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How does Blue Horizons be4 manage to achieve such low chamber pressure compared to the SpaceX raptor engine or the shuttle main engines and yet still deliver more thrust given it does not appear to be significantly bigger ?

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    $\begingroup$ You should compare not only chamber pressure but also nozzle diameter (at the narrow point), exhaust velocity and propellant mass flow. $\endgroup$ – Uwe Apr 4 '18 at 15:55
  • $\begingroup$ You might note that chamber pressure does not appear in the thrust equation. grc.nasa.gov/www/k-12/airplane/rockth.html $\endgroup$ – Organic Marble Apr 4 '18 at 20:46
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As Organic Marble suggests the thrust equation

F = dm/dt * v_e + (p_e - p_a) * A_e

somewhat hides the chamber pressure; it is only relevant for v_e. (In this formular a is for ambient, e = nozzle exit)

Being heat engines rocket engine are limited by Carnot's theorem which basically says that the maximum efficiency (how much of the energy stored in the fuel can actually be used) is:

eta = 1 - T_min / T_max

In the case of a rocket engine the T_max is always the chamber temperature and T_min is the temperature of the exhaust gas. The chamber temperature is relatively fixed by the flame temperature of our fuel and only slightly increases with the chamber pressure (but that additional energy is gained from the pumps so this is more or less a zero sum game).

What we can do is reduce T_min namely the exhaust temperature. In an optimal isentropic process:

T_min / T_max = (p_min / p_max)^((kappa - 1) / kappa)

where kappa is the heat capacity ratio.

That means we can increase our efficiency if we reduce p_min / p_max. Sadly we can't reduce p_min = p_e below the ambient pressure p_a simply using a bigger nozzle because we would face flow seperation and a performance degradation as seen in the second part of the thrust equation (p_e - p_a) * A_e. The optimal exhaust pressure is exactly equal to the ambient pressure.

In order to achieve small ratios we can therefore do only one thing in high ambient pressures: increase the chamber pressure. But working out the numbers there isn't that much gain (asuming p_a = 0.5 bar which is true after the first 5km, kappa ~ 1.4):

eta_merlin = 1 - (0.05 MPa / 9.72 MPa)^((1.4 - 1) / 1.4) = 0.7781
eta_be4    = 1 - (0.05 MPa / 13.4 MPa)^((1.4 - 1) / 1.4) = 0.7976
eta_raptor = 1 - (0.05 MPa / 25.0 MPa)^((1.4 - 1) / 1.4) = 0.8306

which means Raptor might perform up to ~4% better in the atmosphere but higher chamber pressure adds complexity and results in higher dry mass which reduces payload.

In the end the difference is not as big as one might expect.


The Wikipedia article on isentropic nozzle flow is a good read if you want more details.

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  • $\begingroup$ But what about the combustion chamber volume? Is it possible to reduce the necessary chamber volume by increasing chamber pressure? $\endgroup$ – Uwe Apr 5 '18 at 15:22

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