Tag Archives: velocity

It’s rocket science

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Here are six or so rocket science insights, some simple, some advanced. It’s a fun area of engineering that touches many areas of science and politics. Besides, some people seem to think I’m a rocket scientist.

A basic question I get asked by kids is how a rocket goes up. My answer is it does not go up. That’s mostly an illusion. The majority of the rocket — the fuel — goes down, and only the light shell goes up. People imagine they are seeing the rocket go up. Taken as a whole, fuel and shell, they both go down at 1 G: 9.8 m/s2, 32 ft/sec2.

Because 1 G ofupward acceleration is always lost to gravity, you need more thrust from the rocket engine than the weight of rocket and fuel. This can be difficult at the beginning when the rocket is heaviest. If your engine provides less thrust than the weight of your rocket, your rocket sits on the launch pad, burning. If your thrust is merely twice the weight of the rocket, you waste half of your fuel doing nothing useful, just fighting gravity. The upward acceleration you’ll see, a = F/m -1G where F is the force of the engine, and m is the mass of the rocket shell + whatever fuel is in it. 1G = 9.8m/s is the upward acceleration lost to gravity.  For model rocketry, you want to design a rocket engine so that the upward acceleration, a, is in the range 5-10 G. This range avoids wasting lots of fuel without requiring you to build the rocket too sturdy.

For NASA moon rockets, a = 0.2G approximately at liftoff increasing as fuel was used. The Saturn V rose, rather majestically, into the sky with a rocket structure that had to be only strong enough to support 1.2 times the rocket weight. Higher initial accelerations would have required more structure and bigger engines. As it was the Saturn V was the size of a skyscraper. You want the structure to be light so that the majority of weight is fuel. What makes it tricky is that the acceleration weight has to sit on an engine that gimbals (slants) and runs really hot, about 3000°C. Most engineering projects have fewer constraints than this, and are thus “not rocket science.”

Basic force balance on a rocket going up.

Basic force balance on a rocket going up.

A space rocket has to reach very high, orbital speed if the rocket is to stay up indefinitely, or nearly orbital speed for long-range, military uses. You can calculate the orbital speed by balancing the acceleration of gravity, 9.8 m/s2, against the orbital acceleration of going around the earth, a sphere of 40,000 km in circumference (that’s how the meter was defined). Orbital acceleration, a = v2/r, and r = 40,000,000 m/2π = 6,366,000m. Thus, the speed you need to stay up indefinitely is v=√(6,366,000 x 9.8) = 7900 m/s = 17,800 mph. That’s roughly Mach 35, or 35 times the speed of sound at sea level, (343 m/s). You need some altitude too, just to keep air friction from killing you, but for most missions, the main thing you need is velocity, kinetic energy, not potential energy, as I’ll show below. If your speed exceeds 17,800 m/s, you go higher up, but the stable orbital velocity is lower. The gravity force is lower higher up, and the radius to the earth higher too, but you’re balancing this lower gravity force against v2/r, so v2 has to be reduced to stay stable high up, but higher to get there. This all makes docking space-ships tricky, as I’ll explain also. Rockets are the only way practical to reach Mach 35 or anything near it. No current cannon or gun gets close.

Kinetic energy is a lot more important than potential energy for sending an object into orbit. To get a sense of the comparison, consider a one kg mass at orbital speed, 7900 m/s, and 200 km altitude. For these conditions, the kinetic energy, 1/2mv2 is 31,205 kJ, while the potential energy, mgh, is only 1,960 kJ . The potential energy is thus only 1/16 the kinetic energy.

Not that it’s easy to reach 200 miles altitude, but you can do it with a sophisticated cannon. The Germans did it with “simple”, one stage, V2-style rockets. To reach orbit, you generally need multiple stages. As a way to see this, consider that the energy content of gasoline + oxygen is about 10.5 MJ/kg (10,500 kJ/kg); this is only 1/3 of the kinetic energy of the orbital rocket, but it’s 5 times the potential energy. A fairly efficient gasoline + oxygen powered cannon could not provide orbital kinetic energy since the bullet can move no faster than the explosive vapor. In a rocket this is not a constraint since most of the mass is ejected.

A shell fired at a 45° angle that reaches 200 km altitude would go about 800 km — the distance between North Korea and Japan, or between Iran and Israel. That would require twice as much energy as a shell fired straight up, about 4000 kJ/kg. This is still within the range for a (very large) cannon or a single-stage rocket. For Russia or China to hit the US would take much more: orbital, or near orbital rocketry. To reach the moon, you need more total energy, but less kinetic energy. Moon rockets have taken the approach of first going into orbit, and only later going on. While most of the kinetic energy isn’t lost, it’s likely not the best trajectory in terms of energy use.

