Worked examples — Regenerative cooling — heat flux, coolant flow, pressure drop

---

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The examples below are labelled by cell so you can see the whole map is covered.

Cell	What varies / breaks	Example
A. Balanced series	all three resistances comparable	Ex 1
B. One resistance dominates	gas-side $R_g\gg$ others → it alone sets $q$	Ex 2
C. Degenerate: thick wall	$t_w/k_w$ blows up → conduction chokes flux	Ex 3
D. Degenerate: near-zero wall	$t_w\to 0$ → wall term vanishes, two convections left	Ex 3
E. Coolant $\Delta T_c$ capped	boiling/coking ceiling forces ${\dot{m}}_c$ up	Ex 4
F. Limiting: channel narrows	$d_h\downarrow$ → $h_c\uparrow$ (good) but $\Delta p\uparrow$ (bad)	Ex 5
G. Sign / direction check	is heat entering or leaving? get the arrow right	Ex 6
H. Real-world word problem	"does the engine even have enough propellant?"	Ex 7
I. Exam twist: couple two formulas	flow sets $h_c$ which sets $q$ which sets flow	Ex 8

We will hit every cell.

---

Example 1 — Cell A: balanced series (the baseline)

1. Convert each layer to a specific resistance. $R_g=\frac{1}{h_g}=5.0\times 10^{-5},R_w=\frac{t_w}{k_w}=\frac{0.001}{350}=2.857\times 10^{-6},R_c=\frac{1}{h_c}=2.0\times 10^{-5}$ Why this step? Heat crosses gas-film, metal, coolant-film in series, so like series resistors they simply add. Working in resistances turns "three physics laws" into one addition.
2. Add them. $R=7.286\times 10^{-5}$ m²K/W. Why? Series resistance is the sum; the reciprocal is the overall coefficient $U$.
3. Divide the driving temperature difference. $q=\frac{T_{aw}-T_{co}}{R}=\frac{3000}{7.286\times 10^{-5}}=4.12\times 10^7{\text{W/m}}^2\approx 41{\text{MW/m}}^2$ Why? Same $q$ flows through every layer, so total drop $÷$ total resistance.
4. Back out the hot-wall temperature using only the gas-side drop: $T_{wg}=T_{aw}-q{R}_g=3300-(4.12\times 10^7)(5\times 10^{-5})\approx 1242\text{K}$ Why this step? The metal only "feels" its own surface temperature. That surface sits below the gas by exactly the gas-side drop $q{R}_g$.

---

Example 2 — Cell B: one resistance dominates

1. Recompute $R_g$. $R_g=1/2000=5.0\times 10^{-4}$. Others unchanged: $R_w=2.86\times 10^{-6}$, $R_c=2.0\times 10^{-5}$. Why? Only $h_g$ changed, so only $R_g$ moves.
2. Compare magnitudes. $R_g/R_c=5.0\times 10^{-4}/2.0\times 10^{-5}=25\times$ larger, and $R_g/R_w=5.0\times 10^{-4}/2.86\times 10^{-6}=175\times$ larger. Why this step? When one series resistor towers over the rest, it alone sets the current — this is the Newton's Law of Cooling gas film being the bottleneck.
3. Approximate flux with $R_g$ alone. $q_{\text{approx}}=\frac{3000}{5.0\times 10^{-4}}=6.0\times 10^6{\text{W/m}}^2=6{\text{MW/m}}^2$ Why this step? If $R_g$ dwarfs the others, then $R\approx R_g$, so $q\approx \Delta T/R_g$. Dropping the two tiny resistances should barely move the answer — this line tests exactly how good that shortcut is.
4. Exact flux with all three. $R=5.229\times 10^{-4}$, $q=3000/R=5.737\times 10^6$ W/m². Why compare? To see how good "ignore the small ones" is.

---

Example 3 — Cells C & D: thick wall vs vanishing wall

1. Case C, thick wall. $R_w=0.050/350=1.429\times 10^{-4}$. Now $R=R_g+R_w+R_c=5\times 10^{-5}+1.429\times 10^{-4}+2\times 10^{-5}=2.129\times 10^{-4}$. $q_C=3000/2.129\times 10^{-4}=1.409\times 10^7{\text{W/m}}^2\approx 14{\text{MW/m}}^2$ Why this step? A fat wall makes conduction (Fourier's Law of Conduction) the bottleneck; flux collapses to a third of Ex 1's value. Less heat reaches the coolant — but the wall interior gets dangerously hot because the drop $q{R}_w$ is huge.
2. Case D, zero wall. $R_w\to 0$, so $R=R_g+R_c=7.0\times 10^{-5}$. $q_D=3000/7.0\times 10^{-5}=4.286\times 10^7{\text{W/m}}^2\approx 43{\text{MW/m}}^2$ Why? Remove the conduction resistance entirely and only the two films remain — the maximum flux this film pair can pass.

---

Example 4 — Cell E: coolant temperature-rise capped

1. Total power into the jacket. $Q=qA=4.1\times 10^7\times 0.5=2.05\times 10^7$ W. Why? Flux is power per area; multiply by wetted area for total.
2. Energy balance, $200$ K case. ${\dot{m}}_c=\frac{Q}{c_p\Delta T_c}=\frac{2.05\times 10^7}{4200\times 200}=24.4\text{kg/s}$ Why this step? Every joule into the wall must land in the coolant's rising temperature — conservation of energy, no boiling.
3. Stricter cap, $120$ K case. ${\dot{m}}_c^{\prime }=\frac{2.05\times 10^7}{4200\times 120}=40.67\text{kg/s}$ Why? The cap on $\Delta T_c$ is the real constraint; a tighter cap forces a bigger flow to remove the same $Q$.
4. Ratio. ${\dot{m}}_c^{\prime }/{\dot{m}}_c=200/120=1.667$.

