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Unformatted text preview: /V1 5 (let ,HW 7‘“ >olufz‘aﬁ PROBLEM 7.2 KNOWN: Temperature, pressure and velocity of atmospheric air in parallel flow over a plate of
prescribed length and temperature.  I ' i I FIND: (a) Boundary layer thickness, surface shear stress and heat flux at trailing edge, (b) Drag force
and total heat transfer per unit width of plate, and (c) Plot the parameters of part (a) as a function of
distance from the leading edge. ‘ ' r SCHEMATIC: ,.
5""
um=5mls —_> . Ts=75°C
75,0: 25 00 L I
poo= 1 atm x L =1 m ASSUMPTIONS: (1) Critical Reynolds number is 5 x 105 , (2) Flow over top and bottom surface. PROPERTIES: cream Air ('13: 323 K, 1 atm): p = 1.085 kg/rn'3, v = 18.2 x 10'6 nil/8,1: = 0.023
W/mK, Pr = 0.707. ‘ ‘ ' ANALYSIS: (a) Calculate the Reynolds number to determine nature of ﬂow, _ umL _' 5m/sx1m
v 18.2x10'6m2/s Hence, the ﬂow is laminar, and at x = L, . ... rReL' = 2.75x105. 5 = 5LR::;”2 = 5x 1m/ (2.75 ><105)U2 = 9.5mm ‘ ' <
15', = (pui, /2)O.664 Reg”2 = “$1335 m/s)2 0.664/ (2.75 x 105)”2 = 0.0172 N/m2 <
. . m p  t ' Using the appropriate correlation, W, NuL = % = 0.332 Re‘,’2 Prm = 0.332(2.75 x 105) 1/2 (0.707)“3 = 155.1 11L = 155.1 (0.028 W/mK)/lm = 4.34 W/m2  K
Hence, the heat ﬂux is
q;'(L) = hL(TS — Tm) = 4.34W/m2 K(75"C — 25°c) = 217W/m2 ' . <, (b) The drag force per unit plate width is D' = ZLfs,L Where the factor of two is included to account for
both sides of the plate. Hence, from with h ' rm = (pug, /2)1.328 1112;”2 =,(1.085kg/m3/2)(5 m/s)21.328(2.75 x 105)”: = 0.0343 N/m2 ‘
the drag is ‘ ‘ i ' I D’ = 2(1m)0.0343N/1m2 = 0.0686N/m _ ‘ . <
For laminar ﬂow, the average value hL over the distance 0 to L is twice the local value, hL, EL = 2hL = 8.68W/m2 K Continued... PROBLEM 7.2 (cent) The total heat transfer rate per unit width of the plate is
q’ = 2U;de — Tm) = 2(1m)8.6SW/m2 K(75 — 25)°c = 868W/m < (c) Using with the equations of part (a), the boundary layer thickness, surface shear. stress and heat
[flux as a notion of distance form the leading edge were calculated and are plotted below. ‘ 51000
2 f {r 300
End Mower roam/Vi ° Distance from leading edge. x (m) —+ BL thicltness._ deltax ' 10 (mm)
+ Shear stress. tausx ' 10/‘4 (Mm/‘2)
9— Heat iiux. q"x (W/mAE) i: COMMENTS: (1) The velocity boundaryvlayer is very thin at the leading edge and increases with
increasing distance.“ The local shear stress and heat ﬂux are very large near the leading edge and
decrease with increasing distance. The shapes of the two curves are similar. (2) A copy of the IHT Workspace used to generate the above plot is shown below. 1/ Boundary layer thickness, delta delta = 5 * x " Flex A0‘.5 delta_mm = delta " 1000 »
delta_plot = delta_mm ' 10 // Scaling parameter for convenience in plotting I/ Surface shear stress, tausx .
