Heat transfer j p holman 10th edition solution manual

SolutionsManual to accomparry Heat Transfer tenth edition J P Holman southernMethodisttJniversity Chapter I 1-1 LT- (39oox9-022 = (0.2x0.6) 625oc l-2 q _(0.035X85) -_ , 22f85 llt'm2 A 0.13 = g2,3g6 iln.^Z 1-3 , ^ cdT ll q- - _l ?1._ f u c- drc r-ax+b; x=0; dT -kw' nrT r-0.0375 x-0.3, r-A.0625 r = 0.083 3x + 0.0375 tuc|J @+lq3-s4o) rc(0.0g33x*AOW=-l_f \ " = o r=_- ( z o ) ( _ 4 ) rce.og33) o.og,n*+ [ o.o3 lj )*=o ? q-2238 W 1{ q _ ( X - )4 F ^ ^ A ffi-1s08 wl^z 1-5 A- w2 q - -k4nrzar d; (r =4n(2xl0u)(zl 1_ -4nk(ra-r) q-Y ri + 196)_ l l w m -0rt85 mass evaporated = 1'.u11= = e.gl3x l0-s kg/s 199'ooo - s.7az kglday rg ChapterI l:7 q 30+20 4-T* T-m= Zrlk ' httdo !(39) , r - 11.89W /m *arffi+er6rI l-8 Like manykindsof homespunadvice,this is badadvice.Alltypes of heattransfer; conductio4 convection,and radiationvary directly with area.The surfaceareaof the headis much lessthan that of the otherportion of the body andthuswill loselessheat This may be shownexperimentallyby comparingexpozurein cold weatherwearing heavyclothing andno hat,to that wearinga heavyhat andonly undergarments! t-g q _ ( X 0 - 1= 00 )Z Zw l ^ , A 0.05 1-10 Ar-: - KAAT q _ (roxlo-3#)tsoo) = 0.0125m - 1.25cm l-l I 4gzLr z.Ls8- (2.7)(o"l(*)1o.to '\12 )' LT = 0.632"C l-12 q-(5.669x lo-*)t(r o:n)4-6214l = 9g [ A ^2 l-13 =188 ry m- g =(5.669x lo-t)t(t 3n)4 - (698)41 A l-14 - (70)oJ = 704.8 W x 10-r)(+ox0.3 q = (s.669 ,2t(300)4 1-15 il q - (5.669x tO-t)t7n)4 -Q7r4J - t.9t 4x104 wl*' b q -(5.669x lo-r)ten)4 - (Til4I = (5.669xlo-tlrtro )4- Q:R\41 T P = 1K q - - 3w l ^ t Reducedby 44.37o 3' ChaPter1 l-16 q - h A ( T w-T p u i a ) FromTablet-z w h-3soo ;zJc q=(3500)rTd.L(40)=(3500)zr(0.025X3X40)=32987w q = mcoLTpaid \[ - (0-5kg/sX4180J/kg'oC)Af 32,987 LT=15.78oC l-17 hfs = 2257 kJ/kg kJ = 2'37 =2-37kw = 853 kJ/kg) S q = tuhfr= (3-78kg/hr)(2257 * r_z FromTable h_Tsoo $- q=hA(T*-Tpuia) - 100) 2370\{ - (7500x0.3)2(T* T* =965oC 1-18 q= LALT x to4 Btu - hQ3z-zrz)"F ffi- h-1s00 +=85 hr'ft' 17 w oC ^2 ' 1-19 q =oa{t( T)a - (Tz)41 - (298)41 2000W - (5.669x l0-sX0.85X0.006X3)t(n)4 \=t233K 1-20 - 1'489x lOs g q+273)a t0-*Xt000 (5 66gx m' dr4= A l-2r q - d t4 A ' ro^ 06 = (5 -66gx1o-8)74 T - 5 6K Chapter r-22 -Tro\=(5.669 gtl4 -2934)=29.t9w q = oeAl(T1o x to-8;1o.oXazrX0.0af r-23 AT q=lAi= M(Ta-T-) - 4L) (1.4X315 - h(41- 38) 4.025 h-s|14 y m'.oc l-.25 0- T q_(t.6)'#_ ro(r*z_ro) T*, - 35.7oC q -10(35 -10) -2s7 g m2 r-26 lt - 4.5 From Table l-2 y for LT: 30oC mt.oc - 12.15 q - uALT= (4.5X0,3)2(rO) V,' Conduqlion q-kA+ Ar W frfor air = 0.03 - m-oC _ 3.24W q,,,_(o.o3xo.3)2(30) 0^025 1-21 ^r_ (lsoo)(H) - l,2soF 25 7".100-1.25-98.75oF 1t8 700- (l r)(T* - 30) Tw = 93'6oC Chapter r-29 Q=Qconv*Qrad qconv = hA(Tr-T*) FromTablel-2 ft = 180+- m'.oc -30) = 4807 y length =(1S0)z(0.05)0X200 econv