Page 1 PROBLEM 3.1 KNOWN: One-dimensional, plane wall ...

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PROBLEM 3.2KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfacesof a rear window.FIND: (a) Inner and outer window surface temperatures, T s,i and T s,o , and (b) T s,i and T s,o as a function ofthe outside air temperature T ∞,o and for selected values of outer convection coefficient, h o .SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiationeffects, (4) Constant properties.PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K.ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.11 and 3.12,( )$ $T40 C 10 C∞,i − T− −∞,oq′′ = =1 L 1 1 0.004 m 1+ + + +h 2 2o k hi65 W m ⋅K 1.4 W m ⋅ K 30 W m ⋅K50 C2q′′ = = 968W m .20.0154 + 0.0029 + 0.0333 m ⋅K W( )Hence, with q h ( T T )$′′ = i ∞,i − ∞,o, the inner surface temperature is2q′′968W mTs,i = T∞,i − h = 40 C − = 7.7 C2i 30 W m ⋅ K$ $
PROBLEM 3.2 (Cont.)Surface temperatures, Tsi or Tso (C)403020100-10-20-30-30 -25 -20 -15 -10 -5 0Outside air temperature, Tinfo (C)Tsi; ho = 100 W/m^2.KTso; ho = 100 W/m^2.KTsi; ho = 65 W/m^2.KTso; ho = 65 W/m^2.KTsi or Tso; ho = 2 W/m^.KCOMMENTS: (1) The largest resistance is that associated with convection at the inner surface. Thevalues of T s,i and T s,o could be increased by increasing the value of h i .(2) The IHT Thermal Resistance Network Model was used to create a model of the window and generatethe above plot. The Workspace is shown below.// Thermal Resistance Network Model:// The Network:// Heat rates into node j,qij, through thermal resistance Rijq21 = (T2 - T1) / R21q32 = (T3 - T2) / R32q43 = (T4 - T3) / R43// Nodal energy balancesq1 + q21 = 0q2 - q21 + q32 = 0q3 - q32 + q43 = 0q4 - q43 = 0/* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal pointsat which there is no external source of heat. */T1 = Tinfo // Outside air temperature, C//q1 =// Heat rate, WT2 = Tso // Outer surface temperature, Cq2 = 0// Heat rate, W; node 2, no external heat sourceT3 = Tsi// Inner surface temperature, Cq3 = 0// Heat rate, W; node 2, no external heat sourceT4 = Tinfi // Inside air temperature, C//q4 =// Heat rate, W// Thermal Resistances:R21 = 1 / ( ho * As )R32 = L / ( k * As )R43 = 1 / ( hi * As )// Convection thermal resistance, K/W; outer surface// Conduction thermal resistance, K/W; glass// Convection thermal resistance, K/W; inner surface// Other Assigned Variables:Tinfo = -10 // Outside air temperature, Cho = 65// Convection coefficient, W/m^2.K; outer surfaceL = 0.004 // Thickness, m; glassk = 1.4// Thermal conductivity, W/m.K; glassTinfi = 40 // Inside air temperature, Chi = 30// Convection coefficient, W/m^2.K; inner surfaceAs = 1// Cross-sectional area, m^2; unit area
Page 1: PROBLEM 3.1KNOWN: One-dimensional,
Page 5 and 6: T∞,i − Ts,i 1hi0.10= = = 0.846,
Page 7 and 8: PROBLEM 3.4 (Cont.)7000Radiant heat
Page 9 and 10: PROBLEM 3.6KNOWN: Design and operat
Page 11 and 12: PROBLEM 3.7KNOWN: A layer of fatty
Page 13 and 14: PROBLEM 3.8KNOWN: Dimensions of a t
Page 15 and 16: PROBLEM 3.9KNOWN: Thicknesses of th
Page 17 and 18: PROBLEM 3.11KNOWN: Drying oven wall
Page 19 and 20: PROBLEM 3.13KNOWN: Composite wall o
Page 21 and 22: PROBLEM 3.15KNOWN: Dimensions and m
Page 23 and 24: PROBLEM 3.17KNOWN: Total floor spac
Page 25 and 26: PROBLEM 3.18KNOWN: Concrete wall of
Page 27 and 28: (2)(2)(2)Rcdm K/WPROBLEM 3.19 (Cont
Page 29 and 30: and with ′′ ( B B)( c,2 − s,2
Page 31 and 32: PROBLEM 3.22KNOWN: Temperatures and
Page 33 and 34: PROBLEM 3.23 (Cont.)1300Temperature
Page 35 and 36: PROBLEM 3.25KNOWN: Thicknesses and
Page 37 and 38: PROBLEM 3.27KNOWN: Operating condit
Page 39 and 40: PROBLEM 3.28KNOWN: Dimensions, ther
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Page 43 and 44: PROBLEM 3.31KNOWN: Temperature depe
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Page 49 and 50: PROBLEM 3.36KNOWN: Temperature and
Page 51 and 52: PROBLEM 3.37KNOWN: Inner and outer
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PROBLEM 3.38 (Cont.)2000016000Heat
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PROBLEM 3.39 (Cont.)and the heat ga
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PROBLEM 3.40 (Cont.)240Outer surfac
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PROBLEM 3.42KNOWN: Electric current
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PROBLEM 3.43KNOWN: Diameter of elec
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PROBLEM 3.44 (Cont.)(b) From an ene
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PROBLEM 3.45 (Cont.)The heat extrac
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PROBLEM 3.48KNOWN: Saturated steam
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PROBLEM 3.49KNOWN: Temperature and
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PROBLEM 3.50 (Cont.)20002Heat loss,
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PROBLEM 3.51KNOWN: Pipe wall temper
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PROBLEM 3.53KNOWN: Surface temperat
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PROBLEM 3.55KNOWN: Diameter of a sp
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PROBLEM 3.56KNOWN: Sphere of radius
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PROBLEM 3.58KNOWN: Dimensions of sp
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PROBLEM 3.60KNOWN: Inner diameter,
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PROBLEM 3.62KNOWN: Dimensions and m
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PROBLEM 3.63 (Cont.)To maintain T 1
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PROBLEM 3.65KNOWN: Thermal conducti
