Exercises — Transpiration cooling

Effectiveness is how far the wall has been dragged down from gas toward coolant: $\varphi =\frac{T_g-T_w}{T_g-T_c}=\frac{3000-1500}{3000-300}=\frac{1500}{2700}=\mathrm{0.556.}$ Meaning: the wall closed $55.6\%$ of the total possible temperature gap $(T_g-T_c)$. $\varphi =0$ would be no cooling ($T_w=T_g$); $\varphi =1$ would be perfect ($T_w=T_c$).

• Gas side: $\eta h_0=0.5\times 5000=2500\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.
• Coolant side: $G{c}_p=3\times 1200=3600\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

Coolant conductance ($3600$) $>$ gas conductance ($2500$), so the weighted average $T_w$ leans toward the coolant side — the wall stays comfortably cool ($\varphi =3600/6100=0.59$).

Step 1 — gas conductance. $\eta h_0=0.35\times 6000=2100$. Step 2 — coolant conductance. $G{c}_p=2.5\times 1000=2500$. Step 3 — weighted average. $T_w=\frac{2100(3200)+2500(400)}{2100+2500}=\frac{6720000+1000000}{4600}=\frac{7720000}{4600}\approx 1678\mathrm{K}.$ The wall runs at $\approx 1678\mathrm{K}$, less than the melting point of most nickel superalloys — survivable versus an impossible $3200\mathrm{K}$.

$\varphi =\frac{G{c}_p}{\eta h_0+G{c}_p}=\frac{2500}{2100+2500}=\frac{2500}{4600}=\mathrm{0.543.}$ Cross-check with $T_w$ from L2.1: $\frac{3200-1678}{3200-400}=\frac{1522}{2800}=0.544$. ✔ (Tiny gap is rounding of $T_w$.)

Step 1 — target effectiveness. $\varphi =\frac{3000-1000}{3000-300}=\frac{2000}{2700}=0.741$. Step 2 — invert $\varphi$. From $\varphi (\eta h_0+G{c}_p)=G{c}_p$: $G{c}_p=\frac{\varphi }{1-\varphi }\eta h_0=\frac{0.741}{0.259}\times (0.4\times 5000)=2.857\times 2000=5714.$ Step 3 — solve for $G$. $G=5714/1200\approx 4.76\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. Any $G\ge 4.76\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$ holds the wall at or below $1000\mathrm{K}$.

With $\eta h_0=2000$ fixed:

• $G=2$: $G{c}_p=2400$, $T_w=\frac{2000(3000)+2400(300)}{4400}=\frac{6720000}{4400}\approx 1527\mathrm{K}$.
• $G=4$: $G{c}_p=4800$, $T_w=\frac{6000000+1440000}{6800}=\frac{7440000}{6800}\approx 1094\mathrm{K}$.
• $G=8$: $G{c}_p=9600$, $T_w=\frac{6000000+2880000}{11600}=\frac{8880000}{11600}\approx 766\mathrm{K}$.

Drops per doubling: $1527\to 1094$ is $433\mathrm{K}$; $1094\to 766$ is only $328\mathrm{K}$. Each doubling buys less, because $T_w$ is squeezing toward its floor $T_c=300\mathrm{K}$.

(a) Constant $\eta =0.4$. $\eta h_0=2000$, $G{c}_p=3\times 1200=3600$. $T_w=\frac{2000(3000)+3600(300)}{5600}=\frac{6000000+1080000}{5600}=\frac{7080000}{5600}\approx 1264\mathrm{K}.$ (b) Variable $\eta$. $\eta (3)=\frac{1}{1+0.3\times 3}=\frac{1}{1.9}=0.526$. Wait — that is higher than $0.4$, meaning weaker cooling at this $G$. So $\eta h_0=0.526\times 5000=2632$. $T_w=\frac{2632(3000)+3600(300)}{2632+3600}=\frac{7896000+1080000}{6232}=\frac{8976000}{6232}\approx 1440\mathrm{K}.$ Lesson: at $G=3$ this particular model gives less blowing benefit than the flat $\eta =0.4$ assumption, so the real wall is hotter ($1440$ vs $1264\mathrm{K}$). The moral is not "variable $\eta$ is always better" — it is "you must use the actual local $\eta$, and a flattering constant can mislead you either way."

Common terms: $\eta =0.4$, $G{c}_p=3600$.

