16

The hydrogen side of this question has been addressed here: Combination of liquid hydrogen and liquid oxygen -- short answer is that a hydrogen first stage would be lighter, loaded, but much more expensive. Modern "kerosene"-burning rockets like the Falcon 9 use RP-1 (or RG-1 in Russian engines); this is a refined form of kerosene containing a relatively ...


15

Yes they considered the disassosciation of hydrogen According to this source : Previous testing used a maximum temperature of 2,750° K, short of the 3000+° K design temperature for the NCPS. The NTREES facility is designed to test fuel elements and materials in hot flowing hydrogen, reaching pressures up to 1,000 pounds per square inch and ...


15

The implication of the rocket equation is that linear increases in ∆v require exponential increases in mass ratio for a single stage. There's not strictly a maximum delta-v -- if you redo your plot on a log scale, you'll see that it doesn't go vertical. Getting very high mass ratios (much above 10:1) is difficult to do on a single stage, so there is a ...


14

The easiest way is just to think in terms of energy. Using numbers from wikipedia, the mass of a deuterium nucleus is 2.014 daltons, that of a tritium nucleus is 3.016, helium 4 is 4.0026 and a neutron is 1.0087 Thus the net energy production is about 0.019 or very roughly 1/250 of the mass of the products, and in the perfect engine you describe, all of ...


14

I'm guessing that the chemical rocket envelope in the plot encompasses points representing actually-built rocket engines, rather than theoretical ones, hence some of the irregularity of the shape is due to historical accident. 10N is quite small for a chemical rocket engine. Such units are mainly used for attitude control of small spacecraft rather than ...


13

Your question is about the behavior of the Tsiolkovsky rocket equation itself, in the limit of very small final mass (dry mass). Roughly: "is there any limit to delta-v in theory?" Using MathJax: $$ \Delta v=v_e \ln\frac{m_0}{m_f}. $$ If you just look at the velocity ratio and the mass ratio: $$ \frac{\Delta v}{v_e}=\ln\frac{m_0}{m_f}=-\ln\frac{m_f}{m_0}, $...


13

Some of the hydrogen will be disassociated. For the reaction mass that is not dissociated, and passes through the engine in the form of diatomic hydrogen, in addition to the three translational degrees of freedom, heat energy is also put into the vibrational and rotational energy of the hydrogen molecule. So the energy stored is 6/2kT, not 3/2 kT as in the ...


13

If you want to understand how the 'seconds' value fits the greater image, there's this rather contrived definition (which nobody uses because it's contrived and mostly useless but evocative enough.) 0) $I_{sp}$ in seconds is equal to the amount of time a rocket must be fired to use a quantity of propellant with weight (measured at one standard gravity) ...


13

It's even simpler than a German-American disagreement. It's use of ambiguous units. The term "specific X" means the amount of X you can get from a unit mass of something. For instance, in batteries, specific energy means the total amount of energy you can get from one unit mass of battery. As described in the Wikipedia article, *specific impulse" is the ...


12

I guess that it goes approximately like this: assume that the enthalpy change (I'll denote it $\Delta H$) is fully converted to the kinetic energy of the exhaust and the exhaust moves with velocity $v$ relative to the engine. Then we have: $$ mv^2/2 = m \Delta H;$$ $$ v = \sqrt{2\Delta H};$$ $$I_{sp} = \sqrt{2\Delta H}/g.$$ But that is assuming that all ...


12

Ideally, thrust and specific impulse would gradually trade off during flight, but that isn’t generally achievable. Instead, large changes in thrust and ISP are done during staging. It's pretty common for the second stage (or core stage, for a solid booster/liquid sustainer configuration) to start at less than 1:1 thrust-to-weight ratio. That seems like a ...


11

From the ballistics, I estimate the "exhaust velocity" of a urination stream at about 0.5-1.0 m/s2 (i.e. 0.05-0.1 sec specific impulse) though a number of factors can influence that. Assuming a liter of fluid in the bladder massing about 1kg, and a person massing 75kg "dry", we get via Tsiolkovsky a delta-v figure of $$ \sim 0.75 \, ln \, \frac {76} {75} \...


10

You got caught by those silly English units. Your expression, $I_t = I_{sp}g_0 \dot m \Delta t$, works just fine when you use metric units. The mass flow rate, in metric units, is 127 kg/s. The total impulse, in metric units, is $$I_t = (217.5\,\text{s})(9.80665\,\text{m}/\text{s}^2)(127\, \text{kg}/\text{s})(65\,\text{s}) = 1.76\times 10^7\,\text{N}\,\text{...


10

The current launcher with the highest overall system specific impulse is almost certainly the Delta IV Heavy. As far as I can tell from the list of current orbital launchers on Spaceflight 101, it is the only one that uses hydrogen-fueled engines on all stages. All the others use hypergolics or kerosene or solids somewhere in the stack, which all have far ...


9

450-455s Isp is typical of H2/O2; according to the Huzel and Huang data, a hydrogen-beryllium mix combusted with oxygen can hit ~540s. The numbers in that table are for moderate chamber pressure and expansion ratio; higher values are possible. According to Wikipedia: The highest specific impulse for a chemical propellant ever test-fired in a rocket ...


