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7-99 Wind is blowing over the roof of a house

By Jeanne Crawford,2014-07-13 17:22
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7-99 Wind is blowing over the roof of a house

    Chapter 7 External Forced Convection

    7-99 Wind is blowing over the roof of a house. The rate of heat transfer through the roof and the cost of this

    heat loss for 14-h period are to be determined.

    5Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re = 5?10. 3 Air is cran ideal gas with constant properties. 4 The pressure of air is 1 atm.

    Properties Assuming a film temperature of 10?C, the properties of

    air are (Table A-15) T = 100 K sky;Qk0.02439 W/m.?CAir -52V = 60 km/h ?~1.426?10 m/s T = 10?C ?Pr0.7336

    Analysis The Reynolds number is T = 20?C inVL(60?1000/3600) m/s(20 m),(7? Re2.338?10L52~1.426?10 m/s

    which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow. Then the Nusselt number and the heat transfer coefficient are determined to be

    hL0.81/370.81/34Nu(0.037Re871)Pr[0.037(2.338?10)871](0.7336)2.542?10Lk k0.02439 W/m.?C42hNu(2.542?10)31.0 W/m.?CL20 m

    In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal to

    the heat transfer from the roof to the surroundings (by convection and radiation), which must be equal to

    the heat transfer through the roof by conduction. That is,

    ;;;;QQQQ,,, room to roof, conv+radroof, condroof to surroundings, conv+rad

    Taking the inner and outer surface temperatures of the roof to be Tand T, respectively, the quantities s,in s,out

    above can be expressed as

    2244;QhA(TT)A(TT)(5 W/m.?C)(300 m)(20T)?Croom to roof, conv+rad isrooms,insrooms,ins,in 282444,( (0.9)(300 m)(5.67?10 W/m.K)(20273 K)(T273 K)s,in

    TTTTs,ins,outs,ins,out2;(2 W/m.C)(300 m) QkA?roof, conds0.15 mL

    2244;QhA(TT)A(TT)(31.0 W/m.?C)(300 m)(T10)?Croof to surr, conv+rad oss,outsurrss,outsurrs,out 282444,((0.9)(300 m)(5.67?10 W/m.K)(T273 K)(100 K)s,out

    Solving the equations above simultaneously gives

    ;Q28,025 W28.03 kW, T10.6?C, and T3.5?C s,ins,out

    The total amount of natural gas consumption during a 14-hour period is

    ;QQt(28.03 kJ/s)(14?3600 s)?1 therm?total Q15.75 therms?gas?0.850.850.85105,500 kJ??

    Finally, the money lost through the roof during that period is

    Money lost(15.75 therms)($0.60/therm)$9.45

     7-84

    Chapter 7 External Forced Convection

    7-100 Steam is flowing in a stainless steel pipe while air is flowing across the pipe. The rate of heat loss from the steam per unit length of the pipe is to be determined.

    Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The

    pressure of air is 1 atm.

    Properties Assuming a film temperature of 10?C, the properties of air are (Table A-15)

    -52k0.02439 W/m.?C, ~1.426?10 m/s, and Pr0.7336

    Analysis The outer diameter of insulated pipe is D = 4.6+2?3.5=11.6 cm = 0.116 m. The Reynolds number o

    is Steel pipe

    VD(4 m/s)(0.116 m)D = D = 4 cm i14?o Re3.254?1052D = 4.6 cm 2~1.426?10 m/sInsulation The Nusselt number for flow across a cylinder is determined from = 0.3 Do4/55/80.51/3?hD??0.62RePrReoD i?Nu0.31??1/42/3k282,000???10.4/Pr,(;;?? 4/55/8?40.51/34??0.62(3.254?10)(0.7336)3.254?10?? 0.31107.01/4?2/3?282,000??,(;;10.4/0.7336??Steam, 250?C k0.02439 W/m)?C2and hNu(107.0)22.50 W/m)?CoD0.116 mo

    Area of the outer surface of the pipe per m length of the pipe is Air 23?C, 4 m/s ADL(0.116 m)(1 m)0.3644 m oo

    In steady operation, heat transfer from the steam through the pipe and the insulation to the outer surface (by first convection and then conduction) must be equal to the heat transfer from the outer surface to the surroundings (by simultaneous convection and radiation). That is,

