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C r n r T . 4k 2k 4k 2k = = = - + + = - + & && & l l The temperature distribution and the maximum wall temperature (r = ro) are ( ) ( ) 2 2 2 o si i q rq r T r r r n T 4k 2k r = - - + + && l ( ) ( ) 2 2 2 o o w,max o o si i q r rq T T r r r n T 4k 2k r = = - - + + && l where Ts, the inner surface temperature of the wall at the exit, follows from ( ) ( ) ( ) ( ) 2 2 2 2 o oi i s s m,o i i q /4 D D L q D D q h T T D L 4 D p p - - ¢¢ = = = - & & where h is the local convection coefficient at the exit. With ( )D 6 2i 4 m 4 0.1 kg/s Re 7928 D 0.02m 803 10 N s/mp m p - ´ = = = ´ × & the flow is turbulent and, with (L/Di) = (8.87 m/0.02m) = 444 >> (xfd/D) » 10, it is also fully developed. Hence, from the Dittus-Boelter correlation, Eq. 8.60, ( ) ( )4 / 54/5 0.4 0.4 2D i k 0.617 W/m K h 0.023 Re Pr 0.023 7928 5.45 1840 W/m K. D 0.02 m × = = = × Hence, the inner surface temperature of the wall at the exit is ( ) ( ) ( ) ( ) 2 26 32 2 o i s m,o 2i 10 W/m 0.04m 0.02mq D D T T 40 C 48.2 C 4 h D 4 1840 W/m K 0.02m é ù-- ê úë û= + = + = ´ × o o & and ( ) ( ) 6 3 2 2 w,max 10 W/m T 0.02m 0.01m 4 15 W/m K é ù= - -ê úë û´ × ( )26 310 W/m 0.02m 0.02 n 48.2 C 52.4 C. 2 15 W/m K 0.01 + + = ´ × o ol < COMMENTS: The physical situation corresponds to a uniform surface heat flux, and Tm increases linearly with x. In the fully developed region, Ts also increases linearly with x. PROBLEM 8.37 KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient air. Flow rate and inlet temperature of water flow through duct. FIND: (a) Outlet temperature, (b) Pressure drop and pump power requirement, (c) Effect of flow rate and pipe diameter on outlet temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Fully developed flow throughout duct, (3) Negligible pipe wall conduction resistance, (4) Negligible potential energy, kinetic energy and flow work changes for water, (5) Constant properties. PROPERTIES: Table A-6, water ( )mT 360 K :\u2248 3967 kg / m ,\u3c1 = pc 4203 J / kg K,= \u22c5 6324 10µ \u2212= × 2N s / m ,\u22c5 wk 0.674 W / m K, Pr 2.02.= \u22c5 = ANALYSIS: (a) The outlet temperature is given by ( ) ( )m,o m,i pT T T T exp UA / m c\u221e \u221e= + \u2212 \u2212 \ufffd where ( ) ( ) 11tot cnv,w cnd cnv,aUA R R R R \u2212\u2212= = + + ( ) ( ) ( ) 4 cnd ln 1.08w / D ln 1.08 0.30m / 0.15m R 8.75 10 K / W 2 kL 2 1.4 W / m K 100m\u3c0 \u3c0 \u2212 × = = = × \u22c5 ( ) ( ) 11 2 4cnv,aR 4 w L h 4 0.3m 100m 25W / m K 3.33 10 K / W\u2212\u2212 \u2212= = × × × \u22c5 = × With ( ) ( )6 2DRe 4 m / D 4 2 kg / s / 0.15m 324 10 N s / m 52, 400,\u3c0 µ \u3c0 \u2212= = × × × × \u22c5 =\ufffd ( ) ( )4 / 5 0.3 24 / 5 0.3ww fd Dk 0.674 W / m K 0.023h h 0.023 Re Pr 52, 400 2.02 761W / m KD 0.15m \u22c5 × \u2248 = = = \u22c5 ( ) ( ) 11 2 5cnv,w wR D L h 0.15m 100m 761W / m K 2.79 10 K / W\u3c0 \u3c0 \u2212\u2212 \u2212= = × × × \u22c5 = × ( ) 15 4 4UA 2.79 10 8.75 10 3.33 10 K / W 809 W / K\u2212\u2212 \u2212 \u2212\uf8ee \uf8f9= × + × + × =\uf8ef \uf8fa\uf8f0 \uf8fb m,o 809 W / KT 0 C 90 C exp 81.7 C 2kg / s 4203J / kg K \uf8eb \uf8f6 = ° + ° \u2212 = °\uf8ec \uf8f7 × \u22c5\uf8ed \uf8f8 < Continued \u2026.. PROBLEM 8.37 (Cont.) (b) With f = 0.0206 from Fig. 8.3 and 2mu m / D / 4 0.117 m / s,\u3c1\u3c0= =\ufffd ( )232 2 4m 967 kg / m 0.117 m / sup f L 0.0206 100m 91N / m 8.98 10 bars 2D 2 0.15m \u3c1 \u2212\u2206 = = = = × × < With 3 3m / 2.07 10 m / s,\u3c1 \u2212\u2200 = = ×\ufffd \ufffd the pump power requirement is ( )2 3 3P p 91N / m 2.07 10 m / s 0.19 W\u2212= \u2206 \u2200 = × =\ufffd < (c) The effects of varying the flowrate and duct diameter were assessed using the IHT software, and results are shown below. Although cnv,wR , and hence totR , decreases with increasing m\ufffd , thereby increasing UA, the effect is significantly less than that of m\ufffd to the first power, causing the exponential term, ( )pexp UA / m c ,\u2212 \ufffd to approach unity and m,oT to approach m,iT . The effect can alternatively be attributed to a reduction in the residence time of the water in the pipe (um increases with increasing m\ufffd for fixed D). With increasing D for fixed m\ufffd and w, m,oT decreases due to an increase in the residence time, as well as a reduction in the conduction resistance, cndR . COMMENTS: (1) Use of mT 360 K= to evaluate properties of the water for Parts (a) and (b) is reasonable, and iteration is not necessary. (2) The pressure drop and pump power requirement are small. 