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Chip and cooling channel dimensions. Channel flowrate and inlet temperature. Chip temperature. FIND: Water outlet temperature and chip power. SCHEMATIC: ASSUMPTIONS: (1) Negligible kinetic and potential energy changes for water, (2) Uniform channel surface temperature, (3) mT = 300 K, (4) Fully developed flow. PROPERTIES: Table A-6, Water ( mT = 300 K): cp = 4179 J/kg×K, m = 855 ´ 10 -6 kg/s×m, k = 0.613 W/m×K, Pr = 583. ANALYSIS: Using the hydraulic diameter, find the Reynolds number, ( ) ( ) ( ) 2 6 5 h 4 H W 2 50 200 m D 10 m/ m 8 10 m 2 H W 250 m m m m - -´ ´= = = ´ + ( ) ( ) ( ) 4 5 m h 1 h D 12 2 6c 10 kg/s 8 10 m u D m D Re 936. A 50 200 10 m 855 10 kg/s m r m m - - - - ´ = = = = ´ ´ × & Hence, the flow is laminar and, from Table 8.1, NuD = 4.44, so that ( ) 2 D 5h 4.44 0.613 W/m Kk h Nu 34,022 W/m K. D 8 10 m- × = = = × ´ With P = 2(H + W) = 2(250 mm) 10 -6 m/mm = 5 ´10 -4 m, Eq. 8.42b yields 6 2 2 s m,o m,o 4s m,i 1 p T T 350K T PL 5 10 m 34,022 W/m K exp h exp T T 60 K m c 10 kg/s 4179 J/kg K - - æ öæ ö- - ´ ´ ×ç ÷= = - = -ç ÷ç ÷ ç ÷- ´ ×è ø è ø& ( )m,oT 350K 60 K exp 0.407 310 K.= - - = < Hence, from Eq. 8.37, ( ) ( ) ( ) ( )4p m,o m,i 1 p m,o m,iq m c T T Nm c T T 50 10 kg/s 4179 J/kg K 20 K 418 W.-= - = - = ´ × =& & < COMMENTS: (1) The chip heat flux of 418 W/cm 2 is extremely large and the method provides a very efficient means of heat removal from high power chips. However, clogging of the microchannels is a potential problem which could seriously compromise reliability. (2) L/Dh = 125 and 0.05 ReDPr = 272. Hence, fully developed conditions are not realized and h > 34,022. The actual power dissipation is therefore greater than 418 W. PROBLEM 8.89 KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature. Temperature of chip at base of channel. FIND: (a) Water outlet temperature and chip power, (b) Effect of channel width and pitch on power dissipation. SCHEMATIC: ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes for water, (2) Flow may be approximated as fully developed and channel walls as isothermal for purposes of estimating the convection coefficient, (3) One-dimensional conduction along channel side walls, (4) Adiabatic condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in channels, (6) Constant properties. PROPERTIES: Table A-6, Water ( mT = 300K): cp = 4179 J/kg\u22c5K, µ = 855 × 10 -6 kg/s\u22c5m, k = 0.613 W/m\u22c5K, Pr = 5.83. ANALYSIS: (a) The channel sidewalls act as fins, and a unit channel/sidewall combination is shown in schematic (a), where the total number of unit cells corresponds to N = L/S. With N = 50 and L = 10 mm, S = 200 µm and \u3b4 = S \u2013 W = 150 µm. Alternatively, the unit cell may be represented in terms of a single fin of thickness \u3b4, as shown in schematic (b). The thermal resistance of the unit cell may be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (\u3b7ohAt)-1, where At = Af + Ab = L (2 H + W) = 0.01m (4 × 10-4 + 0.5 × 10-4) m = 4.5 × 10-6 m2. With Dh = 4 (H × W)/2 (H + W) = 4 (2 × 10-4m × 0.5 × 10-4m)/2 (2.5 × 10-4m) = 8 × 10-5m, the Reynolds number is ReD = \u3c1um Dh/µ = 1m\ufffd Dh/Acµ = 10 -4 kg/s × 8 × 10-5m/(2 × 10-4m × 0.5 × 10-4m) 855 × 10-6 kg/s\u22c5m = 936. Hence, the flow is laminar, and assuming fully developed conditions throughout a channel with uniform surface temperature, Table 8.1 yields NuD = 4.44. Hence, 2 D 5h k 0.613W / m K 4.44h Nu 34,022 W / m K D 8 10 m\u2212 \u22c5 × = = = \u22c5 × With m = (2h/kch\u3b4)1/2 = (68,044 W/m2\u22c5K/140 W/m\u22c5K × 1.5 × 10-4m)1/2 = 1800 m-1 and mH = 0.36, the fin efficiency is f tanh mH 0.345 0.958 mH 0.36 \u3b7 = = = and the overall surface efficiency is ( ) ( ) 6 f o f 6t A 4.0 10l l l l 0.958 0.963 A 4.5 10 \u3b7 \u3b7 \u2212 \u2212 × = \u2212 \u2212 = \u2212 \u2212 = × The thermal resistance of the unit cell is then Continued \u2026.. PROBLEM 8.89 (Cont.) ( ) ( ) 