When would the density Isp product be an important performance metric of a propellant?

Question

The quote

NASA is testing hydroxylammonium nitrate in space which is expected to perform 50% better than standard propellants and can be allowed to freeze (hydrazine must be kept liquid). LMP-103S has been tested gets about 30%.

used in Why would it be “less bad” for hydroxylammonium nitrate monopropellant to freeze than for hydrazine? and found in this answer to Is the EU really banning “toxic propellants” in 2020? How is that going to work? raised a question about how a monopropellant could be 50% better than hydrazine.

The reply comment links to GPIM AF-M315E Propulsion System which explains that the performance metric is the product of density and Isp.

Question: When would the density Isp product be an important performance metric of a propellant?

Answer 1

From the ECAPS page on LMP-103S performance...

The specific impulse is ≥ 6% higher and the propellant density is 24% higher. As a result, the satellite can either be fitted with a smaller tank, or the mission duration can be extended while retaining the same tank size.

Volume is important because you only have so much room inside a payload fairing. Propellant available for station keeping and maneuvering is a limit on the lifespan of a spacecraft.

Answer 2

When would the density Isp product be an important performance metric of a propellant?

It's always important.

Specific impulse is a measure of impulse provided per mass unit of combusted propellant, which is a great metric for fuel-efficency in a rocket. However, it doesn't take into account the mass of things other than fuel -- like fuel tankage, which has mass generally proportional to its volume.

Hydrogen, for example, looks vastly better than kerosene if you just look at specific impulse, but it is about 1/15 as dense, so some of the specific impulse advantage is lost due to requiring much more tankage volume.

Density-Isp-product isn't a mathematically "correct" metric for comparing propellants -- you'd need a spreadsheet and information about tankage structure, pressure, etc. to actually compute a correct metric -- but it could be useful for identifying interesting candidate propellants before moving onto the spreadsheet phase of the contest.

Answer 3

While it isn't precisely space exploration per se, John Clark mentions this being important in Ignition for missiles, especially in cases when it is desired to meet a performance goal while having a missile meet some imposed or preexisting form factor, and some of the weirdest extremes of his and others' work (done for the Navy) described in Ignition focused on largely doomed attempts at fuels with moderate Isp and very high density.

Modern-day space exploration is often able to make very lightweight tanks (suitable, among other things, for LH2) but there are many times when one does not want a space vehicle to resemble a water balloon, especially if it must fit in a fairing or aeroshell or undergo atmospheric flight.

Answer 4

Oh, I see.

Let's say I am building an ambitious 3U cubesat and I've allocated 1U for fuel and 2U for payload and bus, and I want to select the propellant that will give me the most delta-v.

I don't know Cubesat mass density (kg/U) statistics? but lets say I've got a 3 kg (dry mass $m_D$) 3U cubesat with a 1 liter tank. What is the delta-v as a function of $I_{sp}$ and $\rho$ with a fixed tank volume $V$? Applying tyrannical Tsiolkovsky we get:

$$\Delta v = I_{sp} g_0 \ln \frac{m_D+\rho V}{m_D} = I_{sp} g_0 \ln \left(1 + \frac{\rho V}{m_D}\right).$$

Remember that the log of 1 plus something small is approximately that thing (i.e. $\ln(1+x) \approx x$ for small $x$)

$$\Delta v \approx I_{sp} g_0 \frac{\rho V}{m_D} = I_{sp}\rho \frac{g_0 V}{m_D}.$$

The density specific impulse product $I_{sp}\rho$ is the predictor of total delta-v possible when the tank volume is fixed, and the relationship is simple linear when the propellant mass is a small fraction of the total spacecraft mass.

For flying fuel tanks you'll need to keep the logarithm in the expression and so while close, the highest $I_{sp}\rho$ product monopropellant may not have exactly the highest $\Delta v$. checking now...