Is a kilowatt per square centimeter a typical launch engine's thermal energy flux (density)?

Question

I saw the following image while searching for this answer in Sung Hwan Kim's thesis Germanium-Source Tunnel Field Effect Transistors for Ultra-Low Power Digital Logic It plots power density in Watts per square centimeter. Data points are for microprocessors, but it also includes indicators for 'Hot Plate", "Nuclear Reactor", "Rocket Nozzle", and "Sun's Surface".

The value plotted for the Rocket Nozzle seems to be 1,000 W/cm^2. Is that a sort-of canonical number with many engines used for launch purposes being somewhat similar, or is that an extreme example?

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The Sun

The solar constant is the total electromagnetic radiation power per unit area at 1AU (about 150 million km) and is about 1361 W/m^2. Scale that by $1/r^2$ $to the radius of the Sun (about 0.696 million km) and that's 6300 W/m^2 which agrees nicely with the plot.

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A Rocket Nozzle

I'll work through one example as a proposed way to estimate this.

Merlin engine with the smaller nozzle for first-stage atmospheric operation.

From this answer and an image in this answer I'll call the exit diameter $D$ of 90 centimeters and so radius $R$ of 45 centimeters.

From Wikipedia's Merlin (rocket engine family) I'll use the sea level $I_{SP}$ of 282 seconds and thrust (force) $F$ of 845 kN to get the total mass flow rate.

$$\dot(m) = \frac{F}{v=gI_{SP}} \approx 305 \ \text{kg/s}$$

Start by assuming correct stoichiometry as an approximation CH2 + 1.5O2 → CO2 + H2OI get that 23% of the mass flow rate comes from the CH2 or kerosene.

$$\dot{m_K} \approx 69 \ \text{kg/s}$$

The energy density of kerosene $U$ is about 43 MJ/kg. Add in a fudge factor of 0.8 for incomplete burning, and I get:

$$I = \frac{P=\dot{m}U}{A=\pi R^2} \approx \ 466,000 \ \text{W/cm^2}$$

or 466 times larger than the number in the plot. This means I'm dramatically mis-interpreting something about the plot or I've mad a mistake in my math.

So the exhaust is certainly cooled by the expansion and not all of the chemical energy released in combustion is still present as heat at the exit of the exhaust, hopefully most of it has been converted to directed kinetic energy.

Answer 1

Using Wikipedia as the source, I repeated your calculations for the Space Shuttle Main Engine (SSME) and compared it to the manufacturer's specification and got good agreement. So your method appears valid. The chart in Sung Hwan Kim's thesis must be referring to some old rocket technology such as the V-2.

For SSME with a sea level $I_{SP}$ of 366 seconds and thrust $F$ of 1859kN

$$\dot m = \frac{F}{v=gI_{SP}} \approx 518 \ \text{kg/s}$$

And a stoichiometry of 1/9 part by weight of Hydrogen (16 for O 1x2 for H) I get

$$\dot{m_K} \approx 57.6 \ \text{kg/s}$$

The energy density $U$ of H2 is 142 MJ/kg and the nozzle exit diameter is 230cm. Calculating the energy flux:

$$I = \frac{P=\dot{m}U}{A=\pi R^2} \approx \ 197,000 \ \text{W/cm^2}$$

Now, comparing to the manufacturer's specification of 12 million horsepower (careful, don't run this engine out to Mars!): I convert 12 million HP to Watts and get 8,950 Megawatts. Dividing this by the area of the nozzle:

$$I = \frac{P}{A=\pi R^2} \approx \ 215,000 \ \text{W/cm^2}$$

Though if you note the manufacturer of the SSME says this is at 109% power, not liftoff power as the Wikipedia article indicates. Dividing by 1.09 yields the same answer.

Too tired to calculate for the V-2, but here's the link to the required info

I also add that the comparison to the CPU to engine nozzle exit is not fair. The equivalent of the nozzle exit is the external area of the CPU heat sink. A more fair comparison would be to the throat of the combustion chamber. With an expansion ratio of 77.5 on the SSME, you get to multiply by 77.5 to get the energy flux at the throat, or $15.3 \ \text{MW/cm^2}$.

Those rocket designers are dealing with energy fluxes that exceed that of the surface of the Sun by three orders of magnitude!