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\uf8f4\uf8ef \uf8fa\uf8f0 \uf8fb\uf8f3 \uf8fe ( ) ( ) 11 2 4conv,o o iR h D L 30.9 W / m K 1m 100m 1.03 10 K / W\u3c0 \u3c0 \u2212\u2212 \u2212\u2248 = \u22c5 × × × = × Hence, from Eq. (1) ( ) ( )o i 3 4 4 3 ins ln D / D 2.96 10 3.33 10 1.03 10 K / W 2.52 10 K / W 2 k L\u3c0 \u2212 \u2212 \u2212 \u2212 = × \u2212 × \u2212 × = × ( ) ( )3 2o i insD D exp 2 k L 2.52 10 K / W 1m exp 1.58 10 K / W 0.125 W / m K 100m 1.22m\u3c0 \u2212 \u2212= × × = × × × \u22c5 × = Hence, the minimum insulation thickness is ( )min o it D D / 2 0.11m= \u2212 = < COMMENTS: With Do = 1.22m, use of Di = 1m to evaluate the outer convection coefficient and thermal resistance is a reasonable approximation. However, improved accuracy may be obtained by using the calculated value of Do to determine conditions at the outer surface and iterating on the solution. PROBLEM 8.42 KNOWN: Flow rate, inlet temperature and desired outlet temperature of liquid mercury flowing through a tube of prescribed diameter and surface temperature. FIND: Required tube length and error associated with use of a correlation for moderate to large Pr fluids. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy, potential energy and flow work effects, (4) Fully developed flow. PROPERTIES: Table A-5, Mercury T Km = 350c h: cp = 137.7 J/kg×K, m = 0.1309 ´ 10-2 N×s/m2, k = 9.18 W/m×K, Pr = 0.0196. ANALYSIS: The Reynolds and Peclet numbers are ( )D 2 2 4m 4 0.5 kg/s Re 9727 D 0.05m 0.1309 10 N s/mp m p - ´= = = ´ × & ( )D DPe Re Pr 9727 0.0196 191.= = = Hence, assuming fully developed turbulent flow throughout the tube, it follows from Eq. 8.66 that ( ) ( )0.8 0.8 2Dk 9.18 W/m Kh 5.0 0.025 Pe 5.0 0.025 191 1224 W/m K.D 0.05 m ×= + = + ´ = × From Eq. 8.42a, it follows that ( ) ( ) p o 2i m c 0.5 kg/s 137.7 J/kg KT 450 400 L n n 0.39 m. Dh T 450 3000.05 m 1224 W/m Kp p ×D - = - = - = D -× & l l < If the Dittus-Boelter correlation, Eq. 8.60, is used in place of Eq. 8.66, ( ) ( ) 2 4/5 0.44/5 0.4 2 D k 9.18 W/m K h 0.023 Re Pr 0.023 9727 0.0196 1358 W/m K D 0.05 m ×= = = × and the required tube length is ( ) ( ) p o 2i m c 0.5 kg/s 137.7 J/kg KT 450 400 L n n 0.35 m. Dh T 450 3000.05 m 1358 W/m Kp p ×D -= - = - = D -× & l l < COMMENTS: Such good agreement between results does not occur in general. For example, if ReD = 2 ´ 10 4 , h = 1463 from Eq. 8.66 and 2417 from Eq. 8.60. Large errors are usually associated with using conventional (moderate to large Pr) correlations with liquid metals. PROBLEM 8.43 KNOWN: Surface temperature and diameter of a tube. Velocity and temperature of air in cross flow. Velocity and temperature of air in fully developed internal flow. FIND: Convection heat flux associated with the external and internal flows. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform cylinder surface temperature, (3) Fully developed internal flow. PROPERTIES: Table A-4, Air (298K): n = 15.71 ´ 10-6 m2/s, k = 0.0261 W/m×K, Pr = 0.71. ANALYSIS: For the external and internal flows, 4m D -6 2 VD u D 30 m/s 0.05 m Re 9.55 10 . 15.71 10 m / sn n ´= = = = ´ ´ From the Zhukauskas relation for the external flow, with C = 0.26 and m = 0.6, ( ) ( ) ( ) ( )0.61/4 0.37 1 / 4m n 4D sDNu C Re Pr Pr/Pr 0.26 9.55 10 0.71 1 223.= = ´ = Hence, the convection coefficient and heat flux are 2 D k 0.0261 W/m K h Nu 223 116.4 W/m K D 0.05 m ×= = ´ = × ( ) ( )2 3 2sq h T T 116.4W/m K 100 25 C 8.73 10 W/m .¥¢¢ = - = × - = ´o < Using the Dittus-Boelter correlation, Eq. 8.60, for the internal flow, which is turbulent, ( ) ( )4/5 0.44/5 0.4 4D DNu 0.023 Re Pr 0.023 9.55 10 0.71 193= = ´ = 2 D k 0.0261 W/m K h Nu 193 101 W/m K D 0.05 m ×= = ´ = × and the heat flux is ( ) ( )2 3 2s mq h T T 101 W/m K 100 25 C 7.58 10 W/m .