Lecture 9: Liquid Cooling

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 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 m cp Tc , more for the smaller i engines. Suppose also the fuel only is used as coolant, with a flow rate mF . i i m− mF i m The O F ratio is defined as O F = 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 i QLOSS = mF ccool ∆T ⎛Q m cp Tc ⎜⎜ LOSS ⎝ QTOT i i i (1) i ⎞ mF ccool ∆T ⎟⎟ = ⎠ 1 + OF ⎛Q ∆T = ⎜⎜ LOSS ⎝ QTOT ⎞ ⎛ cp ⎟⎟ ⎜⎜ ⎠ ⎝ Ccool ⎞⎛ O⎞ ⎟⎟ ⎜ 1 + ⎟ Tc F⎠ ⎠⎝ which must be kept within limits. For example, say (2) cp QLOSS =0.02, =1, Q TOT ccool O =4, T =3000K; we obtain ∆T = 0.02 × 1 × 5 × 3000 = 300 K . This may or c F 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 q = hl ( 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: q = hg ( Taw − Twh ) (4) and yet the same flux must cross the wall by conduction: q=k Twh − Twc δ (5) where k is the wall thermal conductivity, and δ its thickness. q ⎧ ⎪ Taw − Twh = h g ⎪ ⎪⎪ δ Re-writing (3)-(5) as ⎨ Twh − Twc = q k ⎪ ⎪ q ⎪ Twc − Tl = h l ⎪⎩ and adding, we obtain ⎛ 1 δ 1⎞ Taw − Tl = ⎜ + + ⎟q; ⎜ hg k hl ⎟ ⎝ ⎠ q= Taw − Tl δ 1 1 + + hg hl k (6) which we can then use to calculate intermediate temperatures from (3), (4), (5). Clearly, what we have done is adding the series “thermal impedances” δ 1 1 , and of the gas boundary layer, the metal, and the liquid. hg k hl Stresses in Cooled Nozzle Walls Ribs carry weak inplane 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 σul steel Fig. 1 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. ε = (1 − ν ) σ (z) E + α ⎡⎣ T ( z ) − T0 ⎤⎦ (1) 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. Both layers now have the same (constant) ε (1 − ν1 ) (1 − ν2 ) σ1 ( z ) E1 σ2 ( z ) E2 ε =0 by definition at assembly + α1 ⎡⎣ T1 ( z ) − T0 ⎤⎦ = ε (2) + α2 ⎡⎣ T2 ( z ) − T0 ⎤⎦ = ε (3) For metals, ν varies little, so take ν1 = ν2 = ν . The temperature T1 ( z ) will be nearly constant (adiabatic outer condition). T2 ( z ) will vary linearly with z, according to −k 2 dT2 =q dz (4) We write (3) at z = 0 , z = t2 and subtract: (1 − ν ) σ2c − σ2h = α2 ⎡⎣ Twh − Twc ⎤⎦ E2 (5) and integrate (4) to q = k2 Twh − Twc t2 (6) so that σ2c − σ2h = α2 E2 t2 q 1 − ν k2 (7) Also (2) reads ε = (1 − ν ) σ1 + α1 ⎣⎡ Tl − T0 ⎦⎤ , which can be combined with E1 ε = (1 − ν ) σ2c + α2 ⎡⎣ Twc − T0 ⎤⎦ E2 ⎛σ σ ⎞ to give (1 − ν ) ⎜ 1 − 2c ⎟ = α2 Twc − α1 Tl − ( α2 − α1 ) T0 E2 ⎠ ⎝ E1 (8) Heat transfer from wall 2 to liquid gives q = hl ( Twc − Tl ) or Twc = Tl + q hl (9b) Substitute into (8) ⎛σ (1 − ν ) ⎜ E1 ⎝ 1 − (9a) σ2c ⎞ q ⎟ = ( α2 − α1 ) ( Tl − T0 ) + α2 E2 ⎠ hl or σ1 σ2c α2 q ( α2 − α1 ) ( Tl − T0 ) − = + E1 E2 1 − ν hl 1−ν (10) 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 : q= Taw − Tl t 1 1 + + 2 hg hl k 2 (11) Equations (7) and (10) relate σ2h , σ2c , σ1 . We need one more equation Force balance Since T2 ( z ) is linear in z, so will σ2 ( z ) . For force calculations, then, we can use the mean value σ2 = σ2c + σ2h 2 (12) The net balance then is 2 σ1 t1 + 2 σ 2 t 2 = PgD + 2Pl t l (13) which is our 3rd equation, together with (7), (10). Notice that, since tl only a minor effect on stresses (except in the ribs). D , Pl has From here, we either solve for σ2h , σ2c , σ1 (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 σ1 = σ1ult,tens. S (14) (S=safety factor ~ 1.5) and ( −σ2h ) = least ⎧⎪σ2ult,comp. S of ⎨ ⎪⎩σ2buckling S (15a) (15b) and use these conditions to determine t1 , t2 . The selection of t2 is complicated by the fact that ( −σ2h ) will decrease with t2 , but so will (quadratically) σ2buckling : For buckling, use a simple clamped-beam formulation: Fc = E2Iπ2 ( l 2)2 where I = = 4π2E2I l2 1 Ht32 . 