Regenerative cooling — heat flux, coolant flow, pressure drop

3.3.28 · Physics › Rocket Propulsion

Intuition Badi picture

Ek rocket engine ke combustion gases 3000–3600 K ke paas hote hain — kisi bhi metal ke survive karne se zyada hot. Phir bhi chamber walls intact rehti hain. KYUN? Kyunki propellant ko jalane se pehle, usse pehle chhote channels ke andar, jo wall ke beech mein hoti hain, pump kiya jaata hai, jahan woh heat soak karta hai. Thanda propellant wall ko cool karta hai injector tak jaate waqt, aur jo heat woh leta hai woh waste nahi hoti — woh wapas flame mein jaati hai (isliye regenerative). Teen quantities decide karti hain ki yeh kaam karega ya nahi:

Heat flux $q$ — kitni tezi se heat wall mein jaati hai (W/m²).
Coolant flow $\overset{m}{˙}_{c}$ — kitni tezi se propellant heat bahar le jaata hai.
Pressure drop $Δ p$ — woh price jo tum coolant ko patli channels mein push karne ke liye pay karte ho.

1. Heat flux: wall kitni zyada "cook" ho rahi hai

YEH KYUN drive karta hai? Heat ko travel karna padta hai: hot gas → wall (gas side) → metal ke through conduction → coolant. Har stage ek resistance hai series mein, jaise ek electrical circuit mein resistors hote hain.

$q$ ko first principles se derive karna

Har interface par Newton's law of cooling se shuru karo aur beech mein Fourier conduction se. Per-area (specific) resistances define karo:

Gas-side convection: $R_{g} = 1/ h_{g}$
Wall conduction: $R_{w} = t_{w} / k_{w}$ (thickness $t_{w}$ , conductivity $k_{w}$ )
Coolant-side convection: $R_{c} = 1/ h_{c}$

COMBINE kaise karein: Steady state mein wahi $q$ har layer se guzarti hai, isliye: $q = h_{g} (T_{a w} - T_{w g}) = \frac{k _{w}}{t _{w}} (T_{w g} - T_{w c}) = h_{c} (T_{w c} - T_{co})$

Yeh step kyun? Steady state ka matlab hai kahin bhi heat pile up nahi hoti, isliye flux layer-to-layer conserved hoti hai. Temperature drops add karo (woh telescope karte hain):

$q = \frac{T _{a w} - T _{co}}{\frac{1}{h _{g}} + \frac{t _{w}}{k _{w}} + \frac{1}{h _{c}}}$

$T_{a w}$ kyun, flame temperature kyun nahi? Wall ke paas fast gas slow ho jaati hai; friction usse aadha reheats kar deta hai, toh effective driving temperature recovery temperature $T_{a w} = T_{c} (\frac{1 + r \frac{γ - 1}{2} M ^{2}}{1 + \frac{γ - 1}{2} M ^{2}})$ hoti hai jisme recovery factor $r \approx Pr^{1/3}$ hota hai. Stagnation se thoda kam, lekin usske kareeb.

2. Coolant flow: heat ko bahar le jaana

Channel wetted area $A$ par total power $Q = q A$ absorb karta hai. Coolant inlet $T_{in}$ se outlet $T_{o u t}$ tak heat hota hai.

Yeh step kyun? Wall mein enter karne wala har joule coolant ke rising temperature mein store hona chahiye (koi phase change assume nahi). Energy in = energy carried out.

Constraint: $T_{o u t}$ coolant ke boiling/coking limit se neeche rehna chahiye, warna film boiling wall ko insulate kar deta hai → burnout. Yeh $Δ T_{c}$ ko cap karta hai, aur $\overset{m}{˙}_{c}$ ko upar force karta hai.

Coolant-side heat transfer coefficient khud flow par depend karta hai Dittus–Boelter correlation ke zariye: $Nu = 0.023 Re^{0.8} Pr^{0.4}, h_{c} = \frac{Nu k _{c}}{d _{h}}$ Yeh kyun matter karta hai: zyada flow → higher Re → higher $h_{c}$ → lower wall temperature. Zyada flow do baar cool karta hai: bada $\overset{m}{˙}_{c}$ AUR bada $h_{c}$ .

3. Pressure drop: patli channels ki cost

Narrow, lambi channels mein coolant push karne mein pressure lagta hai. Woh pressure pumps se supply hona chahiye — zyada $Δ p$ ⇒ bhaari turbopump ⇒ performance penalty.

DERIVE kaise karein (first principles): Ek pressure force $Δ p A_{cs}$ fluid ko push karta hai; wall shear $τ_{w}$ wetted surface $P L$ par resist karta hai. Steady flow mein yeh balance karte hain: $Δ p A_{cs} = τ_{w} P L \Rightarrow Δ p = τ_{w} \frac{P L}{A _{cs}} = τ_{w} \frac{4 L}{d _{h}}$ $d_{h} = 4 A_{cs} / P$ use karke. Phir $f$ define karo $τ_{w} = \frac{f}{4} \cdot \frac{ρ v ^{2}}{2}$ se (empirical dimensionless friction), jisse boxed form milta hai.

$v^{2}$ kyun? Kyunki turbulent wall shear dynamic pressure $\frac{1}{2} ρ v^{2}$ ke saath scale karta hai — woh momentum flux jo wall ko destroy karna padta hai.

