AW1Steve
Posts: 14233
Joined: 3/10/2007 From: USA MeFLDCGuamWSNEIL? Status: offline

quote:
ORIGINAL: janh Already near 34 knots is quite impressive. Something close to or in the low 40s I would still believe. Someone with a good knowledge of hydrodynamics could probably estimate it based on the power transfer of the propellers or power output of the engines/turbines, if that is known, and the hull shape and displacement of the object. I mean to have read about a civilian liner in the 50k BRT range, maybe one of these new modern Caribbean ships, which was said to make short sprints 40+ knots on steam boilers. So it probably is reasonable also for a CVN as well? Out of pure curiosity, I'll give it a quick try. It's grown lengthier than I thought at the beginning, though. And not quite quick. QM, for example, is given with a top speed of ~30 knots using ~4x22 MW power, QE II with 34 knots on 2x44MW, the difference being due to the hulls, bows, propellers/engines specifics . I suspect it is mainly the 30% larger beam of QM that cost her those 4 knots since the propulsion details are probably less critical as long as the power transferred is effectively the same. The beam enters one of the three friction parameters that can be used to estimate the ships "Cw" friction value, and that is proportional to the power requirement P to reach a given speed (ignoring a tiny few other factors...). More exactly what enters is the wetted area of the hull, but for simple forms that's still approximately right. Another thing to keep in mind, and every biker or speedskater would know this too well, is that the friction Cw typically also goes up nonlinearly with the speed v since one usually is in turbulent conditions, so it rises more like v^2 (so Cw ~ A * v^2, where a depends largely on the object yet also the medium it is in). So 1 knot more at 30 will require much much less power than 1 more at 35. With P = Cw * v ~= A * v^3 where A(QM) = 1.3 A(QE) you actually get quite close to the relative P requirements for 30 and 34 knots. So that works roughly. Quite surprising, though, as frictional resistance is augmented by other contributions that become equally important at high speeds. Both hulls, however, have significantly smaller drafts than those of a CVN, so their Cw value will be sizably lower. Assuming it is accurate that Enterprise has four steam turbines with an maximum output of 210 MW total, this power is probably fully covered by the reactors, and transferred with similar efficiacy as QM to the propellers, the question comes down how much the friction goes up with the much deeper draft and larger displacement? The beam is about the same as QM, 41m, the length along waterline also about the same ~317m, but the draft 20% deeper at 12m. There are some approximations for areas of typical hull shapes, some very simple like A = const * sqrt(V * L), but for a really simple estimate both hulls can be treated as the same. So with only a change of the draft, which enters A over the length and width, A is larger by less than 1.2, more like 1.1. With P(QM) = 88 MW ~= A(QM) * v_max(QM)^3 and v_max(QM) = 30 knots, this gives: P(E) = 210 MW = 2.4 * P(QM) = 1.2 * A(QM) * v_max(E)^3 Putting both together: v_max(E) = (2.4/1.2)^(1/3) * v_max(QM) = 38 knots. Surely someone can do it more accurately, but it sounds reasonable so far. Since there is two other resistance components that grow very fast and important at speeds beyond 20 knots, this is probably an overestimate. Maybe Big E has a slightly better efficiacy, or advantages in hull shape, coating paints or other factors, but that is likely to be more than eaten up by the inreasing additional friction terms at speeds above 30 knots. If the hull area were only larger by "only" 1.1, it would be 39 knots, and 37 for 1.3. If the efficiacy lower or higher, say 180 MW or 250 MW effective output, it would be slightly below 37 or 41 knots. So going over 40 knots, even ignoring additional friciton, might take almost 20% power, that's quite a bit. I guess I will settle with a maximum speed of 38 then. BTW: Searching for some info on this, I found this "old" (1952) but very interesting publication from the US Naval Institute. https://darchive.mblwhoilibrary.org/bitstream/handle/1912/191/chapter%202.pdf?sequence=9 Check out how sizable the speed loss/increase of propulsion/shaft power need is depending on sea time since the last dock visit. Even after "only" 300 days, a DD would typically loose 5 knots on its top speed, and the capital ship Tennessee also some 20% after 12 months. Quite large these problems. Probably a reason why they spent so much time in yards. I never tried to keep a DD at sea long enough to accumulate enough SYS damage to loose 5 knots top. Any idea? The Enterprise hull is longer and thinner than a Nimitz. That might slightly throw off the calculations.
_____________________________
VP92 sig banner
