Question bank — Regenerative cooling — heat flux, coolant flow, pressure drop

False. Thickness $t_w$ raises the wall resistance $R_w=t_w/k_w$, and a bigger resistance means a bigger temperature drop across the metal, so the gas-side face $T_{wg}$ gets hotter. Thin, high-$k$ copper is the fix.

True. In steady state the same heat flux $q$ threads through all three layers one after another, exactly like current through series resistors, so their per-area resistances sum.

False. Darcy–Weisbach gives $\Delta p\propto v^2$, so doubling $v$ quadruples $\Delta p$. Velocity is the expensive knob.

True. More flow raises ${\dot{m}}_c$ (carries heat away faster, $Q={\dot{m}}_c{c}_p\Delta T_c$) and raises $h_c$ via Dittus–Boelter ($h_c\propto {\text{Re}}^{0.8}$), lowering wall temperature.

False. The coolant is the propellant; the heat it soaked up goes back into the combustion chamber with it. That recycling is exactly what "regenerative" means.

False (in this design). A vapour film insulates the wall (film boiling), collapsing $h_c$ and spiking $T_{wg}$ toward burnout. This is why $\Delta T_c$ is capped below the boiling/coking limit.

True (for $M>0$). Friction reheats the near-wall gas only partway back to stagnation, and the recovery factor $r={\text{Pr}}^{1/3}<1$ for gases, so $T_{aw}<T_c$. See Adiabatic Wall Temperature and Recovery Factor.

False. $q$ is a density (W/m²) — heat per unit wall area. $Q=qA$ is the total power (W) over the wetted area $A$. Confusing them scrambles the coolant flow calculation.

Error: the driving temperature is the recovery temperature $T_{aw}$, not $T_c$. The gas moves fast, so only $T_{aw}<T_c$ actually pushes heat into the wall; using $T_c$ overestimates $q$ and wastes coolant.

Error: high conductivity makes $R_w=t_w/k_w$ tiny, so it is the smallest resistance. The gas-side convection $R_g=1/h_g$ is usually the largest and controls the flux.

Error: the energy balance uses the temperature rise $\Delta T_c=T_{out}-T_{in}$, not $T_{out}$ alone. Coolant already enters warm at $T_{in}$; only the rise stores absorbed heat.

Error: short channels lower $\Delta p$ but reduce wetted area $A$, giving less room to dump heat, and may force higher $v$ (which raises $\Delta p$ back up as $v^2$). It's a trade-off, not a free win.

Correct, not an error — the drops telescope: $(T_{aw}-T_{wg})+(T_{wg}-T_{wc})+(T_{wc}-T_{co})=T_{aw}-T_{co}$. (Trap: check you didn't blink and call this wrong.)

Error: $\Delta p\propto 1/d_h$, so larger $d_h$ lowers $\Delta p$. Larger channels are gentler on the pump — but they slow the flow and weaken $h_c$, the other side of the trade.

Error: you add resistances, then invert: $U=(1/h_g+t_w/k_w+1/h_c{)}^{-1}$. Adding the conductances directly is dimensionally and physically wrong.

Turbulent wall friction has to destroy the fluid's momentum flux, which scales with dynamic pressure $\frac{1}{2}\rho v^2$. The dimensionless $f$ packages the messy geometry/turbulence into one measured number.

A separate coolant would be dead mass you must carry and cannot burn. Using propellant means the heat is recycled into the flame and no extra tank is needed — mass efficiency is everything in rockets.

It typically has the largest per-area resistance $R_g=1/h_g$ (gas convection is a poor heat mover compared to a dense liquid coolant). In series, the biggest resistance dominates the total, so it sets the flux.

Regen cooling only works if you have that much propellant streaming past. If cooling demands more mass flow than the engine burns, the wall cannot be cooled by regen alone — you'd need Film Cooling or Ablative Cooling to help.

Smaller $d_h$ raises velocity, boosting $h_c$ (better cooling), but $\Delta p\propto v^2/d_h$ explodes (worse pump cost). Cooling and pumping pull in opposite directions on channel size.

Steady state means no heat accumulates anywhere; energy in equals energy out at every layer. If flux differed between layers, temperature would keep changing — not steady. This conservation is what lets resistances telescope.

The coolant resistance $R_c=1/h_c\to 0$ drops out. The flux is then capped by the remaining gas and wall resistances — you cannot cool below what those two allow no matter how good the coolant is.

Then $R_w=t_w/k_w\to 0$ and $T_{wg}=T_{wc}$ — no temperature drop across the metal. Flux is set entirely by the two convection resistances $R_g$ and $R_c$.

The coolant-side driving difference $T_{wc}-T_{co}\to 0$, so that stage carries no heat — the coolant can no longer absorb energy and the wall overheats. Cool inlet temperature is essential.

$\Delta p\to 0$ — a vanishingly short channel has no wall to rub against. But wetted area $A\to 0$ too, so no heat is removed; the zero-pressure "win" is useless.

The velocity-recovery correction vanishes and $T_{aw}\to T_c$: with no motion there's no slowdown-reheating distinction, so the recovery temperature collapses onto the flame temperature.

The full stagnation temperature — all the kinetic energy of the near-wall gas is recovered as heat. Real gases have $r={\text{Pr}}^{1/3}<1$, so $T_{aw}$ sits just below stagnation.