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Modelling Workshop Fluid Mechanics 1: Solution

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The Modelling Workshop Fluid Mechanics 1 Solution of the Modelling Course of Industrial Design of the TU Delft

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Transcript of "Modelling Workshop Fluid Mechanics 1: Solution"

1. 1. P-W-F-1: Fill your glass using Schlenkerla beer barrel Analysis Model the system > restart : with plots : mass flow rate in the orifice > q d t/Cd\$Ah\$rho\$sqrt 2\$g\$h t ; q := t/Cd Ah ρ 2gh t Addtional in this file, not useful in calculation: The resistancts of the orifice (3.1)
2. 2. > #Rd 1 2\$rho\$C2\$A2 d h The consevation of mass: Mass flow rate in the barrel = mass flow rate in the orifice > ode d rho\$diff h t , t \$Ab =K t q d ode := ρ h t dt Ab = K d Ah ρ C 2 gh t (3.2) Solve the problem parameters > Ab d 3.14\$0.32 : g d 9.8 : Cd d 0.6 : rho d 1060 : Ah d 3.14\$r2 : Initial Maximum height of the beer - 7cm (height of the hose) > h0Max d 0.225 K0.07 Ab h0Max := 0.7261783440 (4.1) Initial conditions, let's suppose h0 is the initial height since the drum is not always full > ics d h 0 = h0; ics := h 0 = h0 (4.2) review the ode > ode 299.5560 d h t dt =K 6251.724070 r2 2 h t (4.3) solve > dsol d dsolve ode, ics , h t ; 625172407 2 390840538474173649 4 2 r 2 t h0 Ch0 C r t dsol := h t = K 29955600 1794675942720000 change the solution to a function named h with parameter r and time t (4.4) > hh d rhs dsol : h d unapply hh, t ; 390840538474173649 4 2 625172407 2 r 2 t h0 Ch0 C r t h := t/K 1794675942720000 29955600 now the flow rate (4.5) > q t (4 6)
3. 3. 1997.040 r2 K 409.0513686 r2 h0 C19.6 h0 C4268.444442 r4 t2 2 t (4.6) every 2.5 second how much beer will pass the hose > beer d tt/ int q t , t = tt ..tt C2.5 rho tt C 2.5 q t dt beer := tt/ tt ρ (4.7) Green line - Congraluations, you are qualified for workshop PW-F-1 Here you get the full grade of this workshop, but don't stop... Evaluation - Start from the maximum height - time dependent evaluation - at which time it can fill 1 liter in 2.5 seconds View the height of beers of the system (refering to time from 0 to 60, radius from 0.5cm to 2 cm) > h0 d h0Max; h0 := 0.7261783440 > plot3d h t , t = 0 ..60, r = 0.005 ..0.02, axes = boxed ; (6.1)
4. 4. View the flow rate of the system (time from 0 to 60, radius from 0.5cm to 2 cm) > plot3d q t , t = 0 ..60, r = 0.005 ..0.02, axes = boxed ;
5. 5. View the beer the system can offer with in 2.5 seconds (time from 0 to 60, radius from 0.5cm to 2 cm) > plot3d beer tt , tt = 0 ..60, r = 0.005 ..0.01, axes = boxed ;
6. 6. Now we are using 2D plots to plot several profile of theabove figue in one picture following standard tube size Tube 3/8 inch 1/2 inch 3/4 inch 1 inch divided by 2 for radius, then change from inch to meters > change d 0.0254; change := 0.0254 3 1 1 1 3 1 1 > tube d \$change\$ , \$change\$ , \$change\$ , change\$ ; 8 2 2 2 4 2 2 tube := 0.004762500000, 0.006350000000, 0.009525000000, 0.01270000000 > > r d tube 1 ; pl 1 d plot beer tt , tt = 0 ..60, color = blue, style = line : (6.2) (6.3) (6.4)
