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Multidisciplinary
Design, Analysis, and
Optimization Branch
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EDUCATIONAL ACTIVITIES: THE NASA AEROQUIZ
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Week of 7/2/01:
The United States is celebrating its Independence Day this week,
so it's time for an Aeroquiz question involving the US flag!
Q:
The late 1960s were turbulent times, and burning United States flags
was unfortunately commonplace. Indeed, NASA in those days took
extraordinary precautions to prevent several extraordinary US flags from
being burned. What were these flags and why were they so important?
A:
The Flags are those that are "flying" on the moon, having been erected
during the Apollo missions. The only place available to stow the flags
on the already-cramped lunar landers was on the ladder -- right next to the
engine exhaust! The housing, made of a stainless steel outer case
separated from an aluminum layer by Thermoflex insulation, had to
protect the flag from the 2000F temperatures that the ladder would
experience during descent. The flag itself could only withstand 300F
before damage ocurred.
Of course, the flag raising issue was much more than a technical
challenge: politics were involved!
Congratulations to Jim Griffin.
Week of 7/9/01:
Fellow NASA Glenn Engineer Chris Miller and I were at the
water cooler one day and came up with this week's question.
It's more complex than my usual Aeroquiz
questions, and it requires more than a little knowledge of math and
acoustics, but
the answer is really interesting!
Q:
"Hey, Chris, ever
wonder how much more powerful a big jet engine could
be if it didn't make all
that noise? I wonder how much power is
wasted through acoustics?"
"Well, Jeff, I think a couple of guys like us should be able
to figure that out." Chris drew a
mechanical pencil from his pocket
protector and began scribbling on the back of
an envelope. Luckily,
NASA has
envelopes strewn everywhere for occasions like these.
"Let's see, we'll have to make some assumptions. Let's consider a
Boeing 777 with two big
GE90 engines.
"Each making roughly 50 megawatts of useful power."
"And a 777 at
full throttle makes about 94 dB at seventeen hundred feet."
"And that's not bad, even by today's standards. Why don't you make it easier
and assume that it's a
uniform, omnidirectionally-radiating noise source?"
"Quiet, I'm on a roll now. Let's use standard atmospheric properties....
Oh. Well, now, that's certainly interesting."
What's so interesting?
Hints:
Sound Pressure Level = SPL = 10 log10(P2/Pref2),
where P2 is the root mean square sound wave pressure and Pref
= 20 mPa;
Sound Intensity Level = IL = 10 log10(I/Iref),
where I is the sound intensity and Iref =
10-12 W/m2
A:
I find that only about 2.1 kW goes into making that beautiful 94 dB
roar at 1700 ft. What's interesting is how incredibly small a
quantity that is compared to the useful energy produced by the
engines (roughly 0.002 percent, virtually nothing).
Also, I would speculate that this is why it is so difficult to
reduce noise emissions without severely impinging on useful energy
provided by the engine. If you assume a direct relationship between
the energy going to noise and useful energy produced, only a 1 db
reduction at 1700 feet, seems to require that the useful energy
provided be reduced to around 80 megawatts, a 20 percent reduction!
To further hinder the situation, aircraft engines are not
omnidirectional noise sources, but are louder when viewed from
certain angles, which means the component of energy going to
produce sound is probably even smaller than I figured.
Congratulations to "Ross."
Ross' answer differs from mine a by about a factor of four, but no one
else came anywhere close! In any case, everyone can check my math below:
The trick is to
relate the known decibel rating at 1700 ft with the
central acoustic power
source, which is radiating uniformly at some
unknown intensity.
The sound intensity can be related to the rms pressure in
the acoustic far field like this:
I = P2/rc,
where r is the ambient air
density (about 1.22 kg/m3) and c is the ambient speed of sound (about
343 m/s).
For a noise source W (in Watts) radiating uniformly
and omnidirectionally at radius r (in meters), I = W/4pr2 (W/m2).
Arranging terms, we get P2 = rcI = rcW/4pr2.
Now write SPL in terms of W:
SPL = 10 log10(Wrc/4pr2Pref2)
= 10 log10(1.05x1012 W/4pr2) = 120.2 + 10 log10(W/4pr2)
Solving for W, we get:
W = 4pr2
10(SPL-120.2)/10, where SPL is in dB, r is in
meters, and W is in Watts.
Plugging in 94 dB and 518 m (1700 ft), we get W = 8089 Watts.
So out of 100 megawatts of power generated by the twin GE90
engines, only about 8100 Watts ends up as noise! The acoustic power
is only about 0.008% of the overall power
being generated!
- The Aeroquiz Editor
Week of 7/16/01:
Second in a series of GE90-related questions!
Q:
Last week, we showed that the noise from a Boeing 777 (about 94 dB at 1700
feet) is generated by only about 8100 Watts of acoustic power. Surprisingly,
this amounts to only 0.008% of the total useful power made by the 777's twin
GE90 engines! One of the goals of NASA's noise reduction technology programs
is to "contain" aircraft noise within airport boundaries. How much would
acoustic power need to be reduced to limit noise to only 55 dB at a
typical 1700-foot airport boundary?
No one got the correct answer! The question stands another week!
- The Aeroquiz Editor
Week of 7/23/01:
Second in a series of GE90-related questions!
Q:
Last week, we showed that the noise from a Boeing 777 (about 94 dB at 1700
feet) is generated by only about 8100 Watts of acoustic power. Surprisingly,
this amounts to only 0.008% of the total useful power made by the 777's twin
GE90 engines! One of the goals of NASA's noise reduction technology programs
is to "contain" aircraft noise within airport boundaries. How much would
acoustic power need to be reduced to limit noise to only 55 dB at a
typical 1700-foot airport boundary?
A:
Using the equation developed from last week's answer, I calculate
the acoustic power needs to be reduced from 8100 Watts to about 1 Watt!
Congratulations to "glfras."
The equation from last week is W = 4pr2
10(SPL-120.2)/10, where SPL is in dB, r is in
meters, and W is in Watts. Although this is a simplistic calculation and many
effects are ignored, it nicely illustrates how enormous the technical challenge
is for aircraft noise reduction!
- The Aeroquiz Editor
Week of 7/23/01:
Third in a series of GE90-related questions!
Q:
In 1994 there were 107 nuclear power plants generating 96,637 MW of
power. How many GE90 engines are needed to match the power output of the
average nuclear plant?
A:
Unless this is a trick question, it appears that the answer lies
in the question itself. The question states that 107 nuclear plants
generate 96,637 MW. This means an average of 903 MW each. It should
be noted that this is 903 MW of electrical (useful) power. The
question for the week of 7/9/01 tells us that a GE90 engine generates
"roughly 50 MW of useful power".
Therefore, it would take only 18 GE90 engines to produce
the (useful) power output of of the average nuclear unit!
For the record, the typical power plant steam cycle has an efficiency
of approximately one-third. This means that the reactor core of the
average nuclear power plant generates approximately 2700 MW of thermal
power, with 1800 MW rejected as waste heat, leaving 900 MW of electrical
power to be delivered to the consumers.
Congratulations to A. R. Nies.
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