The force produced by a rocket is equal to the rate of mass shot out times its velocity. F = ∆(mv). To get a lot of force for each bit of fuel, you want the gas exit velocity to be as fast as possible. A typical maximum is about 2,500 m/s. Mach 10, for a gasoline – oxygen engine. The acceleration of the rocket itself is this ∆mv force divided by the total remaining mass in the rocket (rocket shell plus remaining fuel) minus 1 (gravity). Thus, if the exhaust from a rocket leaves at 2,500 m/s, and you want the rocket to accelerate upward at an average of 10 G, you must exhaust fast enough to develop 10 G, 98 m/s2. The rate of mass exhaust is the average mass of the rocket times 98/2500 = .0392/second. That is, about 3.92% of the rocket mass must be ejected each second. Assuming that the fuel for your first stage engine is less than 80% of the total mass, the first stage will flare-out in about 20 seconds. Typically, the acceleration at the end of the 20 burn is much greater than at the beginning since the rocket gets lighter as fuel is burnt. This was the case with the Apollo missions. The Saturn V started up at 0.5G but reached a maximum of 4G by the time most of the fuel was used.

If you have a good math background, you can develop a differential equation for the relation between fuel consumption and altitude or final speed. This is readily done if you know calculous, or reasonably done if you use differential methods. By either method, it turns out that, for no air friction or gravity resistance, you will reach the same speed as the exhaust when 64% of the rocket mass is exhausted. In the real world, your rocket will have to exhaust 75 or 80% of its mass as first stage fuel to reach a final speed of 2,500 m/s. This is less than 1/3 orbital speed, and reaching it requires that the rest of your rocket mass: the engine, 2nd stage, payload, and any spare fuel to handle descent (Elon Musk’s approach) must weigh less than 20-25% of the original weight of the rocket on the launch pad. This gasoline and oxygen is expensive, but not horribly so if you can reuse the rocket; that’s the motivation for NASA’s and SpaceX’s work on reusable rockets. Most orbital rocket designs require three stages to accelerate to the 7900 m/s orbital speed calculated above. The second stage is dropped from high altitude and almost invariably lost. If you can set-up and solve the differential equation above, a career in science may be for you.

Now, you might wonder about the exhaust speed I’ve been using, 2500 m/s. You’ll typically want a speed at lest this high as it’s associated with a high value of thrust-seconds per weight of fuel. Thrust seconds pre weight is called specific impulse, SI, SI = lb-seconds of thrust/lb of fuel. This approximately equals speed of exhaust (m/s) divided by 9.8 m/s2. For a high molecular weight burn it’s not easy to reach gas speed much above 2500, or values of SI much above 250, but you can get high thrust since thrust is related to momentum transfer. High thrust is why US and Russian engines typically use gasoline + oxygen. The heat of combustion of gasoline is 42 MJ/kg, but burning a kg of gasoline requires roughly 2.5 kg of oxygen. Thus, for a rocket fueled by gasoline + oxygen, the heat of combustion per kg is 42/3.5 = 12,000,000 J/kg. A typical rocket engine is 30% efficient (V2 efficiency was lower, Saturn V higher). Per corrected unit of fuel+oxygen mass, 1/2 v2 = .3 x 12,000,000; v =√7,200,000 = 2680 m/s. Adding some mass for the engine and fuel tanks, the specific impulse for this engine will be, about 250 s. This is fairly typical. Higher exhaust speeds have been achieved with hydrogen fuel, it has a higher combustion energy per weight. It is also possible to increase the engine efficiency; the Saturn V, stage 2 efficiency was nearly 50%, but the thrust was low. The sources of inefficiency include inefficiencies in compression, incomplete combustion, friction flows in the engine, and back-pressure of the atmosphere. If you can make a reliable, high efficiency engine with good lift, a career in engineering may be for you. A yet bigger challenge is doing this at a reasonable cost.

At an average acceleration of 5G = 49 m/s2 and a first stage that reaches 2500 m/s, you’ll find that the first stage burns out after 51 seconds. If the rocket were going straight up (bad idea), you’d find you are at an altitude of about 63.7 km. A better idea would be an average trajectory of 30°, leaving you at an altitude of 32 km or so. At that altitude you can expect to have far less air friction, and you can expect the second stage engine to be more efficient. It seems to me, you may want to wait another 10 seconds before firing the second stage: you’ll be 12 km higher up and it seems to me that the benefit of this will be significant. I notice that space launches wait a few seconds before firing their second stage.

