Space Shuttle Launch and Sound Suppression

The Space Shuttle’s first flight (STS-1) in 1981 was considered a great success as almost all the technical and scientific goals were achieved. However, post flight analysis showed one potentially fatal problem: 16 heat shield tiles had been destroyed and another 148 damaged. How did that happen? The culprit was quickly determined to be sound. During launch the shuttle’s main engine and the SRBs (Solid Rocket Boosters) produce intense sound waves which cause strong vibrations. A sound suppression system was needed to protect the shuttle from acoustically induced damage such as cracks and mechanical fatigue. But how do you suppress the sound coming from a jet engine?

Let’s take a step back. What is the source of this sound? When the hot exhaust gas meets the ambient air, mixing occurs. This leads to the formation of a large number of eddies. The small-scale eddies close to the engine are responsible for high frequency noise, while the large-scale eddies that appear downstream cause intense low-frequency noise. Lighthill showed that the power P (in W) of the sound increases with the jet velocity v (in m/s) and the size s (in m) of the eddies:

P = K * D * c-5 * s2 * v8

with K being a constant, D the exhaust gas density and c the speed of sound. Note the extremely strong dependence of acoustic power on jet velocity: if you double the velocity, the power increases by a factor of 256. Such a strong relationship is very unusual in physics. The dependence on eddy size is also significant, doubling the size leads to a quadrupling in power. The formula tells us what we must do to effectively suppress sound: reduce jet velocity and the size of the eddies. Water injection into the exhaust gas achieves both. The water droplets absorb kinetic energy from the gas molecules, thus slowing them down. At the same time, the water breaks down the eddies.

During the second Space Shuttle launch (STS-2) a water injection system was used to suppress potentially catastrophic acoustic vibrations. This proved to be successful, it reduced the sound level by 10 – 20 dB (depending on location), and accordingly was used during every launch since then. But large amounts of water are needed to accomplish this reduction. The tank at the launch pad holds about 300,000 gallons. The flow starts at T minus 6.6 seconds and last for about 20 seconds. The peak flow rate is roughly 15,000 gallons per seconds. That’s a lot of water!

The video below shows a test run of the sound suppression system:

Sources and further reading:

Click to access art09.pdf


Click to access CAE_XUYue_Investigation-of-Flow-Control-with-Fluidic-injection-for-Jet-Noise-Reduction.pdf

Physics (And The Formula That Got Me Hooked)

A long time ago, in my teen years, this was the formula that got me hooked on physics. Why? I can’t say for sure. I guess I was very surprised that you could calculate something like this so easily. So with some nostalgia, I present another great formula from the field of physics. It will be a continuation of and a last section on energy.

To heat something, you need a certain amount of energy E (in J). How much exactly? To compute this we require three inputs: the mass m (in kg) of the object we want to heat, the temperature difference T (in °C) between initial and final state and the so called specific heat c (in J per kg °C) of the material that is heated. The relationship is quite simple:

E = c · m · T

If you double any of the input quantities, the energy required for heating will double as well. A very helpful addition to problems involving heating is this formula:

E = P · t

with P (in watt = W = J/s) being the power of the device that delivers heat and t (in s) the duration of the heat delivery.


The specific heat of water is c = 4200 J per kg °C. How much energy do you need to heat m = 1 kg of water from room temperature (20 °C) to its boiling point (100 °C)? Note that the temperature difference between initial and final state is T = 80 °C. So we have all the quantities we need.

E = 4200 · 1 · 80 = 336,000 J

Additional question: How long will it take a water heater with an output of 2000 W to accomplish this? Let’s set up an equation for this using the second formula:

336,000 = 2000 · t

t ≈ 168 s ≈ 3 minutes


We put m = 1 kg of water (c = 4200 J per kg °C) in one container and m = 1 kg of sand (c = 290 J per kg °C) in another next to it. This will serve as an artificial beach. Using a heater we add 10,000 J of heat to each container. By what temperature will the water and the sand be raised?

Let’s turn to the water. From the given data and the great formula we can set up this equation:

10,000 = 4200 · 1 · T

T ≈ 2.4 °C

So the water temperature will be raised by 2.4 °C. What about the sand? It also receives 10,000 J.

10,000 = 290 · 1 · T

T ≈ 34.5 °C

So sand (or any ground in general) will heat up much stronger than water. In other words: the temperature of ground reacts quite strongly to changes in energy input while water is rather sluggish. This explains why the climate near oceans is milder than inland, that is, why the summers are less hot and the winters less cold. The water efficiently dampens the changes in temperature.

It also explains the land-sea-breeze phenomenon (seen in the image below). During the day, the sun’s energy will cause the ground to be hotter than the water. The air above the ground rises, leading to cooler air flowing from the ocean to the land. At night, due to the lack of the sun’s power, the situation reverses. The ground cools off quickly and now it’s the air above the water that rises.


I hope this formula got you hooked as well. It’s simple, useful and can explain quite a lot of physics at the same time. It doesn’t get any better than this. Now it’s time to leave the concept of energy and turn to other topics.

This was an excerpt from my Kindle ebook: Great Formulas Explained – Physics, Mathematics, Economics. For another interesting physics quicky, check out: Intensity (or: How Much Power Will Burst Your Eardrums?).