This page is a self-testing ladder. Each problem is stated cleanly, then a collapsible solution is hidden inside a [!recall]- callout — try it first, then open. Difficulty climbs from L1 Recognition (do you know the definitions?) up to L5 Mastery (can you weave everything together?). Every numeric answer is machine-checked.

Prerequisites you should have from the parent Propellant Properties — Density, Freezing Point, Toxicity, Storability and its neighbours: Rocket Equation, Specific Impulse, Boil-off Losses, Cryogenic Propellants, Hypergolic Propellants, Tank Design, Methalox.

Symbols we will use (all earned before use)

Definition The cast of characters

$ρ$ (rho) — density, mass packed into one unit of volume. Units g/cm³ or kg/m³. Picture: how many kilograms fit in a box one metre on a side.
$V$ — volume the propellant occupies (m³). Picture: the size of the tank you must build.
$m$ — a mass in kilograms.
$T_{f}$ — freezing point: the temperature at which the liquid turns solid and stops flowing.
$\dot{Q}$ — heat leak rate: joules of heat sneaking into a cold tank every second (the dot means "per second").
$L_{v}$ — latent heat of vaporization: energy needed to boil 1 kg of liquid into gas (J/kg).
$\overset{m}{˙}_{boil}$ — boil-off rate: kilograms lost to evaporation per second.
$I_{s p}$ — specific impulse: seconds of thrust you get per unit weight-flow of propellant; higher = more efficient. See Specific Impulse.
$Δ v$ (delta-vee) — the change in speed a rocket can produce; the currency of space travel.
$g_{0} = 9.81 m/s^{2}$ — standard gravity, a fixed conversion constant.

Two formulas we lean on repeatedly, both from the parent note:

L1 — Recognition

Exercise 1.1

State the four engineering constraints on propellant selection listed in the parent note, and for each write one sentence on why it matters.

Recall Solution

Density $ρ$ — mass per volume. Higher density → smaller, lighter tanks → better structural efficiency.
Freezing point $T_{f}$ — below it the propellant solidifies and cannot flow, so the engine cannot start.
Toxicity — how much exposure harms people; drives ground-handling cost and crew safety.
Storability — whether the propellant can sit in a tank for months/years without boiling off or self-igniting.

Exercise 1.2

A propellant has density $ρ = 0.82 g/cm^{3}$ . Convert this to $kg/m^{3}$ .

Recall Solution

What we do: convert units. Why: tank-volume formulas use SI (kg, m³). $1 g/cm^{3} = 1000 kg/m^{3}$ because $1 g = 1 0^{- 3} kg$ and $1 cm^{3} = 1 0^{- 6} m^{3}$ , so $\frac{1 0 ^{- 3}}{1 0 ^{- 6}} = 1 0^{3}$ . $ρ = 0.82 \times 1000 = 820 kg/m^{3}$

Exercise 1.3

Which of these two toxicity numbers means "more dangerous": a lower TLV-TWA or a higher one? Explain in one line.

Recall Solution

A lower TLV-TWA is more dangerous. TLV-TWA is the safe concentration you may breathe for an 8-hour day — the smaller that safe limit, the tinier the amount that already harms you. Hydrazine's $0.01$ ppm vs RP-1's $200$ ppm means hydrazine is ~20,000× more restrictive.

L2 — Application

Exercise 2.1

You must store 10,000 kg of propellant. Compute the required tank volume for:

RP-1: $ρ = 820 kg/m^{3}$
Liquid hydrogen (LH₂): $ρ = 71 kg/m^{3}$

Then state how many times bigger the LH₂ tank is.

Recall Solution

What we do: apply $V = m / ρ$ . Why: the tank must physically contain the whole mass, and its size is set purely by density once the mass is fixed. $V_{RP-1} = \frac{10 , 000}{820} = 12.2 m^{3}$ $V_{LH_{2}} = \frac{10 , 000}{71} = 140.8 m^{3}$ Ratio: $\frac{V _{LH_{2}}}{V _{RP-1}} = \frac{140.8}{12.2} \approx 11.5$ The hydrogen tank is about 11–12× larger for the same mass — see the figure below.