9

One reason for this preference is that everyone agrees on seconds as a unit for measuring time, whereas we still have multiple conventional units in use for measuring speed (m/s, ft/s). By dividing by g0, in whatever conventional units you're used to working with (9.8 m/s2 or 32 ft/s2) you produce a value that anyone can work with, no matter which system ...


9

In simplest terms it is just the thrust produced divided by the propellant flow rate. "How much thrust am I getting for the propellant I am expending?" So bigger is better - you are getting more thrust for the same propellant flow rate. Bigger numbers = more efficient engine = extracting more thrust from the same amount of propellant. It is a key ...


8

Corroborating Russell Borogove's answer, some Stennis test data I have from 1987 on three different SSMEs shows a small drop in Isp with power level. From 109% to 100% the Isp dropped about 0.08%. I can't find data at lower power levels but my recollection is that the trend continued, with a small degradation in Isp as you throttled down. For reference, ...


8

It's surprisingly difficult to find a good answer to this question. Generally, the rated full power level is where the engine is going to be most efficient. According to Sutton's "Rocket Propulsion Elements", typical deep-throttling engines suffer between 1.5% and 9% reduction in specific impulse (fuel efficiency) at low power levels. It mentions an ...


8

$$I_{sp}=\dfrac{I_{sp1}\dot{m}_1+I_{sp2}\dot{m}_2+{...}}{\dot{m}_1+\dot{m}_2+{...}}$$ So each $I_{sp}$ is simply weighted by its fraction of the total mass flow rate. This extends to any number of $I_{sp}$'s.


8

The historical NASA document "SPACE HANDBOOK: ASTRONAUTICS AND ITS APPLICATIONS" has a useful table, which I will partially reproduce here: TABLE 1.-Specific impulse of some typical chemical propellants Low-energy monopropellants________________________ 160 to 190. High-energy monopropellants: Nitromethane_______________________________ 190 to ...


8

Some resources (Wikipedia, Astronautix) give 312 seconds for the R-4D, but I think that's for the large-nozzle modern version. Marquardt's "Apollo SM-LM RCS Engine Development Program Summary Report" gives the nominal specific impulse as 280 seconds, and gives figures for flown engines on Lunar Orbiter I-V as 276.1 to 279.5 seconds. The Lunar Orbiters, the ...


7

Suppose we have our choices of particles, we can convert all applied energy to exhaust velocity, and the energy is divided amongst the same number of particles whether they have an atom mass of 1 or 131. (Note well: Those are some big assumptions.) The velocity of an exhaust particle is $v=\sqrt{2E/m}$ where $E$ is the energy applied to an individual ...


7

Figure 5 on page 23 of this paper shows the Isp of tri-propellants including iron, where the curve goes out to 95% Fe and 5% H2. Just eyeballing it, the curves might converge at around 190 seconds. On the other hand, the curve might go to hell at 100%, with no light hydrogen for the heat of the burning iron to expel. Note that even at 5% H2 by weight, there ...


7

Since specific impulse and exhaust velocity are directly related via $$I_{SP}=\frac{V_e}{g_0}$$ anything that increases the exhaust velocity necessarily increases the specific impulse. The issue is: do you gain anything from it? That depends on what "gain" you're looking for. Rocket engines of any type are momentum devices. The impulse imparted to the ...


6

The conversion factor isn't the gravitational pull of an arbitrary body, but rather the conversion between pounds as a unit of weight and pounds as a unit of mass -- which is a convention established on Earth, and hence happens to match Earth's gravitational pull. Some rocket scientists work in feet and pounds, others work in meters and kilograms. Exhaust ...


6

The theoretical limit is set by the specific energy of the reaction of combustion of the propellant. Knowing specific energy $e$ of given substance, we can put a cap on obtainable specific impulse $I_{sp}$ by assuming 100% of conversion of chemical energy to kinetic energy. $$ I_{sp} = {v_e \over g_0} $$ $$ E_{chem} = e m \geqslant E_k = {1 \over 2 }{ m ...


6

No, that is not quite right. Let's first state and describe the Tsiolkovsky Rocket Equation: $\displaystyle \Delta v = V_e \times \ln(\frac{m_i}{m_f})$ $\Delta v$ is delta v, the change in velocity in km/s $V_e$ is the effective exhaust velocity in km/s (it's another way of measuring specific impulse) $\ln()$ is just the natural logarithm, or log base e (...


6

How do I get Specific impulse in (s)? First let's look at the units. Always look at the units first. Specific impulse 2314 N·s/kg A Newton has units of kg m / s^2, so that is kg m s / s^2 kg which is m/s, and 2314 m/s is a reasonable value for an effective exhaust velocity. According to the Wikipedia subsection Specific impulse as a speed (effective ...


6

It would be very hard to pinpoint a specific unique condition, but there's a bunch of considerations coming into play that decide it. switching to vacuum-optimized bell nozzle. This alone gives you a significant ISp boost - which is always welcome, but comes at a cost - the nozzle is big, you can't fit a whole lot of engines or combustion chambers with ...


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