0 1 2 3 4 5 Ma s s flow ra te , m d o t(kg /s ) 6 0 6 5 7 0 7 5 8 0 8 5 9 0 O u tle t t e m pe ra tu re , Tm o (C ) 0 .0 5 0 .1 0 .1 5 0 .2 0 .2 5 D u ct d ia m e te r, D (m ) 7 5 7 8 8 1 8 4 8 7 9 0 Ou tle t t e m pe ra tu re , Tm o (C ) PROBLEM 8.38 KNOWN: Water flow through a thick-walled tube immersed in a well stirred, hot reaction tank maintained at 85°C; conduction thermal resistance of the tube wall based upon the inner surface area is 2cdR 0.002 m K / W.\u2032\u2032 = \u22c5 FIND: (a) The outlet temperature of the process fluid, Tm,o; assume, and then justify, fully developed flow and thermal conditions within the tube; and (b) Do you expect Tm,o to increase or decrease if the combined thermal entry condition exists within the tube? Estimate the outlet temperature of the process fluid for this condition. SCHEMATIC: ASSUMPTIONS: (1) Flow is fully developed, part (a), (2) Constant properties, (3) Negligible kinetic and potential energy changes and flow work, and (4) Constant wall temperature heating. PROPERTIES: Table A-6, Water (Tm = (Tm,o + Tm,i)/2 = 337 K): cp = 4187 J/kg\u22c5K, µ = 4.415 × 10-4 N\u22c5s/m2, k = 0.6574 W/m\u22c5K, Pr = 2.80; (Ts = 358 K): µs = 3.316 × 10-4 N\u22c5s/m2. ANALYSIS: (a) The outlet temperature is determined from the rate equation, Eq. 8.46a, written as s m,o s s m,i p T T UA exp T T m c \uf8eb \uf8f6 \u2212 = \u2212\uf8ec \uf8f7\uf8ec \uf8f7 \u2212 \uf8ed \uf8f8\ufffd (1) where the overall coefficient, based upon the inner surface area of the tube is expressed in terms of the convection and conduction thermal resistances, cd,i 1 1 R U h \u2032\u2032= + (2) To estimate h, begin by characterizing the flow DRe 4 m / D\u3c0 µ= \ufffd (3) ( ) 4 2DRe 4 33/ 3600 kg / s / 0.012 m 4.415 10 N s / m 2210\u3c0 \u2212= × × × \u22c5 = Consider the flow as laminar, and assuming fully developed conditions, estimate h with the correlation of Eq. 8.55, DNu hD / k 3.66= = (4) 2h 3.66 0.6574 W / m K / 0.012 m 201 W / m K= × \u22c5 = \u22c5 From Eq. (2), 12 2 2U 1/ 201 W / m K 0.002 m K / W 143.1 W / m K \u2212\uf8ee \uf8f9 = \u22c5 + \u22c5 = \u22c5\uf8ef \uf8fa\uf8f0 \uf8fb and from Eq. (1), with As = \u3c0DL, calculate Tm,o. Continued \u2026.. PROBLEM 8.38 (Cont.) 2 m,o85 T 143.1 W / m K 0.012 m 8 m exp 85 20 33/ 3600 kg / s 4187 J / kg K \u3c0\uf8eb \uf8f6\u2212 \u22c5 × × ×\uf8ec \uf8f7= \u2212\uf8ec \uf8f7 \u2212 × \u22c5\uf8ed \uf8f8 m,oT 64 C= ° < Fully developed flow and thermal conditions are justified if the tube length is much greater than the fully developed length xfd,t. From Eq. 8.23, fd,t D x 0.05 Re Pr D = fd,tx 0.012 m 0.05 221.0 2.41 3.20 m= × × × = That is, the length is only twice that required to reach fully developed conditions. (b) Considering combined entry length conditions, estimate the convection coefficient using the Sieder-Tate correlation, Eq. 8.56, ( ) ( )D D 2/3 D 0.0668 D / L Re Pr Nu 3.66 1 0.04 D / L Re Pr = + \uf8ee \uf8f9+ \uf8f0 \uf8fb (5) substituting numerical values, find D 2Nu 4.05 h 222 W / m K= = \u22c5 which is a 10% increase over the fully developed analysis result. Using the foregoing relations, find 2 m,oU 154 W / m K T 65.5 C= \u22c5 = ° < COMMENTS: (1) The thermophysical properties for the fully developed correlation are evaluated at the mean fluid temperature Tm = (Tm,o + Tm,i)/2. The values are shown above in the properties section. (2) For the Sieder-Tate correlation, the properties are also evaluated at Tm, except for µs, which is evaluated at Ts. (3) For this case where the tube length is about twice xfd,t, the average heat transfer