11 2 6 2t,o o tR h A 0.963 34,022 W / m K 4.5 10 m 6.78K / W\u3b7 \u2212\u2212 \u2212= = × \u22c5 × × = The outlet temperature follows from Eq. (8.46b), ( ) ( ) p t,o l exp 350K 60Km,o s s m,i m c R T T T T \uf8eb \uf8f6 \u2212 = \u2212 ×\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8 = \u2212 \u2212 \ufffd 4 1 exp 307.8K 10 kg / s 4179J / kg K 6.78K / W\u2212 \uf8eb \uf8f6\uf8ec \uf8f7\u2212 =\uf8ec \uf8f7× \u22c5 ×\uf8ed \uf8f8 < The heat rate per channel is then ( ) ( )41 1 p m,o m,iq m c T T 10 kg / s 4179 J / kg K 17.8K 7.46 W\u2212= \u2212 = × \u22c5 =\ufffd and the chip power dissipation is lq Nq 50 7.46 W 373W= = × = < (b) The foregoing result indicates significant heat transfer from the channel side walls due to the large value of \u3b7f. If the pitch is reduced by a factor of 2 (S = 100 µm), we obtain 1S 100 m, W 50 m, 50 m, N 100 : q 7.04 W, q 704 Wµ µ \u3b4 µ= = = = = = < Hence, although there is a reduction in \u3b7f due to the reduction in \u3b4 (\u3b7f = 0.89) and therefore a slight reduction in the value of ql, the effect is more than compensated by the increase in the number of channels. Additional benefit may be derived by further reducing the pitch to whatever minimum value of \u3b4 is imposed by manufacturing or structural limitations. There would also be an advantage to increasing the channel hydraulic diameter and or flowrate, such that turbulent flow is achieved with a correspondingly larger value of h. COMMENTS: (1) Because electronic devices fail by contact with a polar fluid such as water, great care would have to be taken to hermetically seal the devices from the coolant channels. In lieu of water, a dielectric fluid could be used, thereby permitting contact between the fluid and the electronics. However, all such fluids, such as air, are less effective as coolants. (2) With L/Dh = 125 and L/Dh)fd \u2248 0.05 ReD Pr = 273, fully developed flow is not achieved and the value of h = hfd underestimates the actual value of h in the channel. The coefficient is also underestimated by using a Nusselt number that presumes heat transfer from all four (rather than three) surfaces of a channel. PROBLEM 8.90 KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature. Temperature of chip at base of channel. FIND: (a) Outlet temperature and chip power dissipation for dielectric liquid, (b) Outlet temperature and chip power dissipation for air. SCHEMATIC: ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes, (2) Flow may be approximated as fully developed and channel walls as isothermal for purposes of estimating the convection coefficient, (3) One-dimensional conduction along the channel side walls, (4) Adiabatic condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in channels, (6) Constant properties. PROPERTIES: Prescribed. Dielectric liquid: cp = 1050 J/kg\u22c5K, k = 0.065 W/m\u22c5K, µ = 0.0012 N\u22c5s/m2, Pr = 15. Air: cp = 1007 J/kg\u22c5K, k = 0.0263 W/m\u22c5K, µ = 185 × 10 -7 N\u22c5s/m2, Pr = 0.707. ANALYSIS: (a) The channel side walls act as fins, and a unit channel/sidewall combination is shown in schematic (a), where \u3b4 = S \u2013 W = 150 µm. Alternatively, the unit cell may be represented in terms of a single fin of thickness \u3b4, as shown in schematic (b). The thermal resistance of the unit cell may be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (\u3b7o h At)-1, where At = Af + Ab = L (2 H + W) = 4.5 × 10-6 m2. With Ac = H × W = 10-8 m2 and Dh = 4 Ac/2(H + W) = 8 × 10- 5 m, the Reynolds number is ReD = \u3c1umDh/µ = 1m\ufffd Dh/Acµ = 667. Hence, the flow is laminar, and assuming fully developed conditions throughout a channel with uniform surface temperature, Table 8.1 yields NuD = 4.44. Hence, 2D 5h k 0.065 W / m K 4.44h Nu 3608 W / m K D 8 10 m\u2212 \u22c5 × = = = \u22c5 × With m = (2 h/kch \u3b4)1/2 = 586 m-1 and mH = 0.117, the fin efficiency is f tanh mH 0.1167 0.995 mH 0.117 \u3b7 =