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HWS_CH6_EGN3358_Sleiti

HWS_CH6_EGN3358_Sleiti - P BLEM-1 M Air at 30° C and...

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Unformatted text preview: P BLEM -1 . M Air at 30° C and atmospheric pressure flows past a ﬂat plate at 15 m/s. The transition Reynolds number is 0.3 x 105. EIHD: What is the thickness of the boundary layer and the wall shear stress at a point 0.6 m from the leading edge of the plate? H AT] ' U = 15 m/s “a T0" = 30 C 5 I . x = 0.6 m . l ASSUMPTIQNS: ERQPERTIES; Air at 30°C from Table A-8; v = 16.01 x 10-6 mzls p = 1.164 kg/m3 A ALY I ' Rex = (U-x)lv = [(15.0)-(0.60)]l(6.01 x 10'6) = 5.621 x 105 Since Rex > 3 x 105 the bomdary-layer is turbulent. From Table 6-1: 1 -_l. 5}: 0.371{Rex]'; ; 2: 0.371{5.621 x 105] 5 = 2.627 x 1o2 8 = (0.60)-(2.627 x 10-2) = 1.576 x 10-2 ¢=¢=¢=¢=¢=<= Res = U-5/v = (15.0)-(l.576 x 10-2)/(16.01 x 10-6)= 1.477 x 104 I“, = 0.0225-p'U2'[R65]-1f4=(0.0225)~(1.1644)-[l.477 x 10“)“4 Tw = 0.5347 Nt’l‘ﬂ2 ¢=¢=¢=¢=¢=¢= PRQBLEM 6-9 [(55 QWN; Air ﬂows pasta smooth ﬂat plate which is parallel to the ﬂow. El ND ; Determine the ratio of the friction drag over the front half of the plate. x = O to x = U2, to the friction drag of the entire plate if: (a) The flow is laminar over the entire plate. (b) The flow is turbulent over the entire plate 'and ReL < 107. H AT] ' U #- I I H———L—-—-—>-l ASSUMPTIQNS; R PERTIE ' ANALYSIS; (a) Flow is completely laminar: Leading one-half; x = U2 ; Rex = U-U(v-2) ; A = b-L/2 G = 1.32:3{Re,.]-”2 From Table 6-1 U. L 1/2 _ _C'}oA-p-U2 _ U 1/2 .ZU b-L Cf=l.:328‘[v—-2] ,Df;2—-—--——-2———l.-328 'v—zJ' 2 ‘2 . . _ P-U2 ”11/2 L112 . Dm _ 1.328 2 (U) M2) Full Length: x = L; Rex = U-Uv ; A = b-L _ _ "1&1“? . _E}A-p-U2 _ [U- LII/2p. .U2 Cr— 1.328[ V 2 , Dm— % _ 1.328 v 2 b L p-U 1/2 _ 1I2 I2 _ Dr = 1-328 2 '(ﬁ‘) 'b'Luz " ._f£r2=(L) (l) =?1f_ 0'70? ¢=¢=¢=¢=¢=<= (b) Flow is completely turbulent: Leading one-half; x = U2; Rex = U-lJ(v-2) ; A = b-U2 __ 1Is Cr = 0.074-[Red'1’5 From Table 6-1; Cf_ — 0. 074{%—- 3‘] _ C_r-A-p-U2 U- L- 4’5 p_- -U2 __-___b L_ p U2 AL US L 415 D‘Q'T=Om4[v2 2 2 LOO-M 2 'bi ] (— Full Length: x- — L; Rex = U-L/‘v; A: b L E ! 1 _ us [/5 2 . 2 Cf: m374[Uv——'I: mm: 0.074{U—'—VL] LE --—bL — 0074'” E 0%]“5 -(L IS _' Dm_ = (Lr15(l)415_ l —O. 57 4 Dr 2 L 2 4/5 ¢=c=c=¢=¢=¢= LL________________ ERQBLEM 6-815 -. w A baseball pitcher is clocked throwing a fast ball at 90 mph when the air temperature is 60°F. ' HELL- If the diameter of the baseball is 2.80 in.. calculate the drag force on the baseball assuming the effects of surface roughness to be negligible, i.e. a new baseball. 512HEMATIQ; MP I ' (l) The baseball is traveling at a constant speed so the flow past it is steady-state. (2) The baseball is new and therefore its surface is assumed to be smooth. PRQPERTIES: Air @ 60 'F from Table 3-2 p = 0.07633 lbm/ft3 v = 0.5692 ft2 [hr = 1.58 x 10‘4ft21's g = 32.17 {:152 LY I ' U = 90 mph (5280 fu’rnile)/(3600 s/hr) = 132 His Red = U-dfv = l32-(2.8l12)l(l.581x104): 1.948 x 105 C1) = 0.4 (Figure 6-15a for a smooth sphere) D = C1) 13-02 -A12 = 0.4-(0.07633l32.l7)-(132)2 -n.