16.512, Rocket Propulsion Prof. Manuel Martinez-Sanchez Lecture 9: Liquid Cooling

Cooling of Liquid Propellant Rockets

We consider only bi-propellant liquid rockets, since monopropellants tend to be small and operate at lower temperatures. In a bi-propellant rocket, both the oxidizer and the fuel streams are in principle available for cooling the most exposed parts of the chamber and nozzle prior to being injected. This is called “regenerative cooling”, because the heat loss from the gas is recovered (“regenerated”) into the liquid, so no heat escapes. This is not to say no thermodynamic loss is incurred, though (heat is transferred from very hot gas to cool liquid, which implies irreversibility and loss of work potential).

Of the two streams, the fuel is normally used for cooling. This is for two reasons:

(a) Fuels tend to have higher specific heats, so more heat is removed for a given ∆T of the coolant, and

(b) Leakage from an oxidizer stream into the normally fuel-rich combustion gas can produce a local flame that can be catastrophic, whereas leakage from a fuel line into the same fuel-rich gas is inert. In addition, exposing hot metal to oxygen or strong oxidants always carries some risk of accelerated chemical attack, or even ignition. Some exceptions do exist where oxidizers are used for cooling, though.

A typical arrangement is as shown below (Figure 1).

The fuel at high pressure from the fuel pump (FP) is sent through a series of narrow passages carved into the nozzle and chamber walls, picks up the wall heat flux from the gas, and is delivered eventually to the injector manifold. Since the nozzle region

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 1 of 12 is the most thermally loaded one, often the coolant flow is split, with one part entering at the nozzle exit and another providing extra cool fluid by entering just downstream of the throat.

A typical construction for the cooling channels is shown in Fig. 2.

The load-bearing part of the structure is milled longitudinally with channels of varying depth and width (to obtain varying liquid velocity), and a high thermal conductivity thin layer of a Copper alloy is then brazed on the inside.

Design Considerations

Two aspects need to be verified in the design of the cooling system:

(a) The coolant should have sufficient thermal capacity to absorb the heat load without exceeding some critical temperature, which may be a chemical decomposition limit (thermal cracking for hydrocarbons) or the boiling point (although, with care, boiling can be sometimes tolerated or exploited for its strong heat absorption properties).

Suppose QLOSS is the calculated total heat loss from the gas. As seen in a i previous lecture, this amounts to 1-3% of mcpTc, more for the smaller i engines. Suppose also the fuel only is used as coolant, with a flow rate mF . ii i ii i mm− F m The OF ratio is defined as OF=mox mF =, and so mF = . If i 1 + O mF F

the liquid fuel has a specific heat ccool , its temperature rise ∆T from inlet to exit of the cooling circuit will be given by

i QmLOSS =F ccool∆T (1)

i i ⎛⎞QLOSS mF mcpTc ⎜⎟=∆ccool T ⎜⎟Q 1 + O ⎝⎠TOT F

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 2 of 12

⎛⎞Q ⎛⎞cp ⎛O ⎞ ∆=T1LOSS + T (2) ⎜⎟⎜⎟⎜⎟c ⎝⎠QCTOT ⎝⎠cool ⎝⎠F

Q c which must be kept within limits. For example, say LOSS =0.02, p =1, QTOT ccool O =4, T =3000K; we obtain ∆T =×0.02 1 ×5 ×3000 =300 K . This may or F c may not be acceptable; for a cryogenic coolant it would most likely be, but a hydrocarbon fuel, exiting the pump at 300K will then leave the cooling circuit at 600K, probably too high for chemical stability.

(b) The local cooling rate at the most exposed location (the throat) must be sufficient to avoid decomposition or boiling even at the contact point of the liquid with the wall. The thermal situation in a cut through the front wall of the cooling passages is as schematizes in Fig. 3.

The Taw temperature is shown dashed because, as we know it is not the actual gas temperature outside the gas boundary layer, but is the one driving

heat. The liquid bulk is at a temperature Tl , which is below that of the wetted

wall ( Twc ), because heat has to be driven through according to

qh=l(Twc −Tl) (3)

where hl is the liquid-side film coefficient, that can be calculated, for instance, from Bartz formula using liquid properties. The same heat flux is supplied from the gas through the gas-side film coefficient:

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 3 of 12 qh=−ga()TwTwh (4)

and yet the same flux must cross the wall by conduction:

T− T qk= wh wc (5) δ

where k is the wall thermal conductivity, and δ its thickness.

⎧ q TT−= ⎪ aw wh h ⎪ g ⎪ δ Re-writing (3)-(5) as ⎨TTwh −wc =q ⎪ k ⎪ q TT−= ⎪ wc l h ⎩⎪ l

and adding, we obtain

⎛⎞11δ TT−=⎜⎟++ q; aw l ⎜⎟ ⎝⎠hkg hl

T− T q = aw l (6) 11δ ++ hhg l k

which we can then use to calculate intermediate temperatures from (3), (4), (5). Clearly, what we have done is adding the series “thermal impedances” 1 δ 1 , and of the gas boundary layer, the metal, and the liquid. hg k hl

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 4 of 12 Stresses in Cooled Nozzle Walls

Ribs carry weak in- plane stress.

