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at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of the glass is constant. 4 Heat transfer by radiation is negligible. Properties The thermal conductivity of the glass is given to be kglass = 0.78 W/m\u22c5°C. Analysis Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum is zero (no medium to conduct heat) and thus its thermal resistance is zero. Therefore, if radiation is disregarded, the heat transfer through the window will be zero. Then the answer of this problem is zero since the problem states to disregard radiation. Discussion In reality, heat will be transferred between the glasses by radiation. We do not know the inner surface temperatures of windows. In order to determine radiation heat resistance we assume them to be 5°C and 15°C, respectively, and take the emissivity to be 1. Then individual resistances are Vacuum R1 Rrad R3 Ro Ri T\u221e1 T\u221e2 2m 4.2m) 2(m) 2.1( =×=A C/W 1426.0 0167.00810.0)0016.0(20417.02 C/W 0167.0 )m 4.2(C). W/m25( 11 C/W 0810.0 ]278288][278288)[m 4.2().K W/m1067.5(1 1 ))(( 1 C/W 0016.0 )m 4.2(C) W/m.78.0( m 003.0 C/W 0417.0 )m 4.2(C). W/m10( 11 2,11, 22 2 2,o 3222428 22 2 1 1 glass31 22 1 1,i °= +++=+++= °=°=== °= ++×= ++= °=°==== °=°=== \u2212 convradconvtotal conv surrssurrs rad conv RRRRR Ah RR K TTTTA R Ak L RRR Ah RR \u3b5\u3c3 The steady rate of heat transfer through window glass then becomes W203=° °\u2212\u2212=\u2212= \u221e\u221e C/W1426.0 C)]5(24[21 totalR TT Q& The inner surface temperature of the window glass can be determined from C15.5°=°\u2212°=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e C/W) W)(0.0417203(C241,11 1, 11 conv conv RQTT R TT Q && Similarly, the inner surface temperatures of the glasses are calculated to be 15.2 and -1.2°C (we had assumed them to be 15 and 5°C when determining the radiation resistance). We can improve the result obtained by reevaluating the radiation resistance and repeating the calculations. 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. 3-6 3-22 EES Prob. 3-20 is reconsidered. The rate of heat transfer through the window as a function of the width of air space is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" A=1.2*2 [m^2] L_glass=3 [mm] k_glass=0.78 [W/m-C] L_air=12 [mm] T_infinity_1=24 [C] T_infinity_2=-5 [C] h_1=10 [W/m^2-C] h_2=25 [W/m^2-C] "PROPERTIES" k_air=conductivity(Air,T=25) "ANALYSIS" R_conv_1=1/(h_1*A) R_glass=(L_glass*Convert(mm, m))/(k_glass*A) R_air=(L_air*Convert(mm, m))/(k_air*A) R_conv_2=1/(h_2*A) R_total=R_conv_1+2*R_glass+R_air+R_conv_2 Q_dot=(T_infinity_1-T_infinity_2)/R_total Lair [mm] Q [W] 2 307.8 4 228.6 6 181.8 8 150.9 10 129 12 112.6 14 99.93 16 89.82 18 81.57 20 74.7 2 4 6 8 10 12 14 16 18 20 50 100 150 200 250 300 350 Lair [mm] Q [ W ] 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. 3-7 3-23E The inner and outer surfaces of the walls of an electrically heated house remain at specified temperatures during a winter day. The amount of heat lost from the house that day and its cost are to be determined. Assumptions 1 Heat transfer through the walls is steady since the surface temperatures of the walls remain constant at the specified values during the time period considered. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of the walls is constant. Properties The thermal conductivity of the brick wall is given to be k = 0.40 Btu/h\u22c5ft\u22c5°F. Analysis We consider heat loss through the walls only. The total heat transfer area is 2ft 1530)935950(2 =×+×=A The rate of heat loss during the daytime is Btu/h 6120 ft 1 F)4555( )ft F)(1530ftBtu/h 40.0( 221day =°\u2212°\u22c5\u22c5=\u2212= L TT kAQ& T2T1 L Q& Wall The rate of heat loss during nighttime is Btu/h 240,12 ft 1 C)3555()ft F)(1530ftBtu/h 40.0( 2 21 night =°\u2212°\u22c5\u22c5= \u2212= L TT kAQ& The amount of heat loss from the house that night will be Btu 232,560= +=+=\u394=\u23af\u2192\u23af\u394= )Btu/h 240,12(h) 14()Btu/h 6120(h) 10(1410 nightday QQtQQt QQ &&&& Then the cost of this heat loss for that day becomes $6.13== kWh)/09.0)($kWh 3412/560,232(Cost 3-24 A cylindrical resistor on a circuit board dissipates 0.15 W of power steadily in a specified environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the resistor. Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is Wh3.6==\u394= h) W)(2415.0(tQQ & Resistor 0.15 W Q& (b) The heat flux on the surface of the resistor is 2 22 m 000127.0m) m)(0.012 003.0( 4 m) 003.0( 2 4 2 =+=+= \u3c0\u3c0\u3c0\u3c0 DLDAs 2 W/m1179=== 2m 000127.0 W15.0 sA Qq & & (c) The surface temperature of the resistor can be determined from C171°=°\u22c5+°=+=\u23af\u2192\u23af\u2212= \u221e\u221e )m 7C)(0.00012 W/m(9 W15.0C40)( 22 s sss hA QTTTThAQ && 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. 3-8 3-25 A power transistor dissipates 0.2 W of power steadily in a specified environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the transistor. Analysis (a) The amount of heat this transistor dissipates during a 24-hour period is Air, 30°C Power Transistor 0.2 W kWh 0.0048===\u394= Wh8.4h) W)(242.0(tQQ & (b) The heat flux on the surface of the transistor is 2 2 2 m 0001021.0m) m)(0.004 005.0( 4 m) 005.0( 2 4 2 =+= += \u3c0\u3c0 \u3c0\u3c0 DLDAs 2 W/m1959=== 2m 0001021.0 W2.0 sA Qq & & (c) The surface temperature of the transistor can be determined from C139°=°\u22c5+°=+=\u23af\u2192\u23af\u2212= \u221e\u221e )m 21C)(0.00010 W/m(18 W2.0C30)( 22 s sss hA QTTTThAQ && 3-26 A circuit board houses 100 chips, each dissipating 0.06 W. The surface heat flux, the surface temperature of the chips, and the thermal resistance between the surface of the board and the cooling medium are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer from the back surface of the board is negligible. 2 Heat is transferred uniformly from the entire front surface. Analysis (a) The heat flux on the surface of the circuit board is Chips Ts Q& T\u221e 2m 0216.0m) m)(0.18 12.0( ==sA 2 W/m278=×== 2m 0216.0 W)06.0100( sA Qq & & (b) The surface temperature of the chips is C67.8°=°\u22c5 ×°=+= \u2212= \u221e \u221e )m 0216.0)(C W/m10( W)06.0100(+C40 )( 22 s s ss hA QTT TThAQ & & (c) The thermal resistance is C/W4.63°=°\u22c5== )m 0216.0(C) W/m10( 11 22 s conv hA R 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. 3-9 3-27 A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding air and surfaces. For a given deep body temperature, the outer skin temperature is to