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

Ek convective interface resistance contribute karta hai $R=1/h$ — kyun? Kyunki Newton's law of cooling kehta hai $q=h\Delta T$, toh $\Delta T=q/h=q\cdot (1/h)$; "$\Delta T=q\cdot R$" se compare karo toh resistance coefficient ka reciprocal hai. Dekho Newton's Law of Cooling. $R_g=\frac{1}{h_g}=\frac{1}{2.5\times 10^4}=4.0\times 10^{-5}{\text{m}}^2\text{K/W}$ Yeh gas-side convective stage ko represent karta hai — yaani hot combustion gas se wall surface mein heat cross karna.

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

Teen series resistances ko sum karo (kyun series? steady state mein same $q$ har layer se guzarta hai, toh unke temperature drops add hote hain): $R_g=\frac{1}{1.8\times 10^4}=5.556\times 10^{-5},R_w=\frac{0.0008}{350}=2.286\times 10^{-6},R_c=\frac{1}{4.0\times 10^4}=2.5\times 10^{-5}$ $R=R_g+R_w+R_c=8.284\times 10^{-5}{\text{m}}^2\text{K/W}$ $U=\frac{1}{R}=1.207\times 10^4{\text{W/m}}^2\text{K}$ $q=U(T_{aw}-T_{co})=1.207\times 10^4\times (3200-350)=3.44\times 10^7{\text{W/m}}^2\approx 34.4{\text{MW/m}}^2$

Gas-side interface drop hai $q{R}_g$ (kyun: sirf gas-side resistance $T_{aw}$ aur $T_{wg}$ ke beech hai): $T_{wg}=T_{aw}-q{R}_g=3200-(3.44\times 10^7)(5.556\times 10^{-5})=3200-1912=1288\text{K}$ $T_{wg}\approx 1288$ K above hai ~1000 K soften limit se → as-is safe nahi hai. Design fix: $h_c$ badhao (faster coolant) ya $R_g$ ghataao taaki $T_{wg}$ neeche aaye.

$Q=qA=3.44\times 10^7\times 0.4=1.376\times 10^7\text{W}$ Energy balance (wall mein jaane wala har joule coolant ke badhte temperature mein store hota hai): ${\dot{m}}_c=\frac{Q}{c_p\Delta T_c}=\frac{1.376\times 10^7}{4200\times 180}=18.2\text{kg/s}$

Total $R=8.284\times 10^{-5}$. $\frac{R_g}{R}=\frac{5.556}{8.284}=67.1\%,\frac{R_c}{R}=\frac{2.5}{8.284}=30.2\%,\frac{R_w}{R}=\frac{0.2286}{8.284}=2.76\%$ Gas side dominate karta hai (≈67 %). Yeh parent ka claim confirm karta hai: sabse hot interface ki largest resistance hoti hai aur sab kuch control karta hai. Copper wall khud 3 % se kam contribute karta hai — proof hai ki thin, high-$k$ walls heat ko barely resist karti hain.

(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 $=\Delta p_2/\Delta p_1=60/15=4$. $v$ double karne se $\Delta p$ quadruple ho jaata hai kyunki $\Delta p\propto v^2$. Yeh cruel trade-off hai — dekho Darcy-Weisbach Equation.

$h_c\propto v^{0.8}$, toh $v$ double karne par: $\frac{h_{c,2}}{h_{c,1}}=2^{0.8}=1.741$ Toh $h_c$ sirf ~74 % badhta hai, jabki $\Delta p$ 300 % badha (×4). Cooling modestly improve hoti hai; pump cost explode ho jaata hai. Isliye zyada flow free nahi hai — aur analysis ke liye dekho 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, $N=100$ channels of area $A_{cs}$ mein split: $v=\frac{{\dot{m}}_c/\rho }{N{A}_{cs}}=\frac{0.03064}{100\times 4\times 10^{-6}}=\frac{0.03064}{4\times 10^{-4}}=76.6\text{m/s}$ (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 kaam karti hai ($q$, ${\dot{m}}_c$ reasonable hain), lekin jacket pressure drop ka 141 bar brutal hai — pump ko chamber pressure ke upar woh bhi supply karna hoga. Ek designer channels wide karega ya unki sankhya badhayega $v$ ghataane ke liye.

