6th Homework Solution

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

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

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.

Page1 / 7

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

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online