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Modelling Workshop Thermodynamics 2: Solution
 

Modelling Workshop Thermodynamics 2: Solution

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The Modelling Workshop Thermodynamics 2 Solution of the Modelling Course of Industrial Design of the TU Delft

The Modelling Workshop Thermodynamics 2 Solution of the Modelling Course of Industrial Design of the TU Delft

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    Modelling Workshop Thermodynamics 2: Solution Modelling Workshop Thermodynamics 2: Solution Document Transcript

    • P-W-T-2: The Coke Analysis Model The thermal resistance network > restart; Convection between air & side; Conduction of half of thickness of the side;Convection between side & coke L1 1 2 1 > R1 d ; R2 d ; R3 d R2; R4 d ; hOut$A1 K1$A1 hIn$A1 1 R1 := hOut A1 1 L1 R2 := 2 K1 A1
    • 1 L1 2 K1 A1 1 R4 := (3.1.1) hIn A1 Convection between air & lid; Conduction of half of thickness of the lid;Convection between lid & coke L2 1 2 1 > R5 d ; R6 d ; R7 d R6; R8 d ; hOut$A2 K2$A2 hIn$A2 1 R5 := hOut A2 1 L2 R6 := 2 K2 A2 1 L2 R7 := 2 K2 A2 1 R8 := (3.1.2) hIn A2 R3 := air heat the side, then heat coke Tair KTL1 t TL1 t KT t = mL1$cL1$diff TL1 t , t C R1 CR2 R3 CR4 Tair KTL1 t d TL1 t KT t equ1 := = mL1 cL1 TL1 t C 1 1 L1 dt 1 L1 1 C C hOut A1 2 K1 A1 2 K1 A1 hIn A1 > equ1 d (3.2.1) air heat the lid, then heat coke Tair KTL2 t TL2 t KT t = mL2$cL2$diff TL2 t , t C R5 CR6 R7 CR8 Tair KTL2 t d TL2 t KT t equ2 := = mL2 cL2 TL2 t C 1 dt 1 1 L2 L2 1 C C hOut A2 2 K2 A2 2 K2 A2 hIn A2 > equ2 d (3.3.1) Heat from the side & the lid heat coke Heat from water to bottle, then part is absorded by bottle, part goes to the air TL1 t KT t TL2 t KT t C = mCoke$cCoke$diff T t , t ; R3 CR4 R7 CR8 TL1 t KT t TL2 t KT t d equ3 := C = mCoke cCoke T t 1 1 L1 1 L2 1 dt C C 2 K1 A1 2 K2 A2 hIn A1 hIn A2 > equ3 d (3.4.1)
    • Solving 1. The cup is a McDonald’s® medium cup with a lid; the bottom of the cup is insulated; 2. The cup is made of paper, the specific heat capacity of paper is 1336 J/(kg·K), the thermal conductivity is 0.05 W/(m·K), the density is 1000 kg/m3; the thickness of the cup is 0.3 mm; and the mass of the cup is about 10 g; 3. The lid is made of Polypropylene, the specific heat capacity of polypropylene is 1500 J/(kg·K), the thermal conductivity is 0.3 W/(m·K); the thickness of the lid is about 0.11 mm, the mass of the lid is about 2 g; 4. The cup is fully filled with 470 ml of coke; The density of the coke is 1000 kg/m3, the specific heat capacity of the coke is 4181 J/(kg·K), the initial temperature of the coke is 3°C; 5. The contact area between the coke and the sides of the cup and the contact area between the sides of the cup and the air outside are nearly same, both are 0.025 m2; 6. The contact area between the coke and the lid and the contact area between the lid and the air outside are nearly same, both are 0.005 m2; 7. The heat transfer coefficient between the coke and the cup and the heat transfer coefficient between the coke and the lid are the same, both are 500 W/(m2·K); 8. The heat transfer coefficient between the air and outside of the cup & lid is 50 W/(m2·K); 9. The environment (air) temperature is 20°C. Assign Numerical Values Assign the parameters from choices > Tair d 20; Tair := 20 > mL1 d 0.01; cL1 d 1336; L1 d 0.0003; K1 d 0.05; A1 d 0.025 mL1 := 0.01 cL1 := 1336 L1 := 0.0003 K1 := 0.05 A1 := 0.025 > mL2 d 0.002; cL2 d 1500; L2 d 0.00011; K2 d 0.3; A2 d 0.005; mL2 := 0.002 cL2 := 1500 L2 := 0.00011 K2 := 0.3 A2 := 0.005 > mCoke d 0.47; cCoke d 4181; hOut d 50; hIn d 500; mCoke := 0.47 cCoke := 4181 hOut := 50 hIn := 500 Review 3 differential equations before solving (to check errors) (4.1.1) (4.1.2) (4.1.3) (4.1.4)
