16.50 Lecture 15 Subject: Ablative cooling By ablation we mean the recession of a surface due to heating, usually by a hot gas. It is the key process for a) Re-entry heat shields b) Solid propellant nozzles c) Rocket case insulation d) Fire-proofing skyscrapers' structures Consider a hot gas flowing over a surface which can 1) evaporate, and ) whose vapor can react with the external flow A heat balance at the surface gives q s = q w - (ρv) w Δh w (1) where the heat of ablation, Δh w =h w h s, in J/kg, can include a heat of vaporization and decomposition and q w is the heat flux from the fluid boundary layer. Let us assume that it is useful to write q w = ρ e (H e + ue - Hw) where H w is the total enthalpy for the wall material at the wall temperature, and H e =c pg T ge is the gas specific enthalpy at the temperature T ge just outside the boundary layer. Notice that this generalizes our previous expression for heat transfer by replacing the total enthalpy for the static enthalpy; this is of general validity for high-speed flows. By energy conservation in the core flow, u H e + e = c pgt c. 1
Then the heat transfer to the solid is ue s q s = ρ e (H e + - H w ) - (ρv) w (h w - h ) Since q s is the heat flux available to heat the wall, it is clear that the evaporation reduces the wall heat flux, i.e. the heat flux into the solid. Now what determines (ρv) w? First note that (ρv) w = rρ s where r is our "recession rate". Consider a "thermal wave" propagating into the solid. We will now work in the receding frame, in which we can assume a steady situation. If the recession rate is r(m/s), we see solid material moving at! y = "r, and convecting a heat flux. Then, if there is no local heat generation, d dt (!k s! r" s T ) = 0 dy dy dt which integrates to k s + r T = const. = r! st w, or dy " d(t! Tw" ) r ks + (T! T w" ) = o (! =, heat diffusivity, in m /s) dy # " s T! T w" = (T w! T w )e! # y " dt " r The heat flux at at y = o is q s =!k s $ = k s (T w! T w& ), giving dy # y= o % r q s = r(t w " T w# ), or r q = s (T w " T w# ) So returning to our expression for the heat flux to the wall, r (T w " T w# ) q s = c pg (T c " T w ) " r#h w! c pg (T c " T w )! r c (T " T ) s pg c w r =, = (T w " T w# ) + $hw (T w " T w# ) + $h w %& You may compare this to Sutton pg. 510 Eq. 15-6 Now how does (ρv) w influence? Physically, we know that the vapor comes off the wall with u=0, so it will tend to retard the flow near the wall.
Quantitatively, Lees says, for small surface blowoff effect, B 1. 8,30 (ρv) w = c!! 500 J / Kg / K, B = o e B!1 pg 0. 0 ρ e o while Sutton quotes Lees as giving for the larger blowoffs = 1.7 B -.77 5 < B < 100 o These give the same trends, but somewhat different numbers. We adopt here Lees r B formulation: We had =", so B =! =! B o e "1 or e B!1 =" # B =!n(1 +") so finally, +- % c pg (T c " T w ) (/- r =,!n '1 + *0 o -. & (T w " Tw# ) + $h w )-1 (1) and then % c (T " T ) ( q pg c w s sr) ( T w " T w# ) = 0 (T w " T w# )!n '1+ * & (T w w# ) + $h w ) (1 ) m! p At the throat, ( ue ) = = c t * w, and hence r, is maximum. A t c Numerically, one often sees!h w >> (T w " T w# ), and also! <<1. This would leave us with the approximations c pg (Tc # T! r " w ) s o $h w ()! o (3) and going back to the surface heat balance, q s! o, i.e., to the first order, the heat does not penetrate below the ablating layer, which is as intended. 3
Equations (1) or () can be used to select the proper thickness of the ablative (sacrificial) layer, once the burn time is given:! Abl = r t b. Now let us look at some numerical examples, to show the magnitudes of the terms: Joule A. For carbon, Hf g = 17 kcal/gmole=60 x 10 6 kg 17000 cal h w - h s = 1 = 15,000 g 1. 8, 30 cpg!! 500 J / Kg / K, ρ s 000 kg / m 3 kg / m 3 0. 0 and assume T c -T w =1,000K. p Now ρ e =! " 0.65 RT p RT 10 5 (p c atm) kg 58 (p c atm) 8.3x10 3 m s (3000) 0 Taking p c =100 atm and =0.001, we then calculate,500 *1,000 r!,000 " 60 "10 6 " 58 "100 " 0.001 = 1. "10 #4 m / s = 0.1 mm / s So in 100 sec., the change in the surface is 1 mm = 1. cm. This is of the right magnitude, but most important is to see what it depends on. B. For rubber h w -h s 10,000 cal/gm mole. For an approximate empirical formula C 10 H 0, this gives per unit mass 10000 10000 cal Joule = 70 80,000 10 + 0 140 g kg Also c p 1 cal/g 4,000 Joule/kg K So now we would get about 100 times the regression rate for carbon. In this case though, if the rubber is used in the casing of a solid rocket engine then ρ e << 1 (ρ e ) throat, maybe by 100, so the net regression may end up being similar. 4
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