As a final bit, I’d mentioned that docking a rocket with a space station is difficult, in part, because docking requires an increase in angular speed, w, but this generally goes along with a decrease in altitude; a counter-intuitive outcome. Setting the acceleration due to gravity equal to the angular acceleration, we find GM/r2 = w2r, where G is the gravitational constant, and M is the mass or the earth. Rearranging, we find that w2  = GM/r3. For high angular speed, you need small r: a low altitude. When we first went to dock a space-ship, in the early 60s, we had not realized this. When the astronauts fired the engines to dock, they found that they’d accelerate in velocity, but not in angular speed: v = wr. The faster they went, the higher up they went, but the lower the angular speed got: the fewer the orbits per day. Eventually they realized that, to dock with another ship or a space-station that is in front of you, you do not accelerate, but decelerate. When you decelerate you lose altitude and gain angular speed: you catch up with the station, but at a lower altitude. Your next step is to angle your ship near-radially to the earth, and accelerate by firing engines to the side till you dock. Like much of orbital rocketry, it’s simple, but not intuitive or easy.

Robert Buxbaum, August 12, 2015. A cannon that could reach from North Korea to Japan, say, would have to be on the order of 10 km long, running along the slope of a mountain. Even at that length, the shell would have to fire at 450 G, or so, and reach a speed about 3000 m/s, or 1/3 orbital.

The speed of sound, Buxbaum’s correction

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Ernst Mach showed that sound must travel at a particular speed through any material, one determined by the conservation of energy and of entropy. At room temperature and 1 atm, that speed is theoretically predicted to be 343 m/s. For a wave to move at any other speed, either the laws of energy conservation would have to fail, or ∆S ≠ 0 and the wave would die out. This is the only speed where you could say there is a traveling wave, and experimentally, this is found to be the speed of sound in air, to good accuracy.

Still, it strikes me that Mach’s assumptions may have been too restrictive for short-distance sound waves. Perhaps there is room for other sound speeds if you allow ∆S > 0, and consider sound that travels short distances and dies out far from the source. Waves at these, other speeds might affect music appreciation, or headphone design. As these waves were never treated in my thermodynamics textbooks, I wondered if I could derive their speed in any nice way, and if they were faster or slower than the main wave? (If I can’t use this blog to re-think my college studies, what good is it?)

I t can help to think of a shock-wave of sound wave moving down a constant area tube of still are at speed u, with us moving along at the same speed as the wave. In this view, the wave appears stationary, but there is a wind of speed u approaching it from the left.

Imagine the sound-wave moving to the right, down a constant area tube at speed u, with us moving along at the same speed. Thus, the wave appears stationary, with a wind of speed u from the right.

As a first step to trying to re-imagine Mach’s calculation, here is one way to derive the original, for ∆S = 0, speed of sound: I showed in a previous post that the entropy change for compression can be imagines to have two parts, a pressure part at constant temperature: dS/dV at constant T = dP/dT at constant V. This part equals R/V for an ideal gas. There is also a temperature at constant volume part of the entropy change: dS/dT at constant V = Cv/T. Dividing the two equations, we find that, at constant entropy, dT/dV = RT/CvV= P/Cv. For a case where ∆S>0, dT/dV > P/Cv.

Now lets look at the conservation of mechanical energy. A compression wave gives off a certain amount of mechanical energy, or work on expansion, and this work accelerates the gas within the wave. For an ideal gas the internal energy of the gas is stored only in its temperature. Lets now consider a sound wave going down a tube flow left to right, and lets our reference plane along the wave at the same speed so the wave seems to sit still while a flow of gas moves toward it from the right at the speed of the sound wave, u. For this flow system energy is concerned though no heat is removed, and no useful work is done. Thus, any change in enthalpy only results in a change in kinetic energy. dH = -d(u2)/2 = u du, where H here is a per-mass enthalpy (enthalpy per kg).

dH = TdS + VdP. This can be rearranged to read, TdS = dH -VdP = -u du – VdP.