    ;;;QQQ,, pipe and insulationsurface to surroundings

    Using the thermal resistance network, heat transfer from the steam to the outer surface is expressed as

    11R0.0995 ?C/Wconvi,2hA(80 W/m.?C)(0.04 m)(1 m),(ii

    rrln(/)ln(2.3/2)21R 0.0015 ?C/W pipekL22(15 W/m.?C)(1 m)

    rrln(/)ln(5.8/2.3)32R3.874 ?C/WinsulationkL22(0.038 W/m.?C)(1 m)

    (250)?CTTT1?ss;and Qpipe and ins(0.09950.00153.874) ?C/WRRR,convipipeinsulation

    Heat transfer from the outer surface can be expressed as

    2244;QhA(TT)A(TT)(22.50 W/m.?C)(0.3644 m)(T3)?Csurface to surr, conv+rad oossurrossurrs 282444,( (0.3)(0.3644 m)(5.67?10 W/m.K)(T273 K)(3273 K)s

    Solving the two equations above simultaneously, the surface temperature and the heat transfer rate per m length of the pipe are determined to be

    ;T9.9?C and Q60.4 W (per m length) s

     7-85

    Chapter 7 External Forced Convection

7-101 A spherical tank filled with liquid nitrogen is exposed to winds. The rate of evaporation of the liquid

    nitrogen due to heat transfer from the air is to be determined for three cases.

    Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant properties. 4 The pressure of air is 1 atm.

    Properties The properties of air at 1 atm pressure and the free stream temperature of 20?C are (Table A-15)

    0.02514 W/m.?Ck

    -52Insulation ~1.516?10 m/s

    51.825?10 kg/m.s ?

    65.023?10 kg/m.ss, Do@196 ?CWind 20?C Pr0.7309D i40 km/h Analysis (a) When there is no insulation, D = D = 4 m, i

    and the Reynolds number is

    VD(40?1000/3600) m/s(4 m),(6? Re2.932?1052~1.516?10 m/s

    Nitrogen tank The Nusselt number is determined from -196?C 1/4??hD0.52/30.4??20.4Re0.06RePr,(Nu?ks?? 1/45??1.825?1060.562/30.4?,( 20.4(2.932?10)0.06(2.932?10)(0.7309)23336?5.023?10??

    k0.02514 W/m.?C2and hNu(2333)14.66 W/m.?CD4 m

    The rate of heat transfer to the liquid nitrogen is 2;()()()QhATThDTTss?s? 22 (14.66 W/m.?C)[(4 m)],((20(196) ?C159,200 W

    The rate of evaporation of liquid nitrogen then becomes

    ;Q159.2 kJ/s;;;Qmhm ((?0.804 kg/sifh198 kJ/kgif

    (b) Note that after insulation the outer surface temperature and diameter will change. Therefore we need to

    evaluate dynamic viscosity at a new surface temperature which we will assume to be -100?C. At -100?C, 51.189?10 kg/m.s. Noting that D = D = 4.1 m, the Nusselt number becomes 0

    VD(40?1000/3600) m/s(4.1 m),(6?Re3.005?10 52~1.516?10 m/s

    1/4??hD0.52/30.4??20.4Re0.06RePr,(Nu?ks?? 1/45??1.825?1060.562/30.4?,( 20.4(3.005?10)0.06(3.005?10)(0.7309)19105?1.189?10??

    k0.02514 W/m.?C2and hNu(1910)11.71 W/m.?CD4.1 m

    The rate of heat transfer to the liquid nitrogen is

     7-86

    Chapter 7 External Forced Convection

    222(4.1 m)52.81 mADs

    TTTT,tan,tan?sk?sk;Q rr1RR21convinsulation4krrhA12s

    [20(196)]?C7361 W(2.052) m1224(0.035 W/m.?C)(2.05 m)(2 m)(11.71 W/m.?C)(52.81 m)

    The rate of evaporation of liquid nitrogen then becomes

    ;Q7.361 kJ/s;;;Qmhm ((?0.0372 kg/sifh198 kJ/kgif

    (c) We use the dynamic viscosity value at the new estimated surface temperature of 0?C to be 51.729?10 kg/m.s. Noting that D = D = 4.04 m in this case, the Nusselt number becomes 0

