# CSU MECH 344 - HW9 - Solutions (9 pages)

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## HW9 - Solutions

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## HW9 - Solutions

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Mech 344 - Heat and Mass Transfer
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PROBLEM 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 T1 Th i Tc o 80 C and T2 Th o Tc i 120 C Hence T1m 120 80 C ln 120 80 98 7 C and L q 3000 W 0 968 m D U T1m 0 02 m 500 W m2 K 98 7o C b With q Cc Tc o Tc i the following ratio may be formed in terms of the design and 3 year conditions Cc Tc o Tc i q 60o C q3 1 333 oC Cc Tc o Tc i 45 3 Hence q3 q 1 33 3000 W 1 333 2250 W Having determined the ratio of heat rates it follows that Ch Th i Th o q 20o C q3 1 333 oC T Ch Th i Th o 160 h o 3 3 Hence Th o 3 160o C 20o C 1 333 145o C With Tlm 3 125 95 ln 125 95 109 3o C U3 q3 2250 W 338 W m 2 K o DL T 1m 3 0 02 m 0 968 m 109 3 C Continued PROBLEM 11 18 Cont With U 1 h i 1 h o R f c 1 and U3 1 h i 1 h o R f c 1 1 1 1 1 2 4 2 m K W 9 59 10 m K W U3 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 3 kg m Water 1000 Oil 800 cp J kg K 4200 1900 2 m s k W m K 7 7 10 0 64 4 7 5 1 10 0 134 140 Pr ANALYSIS a With the outlet temperature Tc o 60 C from an overall energy balance on the hot oil fluid find h ch Th i Th o 0 1 kg s 1900 J kg K 100 60 C 7600 W q m From an energy balance on the cold water fluid find c cc 30 C 7600 W 0 1 kg s 4200 J kg K 48 1 C Tc o Tc i q m b Using the LMTD method the length of the CF heat exchanger follows from q UA Tlm CF U DL Tlm CF where Tlm CF L q U D Tlm CF 60 30 C 100 48 1 C 40 0 C T1 T2 ln T1 T2 ln 30 51 9 L 7600 W 60 W m 2 K 0 025 m 40 0 C 40 3m COMMENTS Using the NTU method find Cmin Ch 190 W K and Cmax Cc 420 W K Hence q max Cmin Th i Tc i 190 W K 100 30 K 13 300 W and q qmax 0 571 With Cr Cmin Cmax 0 452 and using Eq 11 29b NTU 1 UA 1 1 0 571 1 ln ln 1 00 Cmin Cr 1 Cr 1 0 452 1 0 571 0 452 1 so that with A DL find L 40 3 m PROBLEM 11 24 KNOWN Counterflow concentric tube heat exchanger undergoing test after service for an extended 2 period of time surface area of 5 m and design value for the overall heat transfer coefficient of 2 Ud 38 W m 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 Tc 304 K assuming Tc o 36oC 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 h c h Th i Th o 01 q m kg s 2166 J kg K 110 66 K 9530 W d b i g Energy balance on the cold fluid d i ccc Tc o Tc i q m find Tc o 36 4o C Rate equation q UA Tln CF Tln CF dTh i Tc o i dTh o Tc i i b110 36 4go C b66 25go C 55 7o C ln 73 6 41 0 ln dTh i Tc o i dTh o Tc i i 9530 W U 5 m2 55 7o C U 34 2 W m2 K Overall resistance including fouling factor U 1 1 Ud R f 34 2 W m2 K 1 1 38 W m2 K R f R f 0 0029 m2 K W 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 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 ln E 1 E l C 19 300 W K ln 0 171 A min 12 03m2 1 2 2 U 1 414 2000 W m K l C2r With 0 50 the heat recovery rate is then q Cmin Th i Tc i 627 000 W and the outlet temperatures are Tc o Tc i q 627 000 W 5 C 37 5 C Cc 19 300 W K Th o Th i q 627 000 W 70 C 37 5 C Ch 19 300 W K …

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