Figure — Propellant properties — density, freezing point, toxicity, storability

Exercise 2.2

A cryogenic tank leaks heat at $\dot{Q} = 500 W$ . The propellant is LH₂ with $L_{v} = 445 kJ/kg$ . Find the boil-off rate in kg/day.

Recall Solution

What we do: every joule that leaks in must go somewhere; here it boils liquid into gas. Why divide by $L_{v}$ ? Because $L_{v}$ is the "price in joules" to vaporize 1 kg, so joules-per-second ÷ joules-per-kg = kg-per-second. $\overset{m}{˙}_{boil} = \frac{Q ˙}{L _{v}} = \frac{500 J/s}{445 , 000 J/kg} = 1.124 \times 1 0^{- 3} kg/s$ Convert to per day ( $86, 400$ s in a day): $\overset{m}{˙}_{boil} = 1.124 \times 1 0^{- 3} \times 86, 400 \approx 97.1 kg/day$ See Boil-off Losses.

Exercise 2.3

NTO has $T_{f} = - 11. 2^{\circ} C$ . A satellite in Earth's shadow cools to $- 15 0^{\circ} C$ . Will the NTO in an unheated line freeze? By how many degrees is it above or below its freezing point at that temperature?

Recall Solution

What we do: compare the environment temperature to $T_{f}$ . Why: freezing happens when the propellant's own temperature falls to or below $T_{f}$ . Environment $- 15 0^{\circ} C$ is below $T_{f} = - 11. 2^{\circ} C$ : $- 150 - (- 11.2) = - 138. 8^{\circ} C$ So the propellant would sit 138.8 °C below its freezing point → it freezes solid and blocks the line. NTO needs line heaters in shadow. (This is exactly why Hypergolic Propellants systems still require thermal management, just far less extreme than Cryogenic Propellants.)

L3 — Analysis

Exercise 3.1

Two propellants store the same mass $m$ . Propellant A has density $ρ_{A}$ , propellant B has $ρ_{B} = 2 ρ_{A}$ . The parent note derived $m_{tank} \propto V$ . By what factor does B's tank mass differ from A's?

Recall Solution

What we do: chain two proportionalities. Why: the note showed tank mass scales linearly with tank volume, and volume $= m / ρ$ . $m_{tank} \propto V = \frac{m}{ρ} \Rightarrow \frac{m _{tank, B}}{m _{tank, A}} = \frac{ρ _{A}}{ρ _{B}} = \frac{ρ _{A}}{2 ρ _{A}} = \frac{1}{2}$ B's tank is half the mass of A's. Doubling density halves tank mass — a direct win for Tank Design.

Exercise 3.2

A tank's insulation is upgraded, halving heat leak from $\dot{Q} = 500 W$ to $250 W$ . But the extra multi-layer insulation adds $120 kg$ of mass. Over a 10-day mission, does the insulation save net mass, given LH₂ with $L_{v} = 445 kJ/kg$ ?

Recall Solution

What we do: compare propellant saved against insulation added. Why: insulation is only worth it if boil-off saved exceeds the insulation's own mass. Boil-off at 500 W over 10 days (from Ex 2.2, $97.1$ kg/day): $m_{lost, before} = 97.1 \times 10 = 971 kg$ Halving $\dot{Q}$ halves boil-off (it's linear in $\dot{Q}$ ): $m_{lost, after} = 485.5 kg$ Propellant saved: $Δ m_{saved} = 971 - 485.5 = 485.5 kg$ Insulation cost: $120$ kg. Net benefit: $485.5 - 120 = + 365.5 kg saved$ Yes — for a 10-day mission the insulation pays for itself over 3× over. (For a short mission it might not; the break-even is when saved boil-off = 120 kg.)

Exercise 3.3

Find the mission duration at which the insulation upgrade in Ex 3.2 exactly breaks even (saved boil-off equals the 120 kg penalty).