• Chamber: $\eta h_0=0.4\times 4000=1600$. $T_w=\frac{1600(3000)+3600(300)}{5200}=\frac{4800000+1080000}{5200}=\frac{5880000}{5200}\approx 1131\mathrm{K}.$
• Throat: $\eta h_0=0.4\times 9000=3600$. $T_w=\frac{3600(3000)+3600(300)}{7200}=\frac{10800000+1080000}{7200}=\frac{11880000}{7200}=1650\mathrm{K}.$

Comment: with uniform coolant the throat runs $\sim 519\mathrm{K}$ hotter than the chamber. This is exactly why designers inject more coolant at the throat — the heat flux is worst precisely where the geometry is tightest.

(a) Target $\varphi =\frac{3000-1200}{3000-300}=\frac{1800}{2700}=0.667$. $G{c}_p=\frac{\varphi }{1-\varphi }\eta h_0=\frac{0.667}{0.333}\times (0.4\times 9000)=2\times 3600=7200.$ $G=7200/1200=6\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. (b) ${\dot{m}}_c=GA=6\times 0.02=0.12\mathrm{kg}{\mathrm{s}}^{-1}$. (c) Fraction $=\frac{{\dot{m}}_c}{\dot{m}}=\frac{0.12}{50}=0.0024=0.24\%$. Reading: only about a quarter-percent of flow protects the throat — a good deal, but this dumped coolant produces little thrust, nudging Specific Impulse down. Minimize $G$ to the target, no more.

Step 1 — target effectiveness. $\varphi =\frac{3000-1000}{3000-300}=\frac{2000}{2700}=0.741$. Step 2 — write the balance in $G$. With $\eta h_0=\frac{5000}{1+0.3G}$ and $G{c}_p=1200G$: $\frac{\varphi }{1-\varphi }=\frac{G{c}_p}{\eta h_0}\Rightarrow \frac{0.741}{0.259}=\frac{1200G(1+0.3G)}{5000}.$ Left side $=2.857$. So $1200G(1+0.3G)=2.857\times 5000=14286$. Step 3 — quadratic. $360G^2+1200G-14286=0$, i.e. $G^2+3.333G-39.68=0$. $G=\frac{-3.333+\sqrt{3.333^2+4\times 39.68}}{2}=\frac{-3.333+\sqrt{11.11+158.73}}{2}=\frac{-3.333+\sqrt{169.84}}{2}=\frac{-3.333+13.03}{2}\approx 4.85\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}.$ Compare: with constant $\eta =0.4$ (L3.1) we needed $G\approx 4.76$. Here the falling blowing benefit means we need slightly more coolant, $G\approx 4.85$ — a small but honest correction from modelling $\eta (G)$.

• (a) $G\to 0$ (no coolant): $G{c}_p\to 0$, so $\varphi =\frac{0}{\eta h_0}=0$ and $T_w\to T_g$. Wall as hot as the gas — no protection at all.
• (b) $G\to \infty$ (flooded with coolant): $\frac{\eta h_0}{G{c}_p}\to 0$, so $\varphi \to 1$ and $T_w\to T_c$. Wall as cold as inlet coolant — perfect but propellant-hungry.
• (c) $\eta \to 0$ (blanket so thick almost no heat gets through): $\varphi \to \frac{G{c}_p}{G{c}_p}=1$, $T_w\to T_c$. Same perfect limit, reached via blocking rather than absorbing.
• (d) $T_c=T_g$: the driving gap $T_g-T_c=0$, so $\varphi$ is undefined (0/0) but $T_w=T_g=T_c$ trivially — there is simply no temperature difference to fight. Cooling is meaningless.

These four corners bracket every real operating point: the wall always lies between $T_c$ and $T_g$.

Compute $T_w$ from conductances ::: $T_w=\frac{\eta h_0{T}_g+G{c}_p{T}_c}{\eta h_0+G{c}_p}$ Invert for required coolant ::: $G{c}_p=\frac{\varphi }{1-\varphi }\eta h_0$ Why doubling $G$ doesn't halve $T_w$ ::: $T_w$ is bounded below by $T_c$; returns diminish What happens at the throat ::: $h_0$ spikes, so $T_w$ rises unless local $G$ is raised Cost of extra coolant ::: dumped coolant lowers specific impulse