations on page 12of the CAP 698. I cannot get any of the options exactly, but option b is the closest
