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are , the single equivalent heat transfer coefficient is h hconv rad and radconveqv hhh += when the medium and the surrounding surfaces are at the same temperature. Then the equivalent thermal resistance will be . )/(1 AhR eqveqv = 3-9C The thermal resistance network associated with a five-layer composite wall involves five single-layer resistances connected in series. 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. 3-2 3-10C Once the rate of heat transfer Q is known, the temperature drop across any layer can be determined by multiplying heat transfer rate by the thermal resistance across that layer, & layerlayer RQT &=\u394 3-11C The temperature of each surface in this case can be determined from )(/)( )(/)( 22222222 11111111 \u221e\u2212\u221e\u221e\u2212\u221e \u2212\u221e\u221e\u2212\u221e\u221e +=\u23af\u2192\u23af\u2212= \u2212=\u23af\u2192\u23af\u2212= ssss ssss RQTTRTTQ RQTTRTTQ && && where is the thermal resistance between the environment iR \u2212\u221e \u221e and surface i. 3-12C Yes, it is. 3-13C The window glass which consists of two 4 mm thick glass sheets pressed tightly against each other will probably have thermal contact resistance which serves as an additional thermal resistance to heat transfer through window, and thus the heat transfer rate will be smaller relative to the one which consists of a single 8 mm thick glass sheet. 3-14C Convection heat transfer through the wall is expressed as . In steady heat transfer, heat transfer rate to the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer coefficient three times that of the inner surface will experience three times smaller temperature drop compared to the inner surface. Therefore, at the outer surface, the temperature will be closer to the surrounding air temperature. )( \u221e\u2212= TThAQ ss& 3-15C The new design introduces the thermal resistance of the copper layer in addition to the thermal resistance of the aluminum which has the same value for both designs. Therefore, the new design will be a poorer conductor of heat. 3-16C The blanket will introduce additional resistance to heat transfer and slow down the heat gain of the drink wrapped in a blanket. Therefore, the drink left on a table will warm up faster. 3-17 The two surfaces of a wall are maintained at specified temperatures. The rate of heat loss through the wall is to be determined. Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain constant 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 is constant. 2°C14°C L= 0.3 m Q& Wall Properties The thermal conductivity is given to be k = 0.8 W/m\u22c5°C. Analysis The surface area of the wall and the rate of heat loss through the wall are 2m 18m) 6(m) 3( =×=A W576=°\u2212°\u22c5=\u2212= m 3.0 C)214( )m C)(18 W/m8.0( 221 L TT kAQ& 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. 3-3 3-18 A double-pane window is considered. The rate of heat loss through the window and the temperature difference across the largest thermal resistance are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer coefficients are constant. Properties The thermal conductivities of glass and air are given to be 0.78 W/m\u22c5K and 0.025 W/m\u22c5K, respectively. Analysis (a) The rate of heat transfer through the window is determined to be [ ] [ ] W210=++++ °×= °\u22c5+°\u22c5+°\u22c5+°\u22c5+°\u22c5 °×= ++++ \u394= 05.0000513.02.0000513.0025.0 C(-20)-20)m 5.11( C W/m20 1 C W/m78.0 m 004.0 C W/m025.0 m 005.0 C W/m78.0 m 004.0 C W/m40 1 C(-20)-20)m 5.11( 11 2 22 2 og g a a g g i hk L k L k L h TAQ& (b) Noting that the largest resistance is through the air gap, the temperature difference across the air gap is determined from C28°=×°\u22c5===\u394 )m 5.11(C) W/m025.0( m 005.0 W)210( 2Ak L QRQT a a aa && 3-19 The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through the window and the inner surface temperature are to be determined. Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant 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 is constant. 4 Heat transfer by radiation is negligible. Properties The thermal conductivity of the glass is given to be k = 0.78 W/m\u22c5°C. Analysis The area of the window and the individual resistances are 2m 4.2m) 2(m) 2.1( =×=A C/W06155.001667.000321.004167.0 C/W 01667.0 )m 4.2(C). W/m25( 11 C/W 00321.0 )m 4.2(C) W/m.78.0( m 006.0 C/W 04167.0 )m 4.2(C). W/m10( 11 2,1, 22 2 2,o 2 1 glass 22 1 1,i °=++= ++= °=°=== °=°== °=°=== convglassconvtotal conv conv RRRR Ah RR Ak LR Ah RR T1 Q& Glass Ri Rglass Ro T\u221e1 T\u221e2 L The steady rate of heat transfer through window glass is then W471=° °\u2212\u2212=\u2212= \u221e\u221e C/W 06155.0 C)]5(24[21 totalR TT Q& The inner surface temperature of the window glass can be determined from C4.4°=°\u2212°=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e C/W) 7 W)(0.0416471(C241,11 1, 11 conv conv RQTT R TT Q && 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. 3-4 3-20 A double-pane window consists of two layers of glass separated by a stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined. Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant 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 conductivities of the glass and air are constant. 4 Heat transfer by radiation is negligible. Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/m\u22c5°C and kair = 0.026 W/m\u22c5°C. Air R1 R2 R3 Ro Ri T\u221e1 T\u221e2 Analysis The area of the window and the individual resistances are 2m 4.2m) 2(m) 2.1( =×=A C/W 2539.0 0167.01923.0)0016.0(20417.02 C/W 0167.0 )m 4.2(C). W/m25( 11 C/W 1923.0 )m 4.2(C) W/m.026.0( m 012.0 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 2,211, o 2o2 2 2,o 2 2 2 2 2 1 1 glass31 22 1 1,i °= +++=+++= ==== °=°=== °=°==== °=°=== convconvtotal conv air conv RRRRR Ah RR Ak LRR Ak LRRR Ah RR The steady rate of heat transfer through window glass then becomes W114=° °\u2212\u2212=\u2212= \u221e\u221e C/W2539.0 C)]5(24[21 totalR TT Q& The inner surface temperature of the window glass can be determined from C19.2°°\u2212=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e =C/W) W)(0.0417114(C24o1,11 1, 11 conv conv RQTT R TT Q && 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. 3-5 3-21 A double-pane window consists of two layers of glass separated by an evacuated space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined. Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant