ROCKET PROPULSION Supplement #1 |

Rocket Nozzle Design: Optimizing Expansion for Maximum Thrust

A rocket engine is a device in which propellants are burned in a combustion chamber and the resulting high pressure gases are expanded through a specially shaped nozzle to produce thrust. The function of the nozzle is to convert the chemical-thermal energy generated in the combustion chamber into kinetic energy. The nozzle converts the slow moving, high pressure, high temperature gas in the combustion chamber into high velocity gas of lower pressure and temperature. Gas velocities from 2 to 4.5 kilometers per second can be obtained in rocket nozzles. The nozzles which perform this feat are called DeLaval nozzles (after the inventor) and consist of a convergent and divergent section. The minimum flow area between the convergent and divergent section is called the nozzle throat. The flow area at the end of the divergent section is called the nozzle exit area.

Hot exhaust gases expand in the diverging section of the nozzle. The pressure of these gases will decrease as energy is used to accelerate the gas to high velocity. The nozzle is usually made long enough (or the exit area great enough) such that the pressure in the combustion chamber is reduced at the nozzle exit to the pressure existing outside the nozzle. It is under this condition that thrust is maximum and the nozzle is said to be adapted, also called optimum or correct expansion. To understand this we must examine the basic thrust equation:

F = q × Ve + (Pe - Pa) × Ae where F = Thrust q = Propellant mass flow rate Ve = Velocity of exhaust gases Pe = Pressure at nozzle exit Pa = Ambient pressure Ae = Area of nozzle exitThe product

Let us now consider an example. Assume we have a rocket engine equipped with an extendible
nozzle. The engine is test fired in an environment with a constant ambient pressure. During the
burn, the nozzle is extended from its fully retracted position to its fully extended position.
At some point between fully retracted and fully extended *Pe=Pa* (see figure below).

As we extend the nozzle, the momentum thrust increases as *Ve* increases. At the same
time the pressure thrust decreases as *Pe* decreases. The increase in momentum thrust is
greater than the decrease in pressure thrust, thus the total thrust of the engine increases as
we approach the condition *Pe=Pa*. As we continue to extend to nozzle the situation changes
slightly. Now the pressure thrust changes in magnitude more rapidly than the momentum thrust,
thus the total thrust begins to decrease.

Let's now apply some numbers to our example and run through the calculations to prove that this is true. Assume our rocket engine operates under the following conditions:

q = Propellant mass flow rate = 100 kg/s k = Specific heat ratio = 1.20 M = Exhaust gas molecular weight = 24 Tc = Combustion chamber temperature = 3600 K Pc = Combustion chamber pressure = 5 MPa Pa = Ambient pressure = 0.05 MPa

If the nozzle is properly adapted to the operating conditions we have *Pe=Pa*, or
*Pe*=0.05 MPa.

The gas pressure and temperature at the nozzle throat is less than in the combustion chamber due to the loss of thermal energy in accelerating the gas to the local speed of sound at the throat. Therefore, we calculate the pressure and temperature at the nozzle throat,

Pt = Pc × [1 + (k - 1) / 2]^{-k/(k-1)}Pt = 5 × [1 + (1.20 - 1) / 2]^{-1.20/(1.20-1)}Pt = 2.82 MPa = 2.82x10^{6}N/m^{2}Tt = Tc × [1 / (1 + (k - 1) / 2)] Tt = 3,600 × [1 / (1 + (1.20 - 1) / 2)] Tt = 3,273 K

The area at the nozzle throat is given by

At = (q / Pt) × SQRT[ (R' × Tt) / (M × k) ] At = (100 / 2.82x10^{6}) × SQRT[ (8,314 × 3,273) / (24 × 1.20) ] At = 0.0345 m^{2}

The hot gases must now be expanded in the diverging section of the nozzle to obtain maximum thrust. The Mach number at the nozzle exit is given by

Nm^{2}= (2 / (k - 1)) × [(Pc / Pa)^{(k-1)/k}- 1] Nm^{2}= (2 / (1.20 - 1)) × [(5 / 0.05)^{(1.20-1)/1.20}- 1] Nm^{2}= 11.54 Nm = (11.54)^{1/2}= 3.40

The nozzle exit area corresponding to the exit Mach number is given by

Ae = (At / Nm) × [(1 + (k - 1) / 2 × Nm^{2})/((k + 1) / 2)]^{(k+1)/(2(k-1))}Ae = (0.0345 / 3.40) × [(1 + (1.20 - 1) / 2 × 11.54)/((1.20 + 1) / 2)]^{(1.20+1)/(2(1.20-1))}Ae = 0.409 m^{2}

The velocity of the exhaust gases at the nozzle exit is given by

Ve = SQRT[ (2 × k / (k - 1)) × (R' × Tc / M) × (1 - (Pe / Pc)^{(k-1)/k}) ] Ve = SQRT[ (2 × 1.20 / (1.20 - 1)) × (8,314 × 3,600 / 24) × (1 - (0.05 / 5)^{(1.20-1)/1.20}) ] Ve = 2,832 m/s

Finally, we calculate the thrust,

F = q × Ve + (Pe - Pa) × Ae F = 100 × 2,832 + (0.05x10^{6}- 0.05x10^{6}) × 0.409 F = 283,200 N

Let's now consider what happens when the nozzle is under-extended, that is *Pe>Pa*. If we
assume *Pe=Pa* × 2, we have

Pe = 0.05 × 2 = 0.10 MPa At = 0.0345 m^{2}Nm^{2}= (2 / (1.20 - 1)) × [(5 / 0.10)^{(1.20-1)/1.20}- 1] Nm^{2}= 9.19 Nm = (9.19)^{1/2}= 3.03 Ae = (0.0345 / 3.03) × [(1 + (1.20 - 1) / 2 × 9.19)/((1.20 + 1) / 2)]^{(1.20+1)/(2(1.20-1))}Ae = 0.243 m^{2}Ve = SQRT[ (2 × 1.20 / (1.20 - 1)) × (8,314 × 3,600 / 24) × (1 - (0.10 / 5)^{(1.20-1)/1.20}) ] Ve = 2,677 m/s F = 100 × 2,677 + (0.10x10^{6}- 0.05x10^{6}) × 0.243 F = 279,850 N

Now we consider the over-extended condition, that is *Pe<Pa*. If we assume
*Pe=Pa* / 2, we have

Pe = 0.05 / 2 = 0.025 MPa At = 0.0345 mWe see that both the under-extended and over-extended nozzles produce thrusts less than that produced when the condition^{2}Nm^{2}= (2 / (1.20 - 1)) × [(5 / 0.025)^{(1.20-1)/1.20}- 1] Nm^{2}= 14.18 Nm = (14.18)^{1/2}= 3.77 Ae = (0.0345 / 3.77) × [(1 + (1.20 - 1) / 2 × 14.18)/((1.20 + 1) / 2)]^{(1.20+1)/(2(1.20-1))}Ae = 0.696 m^{2}Ve = SQRT[ (2 × 1.20 / (1.20 - 1)) × (8,314 × 3,600 / 24) × (1 - (0.025 / 5)^{(1.20-1)/1.20}) ] Ve = 2,963 m/s F = 100 × 2,963 + (0.025x10^{6}- 0.05x10^{6}) × 0.696 F = 278,900 N

As can be easily seen, thrust is maximum when *Pa/Pe*=1, or when *Pe=Pa*.

By Robert A. Braeunig, 1998.