Modelling Workshop Thermodynamics 1: Solution
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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

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 Modelling Workshop Thermodynamics 1: Solution Document Transcript

  • P-W-T-1: The Spa mineral water Analysis Model The thermal resistance network > restart; Convection between water & bottle 1 > R1 d ; h1$A R1 := Conduction of half of thickness of the bottle 1 h1 A (3.1.1)
  • L 2 > R2 d ; R3 d R2; K$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)
  • > 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)
  • 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?
  • 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$diff TWater1 t , t = TWater1 t KTair ; R1 CR4 d TWater1 t = 0.7741935482 TWater1 t K3.096774193 dt > ics d TWater1 0 = 20; ics := TWater1 0 = 20 > sol d dsolve equ3, ics , TWater1 t , numeric, output = listprocedure, maxfun = 500000 sol := t = proc t ... end proc, TWater1 t = proc t ... end proc equ3 := K 6271.5 (7.1) (7.2) (7.3) > TWater1 d rhs sol 2 ; TWater1 := proc t ... end proc (7.4) See the difference (red-original model, green-new model) > plot TWater t , TWater1 t , t = 0 ..36000, numpoints = 100, axis = gridlines = 10, color = blue , color = red, green ;
  • 20 18 16 14 12 10 8 6 0 5000 10000 15000 20000 25000 30000 35000 t Hours take to cool water to 6 degree > TimeInSeconds d fsolve TWater1 t = 6, t, t = 15000 ..20000 ; TimeInSeconds := 16844.89898 fsolve TWater1 t = 6, t, t = 15000 ..20000 > TimeInHours d ; 3600 TimeInHours := 4.679138606 (7.5) (7.6) You can also solve equ3 by hand by seperation of variables Here is the manual procedures equ3 can be written as > a$diff TWater1 t , t = TWater1 t Cb d a TWater1 t dt where = TWater1 t Cb (8.1)
  • 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)