Nuclear thermal propulsion — NTR Isp ~900 s concept: Exercises

Parent topic: NTR concept note.

The symbols this page uses (define before you compute)

Definition Every letter in the master formula, in plain words

$v_{e}$ — exhaust velocity (m/s): how fast the gas leaves the nozzle. This is the number we chase; bigger $v_{e}$ = better rocket.
$T_{c}$ — combustion-chamber temperature (K): the temperature of the propellant inside the chamber, before it accelerates. In an NTR the reactor heats the gas to this $T_{c}$ . "Hot chamber" means big $T_{c}$ .
$M$ — molar mass of the propellant (kg/mol): the mass of one mole of the gas. Hydrogen is $M = 2 \times 1 0^{- 3}$ kg/mol (feather-light); steam is $18 \times 1 0^{- 3}$ . Small $M$ = light molecule.
$γ$ — the ratio of specific heats $γ = c_{p} / c_{v}$ (dimensionless, no units). It measures how "springy" the gas is when it expands — how much of its heat converts into directed motion. For a diatomic gas like H₂, $γ \approx 1.4$ ; for a triatomic gas like steam/CH₄, $γ \approx 1.3$ .
$R = 8.314$ J/mol·K — the universal gas constant (same for every gas).
$g_{0} = 9.81$ m/s² — standard gravity, used only to convert $v_{e}$ into "seconds" of $I_{s p}$ .

Formula The master formula — and WHERE the

$\frac{2 γ}{γ - 1}$ comes from $v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}, I_{s p} = \frac{v _{e}}{g _{0}} .$ Why this shape? Start from energy conservation for the flowing gas (see Adiabatic Flow & Enthalpy): the heat energy the gas loses as it cools from $T_{c}$ to the nozzle-exit temperature $T_{e}$ turns into kinetic energy, $\frac{1}{2} v_{e}^{2} = c_{p} (T_{c} - T_{e}) .$ We use enthalpy's $c_{p}$ (not $c_{v}$ ) because a flowing gas also does push-work on itself. Now two substitutions:

Full expansion: a good nozzle cools the gas almost completely, so $T_{e} \to 0$ and $(T_{c} - T_{e}) \to T_{c}$ . This is the best case (an upper bound on $v_{e}$ ).
Write $c_{p}$ per unit mass using $c_{p} = \frac{γ}{γ - 1} \frac{R}{M}$ — this is where $γ$ , $R$ , $M$ enter.

Substitute both: $\frac{1}{2} v_{e}^{2} = \frac{γ}{γ - 1} \frac{R}{M} T_{c}$ . Multiply by 2 and take the root → the factor $\frac{2 γ}{γ - 1}$ is just "2 from the $\frac{1}{2}$ times the specific-heat bundle $\frac{γ}{γ - 1}$ ." It packages the thermodynamic efficiency of turning heat into a straight-line jet.

Level 1 — Recognition

L1.1 — Read the scaling off the formula

Without plugging numbers, state how $v_{e}$ changes if you (a) double the chamber temperature $T_{c}$ , (b) quarter the molar mass $M$ .

Recall Solution

What we use: the fingerprint $v_{e} \propto T_{c} / M$ (everything else in the formula is a constant here). (a) Doubling $T_{c}$ : $v_{e} \to 2 v_{e}$ , i.e. $\times 1.41$ . (b) Quartering $M$ : $v_{e} \to 1/ (1/4) v_{e} = 4 v_{e} = \times 2$ . Why the square root? $v_{e}$ comes from $\frac{1}{2} v_{e}^{2} = (energy per kg)$ , so speed grows as the square root of that energy — never linearly.

L1.2 — Identify the metric

A design team reports "our engine puts out 900 kN of thrust." Does this tell you the $I_{s p}$ ? Explain in one line.

Recall Solution

No. Thrust is $F = \overset{m}{˙} v_{e}$ — it mixes exhaust speed with mass flow $\overset{m}{˙}$ . $I_{s p} = v_{e} / g_{0}$ depends only on exhaust speed. You could get 900 kN by throwing a lot of slow gas (low $I_{s p}$ ) or a little fast gas (high $I_{s p}$ ).

Level 2 — Application

L2.1 — Reproduce the 900 s number

Hydrogen propellant: $T_{c} = 2700$ K, $M = 2 \times 1 0^{- 3}$ kg/mol, $γ = 1.4$ . Find $v_{e}$ and $I_{s p}$ .

