hw5 - PROBLE M 4:? E3332.“ 'Wmfier- 9.5+”: and axi-H...

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Unformatted text preview: PROBLE M 4:? E3332.“ 'Wmfier- 9.5+”: and axi-H a. +anK. Sb+g;cl‘.g+a~. gr-e pcov-§J_§c{.: Erik-Dr" ' Dcflrmmx Mu mu; ‘5’“?! Ara-1* “1' H“-""-‘“" ‘3“‘4 £4‘l-1L). I ’ 999‘? {ms/is' Arse awrmc +t~a~+m rm“ :c.‘~.m3¢. I . . of MN: (janfi‘ncd WH'M‘" +9“ “Iv-K; (gain: 2 _ eficgrgmnéu; i; Gag: WM DATrt: ' 5::qu vh z 2U mfs Alzlilxlli"5}]12 1), z 20 bzu' Tl = OUUEC Liquid V2 : J nals A: z 6 x 10" m" Fig. PM :ANPQLT'SlS‘. 2pm; plum" I‘D—- r—‘H. 1 “C. (TAU? '3) EM¢1M£$MMG MODEL. 1. Tm gen-No! Volume is de‘no‘f'fcl by masked hue. Oh'fM'fi-Jwa . —3 t. . ' . loam zo :- _. NUH'héA-h; kaabl¢A~4)Wl=fiL!'—BL "X 3:” ~13 ' 6‘ O-l‘i‘iém {:3 s Wan U'ZQVfCTL} and. dahm Fran». TOEIQA’LJ - MAE _C<-**53M")('“‘l§) 5.9M «gm—m "‘1: “‘ —.:—““" ' “r” ' i = 0-040; x10 3 Malacg) 5 A Mass {‘q‘t‘e bqfanc: reads) AMCV __ at ‘ 1-,“ M.“- mwfnmedf ban—Jain Nu +anr- decreasrr wa'f'k FM W 4.13 As shown in Fig. 194.13, steam at 80 bar, 440°C, enters a turbine operating at steady state with a volumetric flow rate of 236 m3/min. Twenty percent oi'the entering mass flow exits through a diameter of 0.25 tn at 60 bar, 400°C. The rest exits through a diameter of 1.5 m with a pressure of 0.7 bar and a quality of 90%. Determine the velocity at each exit duct, in 111/5. p, m 80 bar 1' than.” T; m 4400c (AV). =236 [Its/mil] 3 1513 2 0.20m, I; 9.90 pz : bar M 2 1.73 2 bar 400%: D3 = 1.5 m [)3 = 0.25 m Fig. P4. 13 KNOWN: Data are given for steam flowing through a turbine with one inlet and two exits. FIND: Determine the velocity, in m/s, at each exit duct. SCHEMA'I‘IC AND GIVEN DATA: Refer to Fig. 134.13 for schematic and given information. ENGINEERING MODEL: (1) 'i‘he control volume shown in the actiomptmying schematic operates at steady state. (2) The flow at the inlet and each exit are onewtlimensional. ANALYSIS: The mass tlow rate at the inlet and each exit is given by an expression of the form r: = iii—u.)- (o \! Front Table A-4, v. = 0.03742 nil/kg. Thus, PROBLE M 4.13Ccmh‘nmc1) 3 [235-13] [HIE Jill = -._J.. : gosll 0.03 742 .m- S 5 kg m. = 0.20:. z 2] .GE 5 dIfigv : U : ml «A ".2, kg nil} c1! * m. z m, m 0'22 =(i05.1— 21.0)535 a 84.15% S S Rearrange Eq. {1) 10 determine velocity at cxiL 2 and 3. From Table A-4, V2 3 0.04739 [IR/kg. ; \ 3 [21.0 kg” {004730 I: . ) , , A2 0(025 m) s 4 From Table A—3, an); E 0.7 bar and X} t 0.90: :23 -:. 1".) + x161? — 0,. )= 0.001036 + 0.90(2.365 u- 0.001036) = 2.128621:- l 1 g . , .1 . [84.1%][11286In-J T 121303 _ W 3 kg = 1013111- 4 l "a m. ‘ 01.5.01)? b 4 PROF?) 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(9.0,: 4505M} +€¥:Zf§2_)[3?sfizw 243.4 a? “M .= -Jsdxw _ ~I : ,...uu-.___..»m.._._u———r~—--——‘x Hug-'3 thc‘I-‘V LE +H .Udh‘ns: Lym‘ffr‘ ’L W Air enters ll compressor operating at steady static at 14.7 lhl’jitt.2 and MN.' and is eontpressw to a pressure of i5() lhl‘lini As the air passes through the compressor. it is cuuled at :t rate of 1U Btu per lb of air flowing by wzllur cir- cuiated through the compressor casin .'i‘hc volumetric flow rate of the air tit the inict is 5000 ft Ituirt, and the power input R) the compressor is 700 hp. 111:: air behaves as an ideal 3115, there is no stray heztt transfer. and kinetic and potential effects are negligible. Determine (at) the mass flow rate of the air. this, and (h) the tetsipemture of the air at the compressor exit. in “F. EN“?- MODEL.- fi:{‘f:?lb-an-L l. Th4 Cetan vamp“: shown. "fl-:GO'W' {n+mrhe'l’chirafskmdy IL (AV)I=5'D£J0 Ha/m‘m gfi'fe- Coah'vl.) 2. 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This note was uploaded on 04/02/2012 for the course ECS 222 taught by Professor Staff during the Fall '08 term at Syracuse.

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hw5 - PROBLE M 4:? E3332.“ 'Wmfier- 9.5+”: and axi-H...

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