Iterative Method for Heat Transfer Coefficient Determination
Explore an iterative approach to determine the heat transfer coefficient in fluid systems by iteratively guessing the required parameters and calculating the necessary values, leading to an effective solution for this engineering problem.
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Presentation Transcript
Exercise 5.29 Created by Lauren Taylor ('14) Redesigned by Angela Tang ('18) Revised by Rose Yin ( 19) Michelle Quien ( 20) Gavin Batsimm ( 21) Sarah Huang ( 22) Leon Lee ( 23) Austin Vollweiler ( 24) Donovan Cho ( 25)
Heat transfer coefficient Fluid thermal conductivity Fluid viscosity h = 2500 J/m2 s K k = 0.59 J/m s K = 0.001 Pa s = 1000 kg/m3 CP= 4170 J/kg K Fluid density Fluid heat capacity d v = 0.17 m/s CP dv hd = = = Re Pr Nu k k
Soln 1. Guess and Check Iteratively Guess d1 Calculate RePr2/3 using d1 Find Nu on Graph Calculate d2 from Nu We re done! YES! Does d1 = d2? No
Derive d = f1(Nu) 1000 hd = Nu k k d = Nu h 100 Nu J . 0 59 m s K 10 = Nu d J 2500 2 m s K 1 1 10 100 1000 10000 100000 Re Pr2/3
Guess: d = 0.1 m 1000 hd = = Nu 152 k 152 100 Nu Try a lower d 10 1 1 10 100 1000 10000 100000 Re Pr2/3 2 2 P . 0 . 0 1 . 0 17 1000 4170 001 dv C 3 3 2 = = = 4 Re Pr . 6 26 10 3 . 0 001 . 0 59 k
Now guess d = 0.01 m 1000 hd = = Nu 50 k 100 50 = 0.01 0.012 m m d Nu Try a slightly higher d 10 1 1 10 100 1000 10000 100000 Re Pr2/3 2 2 P . 0 . 0 . 0 01 17 1000 4170 001 dv C 3 3 2 = = = 3 Re Pr . 6 26 10 3 . 0 001 . 0 59 k
Try d = 0.013 m 1000 hd = = Nu 55 k 100 55 = 0.013 = 0.013 m m d Nu 001 10 1 1 10 100 1000 10000 100000 Re Pr2/3 2 2 P . 0 . 0 . 0 013 17 1000 4170 dv C 3 3 2 = = = 3 Re Pr . 8 14 10 3 . 0 001 . 0 59 k
Soln 2. Derive functional relationships Derive d=f1(Nu) Derive d=f2(RePr2/3) Find where f1(Nu)=f2(RePr2/3) Figure out d
Soln 2: Derive d = f1(Nu) 1000 hd = Nu k k d = Nu h 100 Nu J . 0 59 m s K 10 = Nu d J 2500 2 m s K 1 1 10 100 1000 10000 100000 Re Pr2/3
Soln 2: Derive d = f2(RePr2/3) 1000 2 P dv C 3 2 = Re Pr 3 k 2 3 k 2 100 = Re Pr d 3 v C P Nu 1 2 k 3 3 2 = Re Pr d 3 2 10 v C 3 P ( ) ( 1000 ) 1 2 0.001 0.59 3 3 2 = Re Pr d 3 ( )( )( ) 2 0.17 4170 3 1 1 10 100 1000 10000 100000 Re Pr2/3 2 = 6 . 1 ( 60 10 m ) Re Pr d 3
Plot f1(Nu) = f2(RePr2/3) 1000 1000 1 2 k k 3 3 2 = = Nu Re Pr d 3 2 h v C 3 P 100 100 55 1 2 h k 3 3 2 = Nu Re Pr 3 2 k v C 3 Nu Nu P 2 = Nu 0068 . 0 Re Pr 3 10 10 y-axis slope x-axis Re Pr2/3 1000 10000 100000 Nu 6.8 68 680 1 1 1 1 10 10 100 100 1000 1000 10000 10000 100000 100000 Re Pr2/3 Re Pr2/3
Plot f1(Nu) = f2(RePr2/3) 1000 1000 hd = = Nu 55 k 100 100 55 = = 0.013 m 1.3 cm d Nu Nu 10 10 1 1 1 1 10 10 100 100 1000 1000 10000 10000 100000 100000 Re Pr2/3 Re Pr2/3
Takeaways: Many ways to solve these problems. We showed: Numerical Guess n Check Analytical Solving Equation for d Understanding of some important dimensionless groups This will be particularly useful for Fluids next semester and Heat and Mass Transfer in junior fall!