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PROBLEM 3.67KNOWN: Spherical tank o
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PROBLEM 3.69KNOWN: Critical and nor
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PROBLEM 3.70 (Cont.)Similarly, with
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PROBLEM 3.72KNOWN: Plane wall with
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PROBLEM 3.73 (Cont.)qT( − LB) = T
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PROBLEM 3.74 (Cont.)Substituting th
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PROBLEM 3.75KNOWN: Composite wall o
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PROBLEM 3.77KNOWN: Geometry and bou
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PROBLEM 3.78KNOWN: Thermal conducti
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PROBLEM 3.78 (Cont.)qL ⎛1 b ⎞Ts
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and from Eq. (3) with x = L and T(L
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PROBLEM 3.80 (Cont.)Surface energy
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PROBLEM 3.82KNOWN: Distribution of
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PROBLEM 3.84KNOWN: Cylindrical shel
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PROBLEM 3.86KNOWN: Diameter, resist
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PROBLEM 3.87KNOWN: Energy generatio
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PROBLEM 3.88KNOWN: Radii and therma
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PROBLEM 3.88 (Cont.)Clearly, the ab
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PROBLEM 3.89 (Cont.)(b) Using the o
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PROBLEM 3.91KNOWN: Materials, dimen
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PROBLEM 3.92KNOWN: Long rod experie
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PROBLEM 3.94KNOWN: Radial distribut
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Henceq2 2() = s,i + ( i − )PROBLE
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PROBLEM 3.96 (Cont.)(b) With the co
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PROBLEM 3.97 (Cont.)(b) The sphere
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PROBLEM 3.98KNOWN: Radius, thicknes
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The boundary conditions are:( ) = w
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PROBLEM 3.101KNOWN: Dimensions of a
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PROBLEM 3.102 (Cont.)Substituting f
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PROBLEM 3.103 (Cont.)Hence, the tem
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PROBLEM 3.105KNOWN: Electric power
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PROBLEM 3.107KNOWN: TC wire leads a
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PROBLEM 3.108KNOWN: Rod (D, k, 2L)
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PROBLEM 3.109KNOWN: Long rod in ove
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( ) θ ( )PROBLEM 3.110 (Cont.)1/2
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PROBLEM 3.111 (Cont.)−( ) 1/21/22
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PROBLEM 3.112 (Cont.)The gradient a
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PROBLEM 3.114KNOWN: Dimensions, end
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PROBLEM 3.116KNOWN: Dimensions and
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PROBLEM 3.117 (Cont.)Continued...dT
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PROBLEM 3.119KNOWN: Long, aluminum
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PROBLEM 3.121KNOWN: Thickness, leng
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PROBLEM 3.122KNOWN: Thickness, leng
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PROBLEM 3.123KNOWN: Length, thickne
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PROBLEM 3.126KNOWN: Pin fin of ther
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PROBLEM 3.128KNOWN: Slender rod of
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PROBLEM 3.130KNOWN: Arrangement of
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PROBLEM 3.131KNOWN: Conditions asso
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PROBLEM 3.132 (Cont.)N t(mm) η f R
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2( ) ( )PROBLEM 3.133 (Cont.)25 0.0
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The fin heat rate is( )( )PROBLEM 3
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PROBLEM 3.135 (CONT.)Heat rate, qt(
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PROBLEM 3.136 (Cont.)With w = 0.25
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where( )( )sinh(mL) + h mk cosh(mL)
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( )PROBLEM 3.138 (Cont.)C1 = 1+ η
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PROBLEM 3.139 (Cont.)Heat generatio
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PROBLEM 3.140 (Cont.)5000Heat rate,
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PROBLEM 3.142KNOWN: Dimensions, bas
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PROBLEM 3.146KNOWN: Dimensions, bas
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PROBLEM 3.147 (Cont.)5 2 3/2( ) 1/2
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PROBLEM 3.148 (Cont.)Once the heat
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PROBLEM 3.149 (Cont.)R′′ t,c to
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PROBLEM 3.151KNOWN: Internal and ex
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PROBLEM 3.152 (Cont.)(c) For the pr
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