6th Homework Solution

# 6th Homework Solution - H W W>0lW/7‘Oﬂ PROBLEM 6.1 I...

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Unformatted text preview: H W W >0lW/7‘Oﬂ PROBLEM 6.1 . I W .KNOWN: Variation of hx with x for laminar ﬂow over a ﬂat plate. FIND: Ratio of average coefﬁcient, hx, to local coefﬁcient, hx, at x. S CHEMATIC: Therma/ boundary layer; 1/- hx= CxJé, where C is a consfarnL ‘ ClélSS ne‘l‘eS ANALYSIS: From W, the average value of hx between 0 and x is —_1 x _g_ x _1/2 hx—Ifo hxdx—xfox dx Ex = 9— 2351/2 = 2035—”2 X Ex = 211x . E Hence, —x = 2 <1 11:: COB/EVENTS: Both the local and average coefﬁcients decrease with increasing distance x from the leading edge as shown in the sketch below. v PROBLEM 6.9 KNOWN: Form of the velocity and temperature proﬁles for flovv over a surface. FIND: Expressions for the friction and convection coefﬁcients. SCHEMATIC: {1017; ——-4> ——{> .62 TIE ANALYSIS: From Sectionﬁﬁﬂ’, the shear stress at the wall is is=u 5;] =MA+2By-3Cy2]y=o=Au- y=o Hence, the friction coefﬁcient has the form, ll:s ZALI. c = = f pug/2 pug. of = 3%1. <1 u.» G ‘I From Section the convection coefﬁcient is h _ —kf(ar/ay)y=o _ —kf [E + 2Fy - 3Gy2]y=o " Ts—T, ‘ D—Tw h: —ka . _ <1 D—T,. a COMMENTS: It is a simple matter to obtain the important surface parameters ﬁ'om knowledge of the corresponding boundary layer proﬁle. However, it is rarely a simple matter to ' determine the form of the proﬁle. ‘ {V} PROBLEM 6.13 ' KNOWN: Air speed and temperature in a wind tunnel. FIND: (a) Minimum plate length to achieve a Reynolds number of 108, (b) Distance from leading edge at which transition would occur. SCHEMATIC: x L x C ASSUMPTIONS: (1) Isothermal conditions, T5 = T”. PROPERTEES: W, Air, (25°C = 298K): v = 15.71x10'6m2/s. the Reynolds number is p. v .. To achieve a Reynolds number of 1x108, the minimum plate length is then _ _ Rexv _ 1x108 (15.71x10'6m2/s) 11.. 50 m/s <1 ‘ Tavemetl; the point of transition corresponds to x _ Remv _ 5x105 (15.71x10'6m2/S) ° 11.. 50 m/s xc = 0.157 m. ' _ ~ <1 COMMENTS: Note that 3‘: _ Rem: L ‘ ReL This expression may be used to quickly establish the location of transition from knowledge of Rem and ReL. is . y ' PROBLEM 6.14 \ i I I )gé/ KNOWN: Transition Reynolds number. Velocity and temperature of atmosphenc air, water, engine oil and mercury ﬂow over a ﬂat plate. FIND: Distance frOm leading edge at which transition occurs for each ﬂuid. SCHEMATIC: ASSUMPTIONS: Transition Reynolds number is Rem = 5x105. PROPERTIES: For the ﬂuids at T = 300K; ' m ‘ ' new“ boat; Fluid Table v(m2/s ) w Var .. Air (1 atm) 15.89x10'6 vat/We“ “44 4 . » j ‘5'“ “\$0257”. Water 0.858x1Cg'6 I 0 i‘ZQ‘iSUO Engine Oil V 550x10‘ a ‘ C 5 6. xii: Mercury é 0.11.3x10"6 I ( ‘43 Fm * r mo cm”) ANALYSIS: From Section 6.3,the point of transition is ' v 5 105 KC" = Rem 1;: = lxm/S V . Substituting appropriate viscosity values, .ﬁnd Fluid xc (m) o“ V 90 dc VOA/“e5 4 Air 7.95 '7" {S I Water 0.43 0‘ 4 5" Oil 275 4 ‘-'+l Mercury ‘ 0.06 ' 0 ram COMMENTS: The distance required to achieve transition increases with increasing v, due to ‘ the effect which viscous forces have on attenuating the instabilities which bring about transition. ' ‘ 4Q}; ‘ T PROBLEM 6.30 ﬁg KNOWN: Local Nusselt number correlation for ﬂow over a roughened, surface. ‘ FIND: Ratio of average heat transfer coefﬁcient to local coefﬁcient. SCHEMATEC: 09 1/3 V “‘f Nux =0.04-Rex' Pr X . ANALYSIS: The local convection coefﬁcient is obtained. from the prescribed correlation, hx = Nux —k— = 0.04 5 Reg-9 Prm . x x 0.9 - 0.9 11,. = 0.04 k [1] 13:1” L a c1 x’0'1 . ‘ v x ' To determine the average heat transfer coefﬁcient for the length zero to x, - _ l _ 1 . _0_1" hx =§Ehx dx—-;C1_f(;x dx 0.9 ‘ - 1 X —o 1 h = — — = 1.11 C x x 0.9. 