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be determined. Assumptions 1 Steady operating conditions exist. 2 The heat transfer coefficient is constant and uniform over the entire exposed surface of the person. 3 The surrounding surfaces are at the same temperature as the indoor air temperature. 4 Heat generation within the 0.5-cm thick outer layer of the tissue is negligible. Qrad Tskin Qconv Properties The thermal conductivity of the tissue near the skin is given to be k = 0.3 W/m\u22c5°C. Analysis The skin temperature can be determined directly from C35.5°=°\u22c5\u2212°=\u2212= \u2212= )m C)(1.7 W/m3.0( m) 005.0 W)(150( C37 21 1 kA LQTT L TT kAQ skin skin & & 3-28 Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface temperature of the bottom of the pan is given. The boiling heat transfer coefficient and the outer surface temperature of the bottom of the pan are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the thickness of the bottom of the pan is small relative to its diameter. 3 The thermal conductivity of the pan is constant. Properties The thermal conductivity of the aluminum pan is given to be k = 237 W/m\u22c5°C. Analysis (a) The boiling heat transfer coefficient is 2 22 m 0491.0 4 m) 25.0( 4 === \u3c0\u3c0DAs 95°C 108°C 600 W 0.5 cm C. W/m1254 2 °=°\u2212=\u2212= \u2212= \u221e \u221e C)95108)(m 0491.0( W800 )( )( 2TTA Qh TThAQ ss ss & & (b) The outer surface temperature of the bottom of the pan is C108.3°=°\u22c5°=+= \u2212= )m C)(0.0491 W/m237( m) 005.0 W)(800(+C108 21,, ,, kA LQTT L TT kAQ innersouters innersouters & & PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-10 3-29E A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal resistance of the wall and its R-value of insulation are to be determined. Assumptions 1 Heat transfer through the wall is one-dimensional. 2 Thermal conductivities are constant. Properties The thermal conductivities are given to be ksheetrock = 0.10 Btu/h\u22c5ft\u22c5°F and kinsulation = 0.020 Btu/h\u22c5ft\u22c5°F. R1 R2 R3 L1 L2 L3 Analysis (a) The surface area of the wall is not given and thus we consider a unit surface area (A = 1 ft2). Then the R-value of insulation of the wall becomes equivalent to its thermal resistance, which is determined from. F.h/Btu.ft 30.34 2 °=+×=+= °=°=== °=°==== 17.29583.022 F.h/Btu.ft 17.29 F)Btu/h.ft. 020.0( ft 12/7 F.h/Btu.ft 583.0 F)Btu/h.ft. 10.0( ft 12/7.0 21 2 2 2 2 2 1 1 31 RRR k L RR k L RRR total fiberglass sheetrock (b) Therefore, this is approximately a R-30 wall in English units. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-11 3-30 The roof of a house with a gas furnace consists of a concrete that is losing heat to the outdoors by radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period are to be determined. Assumptions 1 Steady operating conditions exist. 2 The emissivity and thermal conductivity of the roof are constant. Properties The thermal conductivity of the concrete is given to be k = 2 W/m\u22c5°C. The emissivity of both surfaces of the roof is given to be 0.9. Analysis When the surrounding surface temperature is different than the ambient temperature, the thermal resistances network approach becomes cumbersome in problems that involve radiation. Therefore, we will use a different but intuitive approach. 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), that must be equal to the heat transfer through the roof by conduction. That is, Q& Tsky = 100 K Tair =10°C Tin=20°C L=15 cm rad+conv gs,surroundin toroofcond roof,rad+conv roof, toroom QQQQ &&&& === Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out , respectively, the quantities above can be expressed as [ ]4,44282 , 224 , 4 , rad+conv roof, toroom K) 273(K) 27320()K W/m1067.5)(m 300)(9.0( C))(20m C)(300 W/m5()()( +\u2212+\u22c5×+ °\u2212°\u22c5=\u2212+\u2212= \u2212 ins insinsroominsroomi T TTTATTAhQ \u3c3\u3b5& m 15.0 )m 300)(C W/m2( ,,2,,cond roof, outsinsoutsins TT L TT kAQ \u2212°\u22c5=\u2212=& [ ]44,4282 , 2244 ,, rad+conv surr, toroof K) 100(K) 273()K W/m1067.5)(m 300)(9.0( C)10)(m C)(300 W/m12()()( \u2212+\u22c5×+ °\u2212°\u22c5=\u2212+\u2212= \u2212 outs outssurroutssurroutso T TTTATTAhQ \u3c3\u3b5& Solving the equations above simultaneously gives C1.2and , , out,, °\u2212=°== sins TTQ C7.3 W37,440& The total amount of natural gas consumption during a 14-hour period is therms36.22 kJ 105,500 therm1 80.0 )s 360014)(kJ/s 440.37( 80.080.0 =\u239f\u239f\u23a0 \u239e\u239c\u239c\u239d \u239b×=\u394== tQQQ totalgas & Finally, the money lost through the roof during that period is $26.8== )therm/20.1$ therms)(36.22(lostMoney PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-12 3-31 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined. Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The furnace operates continuously. 4 The given heat transfer coefficient accounts for the radiation effects. Properties The thermal conductivity of the glass wool insulation is given to be k = 0.038 W/m\u22c5°C. Insulation Ro T\u221e Rinsulation Ts L Analysis The rate of heat transfer without insulation is 2m 3m) m)(1.5 2( ==A W1500C)3080)(m 3(C) W/m10()( 22 =°\u2212°\u22c5=\u2212= \u221eTThAQ s& In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be C/W 333.0 W150 C)3080( W150 W150010.0 °=°\u2212=\u394=\u23af\u2192\u23af\u394= =×= Q TR R TQ Q total total & & & and in order to have this thermal resistance, the thickness of insulation must be cm 3.4== °=°+°\u22c5= +=+= m 034.0 C/W 333.0 )m C)(3 W/m.038.0()m C)(3 W/m10( 1 1 222 conv L L kA L hA RRR insulationtotal Noting that heat is saved at a rate of 0.9×1500 = 1350 W and the furnace operates continuously and thus 365×24 = 8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year is therms4.517 kJ 105,500 therm1 h 1 s 3600 0.78 h) kJ/s)(8760 350.1(SavedEnergy =\u239f\u239f\u23a0 \u239e\u239c\u239c\u239d \u239b\u239f\u23a0 \u239e\u239c\u239d \u239b=\u394= Efficiency tQsaved& The money saved is year)(per 1.569$)1.10/therm therms)($4.517(energy) oft Saved)(CosEnergy (savedMoney === The insulation will pay for its cost of $250 in yr 0.44=== $569.1/yr $250 savedMoney spent Money periodPayback which is equal to 5.3 months. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-13 3-32 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to