Pré-visualização50 páginas

aT2 + bT = C1x2 + C2 (e) None of them Answer (c) aT2 + bT = C1x + C2 2-161 Harvested grains, like wheat, undergo a volumetric exothermic reaction while they are being stored. This heat generation causes these grains to spoil or even start fires if not controlled properly. Wheat (k = 0.5 W/m\u22c5K) is stored on the ground (effectively an adiabatic surface) in 5-m thick layers. Air at 20°C contacts the upper surface of this layer of wheat with h = 3 W/m2\u22c5K. The temperature distribution inside this layer is given by 2 0 1 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212=\u2212 \u2212 L x TT TT s s where Ts is the upper surface temperature, T0 is the lower surface temperature, x is measured upwards from the ground, and L is the thickness of the layer. When the temperature of the upper surface is 24oC, what is the temperature of the wheat next to the ground? (a) 39oC (b) 51oC (c) 72oC (d) 84oC (e) 91°C Answer (d) 84oC k=0.5 [W/m-K] h=3 [W/m2-K] L=5[m] Ts=24 [C] Ta=20 [C] To=(h*L/(2*k))*(Ts-Ta)+Ts 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. 2-85 2-162 The conduction equation boundary condition for an adiabatic surface with direction n being normal to the surface is (a) T = 0 (b) dT/dn = 0 (c) d2T/dn2 = 0 (d) d3T/dn3 = 0 (e) -kdT/dn = 1 Answer (b) dT/dn = 0 2-163 Which one of the followings is the correct expression for one-dimensional, steady-state, constant thermal conductivity heat conduction equation for a cylinder with heat generation? (a) t Tce r Trk rr \u2202 \u2202=+\u239f\u23a0 \u239e\u239c\u239d \u239b \u2202 \u2202 \u2202 \u2202 \u3c1gen1 & (b) t T k e r Tr rr \u2202 \u2202=+\u239f\u23a0 \u239e\u239c\u239d \u239b \u2202 \u2202 \u2202 \u2202 \u3b1 11 gen& (c) t T r Tr rr \u2202 \u2202 \u3b1=\u239f\u23a0 \u239e\u239c\u239d \u239b \u2202 \u2202 \u2202 \u2202 11 (d) 01 gen =+\u239f\u23a0 \u239e\u239c\u239d \u239b k e dr dTr dr d r & (e) 0=\u239f\u23a0 \u239e\u239c\u239d \u239b dr dTr dr d Answer (d) 01 gen =+\u239f\u23a0 \u239e\u239c\u239d \u239b k e dr dTr dr d r & 2-164 .... 2-167 Design and Essay Problems KJ 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. 2-33 2-70 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the surface temperatures are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Properties The thermal conductivity is given to be k = 14 W/m\u22c5°C. Analysis (a) Noting that the 85% of the 300 W generated by the strip heater is transferred to the pipe, the heat flux through the outer surface is determined to be 2 22 W/m1.169 m) m)(6 (0.042 W30085.0 2 =×=== \u3c0\u3c0 Lr Q A Q q sss && & Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r direction, the mathematical formulation of this problem can be expressed as 0=\u239f\u23a0 \u239e\u239c\u239d \u239b dr dTr dr d and )]([)( 11 rTThdr rdT k \u2212=\u2212 \u221e sqdr rdT k &=)( 2 (b) Integrating the differential equation once with respect to r gives Heater L=6 m Air, -10°C r r1 r2 1Cdr dTr = Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, r C dr dT 1= 21 ln)( CrCrT += where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r2: k rq Cq r C k ss 2 1 2 1 && =\u2192= r = r1: k rq hr krTC hr krTCCrCTh r C k s 2 1 11 1 12211 1 1 ln=ln )]ln([ & \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212\u2212\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212\u2212=\u2192+\u2212=\u2212 \u221e\u221e\u221e Substituting C1 and C2 into the general solution, the variation of temperature is determined to be \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b ++\u2212=°\u22c5\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b °\u22c5 °\u22c5++°\u2212= \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b ++=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b +\u2212+=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212\u2212+= \u221e\u221e\u221e 61.12ln483.010 C W/m14 m) 04.0)( W/m1.169( m) C)(0.037 W/m30( C W/m14lnC10 lnlnlnlnln)( 1 2 2 1 2 11 1 1 11 1 11 r r r r k rq hr k r rTC hr krrTC hr krTrCrT s & (c) The inner and outer surface temperatures are determined by direct substitution to be Inner surface (r = r1): ( ) C3.91°\u2212=++\u2212=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b ++\u2212= 61.120483.01061.12ln483.010)( 1 1 1 r r rT Outer surface (r = r2): C3.87°\u2212=\u239f\u23a0 \u239e\u239c\u239d \u239b ++\u2212=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b ++\u2212= 61.12 037.0 04.0ln483.01061.12ln483.010)( 1 2 1 r r rT Note that the pipe is essentially isothermal at a temperature of about -3.9°C. 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. 2-34 2-71 EES Prob. 2-70 is reconsidered. The temperature as a function of the radius is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=6 [m] r_1=0.037 [m] r_2=0.04 [m] k=14 [W/m-C] Q_dot=300 [W] T_infinity=-10 [C] h=30 [W/m^2-C] f_loss=0.15 "ANALYSIS" q_dot_s=((1-f_loss)*Q_dot)/A A=2*pi*r_2*L T=T_infinity+(ln(r/r_1)+k/(h*r_1))*(q_dot_s*r_2)/k "Variation of temperature" "r is the parameter to be varied" r [m] T [C] 0.037 3.906 0.03733 3.902 0.03767 3.898 0.038 3.893 0.03833 3.889 0.03867 3.885 0.039 3.881 0.03933 3.877 0.03967 3.873 0.04 3.869 0.037 0.0375 0.038 0.0385 0.039 0.0395 0.04 -3.906 -3.897 -3.888 -3.879 -3.87 r [m] T [C ] 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. 2-35 2-72 A spherical container is subjected to uniform heat flux on the outer surface and specified temperature on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the outer surface temperature, and the maximum rate of hot water supply are to be determined for steady one- dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the mid point. 2 Thermal conductivity is constant. 3 There is no heat generation in the container. Properties The thermal conductivity is given to be k = 1.5 W/m\u22c5°C. The specific heat of water at the average temperature of (100+20)/2 = 60°C is 4.185 kJ/kg\u22c5°C (Table A-9). Analysis (a) Noting that the 90% of the 500 W generated by the strip heater is transferred to the container, the heat flux through the outer surface is determined to be 2 22 22 W/m0.213 m) (0.414 W50090.0 4 =×=== \u3c0\u3c0r Q A Q q sss && & Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r direction, the mathematical formulation of this problem can be expressed as 02 =\u239f\u23a0 \u239e\u239c\u239d \u239b dr dTr dr d and C100)( 11 °== TrT sqdr rdT k &=)( 2 (b) Integrating the differential equation once with respect to r gives 1 2 C dr dTr = Dividing both sides of the equation above by r2 and then integrating, r1 r2 T1k Heater r Insulation 2 1 r C dr dT = 2 1)( C r C rT +\u2212= where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r2: k rq Cq r C k ss 2 2 12 2 1 && =\u2192= r = r1: 1 2 2 1 1 1 122 1 1 11 )( kr rq T r C TCC r C TrT s &+=+=\u2192+\u2212== Substituting C1 and C2 into the general solution, the variation of temperature is determined to be \u239f\u23a0 \u239e\u239c\u239d \u239b \u2212+=°\u22c5\u239f\u23a0 \u239e\u239c\u239d \u239b \u2212+°= \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212+=\u239f\u239f\u23a0