Recall Solution

What we do: set saved mass equal to the penalty and solve for time. Why: break-even is where the two mass terms cancel. Boil-off saved per day $= 97.1 - 48.55 = 48.55 kg/day$ . $48.55 \times t_{days} = 120 \Rightarrow t_{days} = \frac{120}{48.55} \approx 2.47 days$ Missions longer than ~2.5 days benefit; shorter missions are better off carrying the boil-off instead of the insulation.

L4 — Synthesis

Exercise 4.1 — Density–specific-impulse figure of merit

The parent note suggests $ρ \cdot I_{s p}$ as a rough figure of merit for boosters. Compute it for:

RP-1/LOX: bulk $ρ = 1030 kg/m^{3}$ , $I_{s p} = 300 s$
LH₂/LOX: bulk $ρ = 360 kg/m^{3}$ , $I_{s p} = 450 s$

Which wins on this metric, and what does that tell a booster designer?

Recall Solution

What we do: multiply. Why: $ρ I_{s p}$ rewards propellants that are both efficient (high $I_{s p}$ ) and compact (high $ρ$ ) — exactly the two things a first stage cares about. $(ρ I_{s p})_{RP-1} = 1030 \times 300 = 309, 000$ $(ρ I_{s p})_{LH_{2}} = 360 \times 450 = 162, 000$ RP-1/LOX wins by roughly 1.9× on density-impulse. This is why dense kerosene sits in the first stage (Saturn V S-IC, Falcon 9), where compact heavy propellant means small light tanks and high liftoff thrust — while LH₂ is saved for upper stages where raw $I_{s p}$ dominates $Δ v$ .

Exercise 4.2 — $Δ v$ vs storability trade

An upper stage has structure+payload mass $m_{f} = 3000 kg$ and carries $m_{prop} = 7000 kg$ , so $m_{0} = 10, 000$ kg. Compare the $Δ v$ for:

Hypergolic NTO/MMH: $I_{s p} = 320 s$ , storable (0 boil-off)
Methalox: $I_{s p} = 370 s$ , but over a 6-month coast it loses $10%$ of its propellant to boil-off (so only $6300$ kg burns; the lost $700$ kg still counts as dead mass at ignition? No — assume it is vented, so $m_{0}$ drops to $9300$ kg and $m_{f}$ stays $3000$ kg).

Recall Solution

What we do: apply the rocket equation to each case. Why: $Δ v$ is the true currency; storability enters by changing the mass ratio through boil-off.

Hypergolic: $Δ v = 320 \times 9.81 \times ln (\frac{10 , 000}{3000}) = 3139.2 \times ln (3.3333)$ $= 3139.2 \times 1.20397 = 3780 m/s$

Methalox after boil-off ( $m_{0} = 9300$ , $m_{f} = 3000$ ): $Δ v = 370 \times 9.81 \times ln (\frac{9300}{3000}) = 3629.7 \times ln (3.1)$ $= 3629.7 \times 1.13140 = 4106 m/s$

Verdict: even after losing 10% to boil-off, Methalox's higher $I_{s p}$ still delivers ~326 m/s more $Δ v$ . But if the coast were longer (more boil-off) or cryo-cooling mass were added, the storable hypergolic could catch up. This captures the real engineering tension: efficiency vs storability.

L5 — Mastery

Exercise 5.1 — Full trade study

A 6-month Mars-transfer upper stage must deliver $Δ v = 3800 m/s$ with a fixed dry mass (structure + payload, excluding propellant tanks and coolers) of $m_{dry} = 3000 kg$ . You choose between:

Option A — Storable NTO/MMH: $I_{s p} = 320 s$ , bulk $ρ = 1180 kg/m^{3}$ , zero boil-off, no cooler. Tank+plumbing mass $= 8%$ of propellant mass.

Option B — Methalox: $I_{s p} = 370 s$ , bulk $ρ = 830 kg/m^{3}$ , boil-off of $5%$ of loaded propellant over the coast, cryo-cooler mass $= 200 kg$ . Tank+plumbing mass $= 12%$ of loaded propellant mass.

