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walking
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What is wrong with saying that since the speed of fuel goes from 0 to -313m/s (rel to ground) in 1s, then its acceleration is -313m/s^2. This leads to wrong answer according to author's solution but I don't see why.
It's not just the fuel that is ejected !walking said:speed of fuel
The length of time that it takes to accelerate a particular lump of fuel from 0 [big hint, the fuel does not start at a ground-relative speed of 0] to -313 m/sec is utterly irrelevant here. The required acceleration for that lump is also utterly irrelevant here. Irrelevant, that is, unless you can get a handle on the amount of fuel that is being accelerated at a particular point in time.walking said:View attachment 279536
What is wrong with saying that since the speed of fuel goes from 0 to -313m/s (rel to ground) in 1s, then its acceleration is -313m/s^2. This leads to wrong answer according to author's solution but I don't see why.
I was inattentive , yes so any gases it ejects, but still gases does not get an acceleration of 313 m/s^2BvU said:It's not just the fuel that is ejected !
The approach is, perhaps, not complete nonsense, although the execution was borked. If one considers the fuel that is ejected in one second, the center of mass of said fuel begins at +184 m/s (ground relative) and ends at 313 m/s (ground-relative). The center of mass of said diffuse blob of fuel would have an acceleration given by ##\frac{\Delta v}{\Delta t}##. The ##\Delta v## would be 184+313 and the ##\Delta t## would be 1 sec.add314 said:I was inattentive , yes so any gases it ejects, but still gases does not get an acceleration of 313 m/s^2
Sorry about the lack of working and I appreciate everyone's detailed responses. I take down my questions as I read as I have limited internet time. Then I post down when I have written much later on, meaning I don't have time when I post them to actually go through the working for every single problem I have written a question for. I will definitely try to include more detail in my questions as I write them down next time so this problem doesn't occur.haruspex said:In case @jbriggs444 's comment is not quite clear, your calculation of the velocity change of the air is fine but not for the velocity change of the fuel burnt.
And you cannot derive its acceleration by dividing by 1s. You do not know how long it takes to accelerate any given molecule of exhaust. However, the correct analysis is likely to lead to an expression in which this Δv is in the numerator and the given 1s is in the denominator, so depending on how you use this "acceleration" you might still end up with a valid answer.
In short, please show all your working. (You could have done that in less time than the total taken by responders so far! )
Edit: corrected citation of earlier post.
As you have limited internet time, I would recommend to consider rate of change of momentum separetely for the mass of air and for the mass of fuel.walking said:Sorry about the lack of working and I appreciate everyone's detailed responses. I take down my questions as I read as I have limited internet time. Then I post down when I have written much later on, meaning I don't have time when I post them to actually go through the working for every single problem I have written a question for. I will definitely try to include more detail in my questions as I write them down next time so this problem doesn't occur.
I will try my best to include my working out for this problem as soon as possible.
I think this is what the author does but I still don't understand it. I think I am just misunderstanding the problem statement but here is my working out:Lnewqban said:As you have limited internet time, I would recommend to consider rate of change of momentum separetely for the mass of air and for the mass of fuel.
Regarding the change of velocity of the air respect to the engine:
##V_{out}-V_{in}=497-184=313~m/s##
Regarding the change of velocity of the fuel respect to the engine:
##V_{out}-V_{in}=497-0=497~m/s##
What is the change in momentum of the fuel?walking said:2.92(-497+184)
What is the net change in momentum of the air?walking said:70.2(184)
I honestly am still confused. I think it is a problem I am having with understanding the problem statement. I feel that this is one of those problems where if someone could explain to me the PS in detail I would find the problem trivial!haruspex said:What is the change in momentum of the fuel?
What is the net change in momentum of the air?
PS?walking said:I honestly am still confused. I think it is a problem I am having with understanding the problem statement. I feel that this is one of those problems where if someone could explain to me the PS in detail I would find the problem trivial!
Right I get it now. I wasn't taking into account that not only the fuel is ejected (I only realized after re-reading BvU's comment - thanks BvU!) Also for some reason I only just realized that the fuel is actually in the plane rather than somehow converted from the air into fuel!haruspex said:PS?
My questions were not hard:
What mass of fuel is burnt in each second? What was its velocity wrt ground before it was burnt? What was its velocity wrt ground after it was burnt? What was its change in velocity? What was its change in momentum?
Similarly for the air, but starting with its velocity before the plane draws it in.
BvU said:It's not just the fuel that is ejected !
The thrust of a jet plane is the force that propels the plane forward, created by the engines pushing air backwards.
-313m/s^2 is considered wrong because it is an unrealistic and extremely high value for the thrust of a jet plane. This value would result in the plane accelerating at an impossible rate and would not be able to sustain flight.
The typical range of thrust for a jet plane is between 50,000 to 100,000 pounds of force.
The thrust of a jet plane is typically measured in pounds or Newtons of force using a dynamometer or a thrust stand.
The thrust of a jet plane can be affected by factors such as air temperature, air density, altitude, and the condition and power of the engines.