m -Tza) = oeAl(Tta Qrad = (5.669x to-8xo.z)z(0.05X1) (4:a4 -2834) =2 I t" n g th m = 48O7 + 272=5079 Y Qtotat m Most heattransferis by convection 1-30 Q=flconv*Qrad -T*) 4conv = hA(T* FromTablelJ h= 4-5+- m' oC - 20) (2sides) = {conv f;ll]il;r"so -zgla) = o4,(Tta -Tza)= (5.669x to-8X0.gX0.32X323a erad =2 7W 4total= 24'3W +28'7 W = 53 W Convectionandradiationareaboutthe samemagnitude r_3I e = econv* erad= (insulated) 4conv= hA(T*-T*) From Table l-2 h:12 Jm' 'oC - Tzo'),€ = l'0, Tz = 35oC= 308 K Qrad= oeAl(Tl o = h\(Tr - T*) + o€Argt4- rz4) - 3084) o = (12)(4-273)+(s.669x tO-8;1t.0X44 Solutionby iteration: Tt 7.=285K=12"C (2 sides) Chapter1 r-32 - T,r)= (5.669x l0-8X& - T*rn)= l5(T*, - 293) (100X353 ,o ts(Twz- 293)- (5.669x t O-8)[(rg7 - O.LsTw)a- (T.r)a] = = f (T*r) T*, f (Tnr) 320 158.41 350 907.22 3r0 -77.O3 313.3 0.058 = 350K T*r =397- (0.15X313.3) 1-36 h=4'5 w fr (Plate) w (cylinder) h=6.5 + m-."u T* = 2O"C= 293K hA(T -T*) - otA(74 -T*4) Plate x to-8xr4-zgza) 9.5)Q -zg3)=(5.668 T= no realisticvalue(T =247 K, heatgained) Cylinder x to-8xr4 - 2%\ (6.t(r - 293)=(5.668 T =320K= 47"C t-37 The woman is probably correct Her perceivedcomfort is basedon both radiation and convection exchangewith the surroundings.Even though a fan does not blow cool air on her from the refrigerator, her body will radiate to the cold interior and thereby contribute to her feeling of "coolness." r-38 This rs an old story All things being equal, hot water does not freezefaster than cold water The only explanation for the observedfaster cooling is that the refrigerator might be a non-self defrost model which accumulatedan ice layer on the freezing coils Then, when the hot water tray was placed on the ice layer, it melted and reducedthe thermal insulation betweenthe cooling coil and the ice tray ChaPterI 1-39 As in probleml-36, it mustbe observedthat a person'scomfortdependson total by both radiationandconvection.In the heateichangewith the surroundings winter the w-allsof the roomwill presumablybe coolerthanthe room air and increasetheheatlossfrom the bodies.In the sumrn€rthe walls areprobablyhotter than the room air temperatureandtherebyincreasethe heatgain or reducethe heatlossfrom thepeoplein the room l-40 Q=Qconv*Qnd qconv= hA(T* - 7*) = Q)n(l')(6X78- 68)-377 Btu/hr Fot T2= 45oF= 505oR -Tzo) erad=ocA1(T1a - so54) = (0.r714xto-8xo.g)zr(lX6X53g4 = 544 Btu/hr +5M = 921 Btu/hr Qwtat=377 Fot T2= 80oF= 540oR = -36.4 Btu/hr erad=(0.1714x to-8)(o.g)a(1)(6X53845+04) = 377- 36'4= 340'6 Btu/hr Qtotat Conitusion:Radiationplaysa very importantrole in "thermalcomfort." t-41 q ss= T i= O "C=2 K Ta= 22"C Ts= 25"C= 298 K A - (12>(40)= 480 m2 a - 2t 3a')= 60262w - \4 )= (5.668; I 0-8X0.95X480X298 = oeAl(Tsa erad = W {conv= M(To-4)=(10X480X22-0) 105,600 = = 60,262+105,600 165,862W Q,rlvrr For ice 1&= 80 calf9=3.348x tOs ft