tausx = (rho ‘ uinf/‘2 / 2) * 0.664 " Rem0.5
tausx_plot = tausx " 10000 // Scaling parameter for convenience in plotting ll Heat flux, q"x q"x = hx " (Ts  Tinf) NU); = 0.332 ' Rex/‘05 ' Pr"(1/3)
Nux = hx " x / k l/ Reynolds number. '
Rex: uint " x/nu lI Properties Tool: Air
// Alr property functions : From Table A.4
// Units: T(K); 1 atm pressure rho = rho_T("Air".Tf) // Density, kg/mAS
nu = nu_T("Air".Tf) ‘ // Kinematic viscosity, ml‘2/s k = k_T(“Air",Tf) // Thermal conductivity, W/mK
Pr = Pr_T("Air“,Tf) ‘ // Prandti number /I Assigned variables 4 TIM = 25 + 273 , // Airstream temperature, K Ts = 75 + 273 . ll Surface temperature, K uinf = 5 p // Airstream velocity, m/s Tl = 323' // Film temperature, K x =1 _ // Distance from leading edge, m PROBLEM 7.12 . KNOWN: Dimensions and surface temperature of electrically heated strips. Temperature and velocity
of air in parallel ﬂow. ' ‘ FIND: (a) Rate of convection heat transfer from first, fifth and tenth strips as well as from all the Strips,
(b) For air velocities of 2, 5 and 10 m/s, determine the convection heat rates for all the locations of part
(a), and (c) Repeat the calculations. of part (b), but under conditions for which the flow is fully turbulent
over the entire array of strips. ' SCHEMATIC: w=0.2m
. L=25AL=0.25m
' a fq"
um=2mls —> Ts=500°C
, Tw=25°C . , .a .v z . . ~
p=1atm ' 01m number is 5 x 105, (4) Negligible radiation. ' _ PROPERTIES: mm Air (n: 535 K, 1 atm): v = 43.54 x 10’6 mZ/s, k = 0.0429'W/mK, Pr =
0.683. ' .  ' ANALYSIS: (a) The location of transition is determined from 5 43.54 x 106 mz/s xc=5x105l=5x10 I =lO.9m umo . 2 m/s
Since xC >> L = 0.25 m, the air flow is laminar own the entire heater. For the ﬁrst strip, q] = E (AL ><
w_‘)(TS  T”) where h] is obtained from i E; = —k—0.664 Rey2 Pr‘”
AL  1/2 ‘
El =Wx0564 w (0:683)?’3 = 53.23W/m2 K
 0.01m 43.54 x10 m/s . _ _
q1 = 53.8 W/mz K(0.01rn x 0.2 m)(500 — 25)°c = 51.1w ’ < For theﬁfth Strip, C15 = c10—5 "‘ C1044,
Q5 =1"o—5 (SAL X w)(Ts "' T )‘Eo—4(4AL X W)(T5 ‘ Ten) on ‘15 = (530—5 " 4B.0—4 X w)(Ts — Tm) Hence, with x5 = SAL = 0.05 m and x4 = 4AL = 0.04 m, it follows that “EM = 24.1 'W/mgK and EN =
. 26.9 W/mZK and '  q5 = (5 x 24.1— 4 x 26.9)w/m2  K(0.01>< 0.2)m2(500 — 25)K =12.2W. . < . a q L . ‘ . 5 Continued... PROBLEM 7.12 (Cont) For the entire heater, . Hz 1
Eons _—_ £0,664Re‘L/2 pH” = 00429 x 0.664(iﬁ545) (0.683)”3 = 10.75 W/ mz K
‘ ' L 0.25 43.54x 10
and the heat rate over all 25 strips is ‘ ‘
q0_25 = HHS (L x w)(T5 — 1;) = 10.75 w/m2  K(0.25 x 0.2)m2(500 — 25)“c = 255.3w < (b,c) Using the IHT Correlations T001, External F low, for Laminar 0r Mixed F low Conditions, and
llowing the same method of solution as above, the heat rates for the first, fifth, tenth and all the strips u... (In/s) Flow conditio s 51.1 12.1 8.3 . 256‘
80.9 19.1 13.1 404
114 27.0 18.6 572
17.9, 10.6 ‘ 9.1 235
37.3 22.1 6.0 490
64.9 38.5 33.1 r 853 COMMENTS: (1) An alternative ap Vroaeh to evaluating the heat 10 s from a single strip, for example,
strip 5, would take the form q5 = h5(AL W‘)(Ts  T”), where h h,g=4_SAL or h5 z (hx=5AL + hx=4AL)/2. (2) From the tabulated results, note that for b th ﬂow condi ' us, the heat rate for each strip and the entire heater, increases with increasing air velo 'ty. For th ﬂow conditions and for any specified
velocity, the strip heat rates decrease with increa31 ' tance from the leading edge. unexpected behavior. For the um = 5 m/s mud on, the ffect of turbulent ﬂow is to increase the heat