Wall must carry the large hoop

stress due to Pg ,

Pl . In addition, hot side will expand more, forcing back (cold) side to higher tension. So, use high σ steel Fig. 1 ul for back (1) layer. Front (2) layer needs to be good thermal conductor; use Cu or W-Cu alloy (higher strength). Cu has higher expansion coefficient αCu >αsteel , which adds to the effect of higher T, and ends up putting this layer in compression. This can be relieved by hot assembly, so that the Cu is pre-stretched when cold.

Plane strain At any z, within one of the materials.

σ ()z ε=()1T−ν +α⎡()z−T⎤ (1) E ⎣0 ⎦

where T0 can be interpreted as the temperature at which the strain ε is defined to be zero, with zero stress. Since the shape remains planar, ε = constant (at least within the layer).

Write (1) for both layers. We now “assemble” them with a tight fit, but zero stresses, at T0 , which from now on means the assembly temperature. Upon heating or cooling, thermal stresses will arrive, even with no loading or T gradients.

ε =0 by definition at Both layers now have the same (constant) ε assembly

σ ()z 1 ⎡⎤ ()1−ν11+α ⎣T1()z−T0⎦=ε (2) E1

σ ()z 2 ⎡⎤ ()1−ν22+α ⎣T2()z−T0⎦=ε (3) E2

For metals, ν varies little, so take ν12=ν =ν.

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 5 of 12 The temperature Tz1 () will be nearly constant (adiabatic outer condition). Tz2 ( ) will vary linearly with z, according to

dT −k2 =q (4) 2 dz

We write (3) at z0==,zt2 and subtract:

σ−2c σ2h ()1T−ν =α2w⎣⎦⎡⎤h−Twc (5) E2

and integrate (4) to

TTwh − wc qk= 2 (6) t2 so that

α22Et2 σ−2c σ2h = q (7) 1k−ν 2

Also (2) reads

σ1 ε=()1T−ν +α10⎣⎦⎡⎤l −T, which can be combined with E1

σ2c ε=()1T−ν +α2w⎣⎦⎡⎤c−T0 E2

⎛⎞σ1 σ2c to give ()1T−ν ⎜⎟− =α2wc−α1Tl −()α2−α1T0 (8) ⎝⎠EE12

Heat transfer from wall 2 to liquid gives

qh=l(Twc −Tl) (9a)

q or TTwc =l + (9b) hl

Substitute into (8)

⎛⎞σ1 σ2c q ()1T−ν ⎜⎟− =()α21−α (l −T0)+α2 ⎝⎠EE12 hl

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 6 of 12

σασ q ()α−α (T −T ) or 12−=2c +21l 0 (10) EE121−ν hl 1−ν

In all of this, q is taken as a given. It can be calculated from the given Taw , Tl , plus hg , hl and t2 , k2 :

TT− q = aw l (11) 11t ++2 hhg l k2

Equations (7) and (10) relate σσ2h ,,2c σ1 . We need one more equation

Force balance

Since Tz2 ( ) is linear in z, so will σ2 (z) . For force calculations, then, we can use the mean value

σ+σ σ=2c 2h (12) 2 2

The net balance then is

2σ+t 2σt=PD+2Pt (13) 11 2 2 g ll

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 7 of 12

rd which is our 3 equation, together with (7), (10). Notice that, since tDl , Pl has only a minor effect on stresses (except in the ribs).

From here, we either solve for σ2h ,,σσ2c 1 (given geometry, Pg , Pl , q) or, for design, go the reverse route and decide on the geometry for assigned stresses. We will pursue here the second approach.

Design

As noted, σ1 will be positive and high, whereas σ2h (and less so σ2c ) will be negative, and probably high too. We then take the view that

σ=σ S (14) 11ult,tens.

(S=safety factor ~ 1.5) and

⎧σ S ⎪ 2ult,comp. (15a) ()−σ2h =least of ⎨ σ2 S (15b) ⎩⎪ buckling

and use these conditions to determine t,12t. The selection of t2 is complicated by the fact that −σ will decrease with t , but so will (quadratically) σ : ()2h 2 2buckling

For buckling, use a simple clamped-beam formulation:

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 8 of 12

EIππ224 EI F ==22 c 2 2 ()l 2 l

1 where IH= t3 . 12 2

Dividing by A= Ht2 ,

23⎛⎞1 4Eπ Ht 2 22⎜⎟12 π2 ⎛⎞t σ= ⎝⎠σ=2 E (16) 2buckling 2 2buckling ⎜⎟2 l ()Ht2 3 ⎝⎠l

To proceed, start by eliminating σ2c between (7) and (10):

σσ αt αq (α−α) (T −T ) 12−−h 22q =2 +21l 0 EE112 −ν k21−ν hl 1−ν

EE22α 2⎛⎞1 t2ET22(α−α1) ( l −T0) ()−σ2h = − σ1 + ⎜⎟+ q + (17) E112−ν ⎝⎠hl k 1−ν or, recalling (11),

1 t + 2 EEα hk ()α−α ET(−T) ()−σ = − 22σ + 2l 2()TT− + 212l 0 (18) 2h E11 −ν 11t aw l 1−ν 1 ++2 hhg2l k

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 9 of 12 This displays several effects:

σ1 (a) Once σ= ult. is fixed, −σ will increase with t , although weakly, since 1 S 2h 2 t 11 2 , (the Cu wall offers little thermal impedance, compared to the kh2ghl two boundary layers).