1 x Hence, the ratio of the average to local coefﬁcient is hx 1.11C1x’0'1 —=—-‘:OT_= . 11)! Clx ' COMMENTS: Note that N'_‘ux/Nux is also equal to 1.11. Note, however, that _ 1 . ' Nux¢-;J;Nuxdx. t5 PROBLEM 6.29 KNOWN: Experimental measurements of the heat transfer coefﬁcient for a square bar in cross ﬂow. . . FWD: (a) h for the condition when L=1m and V=15m/s, (b) h for the condition when L = 1m and V = 30m/s, (0) Effect of deﬁning a side as the characteristic length. SCHEMATIC: L = 0,51” Resulf's V1 =20m/s El: 50 W/mz-K \(z=15m/5 i=40W/matK ASSUMPTEONS: (1) Functional form in: CRemPrn applies with C, m, n being constants, (.2) Constant properties. ' ANALYSES: (a) For the experiments and the condition L = 1m and V = lSm/s, it follows that , Pr as well as C, m, and n are constants. Hence EL a (VL)"‘ . Using the experimental results, ﬁnd m." Substituting values HiLi _ V1141. m 50x05 _ 20x05 m 5sz “ V21.2 40 xO.5 ' 15 x 0.5 giving m = 0.782. It follows then for L = 1m and V = 15m/s, " 0.782 H—‘H ﬂ 111‘— m—50 W 0‘5 15"”) ~-343W/m2-K < ' 1L Vl'Ll — m2.~1<; x70 20x05 ‘ ' ' (b) For the condition L = 1m and V = 30m/s, ﬁnd ‘ ' 0.782 ' ‘— E L1 ——V'L m — 50 W 0‘5 30 x 1'0 — 59 OW/mZ-K <1 — _L— V1L1 — mz-K XI?)- 20x0.5 _ ‘ ' (c) If the characteristic length were chosen as a side rather than the diagOnal, the value of C would change. However, the coefﬁcients m and u would not change. COMMENTS: (1) The Nusselt number relation with Re, Pr in the form shown will be used frequently in subsequent analysis. (2) ‘It is important to recognize the use of this relation in scaling the value of the-heat transfer coefﬁcient for various characteristic lengths and. velocities. /)( ‘ ‘ PROBLEM 6.37 §< KNOWN: Drag force and air ﬂow conditions associated with a ﬂat plate. FIND: Rate of heat transfer from the plate. ' SCHEMATIC: Um =40 777/5 ___> ASSUMPTIONS: (1) Chilton-Colbum analogy is applicable. PROPERTIES: W4, Air (70°C, 1 atm): p = 1.018 kg/m3, eP = 1009 J/kg-K, Pr: 0.70, v = 20.22x10'6m2/s. an/UQS WE? 1/an ANALYSIS: The rate of heat transfer from the plate is i 0" (a V . q = 2 W) (Ts—Ta) where K may be obtained from the Chilton-Colburn analogy, ﬂail?” (‘7 /i I . _ C _ ‘ — I = .—f = St P125 = h 1312/3 2 p u.” cp E E ‘ 2 ~ _i = .1. _: = i W _ 5.76x1o—4 2 2 p u../2 2 1.018 kg/m (40 m/s) /2 . Hence, — Cf _ h = .5. p u.” cp Pr 2’3 E: 5.76x10-4(1.01gkg/m3) 40m/s (1009J/kg-K) (070)-2/3 E: 30 W/mz-K . ‘ ‘ The heat rate is q = 2(30 W/mz-K) (0.2m)2 (120—20)°c q = 240 w. ‘ ‘ - r . Q COMMENTS: Although the ﬂow is laminar over the entire surface (ReL = umL/v = 40 m/s x 0.2m/20.22><10‘5m2/s = 4.0x105), the pressure gradient is zero and the Chilton-Colbum‘ analogy is applicable to average, as well as local, surface conditions. Note that the only contribution to the drag force is made by the surface shear stress. ...
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## This homework help was uploaded on 04/15/2008 for the course ME 114 taught by Professor Okamoto during the Spring '08 term at San Jose State.

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6th Homework Solution - H W W>0lW/7‘Oﬂ PROBLEM 6.1 I...

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