Exercises — Regenerative cooling — heat flux, coolant flow, pressure drop

A convective interface contributes resistance $R=1/h$ — why? Because Newton's law of cooling says $q=h\Delta T$, so $\Delta T=q/h=q\cdot (1/h)$; comparing to "$\Delta T=q\cdot R$" the resistance is the reciprocal of the coefficient. See Newton's Law of Cooling. $R_g=\frac{1}{h_g}=\frac{1}{2.5\times 10^4}=4.0\times 10^{-5}{\text{m}}^2\text{K/W}$ It represents the gas-side convective stage — heat crossing from hot combustion gas into the wall surface.

• (i) hot gas → wall surface: convection, Newton's Law of Cooling, $q=h_g(T_{aw}-T_{wg})$.
• (ii) inside the metal: conduction, Fourier's Law of Conduction, $q=\frac{k_w}{t_w}(T_{wg}-T_{wc})$.
• (iii) wall → coolant: convection, Newton again, $q=h_c(T_{wc}-T_{co})$; the coefficient $h_c$ comes from the Dittus-Boelter Correlation.

Sum the three series resistances (why series? the same $q$ passes through each layer in steady state, so their temperature drops add): $R_g=\frac{1}{1.8\times 10^4}=5.556\times 10^{-5},R_w=\frac{0.0008}{350}=2.286\times 10^{-6},R_c=\frac{1}{4.0\times 10^4}=2.5\times 10^{-5}$ $R=R_g+R_w+R_c=8.284\times 10^{-5}{\text{m}}^2\text{K/W}$ $U=\frac{1}{R}=1.207\times 10^4{\text{W/m}}^2\text{K}$ $q=U(T_{aw}-T_{co})=1.207\times 10^4\times (3200-350)=3.44\times 10^7{\text{W/m}}^2\approx 34.4{\text{MW/m}}^2$

The gas-side interface drop is $q{R}_g$ (why: only the gas-side resistance stands between $T_{aw}$ and $T_{wg}$): $T_{wg}=T_{aw}-q{R}_g=3200-(3.44\times 10^7)(5.556\times 10^{-5})=3200-1912=1288\text{K}$ $T_{wg}\approx 1288$ K is above the ~1000 K soften limit → not safe as-is. Design fix: raise $h_c$ (faster coolant) or reduce $R_g$ to pull $T_{wg}$ down.

$Q=qA=3.44\times 10^7\times 0.4=1.376\times 10^7\text{W}$ Energy balance (every joule into the wall is stored in the coolant's rising temperature): ${\dot{m}}_c=\frac{Q}{c_p\Delta T_c}=\frac{1.376\times 10^7}{4200\times 180}=18.2\text{kg/s}$

Total $R=8.284\times 10^{-5}$. $\frac{R_g}{R}=\frac{5.556}{8.284}=67.1\%,\frac{R_c}{R}=\frac{2.5}{8.284}=30.2\%,\frac{R_w}{R}=\frac{0.2286}{8.284}=2.76\%$ The gas side dominates (≈67 %). This confirms the parent's claim: the hottest interface has the largest resistance and controls everything. The copper wall itself contributes under 3 % — proof that thin, high-$k$ walls barely resist heat.

(a) $\Delta p_1=f\frac{L}{d_h}\frac{\rho v^2}{2}=0.02\times \frac{0.6}{0.002}\times \frac{800\times 25^2}{2}=0.02\times 300\times 2.5\times 10^5=1.5\times 10^6\text{Pa}=15\text{bar}$ (b) $\Delta p_2=0.02\times 300\times \frac{800\times 50^2}{2}=0.02\times 300\times 10^6=6.0\times 10^6\text{Pa}=60\text{bar}$ Ratio $=\Delta p_2/\Delta p_1=60/15=4$. Doubling $v$ quadruples $\Delta p$ because $\Delta p\propto v^2$. This is the cruel trade-off — see Darcy-Weisbach Equation.

$h_c\propto v^{0.8}$, so doubling $v$ gives: $\frac{h_{c,2}}{h_{c,1}}=2^{0.8}=1.741$ So $h_c$ rises only ~74 %, while $\Delta p$ rose by 300 % (×4). The cooling improves modestly; the pump cost explodes. This is exactly why more flow is not free — analysed further in Turbopump Sizing.

