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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 k1 W / m / K k W 330 3mm mK ⎛⎞ →= ⎜⎟ δ δ= ⎝⎠ ² compared to g 2 W h 50,000 , so OK here). Also, from governing equation 22 TT T ck tt xx ∂∂ ∂∂ =→ = α ∂∂ ρ T ( k c α= , thermal diffusivity , m/ ) 2 s we see that 2 2 x xt , o r , o r t αα 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 ( 1 2 solid graphyte), so 62 1 1.3 10 m /s 710 1100 = × × . The layer “adapts” in () 2 3 6 31 0 t (more like 7 . 0 s e c 1.3 10 × = × 2 1.8sec 4 δ = ). α Treat front layer quasi-statically , i.e., responding instantly to changes in heat flux: 11 wh wc 1 1 kq δ ² t This also means we can lump the thermal resistances of BL and 1 st layer in series: 1 g1 g eff hk h δ + ² 16.512, Rocket Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 2 of 12
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and since 1 g 1 k h δ ± , () 1 gg eff 1 k hh 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 ² (For instance, say Copper, 2 W k3 6 0 mK ³ , with 2 2cm. δ = We now have 2 g 2 eff 2 k W h3 5 0 mK = δ ³ , but 2 2 2 k W 36,000 = δ , so indeed, 2 g eff 2 k h δ ² ). Under these conditions, the heat sink is being “trickle charged” through the high thermal resistance of layer 1. Most likely, heat has time to redistribute internally, so that is nearly uniform 2 T across the layer. We can then write a lumped equation.
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lecture_10 - 16.512, Rocket Propulsion Prof. Manuel...

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