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CSU MECH 344 - HW9 - Solutions

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sm11-018.pdfHW9 Solutions.pdfsm11-024.pdfHW9 SolutionsHW9 SolutionsPROBLEM 11.18 KNOWN: Inner tube diameter (D = 0.02 m) and fluid inlet and outlet temperatures corresponding to design conditions for a counterflow, concentric tube heat exchanger. Overall heat transfer coefficient (U = 500 W/m2⋅K) and desired heat rate (q = 3000 W). Cold fluid outlet temperature after three years of operation. FIND: (a) Required heat exchanger length, (b) Heat rate, hot fluid outlet temperature, overall heat transfer coefficient, and fouling factor after three years. SCHEMATIC: ASSUMPTIONS: (1) Negligible heat loss to the surroundings, (2) Negligible tube wall conduction resistance, (3) Constant properties. ANALYSIS: (a) The tube length needed to achieve the prescribed conditions may be obtained from Eqs. 11.14 and 11.15 where ΔT1 = Th,i - Tc,o = 80°C and ΔT2 = Th,o - Tc,i = 120°C. Hence, ΔT1m = (120 - 80)°C/ln(120/80) = 98.7°C and ()()21mq 3000WL 0.968mDUT0.02m 500W m K 98.7 Cππ== =Δ×⋅×o < (b) With q = Cc(Tc,o - Tc,i), the following ratio may be formed in terms of the design and 3 year conditions. ()()cc,o c,i3cc,o c,i3CT Tq60C1.333qCT T45 C−===−oo Hence, 3q q 1.33 3000W 1.333 2250W== = < Having determined the ratio of heat rates, it follows that ()()hh,i h,o3hh,i h,oh,o(3)3CT Tq20C1.333qCT T160 C T−===−−oo Hence, h,o(3)T 160 C 20 C 1.333 145 C=− =oo o < With ()()lm,3T 125 95 ln 125 95 109.3 CΔ=− =o, ()()()2331m,3q2250WU 338W m KDL T0.02m 0.968m 109.3 Cππ== =⋅Δo < Continued...PROBLEM 11.18 (Cont.) With ()()1ioU1h 1h−⎡⎤=+⎣⎦ and ()()13iof,cU1h1hR−⎡⎤′′=++⎣⎦, 242f,c311 1 1R m K W 9.59 10 m K WU U 338 500−⎛⎞′′=−= − ⋅ = × ⋅⎜⎟⎝⎠ < COMMENTS: Over time fouling will always contribute to a degradation of heat exchanger performance. In practice it is desirable to remove fluid contaminants and to implement a regular maintenance (cleaning) procedure.PROBLEM 11.23 KNOWN: Counterflow concentric tube heat exchanger. FIND: (a) Total heat transfer rate and outlet temperature of the water and (b) Required length. SCHEMATIC: ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible thermal resistance due to tube wall thickness. PROPERTIES: (given): ρ (kg/m3) cp (J/kg⋅K) ν (m2/s) k (W/m⋅K) Pr Water 1000 4200 7 × 10-7 0.64 4.7 Oil 800 1900 1 × 10-5 0.134 140 ANALYSIS: (a) With the outlet temperature, Tc,o = 60°C, from an overall energy balance on the hot (oil) fluid, find ()()hh h,i h,oq m c T T 0.1 kg/s 1900 J / kg K 100 60 C 7600 W.=−=×⋅−°=& < From an energy balance on the cold (water) fluid, find c,o c,i c cT T q / m c 30 C 7600 W / 0.1 kg/s 4200 J / kg K 48.1 C.=+ =°+ × ⋅=°& < (b) Using the LMTD method, the length of the CF heat exchanger follows from ()()lm,CF lm,CF lm,CFqUAT UDLT Lq/UDTππ=Δ = Δ = Δ where ()()()()12lm,CF1260 30 C 100 48.1 CTTT 40.0 Cln T / T ln 30/51.9−°− − °Δ−ΔΔ= = =°ΔΔ ()2L 7600 W / 60 W / m K 0.025m 40.0 C 40.3m.