(2.3/12)2 14 Therefore. D = 0.3535 [bf <=<=<=¢=<=<= MMENT DI U I ° The drag on a new baseball is calculated assuming that the drag coefﬁcient of the baseball can be represented by that for a smooth sphere, Figure 6-15a. At the Reynolds number of the ﬂow, Red = 1.948 x 105 , the flow has not naturally transitioned to turbulent. Hence the larger drag coefﬁcient is used. If the baseball were assumed to be rough then the lower value of drag coefﬁcient would be justiﬁed. 1 i ERQBLEM Q-IBE A well streamlined automobile has a measured drag coefﬁcient of CD = 0.045. A poorly streamlined automobile has a CD = 0.20. if each of these automobiles has an equivalent diameter of 6.0 ft, determine the power required to overcome the air resistance at 60 mph at sea level through air at 60"F. a U.S. Standard Atmosphere, where PSL = 21 16.2 lbplft2 and p = 0.07633 lbmlftz. The speeds of the automobiles are constant with time (steady-state). Air @ 60 "C from Table B-2 p = 0.07633 lbmlft3 P = 2116.21bp'ft2 T = 60 "F g = 32.17 fu's2 NALY I ' 60 mph = 88 Ms Drag = C9 -(pJ'(2-g))-U2-(7t-d2 l4) (Drag)1 = 0.0(115-(0.07633116434)-(88)2 -7t-(6.0)2 I4 = 11.69 lb1r (Drag)2 = 0.20100763376434)-(88)2 -7c-(6.0)2 74 = 51.95 1h; Horsepower = (drag)-(velocity)l550 (Horsepower)1 = 11.69-(88)l550 = 1.870 ¢==¢=¢=<=¢= (Horsepowem = 51.95-(88)!550 = 3312 <=¢=¢=¢=<=<= PR BLE -1 E KNQWN & FIND: Repeat Prob. 6- 18 for Standard Atmosphen’c cenditions at 5000 ft. (Denver, CO) where T = 41.2"F. PIPSL = 0.8321, and pipSL = 0.8616. Compare the power required to overcome air resistance at the two altitudes. SSHEMAILQ: W The speeds of the automobiles are constant with time (steady-state). PRQPERTIES; Air @ 41.2 “F from Table 13-2 pSL = 0.07633 1bm/ft3 PSL = 2116.2 than? TSL = 60 "1: g = 32.17 ft/s2 ANA LYSIS: 60 mph = 88 his Drag = C9 -(p1(2-g))-U2-(zrc.d2 14) Drag = (Drag)SL~(P/PSL) (Dragn = 11.69-(0.8616) = 10.072 (Drag): = 51.95-(O.8616) = 44.76 Horsepower = (drag)-(velocity)1550 (Horsepowerh = 10.072-(88)l550 = 1,611 <=<=¢=¢=¢=¢= (Horsepowem = 44.76-(88)l550 = 7.162 <=<=¢==¢=<= E T DI v A 14% reduction of the drag on each automobile is experienced between sea level and Denver, CO (5,000 ft). This is due to the change in air density between sea level and 5,000 ft. KW N; Water at a velocity of 5 mls flows over a horizontal ﬂat isothermal plate 20 cm long. The temperature of the water is 30 C while that of the surface of the plate 13 60° C. FIND' Calculate the rate of heat transfer per unit width from the upward facing surface of the plate. HE ATIC: Water @ 30°C # b Wtdlh U = 5 mls “A“ A MPTI S: PR PERTIE : ANALYSIS; Use eq. 6-36 H921 2233x10-3 Use eq. 6-35 Nu: VIO- :1 c-_ 1 + 12,7.Et)”2.( pr2f3 _ 1) 1 + 12.7(2233 x 10-3)”2-(5.42m - 1) A -=(AT) -(20.6 x 103)-(0.2)-(60-30) = 123.6 x 103 W/m = 123.6 kW/m ‘=¢=¢=<=¢=¢= ReL- — (U- --L -—p)ltt - ((5)- (0. 2) (995. 7))/(797. 