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# Modelling Workshop Thermodynamics 1: Solution

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

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### Modelling Workshop Thermodynamics 1: Solution

1. 1. P-W-T-1: The Spa mineral water Analysis Model The thermal resistance network > restart; Convection between water & bottle 1 > R1 d ; h1\$A R1 := Conduction of half of thickness of the bottle 1 h1 A (3.1.1)
2. 2. L 2 > R2 d ; R3 d R2; K\$A 1 L 2 KA 1 L R3 := 2 KA R2 := (3.1.2) Convection between bottle & air > R4 d 1 h2\$A R4 := 1 h2 A (3.1.3) Water is cooling down, Heat flow -> Bottle Heat flow to the bottle > equ1 dK mWater\$cWater\$diff equ1 := K mWater cWater TWater t , t = d TWater t dt TWater t KTBottle t ; R1 CR2 TWater t KTBottle t = 1 1 L C h1 A 2 KA (3.2.1) Heat flow-> bottle = Heat absorb by the bottle + Heat flow->Air Heat from water to bottle, then part is absorded by bottle, part goes to the air TWater t KTBottle t = mBottle\$cBottle\$diff TBottle t , t R1 CR2 TBottle t KTair C ; R3 CR4 TWater t KTBottle t d equ2 := = mBottle cBottle TBottle t 1 1 L dt C h1 A 2 KA TBottle t KTair C 1 1 L C h2 A 2 KA > equ2 d Solving Assign Numerical Values Assign the parameters from choices (3.3.1)
3. 3. > mWater d 1.5; cWater d 4181; mBottle d 0.025; cBottle d 1200; h1 d 300; h2 d 10; K d 0.15; A d 0.08; Tair d 4 ; L d 0.0003; mWater := 1.5 cWater := 4181 mBottle := 0.025 cBottle := 1200 h1 := 300 h2 := 10 K := 0.15 A := 0.08 Tair := 4 L := 0.0003 (4.1.1) Review 2 differential equations before solving (to check errors) > equ1 K 6271.5 d TWater t dt = 18.46153846 TWater t K18.46153846 TBottle t (4.2.1) > equ2 18.46153846 TWater t K18.46153846 TBottle t = 30.000 d TBottle t dt (4.2.2) C0.7920792079 TBottle t K3.168316832 Setup initial conditions, bottle & water temperature = 20 > ics d TWater 0 = 20, TBottle 0 = 20; ics := TWater 0 = 20, TBottle 0 = 20 (4.3.1) Simulation : Solve differential Equation in the numerical way > sol d dsolve equ1, equ2, ics , TWater t , TBottle t , numeric, output = listprocedure, maxfun = 500000 ; sol := t = proc t ... end proc, TBottle t = proc t ... end proc, TWater t = proc t (4.4.1) ... end proc Plot Temperature vs time > with plots : > TWater d rhs sol 3 TWater := proc t ... end proc (4.5.1) > TBottle d rhs sol 2 ; (4.5.2)
4. 4. TBottle := proc t ... end proc > plot (4.5.2) TWater t , TBottle t , t = 0 ..36000, numpoints = 100, axis = gridlines = 10, color = blue 20 18 16 14 12 10 8 6 0 5000 10000 15000 20000 25000 30000 35000 t Hours take to cool water to 6 degree > TimeInSeconds d fsolve TWater t = 6, t, t = 15000 ..20000 ; TimeInSeconds := 17248.02097 fsolve TWater t = 6, t, t = 15000 ..20000 ; > TimeInHours d 3600 TimeInHours := 4.791116936 (4.5.3) (4.5.4) Green line - Congraluations, you are qualified for workshop PW-T-1 Here you get the full grade of this workshop, but don't stop... Evaluatuion: What shall we do if we want to cool it faster? Which paprameter in the system is most important 1. Increase the thickness of the bottle?
5. 5. 2. Change to different material; What is the effect of the bottle, can we neglect it? If we neglect the bottle > equ3 dK mWater\$cWater\$diff TWater1 t , t = TWater1 t KTair ; R1 CR4 d TWater1 t = 0.7741935482 TWater1 t K3.096774193 dt > ics d TWater1 0 = 20; ics := TWater1 0 = 20 > sol d dsolve equ3, ics , TWater1 t , numeric, output = listprocedure, maxfun = 500000 sol := t = proc t ... end proc, TWater1 t = proc t ... end proc equ3 := K 6271.5 (7.1) (7.2) (7.3) > TWater1 d rhs sol 2 ; TWater1 := proc t ... end proc (7.4) See the difference (red-original model, green-new model) > plot TWater t , TWater1 t , t = 0 ..36000, numpoints = 100, axis = gridlines = 10, color = blue , color = red, green ;
6. 6. 20 18 16 14 12 10 8 6 0 5000 10000 15000 20000 25000 30000 35000 t Hours take to cool water to 6 degree > TimeInSeconds d fsolve TWater1 t = 6, t, t = 15000 ..20000 ; TimeInSeconds := 16844.89898 fsolve TWater1 t = 6, t, t = 15000 ..20000 > TimeInHours d ; 3600 TimeInHours := 4.679138606 (7.5) (7.6) You can also solve equ3 by hand by seperation of variables Here is the manual procedures equ3 can be written as > a\$diff TWater1 t , t = TWater1 t Cb d a TWater1 t dt where = TWater1 t Cb (8.1)
7. 7. 6271.5 3.096774193 ; b dK ; 0.7741935482 0.7741935482 a := K 8100.687502 b := K 4.000000000 dTWater1 t 1 > = : TWater1 t Cb a Integral both side t > ln TWater1 t Cb = Cc : a > a dK (8.2) t a > TWater1 t Cb = e \$ec : t a > TWater1 t = c1\$e Kb : Initial conditions TWater1(0)=20 0 a > 20 = c1\$e Kb 20 = 1.0 c1 C4.000000000 (8.3) c1 := 16 (8.4) > c1 d 16 The model now is t a > TWater1 t = c1\$e Kb 0.0001234463124 t TWater1 t = 16 eK C4.000000000 Verify > e d 2.71828; e := 2.71828 (8.5) (8.6) t a > TWater1 d unapply c1\$e Kb, t 0.0001234463124 t TWater1 := t/16 2.71828K C4.000000000 > evalf TWater1 TimeInSeconds 6.000004557 So, this is an exam question, even with a pen, you can do it easily :) (8.7) (8.8)