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Wysłany: Nie 1:12, 27 Lut 2011
Temat postu: moncler sale ASU Basic Knowledge Lecture to the he
ASU Basic Knowledge: Lecture to the heart of turbine ...
D. (Or chord length and installation angle) on nozzle performance. (3) The minimum cross-section of high aspect ratio / height and width of the smallest cross-section forming a flow channel exit area, therefore, should meet the traffic requirements of the continuity equation. If we look at exports as a oblique incision approximately right triangle (see Figure 6) available. 6 Ⅳ = tN * z, v · s (1-3O) where t Ⅳ = 2 Ⅳ / pitch left to consider the export of a narrow edge thickness reduction coefficient,
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, generally 0.95 to 0.98. 6 - outflow edge thickness by nozzle exit of the smallest cross-section of each class of traffic on the continuity equation too. GfzN = AN · Ct · pt = IN West N · C1-hZ. . b ~ C1. m (1 = 31) which can get I Ⅳ / ratio, this value is generally recommended 0.5N1.1, 1 so good,
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, when the secondary flow loss can be relatively smaller, in order to meet this requirement, you can adjust the blade number. For the same reason, some textbooks recommend / 6 = o. 30 ~ o. 45. (4) The export side of the thickness of the ) Ⅳ + (7 - a
(1-32) r,
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, = (3 ~ 5) (133) 6 = (O.10N0.25) b +1 (1-34) at this time according to Figure 15 chord the law of cosines in the triangle OAB obtained. that 0 =. +6 a 2COS (90. + port l Ⅳ) =- v / D, '-DNz (I-sin-za, n)-Drsinalt ~ 2 Figure 15 straight circular arc blade (1-35) * 48 * 1992 Cryogenic Technology 5 guillotine following parameters of the nozzle is known, find the flow velocity of the nozzle, losses, export status, and whether the deflection, the export section. Import parameters P0 = 0.9415MPa, T. = 174.2K, the export parameters PI = 0.388MPa, air flow G = 1.501kg / s, the speed coefficient = 0.96, angle of mouth air economic solution l = 15 l ( 1) from the nozzle exit velocity. check the air T__S figure was f. = 4870.5J/mol = 168.18kJ/kg from the passing of P0,
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, and Pl the isentropic Richard f = 3773.23J/mol = 130 .291kJ/kg 『Tony J by (1-15) was C =, / 2 (a ils) = 0.96, / i_ = i.1_ = 264.27m / s (2) nozzle by the type of loss (1-16) get:: (1 a) (f0 a f) = (1-0.96.) (168.18-130.291) = 2.97kJ/kg (3) status by the type nozzle exit (1-17) was lil = $ '15 + Ⅳ = 130.291 +2.97 = 133.261kJ/kg = 3859.24J/mal by the P1, 1 check air T__S figure was T1 = 137K from the P1, T1 Search compression factor was z1 = 0.982 则 nozzle exit air density P1 = P1/z1l = 0.385 × 10/0.982 × 287.2 × 137 = 9.96kg / m ~ (4) deflection angle calculation and more exponent from (1-1
calculation ≈ = K = F Lai Lai .= 1.357 1.4.1 does not consider losses in the nozzle exit Mach number for the three 【,。.:
:. Kz11 , / 1.4 × 0.982 × 287.2 × 13 a 275.281.184282.6-right foot. If after considering the loss for the exit Mach number: M == 7l1:: Type: 1.15. 51 to 111 .357 × 0.982 × the west The results show that a deflection in the oblique incision,
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, according to type (1-2O) calculated deflection-si_n (5 - '+-d): 1.0371STt516 = 0.57. can be ignored deflection effects (5) calculation of the export section of the throat width according to equation (1-31), is to take z Ⅳ = 23, D Ⅳ = 128mm, r Ⅳ = 0.98bs: height according to equation (1-32) calculations. 1G. c1. Pl =! 23 × 4.43 × 264.27 × 9.96 × 10 A = 5.6 × 10 one. m (continuing. under the stresses of work turbo-expander wheel and diffuser)
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