8 x 10 6)— - 1.248 x 106 _6.?08 x103)(0 615):206x103 W/m2 c PRQBLEM ﬁ-Zﬂ For water @ 30"C from Table A-9; k = 0.615 WIm-“C Ll. = 797.3 x 10-6 kglm-s p: 995. 7 kg/m3 Pr = 5.42 61,— _ 4.180 kJ/kg c (turbulent boundary- layer) ‘c? = 0.074{Re._‘|-”5 = 0.074-[1248 x 106]“: 4.466 x 10-3 = 6.708 x 103 (.02) EBQBLEM ﬁ-Zﬁ KNQWN; A fluid at 20° C ﬂows over a 50 cm long ﬂat plate with a velocity of 2 m/s. E IN D '. Detennine the average convection heat transfer coefﬁcient for the following fluids: (:1) Air. (b) Water. W U = 2 mls I ~ “—255Mm—H A MPTI : Uniform surface temperature PBQPERTlEs; Air at 20' c, Table A-8. v = 15.09 x 10*6 1112/3 ' Pr = 0.713 k = 25.64 x 10'3 W!m-°C Water at 20' C. Table A-9. v = 1.004 x 10'6 m2/s Pr = 6.99 k = 0.5996 WIm-"C AMLXSIE: (a) Air Re, =91 539.51% = 66270 a 66.27 x 103 (Laminar now) V 15.09 x 10-46175] Use eq. 6-31. __ -3 h = 0.664% -Pr”3-Re1”2 = 0,664.2_5-§%_’;.1_0_ -0.713“3-(66.27 ><103)”2 = 7.83 FE U-x 2 [11051.0 5 [m] (b) Water Re = ‘ V 1.004x 10061215] Use eq. 6-38. Turbulent Flow; m = 'V Nulzmiw + Nufurbulcm mam“, = 0.664~(Re)”2.(Pr)m = 0.6640 x 106 ”21699)“3 = 1270 eq. 6-30. —— 0.037(ReP'8-(Pr) NU b len =-——-—-—-—-——-—-—-————-— = 6205 .6—37. “" " ' 1 + 2.443-(RerO-l-((Pr)2’3 — 1) eq N1? = 6/0270? + (6205)2 = 6334 i; = FIE-k = 6334-(0.5996) = 7596. 111 .,._¢=¢=¢=¢=¢= L 0.5 m2-'C =——————__— = 996 x 103 = l x 10‘5 (Turbulent ﬂow) PR BLEM -44 KNQWN: A window 0.5 m wide and 0.4 m high has an interior surface temperature of 10° C. A small fan has been installed in order to minimize the danger of the window fogging up. The fan creates a slight upward movement ofair over the window with its velocity being of the order of 1 m/s. FIND ' Estimate the rate of heat loss through the window if the temperature of the air in the room is 20" C. S! :HEMATIQ: = 10°C , 7,, = 20°C 3 [opposing flow] ﬁU = l mls A MPTI N ' PRQPERTIEs: Air @ Tr: (20+10)/2= 15°C (233.15 K) from Table A-8; Pr = 0.715 v = 14.64 x 10'6 m2/s k = 0.02526 W/m “C 13 = 171* = 3.47 x 10-3 UK p = 1.225 kglm3 ANALYSIS: ReL = 95-5 = M = 27.32 x 103 (Laminar flow) V (14.64 x 10-6) .. .3. -3. _ . . a=g914m PF=W=7MGHW V2 (14.64 x 10-5)2 The ratio of GIL [Re2 is 2 0.1 so combined natural and forced convection must be considered in calculating the heat transfer. R The forced convection opposes the natural convection. Use eq. 6—30 to calculate the effect of forced convection. mF = 0.664-(ReL)”2-(Pr)”3 = 0.664-(27.32 x 103)“2-(0.715)m = 93.17 25.6 eq. 6-52 to calculate the effect of natural convection. NuN = 0.68 + 0.67‘[RaL-W(Pr)]m MPr)=[+94—31)9”6]m=ﬂig’lﬁlmwsm [1 + 0.715 m” =0 68 + 0. 67 [(72 66 x 106M 0 343)]”“ = 43. 2 N1? = Nu]: — mm = 198.143 — 48.234”3 = 94.1 H —— h-.(0 40) h“ “= 7* = 94-1 ; E = 5.942 wrmzrc k T(.25 26 x 10 3) Q= 11- A-(A ).=(5 942)“). mm. 10)= 11 33 w <===¢=== MMENT DIS ION: The movement of this air upward by the fan decreases the heat transfer to the window. ...
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HWS_CH6_EGN3358_Sleiti - P BLEM-1 M Air at 30° C and...

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