---

Example 5 — Cell F: narrowing the channel (the cruel trade-off)

1. Baseline $\Delta p$ at $d_h=2$ mm (the blue bar). $\Delta p_1=f\frac{L}{d_h}\frac{\rho v^2}{2}=0.02\cdot \frac{0.6}{0.002}\cdot \frac{800\cdot 30^2}{2}=2.16\times 10^6\text{Pa}=21.6\text{bar}$ Why this step? This is Darcy–Weisbach straight off — pressure lost to wall friction over the channel length.
2. New $\Delta p$ at $d_h=1$ mm, $v=60$ m/s (the pink bar). $\Delta p_2=0.02\cdot \frac{0.6}{0.001}\cdot \frac{800\cdot 60^2}{2}=1.728\times 10^7\text{Pa}=172.8\text{bar}$ Why? Two things worsened at once: $d_h$ halved (factor $2$ in $L/d_h$) and $v$ doubled (factor $4$ in $v^2$).
3. Growth factor (the yellow arrow). $\Delta p_2/\Delta p_1=172.8/21.6=8$. Why this step? $\Delta p\propto v^2/d_h$; halving $d_h$ gives $\times 2$, doubling $v$ gives $\times 4$; together $\times 8$.

---

Example 6 — Cell G: which way does heat go? (sign / direction)

1. Compute the student's expression. $h_g(T_{wg}-T_{aw})=2\times 10^4(1242-3300)=2\times 10^4(-2058)=-4.12\times 10^7{\text{W/m}}^2$ Why this step? To expose the sign. A negative flux with this convention means the flow is opposite to the assumed arrow.
2. Fix the convention. Newton's Law of Cooling for heat into the wall uses hotter-minus-colder in the flow direction: $q=h_g(T_{aw}-T_{wg})>0$. Why? Heat always flows hot → cold. Gas ($3300$ K) is hotter than wall ($1242$ K), so heat enters the wall; the correct expression is positive.
3. Magnitude is identical, only the arrow flipped: $|q|=4.12\times 10^7$ W/m².

---

Example 7 — Cell H: real-world word problem

1. Fuel flow available. ${\dot{m}}_{\text{fuel}}=\frac{1}{3}\times 30=10$ kg/s. Why this step? Only the fuel is routed through cooling channels; oxidizer usually isn't (it would react with the hot metal).
2. Compare to demand. Need $24.4$ kg/s, have $10$ kg/s. Deficit $=24.4-10=14.4$ kg/s. Why? If demand exceeds available coolant, pure regen cooling cannot carry the heat.
3. Design consequence. The shortfall must be met by supplement: Film Cooling (a cool gas layer along the wall) or Ablative Cooling (a sacrificial liner) to cut the heat load $q$ so the required ${\dot{m}}_c$ drops to $\le 10$ kg/s. Why this step? You can't invent fuel; instead you lower the demand.

---

Example 8 — Cell I: exam twist, two formulas coupled

1. $h_c$ scaling. $h_c\propto \text{Nu}\propto {\text{Re}}^{0.8}$. Doubling Re: factor $=2^{0.8}$. $2^{0.8}=1.741$ Why this step? $k_c,d_h,\text{Pr}$ are unchanged, so only the ${\text{Re}}^{0.8}$ term moves; $h_c$ rises by $1.741\times$, less than double — the $0.8$ exponent punishes you.
2. New coolant resistance. $R_c\propto 1/h_c$, so $R_c^{\text{new}}=R_c^{\text{old}}/1.741$. Why? Resistance is the inverse of the transfer coefficient.
3. New total resistance. Old total $R=1$ (normalized), with $R_c^{\text{old}}=0.40$, others $=0.60$. $R^{\text{new}}=0.60+0.40/1.741=0.60+0.2298=0.8298$ Why this step? Only the coolant term shrinks; gas and wall terms are untouched, so improvement is diluted.
4. Flux gain. Since $q\propto 1/R$, the flux grows by the ratio of old total to new total: $\frac{q^{\text{new}}}{q^{\text{old}}}=\frac{R^{\text{old}}}{R^{\text{new}}}=\frac{1}{0.8298}=1.205$ Why this step? Same driving $\Delta T$, so flux is inversely proportional to total resistance; dividing old-by-new gives the improvement factor — a $20.5\%$ gain despite $h_c$ rising $74\%$.

---

Recall Active recall — cover the answers

Ex 1 baseline flux ::: $\approx 41$ MW/m², $T_{wg}\approx 1242$ K Ex 3 which wall gives more flux, thick or none ::: none (zero-thickness), $\approx 43$ vs $14$ MW/m² Ex 4 flow when $\Delta T_c$ goes $200\to 120$ K ::: rises by factor $200/120=1.667$, to $40.7$ kg/s Ex 5 $\Delta p$ factor when $d_h$ halves and $v$ doubles ::: $\times 8$ Ex 6 heat direction when $q$ comes out negative ::: convention backwards; heat still flows gas → wall Ex 8 $h_c$ factor when velocity doubles ::: $2^{0.8}=1.741$, but flux only up $\sim 20\%$