Film Cooling in LF RocketsErin Schmidt

Watch the Heat!

Temperatures in LF rocket engines can reach 3600 K

Heat transfer rates are on the order of 200 MW/m^2

The basic problem is keeping the nozzle and combustion chamber walls from

vaporizing

Rocket Engine Heat Transfer

There are a few classical approaches to dealing with this problem:

Passive cooling -> ablation, radiative cooling, heat sinksActive cooling -> regenerative cooling, film cooling

● LOX-H2 Staged Combustion● 3000 psi Chamber pressure● 2300 kN Thrust Ea.

Late 1960’s technology, but remains the current state-of-the-art

cryogenic engine.The SSME used film cooling

Source: NASA MSFC

RS-25 SSME

Film Cooling

Source: NASA SP-8124, Anon., 1977

Ultimately the thing we care about is Tw

Skipping straight to the solution (similarity solution for B.L. from Schlichting):

(Tw-Ti )/(Tm-Ti )=1-exp[h2αt/k2]erfc[h sqrt(αt)/k]

A Conceptual Model

All of the assumptions implicit in this result are wrong

Big Problems● Flow has 3D wakes due to injector geometry,

lateral flows, separation, and vorticity ● Homogenous reactions within the film● Turbulent mixing within the film● Shear mixing at the film interface● Compressibility-> pressure, density and

temperature gradients● Flow discontinuities (e.g. shock)

For any practical rocket engine the simple model has laughably bad predictive power

Big Problems Cont.

For an isentropic converging-diverging nozzle it is ill-advised to make any of the constant properties assumptions

Besides making the governing equations really complex, this adds several levels of coupling between the governing eqns Source: Huzel and Huang,

1967

Big Problems Cont.

One specific example of coupling is that the reactions in the film are weakly diffusion controlled

But binary diffusion coefficients depend on temperature and pressure (which are changing axially in the isentropic flow). Also the reactions couple the energy and species conservation eqns. because the reactions are not isothermal

Combustion Products

We need to relate T and r. We can do this by balancing enthalpy

Hess’s law states that “if a reaction can be carried out as a series of steps, ∆H for the entire reaction is the sum of all ∆H for each step”...

∆Hrxn= ∑np∆Hf,products- ∑nr∆Hf,reactants

Combustion Products

When a chemical reaction is in equilibrium there are no changes in the molar concentrations of products and reactants with time

For a general reaction aA + bB cC + dD

there is a constant s.t. Kc= [C]c[D]d/[A]a[B]b

Combustion Products

If Kc > 1 equilibrium favors productsIf Kc < 1 equilibrium favors reactants

You can also write Kc in terms of partial pressures

Kc=pCcpD

d/pAapB

b

Equilibrium Constant Method

Example: chemical species at equilibrium2H2(g) + O2(g) 2H2O(g)

a) H2 + ½ O2 H2Ob) ½ O2 + ½ H2 OHc) ½ H2 Hd) ½ O2 O

Equilibrium Constant Method

Equilibrium constants are known functions of temperature! Using the equilibrium constant eqns...

a) Kp,a= pH2O/pH2p1/2O2

b) Kp,b= pOH/p1/2H2p1/2

O2

c) Kp,c= pH/p1/2H2

d)Kp,d= pO/p1/2O2

Equilibrium Constant Method

Solve for pi , but we only have 4 eqns. in 6

unknowns…Using continuity and the ideal gas law:2pH2O+ pOH+ 2pH2+ pH=NH(RT/V)

pH2O+ pOH+ 2pO2+ pO=NO(RT/V)With NH and NO being determined by the concentration gradient

Combustion Temperature

Now we can solve for pi (and thus ni moles of gas per chemical species) ...

We can use ni to get the temperature in the reacting B.L.

Combustion Temperature

Using a simple iterative method:1. Assume a chamber temperature T’ at a

given pressure

2. Determine equilibrium composition of combustion species at T’

3. Calculate:Qavailable=∆Hrxn+ ∑ini,reactants∫TiTrefCp,idT

Combustion Temperature

4. Calculate:

(Use a curve fit equation for Cp , usually found in thermodynamic tables, or use lookups from an enthalpy table in lieu of the integral)

Qrequired= -∑ini,products∫T’TrefCpdT

Combustion Temperature

5. Compare Qavailable and Qrequired ; if the difference is greater than your chosen threshold assume a new T’

6. Iterate until T converges7. Use P, T to compute D with Chapman-Engkog or other

The point is realistic film-cooling simulations accounting

for all these problems are incredibly difficult to do

In fact no one has managed to make one yet

Std. NASA CorrelationBecause analytical and numerical approaches leave much to be desired propulsion engineers rely on empirical correlations. This one is the standard for liquid films:

where,

Reference1. “Liquid Rocket Engine Self-Cooled Combustion Chambers.” 1977.

http://ntrs.nasa.gov/search.jsp?R=19780013268.

2. Huang, D. H., and D. K. Huzel. 1971. “Design of Liquid Propellant

Rocket Engines Second Edition.”

http://ntrs.nasa.gov/search.jsp?R=19710019929.

3. Schlichting, H., and K. Gersten. 2000. Boundary-Layer Theory.

8th edition. Berlin ; New York: Springer.

Thanks!Source: NASA

MSFC