Recall Solution

Prefactor: $\frac{2 γ}{γ - 1} = \frac{2 ( 1.4 )}{0.4} = 7.0$ . Thermal energy per kg: $\frac{R T _{c}}{M} = \frac{8.314 \times 2700}{2 \times 1 0 ^{- 3}} = 1.122 \times 1 0^{7}$ J/kg. $v_{e} = 7.0 \times 1.122 \times 1 0^{7} = 7.86 \times 1 0^{7} \approx 8860 m/s .$ $I_{s p} = \frac{8860}{9.81} \approx 903 s . ✓$

L2.2 — Methane instead of hydrogen

Run the same 2700 K reactor with methane ( $M = 16 \times 1 0^{- 3}$ kg/mol, $γ = 1.3$ ). Find $I_{s p}$ .

Recall Solution

Prefactor: $\frac{2 ( 1.3 )}{0.3} = 8.667$ . $\frac{R T _{c}}{M} = \frac{8.314 \times 2700}{16 \times 1 0 ^{- 3}} = 1.403 \times 1 0^{6}$ J/kg. $v_{e} = 8.667 \times 1.403 \times 1 0^{6} = 1.216 \times 1 0^{7} \approx 3487 m/s .$ $I_{s p} = \frac{3487}{9.81} \approx 355 s .$ Why so much worse than hydrogen? $M$ is 8× larger; the $1/ M$ factor alone costs a factor $8 \approx 2.8$ .

Level 3 — Analysis

L3.1 — Temperature needed to match a target

You want an NTR with hydrogen to hit $I_{s p} = 1000$ s. Ignoring melting limits, what chamber temperature $T_{c}$ is required? ( $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ .)

Recall Solution

Target speed: $v_{e} = I_{s p} g_{0} = 1000 \times 9.81 = 9810$ m/s. Invert the formula for $T_{c}$ (solve $v_{e}^{2} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}$ for $T_{c}$ ): $T_{c} = \frac{v _{e}^{2} M}{\frac{2 γ}{γ - 1} R} = \frac{( 9810 ) ^{2} \times 2 \times 1 0 ^{- 3}}{7.0 \times 8.314} .$ Numerator: $(9810)^{2} = 9.624 \times 1 0^{7}$ ; $\times 2 \times 1 0^{- 3} = 1.925 \times 1 0^{5}$ . Denominator: $7.0 \times 8.314 = 58.2$ . $T_{c} = \frac{1.925 \times 1 0 ^{5}}{58.2} \approx 3308 K .$ Interpretation: 1000 s needs ~3300 K — already above the melt point of solid fuel elements (~2900 K). This is why solid-core NTR stalls near 900 s.

L3.2 — Which lever is stronger?

Starting from the 903 s baseline, you may either raise $T_{c}$ from 2700 K to 3200 K, or switch from a hydrogen mix with effective $M = 3 \times 1 0^{- 3}$ to pure hydrogen $M = 2 \times 1 0^{- 3}$ . Which gives the bigger $I_{s p}$ gain?

Figure — Nuclear thermal propulsion — NTR Isp ~900 s concept

Recall Solution

Since $v_{e} \propto T_{c} / M$ , we compare ratios. The figure above plots each option as a bar of relative exhaust speed against the gray baseline: Temperature lever (orange bar): factor $3200/2700 = 1.185 = 1.089$ → about $+ 8.9%$ . Mass lever (green bar): factor $3/2 = 1.5 = 1.225$ → about $+ 22.5%$ . As the green arrow in the figure highlights, the green bar rises higher than the orange one: the mass lever wins — dropping $M$ from 3 to 2 g/mol beats a 500 K temperature bump. This is the whole moral of the parent note: light molecule > hot reactor.

Level 4 — Synthesis

L4.1 — Feed the $Δ v$ budget

An NTR upper stage has $I_{s p} = 903$ s. Its wet (fuelled) mass is 40 t and dry mass is 15 t. Using the Tsiolkovsky Rocket Equation $Δ v = v_{e} ln (m_{0} / m_{f})$ , find $Δ v$ . Then find what $Δ v$ the same mass ratio gives with a chemical stage at $I_{s p} = 450$ s.

Recall Solution

Effective exhaust speed: $v_{e} = I_{s p} g_{0} = 903 \times 9.81 = 8858$ m/s. Mass ratio: $m_{0} / m_{f} = 40/15 = 2.667$ , so $ln (2.667) = 0.9808$ . $Δ v_{N T R} = 8858 \times 0.9808 \approx 8687 m/s .$ Chemical: $v_{e} = 450 \times 9.81 = 4414.5$ m/s, same $ln$ factor: $Δ v_{c h e m} = 4414.5 \times 0.9808 \approx 4330 m/s .$ Payoff: identical hardware mass ratio, but the NTR delivers almost exactly double the $Δ v$ — because $Δ v$ is linear in $v_{e}$ while $v_{e}$ roughly doubled.

L4.2 — Same $Δ v$ , less fuel

A mission needs $Δ v = 6000$ m/s with a 15 t dry mass. Compare the wet mass required for the NTR ( $I_{s p} = 903$ s) vs the chemical stage ( $I_{s p} = 450$ s).