¢¢ = - = × - = ´o < COMMENTS: Convection effects associated with the two flow conditions are comparable. PROBLEM 8.44 KNOWN: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet temperature. FIND: (a) Water outlet temperature evaluating properties at Tm = 300 K, (b) Repeat calculations using properties evaluated at the appropriate temperature, mT = (Tm,i + Tm,o)/2, and (c) Coolant mean velocities for the range 4 \u2264 L \u2264 7 m which provide the same Tm,o as found in part (b). SCHEMATIC: ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic energy, potential energy and flow work changes. PROPERTIES: Table A.6, Water ( mT = 300 K): \u3c1 = 997 kg/m3, cp = 4179 J/kg\u22c5K, µ = 855 × 10-6 kg/s\u22c5m, k = 0.613 W/m\u22c5K, Pr = 5.83. ANALYSIS: (a) From Equation 8.42b ( ) ( )m,o s s m,i pT T T T exp DL mc h\u3c0\uf8ee \uf8f9= \u2212 \u2212 \u2212\uf8f0 \uf8fb\ufffd . and evaluating properties at mT = 300 K, find ( )3m D 6 997 kg m 1m s 0.0254 mu DRe 29,618 855 10 kg s m \u3c1 µ \u2212 = = = × \u22c5 The flow is turbulent, and since L/D = 197, it is reasonable to assume fully developed flow throughout the tube. Hence, h \u2248 hfd. From the Dittus-Boelter equation, ( ) ( )4 /5 0.44 /5 0.4D DNu 0.023Re Pr 0.023 29,618 5.83 176= = = ( ) ( ) 2Dh Nu k D 176 0.613W m K 0.0254 m 4248 W m K= = \u22c5 = \u22c5 . With ( ) ( ) ( )( )22 3mm u D 4 4 997 kg m 1m s 0.0254m 0.505kg s\u3c1 \u3c0 \u3c0= = =\ufffd . Equation 8.42b yields ( ) ( ) ( )( ) 2 m,o 0.0254m 5m 4248W m K T 350K 60K exp 323K 50 C 0.505kg s 4179J kg K \u3c0\uf8ee \uf8f9\u22c5\uf8ef \uf8fa = \u2212 \u2212 = =\uf8ef \uf8fa \u22c5\uf8ef \uf8fa\uf8f0 \uf8fb $ < (b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with the energy balance and rate equations above, the calculation of part (a) is repeated with mT = (Tm,i + Tm,o)/2 giving these results: mT 307.3K= Tm,o = 51.7°C = 324.7 K < (c) Using the IHT model developed for the part (b) analysis, the coolant mean velocity, um, as a function of tube length L with Tm,o = 51.7°C is calculated and the results plotted below. Continued... PROBLEM 8.44 (Cont.) 4 5 6 7 Tube length, L (m) 0 1 2 3 4 5 6 Ve lo ci ty , u m (m /s) COMMENTS: (1) Using mT = 300 K vs. mT = (Tm,i + Tm,o)/2 = 307 K for this application resulted in a difference of Tm,o = 50°C vs.Tm,o = 51.7°C. While the difference is only 1.7°C, it is good practice to use the proper value for mT . (2) Note that um must be increased markedly with increasing length in order that Tm,o remain fixed. PROBLEM 8.45 KNOWN: Gas turbine vane approximated as a tube of prescribed diameter and length maintained at a known surface temperature. Air inlet temperature and flowrate. FIND: (a) Outlet temperature of the air coolant for the prescribed conditions and (b) Compute and plot the air outlet temperature Tm,o as a function of flow rate, 0.1 \u2264 m\ufffd \u2264 0.6 kg/h. Compare this result with those for vanes having passage diameters of 2 and 4 mm. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes. PROPERTIES: Table A.4, Air (assume mT = 780 K, 1 atm): cp = 1094 J/kg\u22c5K, k = 0.0563 W/m\u22c5K, µ = 363.7 × 10-7 N\u22c5s/m2, Pr = 0.706; (Ts = 650°C = 923 K, 1 atm): µ = 404.2 × 10-7 N\u22c5s/m2. ANALYSIS: (a) For constant wall temperature heating, from Eq. 8.43, s m,o s m,i p T T PLh exp T T mc \uf8eb \uf8f6 \u2212 = \u2212\uf8ec \uf8f7\uf8ec \uf8f7 \u2212 \uf8ed \uf8f8\ufffd (1) where P = \u3c0D. For flow in circular passage, ( ) ( )D 7 2 4 0.18kg h 1 3600s h4mRe 584 D 0.003m 363.7 10 N s m\u3c0 µ \u3c0 \u2212 × = = = × \u22c5 \ufffd . (2) The flow is laminar, and since L/D = 75 mm/3 mm = 25, the Sieder-Tate correlation including combined entry length yields 0.141/3 DD s hD Re PrNu 1.86 k L D µ µ \uf8eb \uf8f6\uf8eb \uf8f6 = = \uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 (3) 0.141/3 7 2 7 0.0563W m K 584 0.706 363.7 10h 1.86 87.5 W m K 0.003m 25 404.2 10 \u2212 \u2212 \uf8eb \uf8f6 \u22c5 × ×\uf8eb \uf8f6 \uf8ec \uf8f7= = \u22c5\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 ×\uf8ed \uf8f8 . Hence, the air outlet temperature is ( ) ( ) ( ) 2 m,o650 T 0.003m 0.075m 87.5W m K exp