    VD(40?1000/3600) m/s(4.04 m),(6? Re2.961?1052~1.516?10 m/s

    1/4??hD0.52/30.4??20.4Re0.06RePr,(Nu?ks?? 1/45??1.825?1060.562/30.4?,( 20.4(2.961?10)0.06(2.961?10)(0.7309)17245?1.729?10??

    k0.02514 W/m.?C2and hNu(1724)10.73 W/m.?CD4.04 m

    The rate of heat transfer to the liquid nitrogen is

    222(4.04 m)51.28 mADs

    TTTT,tan,tan?sk?sk;Q rr1RR21convinsulation4krrhA12s

    [20(196)]?C27.4 W(2.022) m1224(0.00005 W/m.?C)(2.02 m)(2 m)(10.73 W/m.?C)(51.28 m)

    The rate of evaporation of liquid nitrogen then becomes

    ;Q0.0274 kJ/s-4;;;Qmhm ((?1.38?10 kg/sifh198 kJ/kgif

     7-87

    Chapter 7 External Forced Convection 7-102 A spherical tank filled with liquid oxygen is exposed to ambient winds. The rate of evaporation of

    the liquid oxygen due to heat transfer from the air is to be determined for three cases.

    Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant properties. 7 The pressure of air is 1 atm.

    Properties The properties of air at 1 atm pressure and the free stream temperature of 20?C are (Table A-15)

    0.02514 W/m.?Ck

    -52~Insulation 1.516?10 m/s

    5 1.825?10 kg/m.s ?

    56.127?10 kg/m.ss, D@183?CoWind 20?C Pr0.7309D i40 km/h Analysis (a) When there is no insulation, D = D = 4 m, i

    and the Reynolds number is

    VD(40?1000/3600) m/s(4 m),(6? Re2.932?1052~1.516?10 m/s

    Oxygen tank The Nusselt number is determined from

     -183?C 1/4??hD0.52/30.4??20.4Re0.06RePrNu,(?ks?? 1/45??1.825?1060.562/30.4?,( 20.4(2.932?10)0.06(2.932?10)(0.7309)22205?1.05?10??

    k0.02514 W/m.?C2and hNu(2220)13.95 W/m.?CD4 m

    The rate of heat transfer to the liquid oxygen is

    222;,(QhA(TT)h(D)(TT)(13.95 W/m.?C)[(4 m)](20(183) ?C142,372 W ss?s?

    The rate of evaporation of liquid oxygen then becomes

    ;Q142.4 kJ/s;;;Qmhm ((?0.668 kg/sifh213 kJ/kgif

    (b) Note that after insulation the outer surface temperature and diameter will change. Therefore we need to

    evaluate dynamic viscosity at a new surface temperature which we will assume to be -100?C. At -100?C, 51.189?10 kg/m.s. Noting that D = D = 4.1 m, the Nusselt number becomes 0

    VD(40?1000/3600) m/s(4.1 m),(6? Re3.005?1052~1.516?10 m/s

    1/4??hD0.52/30.4??20.4Re0.06RePr,(Nu?ks?? 1/45??1.825?1060.562/30.4?,( 20.4(3.005?10)0.06(3.005?10)(0.7309)19105?1.189?10??

    k0.02514 W/m.?C2and hNu(1910)11.71 W/m.?CD4.1 m

    The rate of heat transfer to the liquid nitrogen is

     7-88

    Chapter 7 External Forced Convection

    222(4.1 m)52.81 mADs

    TTTT,tan,tan?sk?sk;Q rr1RR21convinsulation4krrhA12s

    [20(183)]?C6918 W(2.052) m1224(0.035 W/m.?C)(2.05 m)(2 m)(11.71 W/m.?C)(52.81 m)

    The rate of evaporation of liquid nitrogen then becomes

    ;Q6.918 kJ/s;;;Qmhm ((?0.0325 kg/sifh213 kJ/kgif

    (c) Again we use the dynamic viscosity value at the estimated surface temperature of 0?C to be 51.729?10 kg/m.s. Noting that D = D = 4.04 m in this case, the Nusselt number becomes 0

    VD(40?1000/3600) m/s(4.04 m),(6? Re2.961?1052~1.516?10 m/s

    1/4??hD0.52/30.4??20.4Re0.06RePr,(Nu?ks?? 1/45??1.825?1060.562/30.4?,( 20.4(2.961?10)0.06(2.961?10)(0.713)17245?1.729?10??

