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lecture_10

# lecture_10 - 16.512 Rocket Propulsion Prof Manuel...

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16.512, Rocket Propulsion Prof. Manuel Martinez-Sanchez Lecture 10: Ablative Cooling, Film Cooling Transient Heating of a Slab Typical problem: Uncooled throat of a solid propellant rocket Inner layer retards heat flux to the heat sink. Heat sink’s T gradually rises during firing (60-200 sec). Peak T of heat sink to remain below matl. limit. Back T of heat sink to remain below weakening point for structure. Prototype 1-D problem: Can be solved exactly, or can do transient 1-D numerical computation. But it is useful to look at basic issues first. Thermal conductance of g B.L.=h Thermal conductance of front layer 1 1 k = δ Thermal conductance of layer i i i k = δ ( i δ = thickness, = thermal conductivity) i k 16.512, Rocket Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 1 of 12

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Want layer 1 to have 1 g 1 k h δ ± to protect the rest. (Say, porous, Oriented graphyte, 1 1 2 1 1 k 1W /m/K k W 330 3mm m K = δ δ = ² compared to g 2 W h 50,000 m K , so OK here). Also, from governing equation 2 2 2 2 T T T c k t t x x = = α ρ T ( k c α = ρ , thermal diffusivity , m / ) 2 s we see that 2 2 x x t, or x t , or t α α α . So the layer 1 will “adapt” to its boundary conditions in a time 2 1 1 t δ α . Say, J c 710 KgK ² and 3 Kg 1100 m ² ρ ( 1 2 solid graphyte), so 6 2 1 1.3 10 m /s 710 1100 α = = × × . The layer “adapts” in ( ) 2 3 6 3 10 t (more like 7.0sec 1.3 10 × = × 2 1.8sec 4 δ = ). α Treat front layer quasi-statically , i.e., responding instantly to changes in heat flux: ( ) ( ) ( ) t t 1 1 wh wc 1 1 T T k q δ ² t This also means we can lump the thermal resistances of BL and 1 st layer in series: ( ) 1 g 1 g eff 1 1 h k h δ + ² 16.512, Rocket Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 2 of 12
and since 1 g 1 k h δ ± , ( ) 1 g g eff 1 k h h δ ± For layer 2 (the heat sink), is high (metal) and 2 k ( ) g eff h is now small (thanks to 1 st layer) so, very likely, ( ) 2 g eff 2 k h δ ³ (For instance, say Copper, 2 W k 360 mK ² , with 2 2cm. δ = We now have ( ) 2 g 2 eff 2 k W h 350 m K = δ ² , but 2 2 2 k W 36,000 m K = δ , so indeed, ( ) 2 g eff 2 k h δ ³ ).

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