π=⋅××°= < COMMENTS: Using the ε-NTU method, find Cmin = Ch = 190 W/K and Cmax = Cc = 420 W/K. Hence ()()max min h,i c,iq C T T 190 W / K 100 30 K 13,300 W=−= −= and ε=q/qmax = 0.571. With Cr = Cmin/Cmax = 0.452 and using Eq. 11.29b, min r rUA 1 1 1 0.571 1NTU ln ln 1.00C C 1 C 1 0.452 1 0.571 0.452 1εε⎛⎞−−⎛⎞== = =⎜⎟⎜⎟−− − ×−⎝⎠⎝⎠ so that with A = πDL, find L = 40.3 m.PROBLEM 11.24 KNOWN: Counterflow, concentric tube heat exchanger undergoing test after service for an extended period of time; surface area of 5 m2 and design value for the overall heat transfer coefficient of Ud = 38 W/m2⋅K. FIND: Fouling factor, if any, based upon the test results of engine oil flowing at 0.1 kg/s cooled from 110°C to 66°C by water supplied at 25°C and a flow rate of 0.2 kg/s. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible losses to the surroundings, (3) Constant properties. PROPERTIES: Table A-5, Engine oil (Th = 361 K): c = 2166 J/kg⋅K; Table A-6, Water ()cc,oT 304 K, assuming T 36 C :==o c = 4178 J/kg⋅K. ANALYSIS: For the CF conditions shown in the Schematic, find the heat rate, q, from an energy balance on the hot fluid (oil); the cold fluid outlet temperature, Tc,o, from an energy balance on the cold fluid (water); the overall coefficient U from the rate equation; and a fouling factor, R, by comparison with the design value, Ud. Energy balance on hot fluid q m c T T kg / s 2166 J / kg K 110 66 K 9530 Whh h,i h,o=−=×⋅−=&.dibg01 Energy balance on the cold fluid qmc T T find T Cc c c,o c,i c,o=− =&,.di364o Rate equation qUATn,CF=Δl ΔTTT T TnT T T TC6625Cn 73.6/ 41.0Cn,CFh,i c,o h,o c,ih,i c,o h,o c,iloooll=−−−−−=−−−=didididibgbg/..110 36 4557 9530 557 WU5 m C2=× ×.o U 34.2 W/mK2=⋅ Overall resistance including fouling factor U1/1/U Rdf=+′′ 34 2. W/m K 1/ 1/38 W/m K R22f⋅= ⋅+′′ ′′=⋅R mK/Wf200029. <PROBLEM 11.28 KNOWN: Flow rate, inlet temperatures and overall heat transfer coefficient for a regenerator. Desired regenerator effectiveness. Cost of natural gas. FIND: (a) Heat transfer area required for regenerator and corresponding heat recovery rate and outlet temperatures, (b) Annual energy and fuel cost savings. SCHEMATIC: ASSUMPTIONS: (a) Negligible heat loss to surroundings, (b) Constant properties. PROPERTIES: Problem 11.27, milk: cp = 3860 J/kg⋅K. ANALYSIS: (a) With Cr = 1 and ε = 0.50 for one shell and two tube passes, Eq. 11.30c yields E = 1.414. With Cmin = 5 kg/s × 3860 J/kg⋅K = 19,300 W/K, Eq. 11.30b then yields ()()()()2min1/2 22rln E 1 / E lln 0.171C 19,300W / KA 12.03mU 1.4142000W / m KlC−+⎡⎤⎣⎦=− =− =⋅+ < With ε = 0.50, the heat recovery rate is then ()min h,i c,iq C T T 627,000Wε=−= < and the outlet temperatures are c,o c,icq 627,000WT T 5C 37.5CC 19,300W/ K=+=°+ =° < h,o h,ihq 627,000WT T 70 C 37.5 CC 19,300W / K=−=°− =° < (b) The amount of energy recovered for continuous operation over 365 days is 13E 627,000W 365d/ yr 24h /d 3600s/ h 1.98 10 J / yrΔ= × × × = × The annual fuel savings SA is then 7ngAEC1.98 10 MJ / yr $0.02/ MJS $440,000/ yr0.9ηΔ×××== = < COMMENTS: (1) With Cc = Ch, the temperature changes are the same for the two fluids, (2) A larger effectiveness and hence a


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