We now use conservation of mass to put du into terms of P,V, and T. By conservation of mass, u/V is constant, or d(u/V)= 0. Taking the derivative of this quotient, du/V -u dV/V2= 0. Rearranging this, we get, du = u dV/V (No assumptions about entropy here). Since dH = -u du, we say that u2 dV/V = -dH = -TdS- VdP. It is now common to say that dS = 0 across the sound wave, and thus find that u2 = -V2 (dP/dV) at const S. For an ideal gas, this last derivative equals, PCp/VCv, so the speed of sound, u= √PVCp/Cv with the volume in terms of mass (kg/m3).

The problem comes in where we say that ∆S>0. At this point, I would say that u2 = -V(dH/dV) = VCp dT/dV > PVCp/Cv. Unless, I’ve made a mistake (always possible), I find that there is a small leading, non-adiabatic sound wave that goes ahead of the ordinary sound wave and is experienced only close to the source caused by mechanical energy that becomes degraded to raising T and gives rise more compression than would be expected for iso-entropic waves.

This should have some relevance to headphone design and speaker design since headphones are heard close to the ear, while speakers are heard further away. Meanwhile the recordings are made by microphones right next to the singers or instruments.

Robert E. Buxbaum, August 26, 2014

Heat conduction in insulating blankets, aerogels, space shuttle tiles, etc.

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A lot about heat conduction in insulating blankets can be explained by the ordinary motion of gas molecules. That’s because the thermal conductivity of air (or any likely gas) is much lower than that of glass, alumina, or any likely solid material used for the structure of the blanket. At any temperature, the average kinetic energy of an air molecule is 1/2kT in any direction, or 3/2kT altogether; where k is Boltzman’s constant, and T is absolute temperature, °K. Since kinetic energy equals 1/2 mv2, you find that the average velocity in the x direction must be v = √kT/m = √RT/M. Here m is the mass of the gas molecule in kg, M is the molecular weight also in kg (0.029 kg/mol for air), R is the gas constant 8.29J/mol°C, and v is the molecular velocity in the x direction, in meters/sec. From this equation, you will find that v is quite large under normal circumstances, about 290 m/s (650 mph) for air molecules at ordinary temperatures of 22°C or 295 K. That is, air molecules travel in any fixed direction at roughly the speed of sound, Mach 1 (the average speed including all directions is about √3 as fast, or about 1130 mph).

The distance a molecule will go before hitting another one is a function of the cross-sectional areas of the molecules and their densities in space. Dividing the volume of a mol of gas, 0.0224 m3/mol at “normal conditions” by the number of molecules in the mol (6.02 x10^23) gives an effective volume per molecule at this normal condition: .0224 m3/6.0210^23 = 3.72 x10^-26 m3/molecule at normal temperatures and pressures. Dividing this volume by the molecular cross-section area for collisions (about 1.6 x 10^-19 m2 for air based on an effective diameter of 4.5 Angstroms) gives a free-motion distance of about 0.23×10^-6 m or 0.23µ for air molecules at standard conditions. This distance is small, to be sure, but it is 1000 times the molecular diameter, more or less, and as a result air behaves nearly as an “ideal gas”, one composed of point masses under normal conditions (and most conditions you run into). The distance the molecule travels to or from a given surface will be smaller, 1/√3 of this on average, or about 1.35×10^-7m. This distance will be important when we come to estimate heat transfer rates at the end of this post.

 

Molecular motion of an air molecule (oxygen or nitrogen) as part of heat transfer process; this shows how some of the dimensions work.

Molecular motion of an air molecule (oxygen or nitrogen) as part of heat transfer process; this shows how some of the dimensions work.

The number of molecules hitting per square meter per second is most easily calculated from the transfer of momentum. The pressure at the surface equals the rate of change of momentum of the molecules bouncing off. At atmospheric pressure 103,000 Pa = 103,000 Newtons/m2, the number of molecules bouncing off per second is half this pressure divided by the mass of each molecule times the velocity in the surface direction. The contact rate is thus found to be (1/2) x 103,000 Pa x 6.02^23 molecule/mol /(290 m/s. x .029 kg/mol) = 36,900 x 10^23 molecules/m2sec.