(b) This (σ2h ) (middle term in (18)) arises from the heating of layer 2 relative to 1, due to heat flowing in.

(c) The positive stress σ1 (to counter Pg mostly) relieves this tendency to compress layer 2, and might even reverse it.

(d) The last term in (18) arising from differential expansion coefficients α−21α, −51− could be used as a design aid. If 2=Cu, 1=steel, α=2 2.3 ×10 K , −−51 α=121.4 ×10 K , so α−α1>0 .

Then, if −σ2h is still too high despite σ1 , we could increase T0 , if possible above Tl , to reduce −σ2h . This implies assembly at high temperature.

⎪⎧⎫(18)⎪ We can use ⎨⎬ to construct a plot like Figure 3, and select a viable t2 . Once this ⎩⎭⎪⎪()16 is done, equation (7) gives σ , equation (12) gives σ , and equation (13) gives t . 2c 2 1

Some data:

Material E (Pa ) −1 σult (Pa ) ν ()1 −ν σ d (k ) ZK= ult. o (at 500K) (at 500K) Eα () Cu 0.95 × 1011 2.3 × 10−5 1.1 × 108 0.3 35 St. Steel 302 1.61 × 1011 1.8 × 10−5 4.6 × 108 0.3 111 Ti 1.63 × 1011 1.7 × 10−5 5.4 × 108 0.3 136 Alloy Steel 1.09 × 1011 1.4 × 10−5 5.1 × 108 0.3 234 (SAE x4130)

The last column is a “figure of merit” extracted from equation (18) to give a preliminary rough idea of materials expansion stress. The higher Z, the higher the

∆T to reach σult. in a double strip of this material subject to differential heating ∆T .

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 10 of 12 7 Example P1ga= 00atm 10P (neglect Pe effect) D=0.3m; l=4mm 2 hg = 24000 W / m /K ; 52 h2l =×.7610W/m/K;

k2 = 360 W / m /K

T3aw = 200K,Tl = 400K

5.1 × 108 σ= P ()alloy steel ; E =1.09 ×1011 P ; α=1.4 ×10−−5 K 1 1a1.5 1 a1

8 1.1 × 10 11 −−5 1 −σ2a= PC()u; E2= 0.95× 10 Pa;α2= 2.3× 10 K ult,comp. 1.5

Substituting into (18),

1.303 ×+10−3 t −σ = −2.96 × 1089+ 8.740 × 10 2 + 1.221 × 106( 400 −T ) (19) 2h −3 0 16.30 ×+10 t2

and, from (16)

σ=1.95 ×1016 t2 (t in m.) 22buckling 2

Following are some calculated results:

COMMENTS tm()m 00.2 2 1.1 × 108 88Since −σ = , assembling at T02=−297K σh()Pa5.33 ×10 6.26 ×10 ult,com 1.5 −σ P 0 7.80 × 108 room T is not acceptable for any t . 2abuckling () 0 2

tm2 ()m 00.2 88 T02=−500K σh()Pa2.81 ×10 3.78 ×10 Closer, but still no solution 0 8 −σ2,buckl. ()Pa 7.80 × 10

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 11 of 12 tm2 ()m 0 0.2 0.1 With assembly at 8 88T =700K, a very thin −σ2h ()Pa 0.36 × 10 1.34 ××10 0.85 10 0 88t02 .1mm is T02=−700K σ,buckl.()Pa0 7.80 ×10 1.95 ×10 8 acceptable, but may 27+σ P +×4.48 10 q()w / m =×6.15 10 2c ()a be questionable on 4.40 robustness. Also, σ tm1 ()m 2c is too high

tm2 ()m 0.2 0.3 0 If T0 =800K, is 8 88 −σ2h ()Pa −×0.86 10 0.12 ××10 0.60 10 feasible, then 88t02 = .3mm is T02=−800K σ,buckl.()Pa0 7.8 ×10 1.76 ×10 acceptable. σ also 7 7 2c q6=×.0710 +σ2c ()Pa 9.79 × 10 OK (in tension) 4.48 tm1 ()m

With the assumed l=4mm, buckling is not a problem in any case, but compressive failure is hard to avoid. It may be possible to exceed the elastic limit and go into plastic compressive yield if ductility is high enough to ensure no rupture. But this means no reusability.

16.512, Rocket Propulsion Lecture 9 Prof. Manuel Martinez-Sanchez Page 12 of 12