7. 7. r := 0.004762500000 (6.4) > r d tube 2 ; pl 2 d plot beer tt , tt = 0 ..60, color = yellow, style = line : r := 0.006350000000 > r d tube 3 ; pl 3 d plot beer tt , tt = 0 ..60, color = red, style = line : r := 0.009525000000 > r d tube 4 ; pl 4 d plot beer tt , tt = 0 ..60, color = green, style = line : r := 0.01270000000 > display pl 1 , pl 2 , pl 3 , pl 4 ; 0.0025 0.0020 0.0015 0.0010 0.0005 0 10 20 30 tt 40 50 60 > Evaluation- height dependent evaluation - at which height it can fill 1 liter in 2.5 seconds Here we repeart the calculation except we use the initial height as a parameter as well (6.5) (6.6) (6.7)
8. 8. > restart : > q d t/Cd\$Ah\$rho\$sqrt 2\$g\$h t ; q := t/Cd Ah ρ > Rd 2gh t (7.1) 1 : 2\$rho\$C2\$A2 d h > ode d rho\$diff h t , t \$Ab =K t q d ode := ρ h t dt Ab = K d Ah ρ C 2 gh t (7.2) > Ab d 3.14\$0.32 : g d 9.8 : Cd d 0.6 : rho d 1060 : Ah d 3.14\$r2 : > ics d h 0 = h0; ics := h 0 = h0 (7.3) review the ode > ode 299.5560 d h t dt =K 6251.724070 r2 2 h t (7.4) solve > dsol d dsolve ode, ics , h t ; 625172407 2 390840538474173649 4 2 dsol := h t = K r 2 t h0 Ch0 C r t 29955600 1794675942720000 change the solution to a function named h with parameter r and time t (7.5) > hh d rhs dsol : h d unapply hh, t ; 625172407 2 390840538474173649 4 2 h := t/K r 2 t h0 Ch0 C r t 29955600 1794675942720000 now the flow rate (7.6) > q t 1997.040 r2 > beer d h0/ K 409.0513686 r2 2 t int q t , t = 0 ..0 C2.5 rho h0 C19.6 h0 C4268.444442 r4 t2 (7.7) 2.5 q t dt beer := h0/ 0 ρ > Now we are using 2D plots to plot several profile of the above figure in one picture following (7.8)
9. 9. (standard) tube size. Looking at the results in the first cycle above we saw that that two biggest standard tubes (1/2 and 3/4 inch) we in the neighborhood of the best solution (cyan and yellow lines in the graph of the first cycle). So now we can best look for optimal tube diameters in between these standard diameters to look for the optimal tube diameter. The following tubes are investigated: Tube 1/2 inch (standard) 9/16 inch (might be optimal) 5/8 inch (might be optimal) 11/16 inch (might be optimal) 3/4 inch (standard) divided by 2 for radius > change d 0.0254; change := 0.0254 (7.9) 1 1 9 1 5 1 11 1 3 1 \$change\$ , \$change\$ , \$change\$ , \$change\$ , \$change\$ 2 2 16 2 8 2 16 2 4 2 tube := 0.006350000000, 0.007143750000, 0.007937500000, 0.008731250000, (7.10) 0.009525000000 > r d tube 2 ; r := 0.007143750000 (7.11) > tube d > with plots : > r d tube 1 : pl 1 d plot beer h0 , h0 = 0 ..0.80, color = red, style = line : > r d tube 2 : pl 2 d plot beer h0 , h0 = 0 ..0.80, color = green, style = line : > r d tube 3 : pl 3 d plot beer h0 , h0 = 0 ..0.80, color = blue, style = line : > r d tube 4 : pl 4 d plot beer h0 , h0 = 0 ..0.80, color = yellow, style = line : > r d tube 5 : pl 5 d plot beer h0 , h0 = 0 ..0.80, color = cyan, style = line : > display pl 1 , pl 2 , pl 3 , pl 4 , pl 5 ;
10. 10. 0.0016 0.0014 0.0012 0.0010 0.0008 0.0006 0.0004 0.0002 0 0.1 0.2 0.3 0.4 h0 0.5 0.6 0.7 0.8 Conclusion- 1/2 inch tube can satisfy the demand till 10 cm height (Red line below). 9/16 tube can always satisfy the requirements, so this tube will be the optimal choise. > display pl 1 , pl 2 , pl 3 ;
11. 11. 0.0010 0.0008 0.0006 0.0004 0.0002 0 0.1 0.2 0.3 0.4 h0 0.5 0.6 0.7 0.8 > > Extra information Due to valave resistances and other tube resistances, in reality, the filling speed will be a bit slower and are controlled by the waitor through the valve The important thing here is that we can defined the bounday of the system, e.g. maximum speed Think about, if we need to filled even faster, with the same tube, what shall we do?