Compute $\frac{\gamma -1}{2}{M}^2=\frac{0.2}{2}(0.3{)}^2=0.1\times 0.09=0.009$. $T_{aw}=3600\times \frac{1+0.89\times 0.009}{1+0.009}=3600\times \frac{1.00801}{1.009}=3600\times 0.999118=3596.8\text{K}$ Yah $T_c$ se sirf ~3.2 K neeche hai yahan (chamber mein low Mach → tiny correction). $T_c$ use karne par driving $\Delta T$ aur hence $q$ thodi overestimate hogi, coolant waste hoga — effect high-Mach nozzle throat mein badh jaata hai. Dekho Adiabatic Wall Temperature and Recovery Factor.

(a) Mass flow se velocity: $v=\frac{\dot{m}}{\rho A_{cs}}=\frac{\dot{m}}{\rho w^2}$. Darcy–Weisbach mein $d_h=w$ ke saath substitute karo: $\Delta p=f\frac{L}{w}\frac{\rho v^2}{2}=f\frac{L}{w}\frac{\rho }{2}{\left(\frac{\dot{m}}{\rho w^2}\right)}^2=\frac{fL{\dot{m}}^2}{2\rho w^5}$ Toh $\Delta p\propto w^{-5}$ — channel shrink karne ka ek extremely steep penalty. (b) $w=1$ mm ke saath: $v=\frac{0.25}{800\times (10^{-3}{)}^2}=\frac{0.25}{8\times 10^{-4}}=312.5$ m/s; $\Delta p_{1mm}=\frac{0.02\times 0.5\times 0.25^2}{2\times 800\times (10^{-3}{)}^5}=\frac{6.25\times 10^{-4}}{1.6\times 10^{-12}}=3.906\times 10^8\text{Pa}$ $w=2$ mm ke saath: $\Delta p_{2mm}=\Delta p_{1mm}\times (2{)}^{-5}=3.906\times 10^8/32=1.221\times 10^7$ Pa. Ratio $=32$: kuch half nahi — width double karne se pressure drop 32 ke factor se cut ho jaata hai. (c) Lesson: kyunki $\Delta p\propto w^{-5}$ fixed mass flow par, chhote channels pump load ke liye catastrophic hain. Designers parallel mein kaafi wide channels use karte hain taaki velocity (aur thus $h_c$) cooling ke liye kaafi high rahe aur saath hi $\Delta p$ survivable rahe — parent note ka tug-of-war quantitative ho gaya.

(a) Remove karne wali power: $Q=qA=8.0\times 10^7\times 0.15=1.2\times 10^7$ W. Coking se pehle coolant max power absorb kar sakta hai: $Q_{max}=\dot{m}{c}_p\Delta T_{c,max}=12\times 4200\times 250=1.26\times 10^7$ W. Kyunki $Q=1.2\times 10^7<Q_{max}=1.26\times 10^7$, yeh just fit hota hai — margin sirf $\frac{1.26-1.2}{1.26}=4.8\%$. Dangerously thin: koi bhi flux underestimate coking → film boiling → burnout cause karega. (b) Itne thin margin ke saath designer Film Cooling add karta hai — gas-side wall ke along ek thin coolant film inject karta hai $T_{aw}$ locally kam karne aur $q$ cut karne ke liye — ya, short-burn/expendable engine ke liye, Ablative Cooling jahan ek sacrificial liner char hoti hai aur mass loss se heat carry hoti hai. Dono regen jacket par load kam karte hain, safe margin restore karte hain.

Series resistances sum hoti hain, phir ek baar invert karo ::: $R=1/h_g+t_w/k_w+1/h_c$, aur $U=1/R$ Cooling coefficient ka velocity ke saath scaling ::: $h_c\propto v^{0.8}$ (sub-linear) Pressure-drop ka velocity ke saath scaling ::: $\Delta p\propto v^2$ (super-linear) Pressure-drop ka square channel width ke saath fixed mass flow par scaling ::: $\Delta p\propto w^{-5}$ Gas side flux kyun control karta hai ::: kyunki iska specific resistance sabse bada hota hai (~67 % yahan) Jab regen akela marginal ho, supplement karo ::: Film Cooling ya Ablative Cooling se