    k0.02514 W/m.?C2and hNu(1724)10.73 W/m.?CD4.04 m

    The rate of heat transfer to the liquid nitrogen is

    222(4.04 m)51.28 mADs

    TTTT,tan,tan?sk?sk;Q rr1RR21convinsulation4krrhA12s

    [20(183)]?C25.8 W(2.022) m1224(0.00005 W/m.?C)(2.02 m)(2 m)(10.73 W/m.?C)(51.28 m)

    The rate of evaporation of liquid oxygen then becomes

    ;Q0.0258 kJ/s-4;;;Qmhm ((?1.21?10 kg/sifh213 kJ/kgif

     7-89

    Chapter 7 External Forced Convection

    7-103 A circuit board houses 80 closely spaced logic chips on one side. All the heat generated is conducted

    across the circuit board and is dissipated from the back side of the board to the ambient air, which is forced

    to flow over the surface by a fan. The temperatures on the two sides of the circuit board are to be

    determined.

    5Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re = 5?10. 3 crRadiation effects are negligible. 4 Air is an ideal gas with constant properties. 7 The pressure of air is 1 atm.

    Properties Assuming a film temperature of 40?C, the properties of air are (Table A-15)

    0.02662 W/m.?Ck

    -521.702?10 m/s~

    Pr0.7255

    Analysis The Reynolds number is

    T 1VL(400/60) m/s(0.18 m),(4? Re7.051?10L52~1.702?10 m/sT 2which is less than the critical Reynolds number. Therefore, T =30?C ?the flow is laminar. Using the proper relation for Nusselt 400 m/min number, heat transfer coefficient is determined to be

    hL0.51/340.51/3Nu0.664RePr0.664(7.051?10)(0.7255)158.4Lk k0.02662 W/m.?C2hNu(158.4)23.43 W/m.?CL0.18 m

    The temperatures on the two sides of the circuit board are

    ;Q;()? QhATTTT??s22hAs

    (80?0.06) W 30C?39.48?C2(23.43 W/m.?C)(0.12 m)(0.18 m) ;kAQLs;()?QTTTT1212LkAs

    (800.06 W)(0.003 m)??39.52?C 39.48C(16 W/m.?C)(0.12 m)(0.18 m)

     7-90

    Chapter 7 External Forced Convection

    7-104E The equivalent wind chill temperature of an environment at 10?F at various winds speeds are V = 10 mph: TTVV,????9149140475002030304.(.)(...)equivambient

    ,????,??9149141004750020310..(..(F) mph)+0.30410 mph9F

    T,????,??9149141004750020320..(..(F) mph)+0.30420 mph24.9FV = 20 mph: equiv

    T,????,??9149141004750020330..(..(F) mph)+0.30430 mph33.2FV = 30 mph: equiv

    T,????,??9149141004750020340..(..(F) mph)+0.30440 mph37.7FV = 40 mph: equiv

    In the last 3 cases, the person needs to be concerned about the possibility of freezing.

     7-91

    Chapter 7 External Forced Convection

    7-105E "!PROBLEM 7-105E"

    "ANALYSIS"

    T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*Vel+0.304*sqrt(Vel))

    Vel [mph] T [F] T [F] ambientequiv4 20 19.87 14.67 20 -4.383 25.33 20 -15.05 36 20 -20.57 46.67 20 -23.15 57.33 20 -23.77 68 20 -22.94 78.67 20 -21.01 89.33 20 -18.19 100 20 -14.63 4 40 39.91 14.67 40 22.45 25.33 40 14.77 36 40 10.79 46.67 40 8.935 57.33 40 8.493 68 40 9.086 78.67 40 10.48 89.33 40 12.51 100 40 15.07 4 60 59.94 14.67 60 49.28 25.33 60 44.59 36 60 42.16 46.67 60 41.02 57.33 60 40.75 68 60 41.11 78.67 60 41.96 89.33 60 43.21 100 60 44.77

     7-92

    Chapter 7 External Forced Convection

    60

    5060 F

    40

    30

    2040 F

    10 [F]

    0equivT-10

    20 F

    -20

     -30

    022446688110

    Vel [mph]

    7-106 …. 7-110 Design and Essay Problems

    ;;

     7-93

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