The thermal conductivity is merely this number times the heat capacity transfer per molecule times the distance of the transfer. I will now calculate the heat capacity per molecule from statistical mechanics because I’m used to doing things this way; other people might look up the heat capacity per mol and divide by 6.02 x10^23: For any gas, the heat capacity that derives from kinetic energy is k/2 per molecule in each direction, as mentioned above. Combining the three directions, that’s 3k/2. Air molecules look like dumbbells, though, so they have two rotations that contribute another k/2 of heat capacity each, and they have a vibration that contributes k. We begin with an approximate value for k = 2 cal/mol of molecules per °C; it’s actually 1.987 but I round up to include some electronic effects. Based on this, we calculate the heat capacity of air to be 7 cal/mol°C at constant volume or 1.16 x10^-23 cal/molecule°C. The amount of energy that can transfer to the hot (or cold) wall is this heat capacity times the temperature difference that molecules carry between the wall and their first collision with other gases. The temperature difference carried by air molecules at standard conditions is only 1.35 x10-7 times the temperature difference per meter because the molecules only go that far before colliding with another molecule (remember, I said this number would be important). The thermal conductivity for stagnant air per meter is thus calculated by multiplying the number of molecules times that hit per m2 per second, the distance the molecule travels in meters, and the effective heat capacity per molecule. This would be 36,900 x 10^23  molecules/m2sec x 1.35 x10-7m x 1.16 x10^-23 cal/molecule°C = 0.00578 cal/ms°C or .0241 W/m°C. This value is (pretty exactly) the thermal conductivity of dry air that you find by experiment.

I did all that math, though I already knew the thermal conductivity of air from experiment for a few reasons: to show off the sort of stuff you can do with simple statistical mechanics; to build up skills in case I ever need to know the thermal conductivity of deuterium or iodine gas, or mixtures; and finally, to be able to understand the effects of pressure, temperature and (mainly insulator) geometry — something I might need to design a piece of equipment with, for example, lower thermal heat losses. I find, from my calculation that we should not expect much change in thermal conductivity with gas pressure at near normal conditions; to first order, changes in pressure will change the distance the molecule travels to exactly the same extent that it changes the number of molecules that hit the surface per second. At very low pressures or very small distances, lower pressures will translate to lower conductivity, but for normal-ish pressures and geometries, changes in gas pressure should not affect thermal conductivity — and does not.

I’d predict that temperature would have a larger effect on thermal conductivity, but still not an order-of magnitude large effect. Increasing the temperature increases the distance between collisions in proportion to the absolute temperature, but decreases the number of collisions by the square-root of T since the molecules move faster at high temperature. As a result, increasing T has a √T positive effect on thermal conductivity.

Because neither temperature nor pressure has much effect, you might expect that the thermal conductivity of all air-filed insulating blankets at all normal-ish conditions is more-or-less that of standing air (air without circulation). That is what you find, for the most part; the same 0.024 W/m°C thermal conductivity with standing air, with high-tech, NASA fiber blankets on the space shuttle and with the cheapest styrofoam cups. Wool felt has a thermal conductivity of 0.042 W/m°C, about twice that of air, a not-surprising result given that wool felt is about 1/2 wool and 1/2 air.

Now we can start to understand the most recent class of insulating blankets, those with very fine fibers, or thin layers of fiber (or aluminum or gold). When these are separated by less than 0.2µ you finally decrease the thermal conductivity at room temperature below that for air. These layers decrease the distance traveled between gas collisions, but still leave the same number of collisions with the hot or cold wall; as a result, the smaller the gap below .2µ the lower the thermal conductivity. This happens in aerogels and some space blankets that have very small silica fibers, less than .1µ apart (<100 nm). Aerogels can have much lower thermal conductivities than 0.024 W/m°C, even when filled with air at standard conditions.

In outer space you get lower thermal conductivity without high-tech aerogels because the free path is very long. At these pressures virtually every molecule hits a fiber before it hits another molecule; for even a rough blanket with distant fibers, the fibers bleak up the path of the molecules significantly. Thus, the fibers of the space shuttle (about 10 µ apart) provide far lower thermal conductivity in outer space than on earth. You can get the same benefit in the lab if you put a high vacuum of say 10^-7 atm between glass walls that are 9 mm apart. Without the walls, the air molecules could travel 1.3 µ/10^-7 = 13m before colliding with each other. Since the walls of a typical Dewar are about 0.009 m apart (9 mm) the heat conduction of the Dewar is thus 1/150 (0.7%) as high as for a normal air layer 9mm thick; there is no thermal conductivity of Dewar flasks and vacuum bottles as such, since the amount of heat conducted is independent of gap-distance. Pretty spiffy. I use this knowledge to help with the thermal insulation of some of our hydrogen generators and hydrogen purifiers.

There is another effect that I should mention: black body heat transfer. In many cases black body radiation dominates: it is the reason the tiles are white (or black) and not clear; it is the reason Dewar flasks are mirrored (a mirrored surface provides less black body heat transfer). This post is already too long to do black body radiation justice here, but treat it in more detail in another post.

RE. Buxbaum