Recall Solution

Invert Tsiolkovsky: $m_{0} = m_{f} e^{Δ v / v_{e}}$ . NTR: $v_{e} = 8858$ ; exponent $6000/8858 = 0.6774$ ; $e^{0.6774} = 1.969$ . $m_{0}^{N T R} = 15 \times 1.969 \approx 29.5 t \Rightarrow fuel \approx 14.5 t .$ Chemical: $v_{e} = 4414.5$ ; exponent $6000/4414.5 = 1.359$ ; $e^{1.359} = 3.893$ . $m_{0}^{c h e m} = 15 \times 3.893 \approx 58.4 t \Rightarrow fuel \approx 43.4 t .$ Payoff: the chemical stage needs ~3× the propellant for the same job. High $I_{s p}$ pays off exponentially through the $e^{Δ v / v_{e}}$ — the deeper reason NTRs matter for deep-space transfers.

Level 5 — Mastery

L5.1 — The material ceiling as a design proof

Prove, using the scaling, that a solid-core NTR limited to a wall temperature $T_{ma x} = 2900$ K can never reach $I_{s p} = 1200$ s with hydrogen ( $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ ). Then state what propulsion family you'd switch to and why.

Recall Solution

Recall $T_{e}$ : in the master-formula derivation $T_{e}$ was the nozzle-exit temperature — the temperature of the gas at the very end of the nozzle, after it has expanded and cooled. "Full expansion" means the nozzle cools the gas as much as possible, $T_{e} \to 0$ , which converts the most heat into speed. So setting $T_{e} \to 0$ gives the largest $v_{e}$ any real nozzle at that chamber temperature could reach. Best-case $v_{e}$ at the ceiling (full expansion, $T_{c} = T_{ma x} = 2900$ K): $v_{e}^{ma x} = 7.0 \times \frac{8.314 \times 2900}{2 \times 1 0 ^{- 3}} = 7.0 \times 1.206 \times 1 0^{7} = 8.44 \times 1 0^{7} \approx 9186 m/s .$ $I_{s p}^{ma x} = \frac{9186}{9.81} \approx 936 s .$ Even in the ideal limit the ceiling gives ~936 s, far below 1200 s. To do better you must decouple exhaust speed from a solid melting wall — heat the propellant without a solid touching it. That is the logic of Nuclear Electric Propulsion (ionise + electrostatically accelerate; $I_{s p}$ of thousands of seconds), which trades away thrust for enormous $I_{s p}$ . Why not just gas-core NTR? A gas-core reactor lets the fissile fuel itself be the hot gas (no solid limit), reaching much higher $T_{c}$ — but it is far less mature. The clean argument: solid core ⇒ $T$ capped by melting ⇒ $I_{s p}$ capped near 900 s is unavoidable.

L5.2 — Reverse-engineer the propellant

A future gas-core NTR achieves $I_{s p} = 1500$ s using hydrogen ( $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ ). What chamber temperature does this imply, and comment on feasibility.

Recall Solution

$v_{e} = 1500 \times 9.81 = 14715$ m/s. $T_{c} = \frac{v _{e}^{2} M}{\frac{2 γ}{γ - 1} R} = \frac{( 14715 ) ^{2} \times 2 \times 1 0 ^{- 3}}{7.0 \times 8.314} .$ $(14715)^{2} = 2.166 \times 1 0^{8}$ ; $\times 2 \times 1 0^{- 3} = 4.331 \times 1 0^{5}$ ; $/58.2 = 7442$ K. $T_{c} \approx 7440 K .$ Feasibility: ~7400 K is far above any solid melting point — only a gas-core design (fissioning plasma, no solid wall in contact) could sustain it. This confirms the L5.1 argument: pushing $I_{s p}$ past ~1000 s forces you to abandon the solid-core architecture entirely.

Recall One-line summary of the whole ladder

Recognise $v_{e} \propto T_{c} / M$ (L1) → plug numbers with SI units (L2) → compare levers as ratios (L3) → feed Tsiolkovsky where high $I_{s p}$ pays exponentially (L4) → prove the material ceiling forces you toward Nuclear Electric Propulsion beyond ~1000 s (L5).

Connections

Specific Impulse — every problem optimises this.
Tsiolkovsky Rocket Equation — L4's $Δ v$ payoff.
De Laval Nozzle — the expander turning enthalpy into $v_{e}$ .
Nuclear Fission — the heat source.
Chemical Rocket Propulsion — the ~450 s baseline.
Adiabatic Flow & Enthalpy — where $\frac{1}{2} v_{e}^{2} = c_{p} Δ T$ comes from.
Nuclear Electric Propulsion — where L5 points beyond the solid-core ceiling.