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

Worked examples

Common mistake Steel-manned misconceptions

(a) "Safety ke liye bas wall ko thicker banao." Sahi lagta hai: thicker metal = zyada strong, zyada thermal mass. Reality: thicker wall $R_{w} = t_{w} / k_{w}$ badha deti hai, toh $T_{w g}$ upar jaata hai — zyada hot, safer nahi. Fix: rocket walls thin banai jaati hain (≈1 mm) high- $k$ copper alloys ke saath taaki $R_{w}$ chhota rahe.

(b) "Zyada coolant flow hamesha help karta hai, koi downside nahi." Sahi lagta hai: zyada flow zyada heat carry karta hai aur $h_{c}$ badhata hai. Reality: $Δ p \propto v^{2}$ , toh velocity double karne se pump work chaar guna ho jaata hai. Fix: optimize karo — cooling ke liye kaafi flow, itna nahi ki pump budget toot jaaye.

(c) "Driving temperature ke roop mein flame temperature $T_{c}$ use karo." Sahi lagta hai: woh sabse hot number hai jo available hai. Reality: gas tezi se move kar rahi hai; sirf recovery temperature $T_{a w} < T_{c}$ heat ko wall mein drive karti hai. Fix: $T_{a w} = r$ -corrected value use karo; $q$ overestimate karna coolant waste karta hai.

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek metal cup itni hot hai ki woh melt ho jaaye — jaise blowtorch ke paas. Apna thanda juice peene se pehle, tum juice ko chhote tubes ke through chalate ho jo cup ke around wrapped hain. Thanda juice heat grab karta hai, cup ko cool rakhta hai. Phir tum (ab warm) juice peete ho — kuch waste nahi! Lekin agar tubes bahut patli hain, toh juice bahar karne ke liye tumhe bahut zyada suck karna padta hai. Woh "zyada suck karna" pressure drop hai, aur woh annoying cost hai.

Recall Active recall — answers cover karo

$1/ U$ dene ke liye kaunse resistances add hote hain? ::: gas convection $1/ h_{g}$ , wall conduction $t_{w} / k_{w}$ , coolant convection $1/ h_{c}$ .
Thin walls kyun? ::: $R_{w} = t_{w} / k_{w}$ minimize karne ke liye taaki wall cool rahe.
Zyada velocity better cool kyun karti hai lekin zyada cost kyun karti hai? ::: $h_{c} ↑$ (Re $^{0.8}$ ) lekin $Δ p \propto v^{2}$ .

Flashcards

Heat flux kya hai aur uske units kya hain?

Wall ke unit area per thermal power, W/m². Yeh power ki ek density hai, total power nahi.

Overall heat flux ke liye series-resistance formula likho.

q = (T_{a w} - T_{co}) / (1/ h_{g} + t_{w} / k_{w} + 1/ h_{c})

Flame temperature ki jagah recovery temperature

T_{a w}

kyun use karte hain?

Fast gas wall ke paas sirf aadha reheat hoti hai; effective driving temp

T_{a w} = r

-corrected hoti hai, stagnation se thodi neeche.

Wall ko thicker karne se woh zyada hot kyun hoti hai, safer kyun nahi?

Yeh

R_{w} = t_{w} / k_{w}

badhata hai, gas-side wall temperature

T_{w g} = T_{a w} - q R_{g}

raise karta hai... actually

T_{w c}

raise karta hai; overall higher wall temps.

Required coolant mass flow kaise compute karte hain?

\overset{m}{˙}_{c} = q A / (c_{p} Δ T_{c})

energy balance

Q = \overset{m}{˙}_{c} c_{p} Δ T_{c}

se.

Darcy–Weisbach state karo aur uska origin derive karo.

Δ p = f \frac{L}{d _{h}} \frac{ρ v ^{2}}{2}

; pressure force = wall shear force se,

Δ p A_{cs} = τ_{w} P L

d_{h} = 4 A_{cs} / P

ke saath.

Allowed coolant temperature rise ko kya limit karta hai?

Coolant ka boiling/coking; film boiling insulate karke wall burnout cause karta.

Regenerative cooling ko "regenerative" kyun kehte hain?

Coolant dwara absorbed heat combustion mein wapas jaati hai jab woh propellant jalta hai — energy recover hoti hai.

h_{c}

flow par kaise depend karta hai (Dittus–Boelter)?

N u = 0.023 R e^{0.8} P r^{0.4}

, toh

h_{c} \propto R e^{0.8} \propto v^{0.8}

Design mein central trade-off ek line mein?

Chhote channels better cool karte hain (

h_{c} ↑

) lekin zyada pump work cost karte hain (

Δ p \propto v^{2} / d_{h}

Connections

Newton's Law of Cooling — gas- aur coolant-side convection terms.
Fourier's Law of Conduction — $t_{w} / k_{w}$ wall term.
Dittus-Boelter Correlation — Reynolds & Prandtl se $h_{c}$ set karta hai.
Darcy-Weisbach Equation — coolant channels mein pressure drop.
Adiabatic Wall Temperature and Recovery Factor — real driving temperature.
Turbopump Sizing — jahan $Δ p$ ek mass penalty ban jaata hai.
Film Cooling & Ablative Cooling — alternative wall-protection schemes.