rates for the entire heater and the tenth and ' h strips. For e 11m = 10 m/s, theeffect of turbulent ﬂow is to increase the heat rates at all lOcation . This behavior is a nsequence of low Reynolds number (Rex~
= 2.3 x 10“) at x = 0.25 m with um = 1 m/s. (4) To more fully appreciate ti}? cts due to laminar vs. turbulent w conditions and air Velocity, it is (3) The effect of flow conditions, laminar my ulent ﬂow, on strip heat rates shows some useful to examine the local coe cient as a function of distance from the eading edge. How would you
use the results plotted below t explain heat rate behavior evident in the su mary table above? 100
80 l
. so
40 20 _ Local coefﬁcient. hx (W/mAZK) _ o 0.02 , 0.04 0.05 0.013 0.1 DIstanca irom' the leading edge. x (rn) — ulnt = 2 mls. laminar ilow
—X— ulnt =1 5 mls. laminar tlow
~6— uint = 10 mls. laminar llow
—A— uini = 10 rule. luliy turbulent llow PROBLEM 7.58 KNOWN: Diameter and emissivity of a cylinder placed in an oven with a surface temperature T5 = 650
K. Velocity and temperature of nitrogen in cross flow over the cylinder. ' FIND: (a) Steadystate temperature of the‘cylinder, and (b) Change in nitrogen gas velocity, V, or
change in oven surface temperature, Ts, if you wised to increase the product temperature, Tp, by 25 °C
above the result found in part (a). 3 SCHEMATIC:
Product, TP, 5 = 0.8 31:3C0ven surface ~ ' 7 . ' , "
TS=G5OK I, g <———T.,°=350K
D = 0.005 m " ASSUMPTIONS: (1) Steadystate conditions, (2) Product surfaceis‘diffuse—gray, (3) Oven forms a.
large enclosure about the product, (4) Product length—to—dianieter ratio is >> 1. ' ' PROPERTIES: Nitrogen (Tm = 350 K, 1 atm): v = 20.78 X 10'5 mz/s, k = 0.0293 W/mK, Pr
= 0.711. ‘ " ‘ ANALYSIS: (a) Performing an energy balance for a control surface about the product, p Ein _ Eout : 0 V I qrad _ qconv = O \Uﬂé and substituting the rate equations, ZLLODW , a \Me {/0 > 0* ,7 \A V 00W) .3. 051' From the Zhukauskus correlation, (Pr/PrS :: 1, . . Pr0.37(Pr/Prs)1/4 I and ReD D p v 20.78x10 m /s
find from Table 9.4, c = 0.26 and m = 0.6,
E = _—_—_O'0293 W/m' K 0.26(7218)°‘6(0.711)°‘37 = 2,8W/m2 K. 0.05 m
Substituting numerical values into the rate‘equations,
0.80 x 5.67 x 10'8 W/mz K"(6504 — T;)K“ — 28 w/m2  K(Tp — 350)K = 0
From a trialanderror solution, it follows that
Tp z 520 K. (b) The fore ' equations were used in the IHT Workspace to dete ' Mrogen velocity V or
oven surface tempera TS would be required W to increase by 25°C above the
result of part (a). The exact so ' part (a) considering (Pr/Prs) was found as Continued... PROBLEM 7.33 KNOWN: Operating power of electrical components attached to one side of copper plate.
Contact resistance. Velocity and temperature of water ﬂow on opposite side. FIND: (a) Plate temperature, (b) Component temperature. SCHEMATIC: IV=100 Componenfs
4“ 4’59. —""‘> =25 W
:‘Lwafer/Lrhﬂ I. I , in“; g
wag“ A .. ' ., . Jri.
um=2 m/s
7; =1 7°C ASSUMPTIONS: (1) Steadystate conditions, (2) Constant properties, (3) Negligible heat loss
frOm sides and bottom, (4) Turbulent ﬂow throughout.  PROPERTIES: Water (given): v = 0.9mm“6 m2/s, k = 0.620 W/mK, Pr = 5.2. ANALYSIS: (a) From the convection rate equation, TS=T,.+q/l_1A where q'= ch = 2500 W and A = L2 = 0.04 m2. The convection coefﬁcient is given by the
turbulent ﬂow correlation ' . H: N_uL(k/L) = 0.037Ref/5Pr1/3(k/L) where ReL = (umL/v) = (2 m/sx0.2 m)/o.96x106 m2/s = 4.17x105 and hence E: 0.037(4.17x105)4/5 (5.2)“3 (0.62 W/mpK/o2 m) = 6228 W/mZ‘K.