(a) Resistances: $R_g=5\times 10^{-5}$, $R_w=0.001/350=2.857\times 10^{-6}$, $R_c=2\times 10^{-5}$; $R=7.286\times 10^{-5}$. $q=\frac{3300-300}{7.286\times 10^{-5}}=4.117\times 10^7{\text{W/m}}^2\approx 41{\text{MW/m}}^2$ (b) $Q=qA=4.117\times 10^7\times 0.5=2.059\times 10^7$ W. ${\dot{m}}_c=\frac{Q}{c_p\Delta T_c}=\frac{2.059\times 10^7}{4200\times 200}=24.5\text{kg/s}$ (c) Volume flow $={\dot{m}}_c/\rho =24.5/800=0.03064$ m³/s, split over $N=100$ channels of area $A_{cs}$: $v=\frac{{\dot{m}}_c/\rho }{N{A}_{cs}}=\frac{0.03064}{100\times 4\times 10^{-6}}=\frac{0.03064}{4\times 10^{-4}}=76.6\text{m/s}$ (d) $\Delta p=f\frac{L}{d_h}\frac{\rho v^2}{2}=0.02\times \frac{0.6}{0.002}\times \frac{800\times 76.6^2}{2}=0.02\times 300\times 2.347\times 10^6=1.408\times 10^7\text{Pa}\approx 141\text{bar}$ Verdict: cooling works ($q$, ${\dot{m}}_c$ reasonable), but 141 bar of jacket pressure drop is brutal — the pump must supply that on top of chamber pressure. A designer would widen channels or add more of them to cut $v$.

Compute $\frac{\gamma -1}{2}{M}^2=\frac{0.2}{2}(0.3{)}^2=0.1\times 0.09=0.009$. $T_{aw}=3600\times \frac{1+0.89\times 0.009}{1+0.009}=3600\times \frac{1.00801}{1.009}=3600\times 0.999118=3596.8\text{K}$ It is only ~3.2 K below $T_c$ here (low Mach in the chamber → tiny correction). Using $T_c$ would slightly overestimate the driving $\Delta T$ and hence $q$, wasting coolant — the effect grows in the high-Mach nozzle throat. See Adiabatic Wall Temperature and Recovery Factor.

(a) Velocity from mass flow: $v=\frac{\dot{m}}{\rho A_{cs}}=\frac{\dot{m}}{\rho w^2}$. Substitute into Darcy–Weisbach with $d_h=w$: $\Delta p=f\frac{L}{w}\frac{\rho v^2}{2}=f\frac{L}{w}\frac{\rho }{2}{\left(\frac{\dot{m}}{\rho w^2}\right)}^2=\frac{fL{\dot{m}}^2}{2\rho w^5}$ So $\Delta p\propto w^{-5}$ — an extremely steep penalty for shrinking the channel. (b) With $w=1$ mm: $v=\frac{0.25}{800\times (10^{-3}{)}^2}=\frac{0.25}{8\times 10^{-4}}=312.5$ m/s; $\Delta p_{1mm}=\frac{0.02\times 0.5\times 0.25^2}{2\times 800\times (10^{-3}{)}^5}=\frac{6.25\times 10^{-4}}{1.6\times 10^{-12}}=3.906\times 10^8\text{Pa}$ With $w=2$ mm: $\Delta p_{2mm}=\Delta p_{1mm}\times (2{)}^{-5}=3.906\times 10^8/32=1.221\times 10^7$ Pa. Ratio $=32$: halving nothing — doubling the width cuts pressure drop by a factor of 32. (c) Lesson: because $\Delta p\propto w^{-5}$ at fixed mass flow, small channels are catastrophic for pump load. Designers use many wider channels in parallel to hold velocity (and thus $h_c$) high enough for cooling while keeping $\Delta p$ survivable — the tug-of-war of the parent note made quantitative.

(a) Power to remove: $Q=qA=8.0\times 10^7\times 0.15=1.2\times 10^7$ W. Max power the coolant can absorb before coking: $Q_{max}=\dot{m}{c}_p\Delta T_{c,max}=12\times 4200\times 250=1.26\times 10^7$ W. Since $Q=1.2\times 10^7<Q_{max}=1.26\times 10^7$, it just fits — margin only $\frac{1.26-1.2}{1.26}=4.8\%$. Dangerously thin: any flux underestimate causes coking → film boiling → burnout. (b) With such thin margin a designer adds Film Cooling — injecting a thin coolant film along the gas-side wall to lower $T_{aw}$ locally and cut $q$ — or, for a short-burn/expendable engine, Ablative Cooling where a sacrificial liner chars and carries heat away by mass loss. Both reduce the load the regen jacket must carry, restoring safe margin.

Series resistances sum, then invert once ::: $R=1/h_g+t_w/k_w+1/h_c$, and $U=1/R$ Cooling coefficient scaling with velocity ::: $h_c\propto v^{0.8}$ (sub-linear) Pressure-drop scaling with velocity ::: $\Delta p\propto v^2$ (super-linear) Pressure-drop scaling with square channel width at fixed mass flow ::: $\Delta p\propto w^{-5}$ Why the gas side controls flux ::: it carries the largest specific resistance (~67 % here) When regen alone is marginal, supplement with ::: Film Cooling or Ablative Cooling