12 Dividing by A = Ht2 , σ2buckling ⎛ 1 ⎞ 4π2E2 ⎜ Ht23 ⎟ 12 ⎝ ⎠ = l 2 (Ht2 ) 2 σ2buckling = π2 ⎛ t2 ⎞ ⎜ ⎟ E2 3 ⎝ l ⎠ (16) To proceed, start by eliminating σ2c between (7) and (10): t2 σ1 σ2h α α2 q ( α2 − α1 ) ( Tl − T0 ) q= − − 2 + E1 E2 1 − ν k2 1 − ν hl 1−ν E α E ( −σ2h ) = − E2 σ1 + 1 2− ν2 1 E2 ( α2 − α1 ) ( Tl − T0 ) ⎛1 t ⎞ + 2 ⎟q+ ⎜ 1−ν ⎝ hl k2 ⎠ (17) or, recalling (11), ( −σ2h ) t 1 + 2 ( α2 − α1 ) E2 ( Tl − T0 ) E2 hl k2 α2E2 =− σ1 + Taw − Tl ) + ( t 1 E1 1−ν 1 1−ν + + 2 hg hl k2 (18) This displays several effects: (a) Once σ1 = t2 k2 σ1ult. S is fixed, −σ2h will increase with t2 , although weakly, since 1 1 (the Cu wall offers little thermal impedance, compared to the , hg hl 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 α2 − α1 , could be used as a design aid. If 2=Cu, 1=steel, α2 = 2.3 × 10−5 K −1 , α1 = 1.4 × 10−5 K −1 , so α2 − 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 σ2c , equation (12) gives σ 2 , and equation (13) gives t1 . Some data: ( ) σult ( Pa ) (at 500K) ν 2.3 × 10−5 1.1 × 108 0.3 35 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 (SAE x4130) 1.09 × 1011 1.4 × 10−5 5.1 × 108 0.3 234 Material E ( Pa ) (at 500K) d k −1 Cu 0.95 × 1011 St. Steel 302 Z= (1 − ν ) σult. Eα ( K) o 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 . Example Pg = 100 atm (neglect Pe effect) 107 Pa D=0.3m; l=4mm hg = 24000 W / m2 / K ; hl = 2.76 × 105 W / m2 / K ; k2 = 360 W / m / K Taw = 3200 K , Tl = 400 K σ1 = 5.1 × 108 Pa ( alloy steel) ; E1 = 1.09 × 1011 Pa ; α1 = 1.4 × 10−5 K −1 1.5 −σ2ult,comp. = 1.1 × 108 Pa ( Cu) ; 1.5 E2 = 0.95 × 1011 Pa ; α2 = 2.3 × 10−5 K −1 Substituting into (18), −σ2h = −2.96 × 108 + 8.740 × 109 1.303 × 10−3 + t2 16.30 × 10−3 + t2 + 1.221 × 106 ( 400 − T0 ) (19) and, from (16) σ2buckling = 1.95 × 1016 t22 ( t2 in m.) Following are some calculated results: t2 (mm) T0 = 297 K −σ2h (Pa ) −σ2buckling 0 5.33 × 108 6.26 × 108 7.80 × 108 (Pa ) 0 t2 (mm) T0 = 500 K 0.2 −σ2h (Pa ) −σ2,buckl. (Pa ) 0 COMMENTS 1.1 × 108 , assembling at Since −σult,com = 1.5 room T0 is not acceptable for any t2 . 0.2 8 2.81 × 10 3.78 × 108 0 7.80 × 108 Closer, but still no solution t2 (mm) −σ2h (Pa ) T0 = 700 K ( ) q w / m2 = 6.15 × 107 0 0.2 0.1 8 With assembly at T0 =700K, a very thin 8 0.36 × 108 1.34 × 10 0.85 × 10 −σ2,buckl. (Pa ) 0 t2 7.80 × 108 1.95 × 108 +σ2c (Pa ) +4.48 × 108 4.40 t1 (mm) 0.1mm is acceptable, but may be questionable on robustness. Also, σ2c is too high t2 (mm) −σ2h (Pa ) T0 = 800 K q = 6.07 × 107 t1 (mm) 0.3 8 −0.86 × 10 0.12 × 10 0.60 × 10 −σ2,buckl. (Pa ) 0 +σ2c (Pa ) 0.2 0 8 8 8 7.8 × 10 8 1.76 × 10 9.79 × 107 4.48 If T0 =800K, is feasible, then t2 = 0.3 mm is acceptable. σ2c also OK (in tension) With the assumed l = 4 mm , 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.