For each option, find the loaded propellant mass required to hit $3800 m/s$ , then the tank volume, then the total ignition-stack mass. Which option is lighter overall?

Recall Solution

Strategy. The rocket equation fixes the mass ratio $m_{0} / m_{f}$ ; everything else (tanks, coolers) piles into $m_{f}$ . We solve for propellant, then bolt on the hardware.

Step 1 — required mass ratio. Invert the rocket equation: $\frac{m _{0}}{m _{f}} = exp (\frac{Δ v}{I _{s p} g _{0}})$ Option A: $\frac{3800}{320 \times 9.81} = \frac{3800}{3139.2} = 1.21050$ , so ratio $= e^{1.21050} = 3.3552$ . Option B: $\frac{3800}{370 \times 9.81} = \frac{3800}{3629.7} = 1.04692$ , so ratio $= e^{1.04692} = 2.8493$ .

Step 2 — set up $m_{f}$ and $m_{0}$ for each. Let $m_{p}$ = loaded propellant.

Option A (no boil-off): burned propellant = $m_{p}$ . $m_{f} = m_{dry} + 0.08 m_{p} = 3000 + 0.08 m_{p}$ $m_{0} = m_{f} + m_{p} = 3000 + 1.08 m_{p}$ Ratio condition: $\frac{3000 + 1.08 m _{p}}{3000 + 0.08 m _{p}} = 3.3552$ . Solve: $3000 + 1.08 m_{p} = 3.3552 (3000 + 0.08 m_{p})$ $3000 + 1.08 m_{p} = 10065.6 + 0.268416 m_{p}$ $0.811584 m_{p} = 7065.6 \Rightarrow m_{p} = 8706.4 kg$ Tank mass $= 0.08 \times 8706.4 = 696.5$ kg. $m_{f} = 3696.5$ kg, $m_{0} = 12406.9$ kg. Volume $V_{A} = 8706.4/1180 = 7.38 m^{3}$ . Total ignition stack $m_{0} = 12, 407 kg$ .

Option B (5% boil-off, cooler + tanks are dead mass at ignition): only $0.95 m_{p}$ burns. $m_{f} = m_{dry} + 200 + 0.12 m_{p} + 0.05 m_{p} = 3200 + 0.17 m_{p}$ (the vented $5%$ is gone by burn's end, but at ignition it isn't yet lost — treat the coast boil-off as vented before the burn, so it does not contribute to $m_{0}$ at ignition; loaded propellant available to burn is $0.95 m_{p}$ .) $m_{0}^{(at burn)} = m_{f} + 0.95 m_{p} = 3200 + 1.12 m_{p}$ Ratio: $\frac{3200 + 1.12 m _{p}}{3200 + 0.17 m _{p}} = 2.8493$ . $3200 + 1.12 m_{p} = 2.8493 (3200 + 0.17 m_{p})$ $3200 + 1.12 m_{p} = 9117.76 + 0.484381 m_{p}$ $0.635619 m_{p} = 5917.76 \Rightarrow m_{p} = 9310.7 kg$ Volume $V_{B} = 9310.7/830 = 11.22 m^{3}$ . Loaded-stack mass at launch (before venting) $= m_{dry} + 200 + 0.12 m_{p} + m_{p} = 3200 + 1.12 \times 9310.7 = 13, 628 kg$ .

Step 3 — verdict. Option A loads $8706$ kg propellant into a compact $7.4 m^{3}$ tank, total launch mass $\approx 12, 407$ kg. Option B needs $9311$ kg into a bulky $11.2 m^{3}$ tank, total $\approx 13, 628$ kg. For this dry mass and this coast duration, the storable NTO/MMH (Option A) is lighter and more compact — its zero boil-off and lighter tankage beat methalox's $I_{s p}$ edge once the 5% loss, cooler, and thicker cryo-tanks are counted. The lesson: raw $I_{s p}$ does not decide the winner — the whole system does.