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flow (part a), (4) Negligible kinetic and potential energy changes, (5) Negligible tube wall thermal resistance. PROPERTIES: Table A.5, Glycerin ( mT \u2248 300 K): \u3c1 = 1259.9 kg/m3, cp = 2427 J/kg\u22c5K, µ = 79.9 × 10- 2 N\u22c5s/m2, k = 286 × 10-3 W/m\u22c5K, Pr = 6780. ANALYSIS: (a) From an overall energy balance assuming the maximum temperature rise of the glycerin coolant is 6°C, find the flow rate as ( )p m,o m,iq mc T T= \u2212\ufffd ( ) ( ) 2p m,o m,im q c T T 1000 W 2427 J kg K 6 K 6.87 10 kg s\u2212= \u2212 = \u22c5 = ×\ufffd < From Eq. 8.43, the length of tubing can be determined, ( )s m,o p s m,i T T exp PLh mc T T \u2212 = \u2212 \u2212 \ufffd where P = \u3c0D. For the tube flow, find 2 D 2 2 4m 4 6.87 10 kg sRe 5.47 D 0.020 m 79.9 10 N s m\u3c0 µ \u3c0 \u2212 \u2212 × × = = = × × × \u22c5 \ufffd which implies laminar flow, and if fully developed, D hDNu 3.66 k = = 3 23.66 286 10 W m Kh 52.3W m K 0.020 m \u2212× × \u22c5 = = \u22c5 ( ) ( ) ( )( ) ( )2 2 47 30 C exp 0.020 m 52.3W m K L 6.87 10 kg s 2427 J kg K 47 24 C \u3c0 \u2212 \u2212 = \u2212 × \u22c5 × × × \u22c5 \u2212 \uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb $ $ L = 15.3 m. < The number of turns of the tubing, N, is N = L/(\u3c0D) = (15.3 m)/\u3c0(0.3 m) = 16.2 and hence the spacing S will be S = H/N = 500 mm/16.2 = 30.8 mm. < Continued... PROBLEM 8.29 (Cont.) (b) Parametric calculations were performed using the IHT Correlations Toolpad based on Eq. 8.56 (a thermal entry length condition), and the following results were obtained. 0.05 0.09 0.13 0.17 0.21 0.25 Mass flowrate, mdot(kg/s) 1000 1400 1800 2200 2600 3000 H ea t r at e, q (W ) 0.05 0.09 0.13 0.17 0.21 0.25 Mass flowrate, mdot(kg/s) 25 27 29 31 33 35 O ut le t t em pe ra tu re , T m o(C ) With Ts maintained at 47°C, the maximum allowable transformer power (heat rate) and glycerin outlet temperature increase and decrease, respectively, with increasing \ufffdm . The increase in q is due to an increase in DNu (and hence h ) with increasing ReD. The value of DNu increased from 5.3 to 9.4 with increasing \ufffdm from 0.05 to 0.25 kg/s. COMMENTS: Since ( )1 DDGz L D Re Pr\u2212 = = (15.3 m/0.02 m)/(5.47 × 6780) = 0.0206 < 0.05, entrance length effects are significant, and Eq. 8.56 should be used to determine DNu . PROBLEM 8.30 KNOWN: Diameter and length of copper tubing. Temperature of collector plate to which tubing is soldered. Water inlet temperature and flow rate. FIND: (a) Water outlet temperature and heat rate, (b) Variation of outlet temperature and heat rate with flow rate. Variation of water temperature along tube for the smallest and largest flowrates. SCHEMATIC: ASSUMPTIONS: (1) Straight tube with smooth surface, (2) Negligible kinetic/potential energy and flow work changes, (3) Negligible thermal resistance between plate and tube inner surface, (4) ReD,c = 2300. PROPERTIES: Table A.6, water (assume mT = (Tm,i + Ts)/2 = 47.5°C = 320.5 K): \u3c1 = 986 kg/m3, cp = 4180 J/kg\u22c5K, µ = 577 × 10-6 N\u22c5s/m2, k = 0.640 W/m\u22c5K, Pr = 3.77. Table A.6, water (Ts = 343 K): µs = 400 × 10-6 N\u22c5s/m2. ANALYSIS: (a) For \ufffdm = 0.01 kg/s, ReD = 4 m D\u3c0 µ\ufffd = 4(0.01 kg/s)/\u3c0(0.01 m)577 × 10-6 N\u22c5s/m2 = 2200, in which case the flow may be assumed to be laminar. With fd,tx D \u2248 0.05ReDPr = 0.05(2200)(3.77) = 415 and L/D = 800, the flow is fully developed over approximately 50% of the tube length. With ( ) ( )1/3 0.14D sRe Pr L D µ µ\uf8ee \uf8f9\uf8f0 \uf8fb = 2.30, Eq. 8.57 may therefore be used to compute the average convection coefficient 0.141/3 DD s Re PrNu 1.86 4.27 L D µ µ \uf8eb \uf8f6\uf8eb \uf8f6 = =\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 ( ) ( ) 2Dh k D Nu 4.27 0.640 W m K 0.01m 273W m K= = \u22c5 = \u22c5 From Eq. 8.42b, 2 s m,o s m,i p T T DL 0.01m 8m 273W m K exp h exp T T mc 0.01kg s 4180J kg K \u3c0 \u3c0\uf8eb \uf8f6\uf8eb \uf8f6\u2212 × × × \u22c5\uf8ec \uf8f7= \u2212 = \u2212\uf8ec \uf8f7\uf8ec \uf8f7 \uf8ec \uf8f7 \u2212 × \u22c5\uf8ed \uf8f8 \uf8ed \uf8f8\ufffd ( )m,o s s m,iT T 0.194 T T 70 C 8.7 C 61.3 C= \u2212 \u2212 = \u2212 =$ $ $ < Hence, ( ) ( )( )p m,o m,iq mc T T 0.01kg s 4186 J kg K 36.3K 1519 W= \u2212 = \u22c5 =\ufffd < (b) The IHT Correlations, Rate Equations and Properties Tool Pads were used to determine the parametric variations. The effect of m\ufffd was considered in two steps, the first corresponding to m\ufffd < 0.011 kg/s (ReD < 2300) and the second for m\ufffd > 0.011 kg/s (ReD > 2300). In the first case, Eq. 8.57 was used to determine h , while in the second Eq. 8.60 was used. The effects of m\ufffd are as follows. Continued... PROBLEM 8.30 (Cont.) 0.005 0.006 0.007 0.008 0.009 0.01 0.011 Mass flowrate, mdot(kg/s) 60 61 62 63 64 65 66 67 O ut le t t em pe ra tu re , T m o(C ) Laminar flow (ReD < 2300) 0.01 0.02 0.03 0.04 0.05 Mass flowrate, mdot(kg/s) 69 69.2 69.4 69.6 69.8 70 O ut le t t em pe ra tu re , T m o(C ) Turbulent flow (ReD>2300) 0.005 0.006 0.007 0.008 0.009 0.01 0.011 Mass flowrate, mdot(kg/s) 800 900 1000 1100 1200 1300 1400 1500 1600 1700 H ea t r at e, q (W ) Laminar flow (ReD < 2300) 0.01 0.02 0.03 0.04 0.05 Mass flowrate, mdot(kg/s) 1500 3500 5500 7500 9500 H ea t r at e, q (W ) Turbulent flow (ReD>2300) The outlet temperature decreases with increasing m\ufffd , although the effect is more pronounced for laminar flow. If q were independent of \ufffdm , (Tm,o - Tm,i) would decrease inversely with increasing \ufffdm . In turbulent flow, however, the convection coefficient, and hence the heat rate, increases approximately as 0.8m\ufffd , thereby attenuating the foregoing effect. In laminar flow, q ~ 0.5m\ufffd and this attenuation is not as pronounced. The temperature distributions were computed from Eq. 8.43, with h assumed to be independent of x. For laminar flow ( m\ufffd = 0.005 kg/s), h was based on the entire tube length (L = 8 m) and computed from Eq. 8.57, while for turbulent flow ( m\ufffd = 0.05 kg/s) it was assumed to correspond to the value for fully developed flow and computed from Eq. 8.60. The corresponding temperature distributions are as follows. Continued... PROBLEM 8.30 (Cont.) 0 2 4 6 8 Axial location, x(m) 20 30 40 50 60 70 M ea n te m pe ra tu re , T m (C ) mdot = 0.005 kg/s 0 2 4 6 8 Axial location, x(m) 20 30 40 50 60 70 M ea n te m pe ra tu re , T m (C ) mdot = 0.05 kg/s The more pronounced increase for turbulent flow is due to the much larger value of h (4300 W/m2\u22c5K for \ufffdm = 0.05 kg/s relative to 217 W/m2\u22c5K for m\ufffd = 0.05 kg/s). PROBLEM 8.31 KNOWN: Diameter and surface temperature of ten tubes in an ice bath. Inlet temperature and flowrate per tube. Volume (\u2200) of container and initial volume fraction, fv,i, of ice. FIND: (a) Tube length required to achieve a prescribed air outlet temperature Tm,o and time to completely melt the ice, (b) Effect of mass flowrate on Tm,o and suitable design and operating conditions. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Negligible kinetic/potential energy and flow work changes, (3) Constant properties, (4) Fully developed flow throughout each tube, (5) Negligible tube wall thermal resistance. PROPERTIES: Table A.4, air (assume mT = 292 K): cp = 1007 J/kg\u22c5K, µ = 180.6 × 10-7 N\u22c5s/m2, k = 0.0257 W/m\u22c5K, Pr = 0.709; Ice: \u3c1 = 920 kg/m3, hsf = 3.34 × 105 J/kg. ANALYSIS: (a) With ReD = 4 m\ufffd /\u3c0Dµ = 4(0.01 kg/s)/\u3c0(0.05 m)180.6 × 10-7 N\u22c5s/m2 = 14,100 for m\ufffd = 0.01 kg/s, the flow is turbulent, and from Eq. 8.60, ( ) ( )0.8 0.30.8 0.3D D DNu Nu 0.023Re Pr 0.023 14,100 0.709 43.3= = = = ( ) ( ) 2Dh Nu k D 43.3 0.0257 W m K 0.05m 22.2 W m K= = \u22c5 = \u22c5 With Tm,o = 14°C, the tube length may be obtained from Eq. 8.42b, ( )( ) ( ) 2 s m,o s m,i p 0.05m 22.2 W m K LT T 14 DLh exp exp T T 24 mc 0.01kg s 1007 J kg K \u3c0\u3c0 \uf8ee \uf8f9 \u22c5\uf8eb \uf8f6 \u2212 \u2212 \uf8ef \uf8fa = = \u2212 = \u2212\uf8ec