The plate temperature is then T5 = 17°C + 2500 W/ (6228 W/mzK)(0.04 m)2 = 27°C. < (b) For an individual component, the rate equation in terms of the component’s thermal
resistance can be used to ﬁnd the component temperature, qc = (Tc " Ts)/Rt,c = (Tc "’ Ts)/(R't’,c/Ac)
Tc = T, + qc 'gc/Ac = 27°C + 25 W(2>'<104 mzK/W)/10'4 m2
Tc = 77°C. ' . . ' < COMMENTS: With ReL = 4.17x105, the boundary layer would be laminar over the entire
plate without the boundary layer trip, causing Ts and TO to be appreciably larger. PROBLEM 7.50 KNOWN: Dimensions of chip and pin fin. Chip temperature. Free stream velocity and temperature of
air coolant. FIND: (21) Average pin convection coefficient, (b) Pin heat transfer rate, (0) Total heat rate, (d) Effect of
velocity and pin diameter on total heat rate. ' SCHEMATIC:
25Ds4mm 105Vs40m/s—* L=12mm if Tw= 300 K —>
Tb = 350 K /
ASSUMPTIONS: (1) Steadystate cnditions, (2) Onedimensional conduction in pin, (3) Constant properties, (4) Convection coefficients on pin surface (tip and side) and chip surface‘correspond to single
cylinder in cross ﬂow, (5) Negligible radiation. PROPERTIES: Table A.1, Copper (350 K): k = 399 W/mK; Table A4, Air (Tf =325 K, 1 atm): v =
18.41 x 10'6 m2/s, k = 0.0282 W/rnK,‘ Pr = 0.704. aﬁdm
ANALYSIS: (a) With V = 10 m/S and D = 0.002 m, M _y_l?__ lOm/sx 0.002m _V a f“fl/V
RED _ v _18.41><10'6 mz/s _1087 I " y CC) \1: M
Using the Churchill and Bernstein correlations, Em, ! ' \[ r r p (b.‘)’/ , wt
__ 1/2 1/3 5/3 4/5 NuD=0.3,+%L[1+[ RED ] ] —16.7 0 [1+ (O.4/Pr)2/3]m F = (ﬁnk/D) = (16.7 x 0.0282 w/m . K/0.002 m) ‘= 23 5 Wm: . K < l I I  0 . If ‘
(b) For the flu Whom and ______J \ h I? F rid K: 1/2 (“# 7 3 1/2
M =l hnDknD2/4) _ b = (n/2)[235 w/m  K(0.002 m)’ 399 W/m K] 50K = 2.15W m= (HP/loan)”2 = (4 ><235W/m2 K/399W/mKx 0.002m)”2 = 34.3 m“
mL = 34.3 m'1(0.012 m) = 0.412
(E/mk) = (235W/m2  K/34.3 m" x 399W/n1 K) = 0.0172. The fin heat rate is UAW)‘ _ M sinh mL + (E/mk)cosh mL
.6 if 35’ qf _ costh + (h/mk)sinth
930 f1 W 091 g I h ' ~ b I
i: Jk OY’ 8'5” ; 761mg“ Continued... = 0.868W L n . < PROBLEM 7.50 (Cont) (c) The total heat rate is that from the base and through the fin, q = qb + qf = E(w2 — 71:D2/4)eb + qf = (0.151 + O.868)W =1.019w. <, l (C!) Using t e x en e Surface Model for a Pin Fin with the Correlations Tool Pad or‘a y in '
. t ossﬂow and Properties Tool Pad for 'Air, the following results were generated. / 1.8 . I l“‘ Total heal rate, q(W) Freestream velocity. V(m/s) —' D=2mm ’: Clearly, there is significa? enefit associated with increasing V whic increases the convection coefficient and the total) eat rate. Although the convection Coefficient d\e\cr ases with increasing D, the
increase in the total at transfer surface area is sufﬁcient to yield an increase ' q with increasing D.
The maXimum h t rate is q = 2.77 W for V = 40 m/s and D = 4 mm. ' COMMENTS: Radiation effects should be negligible, although tipwand base convectio oefficients will dif from those calculated in parts (a) and (d). ...
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 Spring '08
 Okamoto
 Heat Transfer

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