    • > equ1 21.73913044 K1.086956522 TL1 t = 13.36 d TL1 t dt C5.000000000 TL1 t (4.2.1) K5.000000000 T t > equ2 4.954582988 K0.2477291494 TL2 t = 3.000 d TL2 t dt C2.290076336 TL2 t (4.2.2) K2.290076336 T t > equ3 5.000000000 TL1 t K7.290076336 T t C2.290076336 TL2 t = 1965.07 d T t dt (4.2.3) Setup initial conditions, coke is 3, cup& lid can be 3 or 20, depdends on your interpreation > ics d T 0 = 3, TL1 0 = 20, TL2 0 = 20; ics := T 0 = 3, TL1 0 = 20, TL2 0 = 20 (4.3.1) Simulation : Solve differential Equation in the numerical way > sol d dsolve equ1, equ2, equ3, ics , T t , TL1 t , TL2 t , numeric, output = listprocedure, maxfun = 500000 ; sol := t = proc t ... end proc, T t = proc t ... end proc, TL1 t = proc t ... end proc, TL2 t = proc t ... end proc (4.4.1) Plot Temperature vs time > with plots : > T d rhs sol 2 T := proc t ... end proc (4.5.1) TL1 := proc t ... end proc (4.5.2) TL2 := proc t ... end proc (4.5.3) > TL1 d rhs sol 3 ; > TL2 d rhs sol 4 ; > plot T t , TL1 t , TL2 t , t = 0 ..1800, axes = boxed, axis = gridlines = 10, color = blue , color = red, green, blue
    • 20 18 16 14 12 10 8 6 4 0 200 400 600 800 1000 1200 1400 1600 1800 t Temperature of coke (15 minutes = 900 seconds) > Temp d T 900 ; Temp := 9.83406853798502 (4.5.4) Green line - Congraluations, you are qualified for workshop PW-T-2 Here you get the full grade of this workshop, but don't stop... Evaluatuion: (Same as previous workshop) Which paprameter in the system is most important if we want to adjust system's performance? What is the effect of the lid and the cup, can we neglect it in the calculation?
    • If we neglect the bottle Tair KTcoke t Tair KTcoke t C = mCoke$cCoke$diff Tcoke t , t ; R1 CR2 CR3 CR4 R5 CR6 CR7 CR8 d equ4 := 22.32808176 K1.116404088 Tcoke t = 1965.07 Tcoke t (7.1) dt > ics d Tcoke 0 = 3; ics := Tcoke 0 = 3 (7.2) > sol d dsolve equ4, ics , Tcoke t , numeric, output = listprocedure, maxfun = 500000 sol := t = proc t ... end proc, Tcoke t = proc t ... end proc (7.3) > equ4 d > Tcoke d rhs sol 2 ; Tcoke := proc t ... end proc See the difference (red-original model, green-new model) > plot T t , Tcoke t , t = 0 ..1800, numpoints = 100, axis = gridlines = 10, color = blue , color = red, green ; 13 12 11 10 9 8 7 6 5 4 3 0 200 400 600 800 1000 1200 1400 1600 1800 t (7.4)
    • > Tcoke 900 9.80495974049290 (7.5) A := 1965.07 (7.6) B := 13.36 (7.7) Why we can do that. the key is: > A d mCoke$cCoke > B d mL1$cL1 > C d mL2$cL2 C := 3.000 A >> (B+C), comparing to the thermal "mass" of the coke, the cup and the lid is neglectable (7.8) >> means 20 times larger for resistance > R1; R2 CR3; R4; R5; R6 CR7; R8 0.8000000000 0.2400000000 0.08000000000 4.000000000 0.07333333334 0.4000000000 We cannot neglect the resistance of the cup & the lid (7.9) You can also solve the question by hand Here is the manual procedures > equ4 d Tair KTcoke t $diff Tcoke t , t ; 1 1 C R1 CR2 CR3 CR4 R5 CR6 CR7 CR8 equ4 := 20 KTcoke t 1.116404088 = 1965.07 d Tcoke t dt = mCoke$cCoke (8.1) where 1 1 C ; b d mCoke$cCoke; R5 CR6 CR7 CR8 R1 CR2 CR3 CR4 a := 1.116404088 b := 1965.07 dTcoke t a > = dt ; 20 KTcoke t b dTcoke t = 0.0005681243355 dt 20 KTcoke t > ad (8.2) (8.3)
    • Integral both side > K 20 KTcoke t ln at b K > 20 KTcoke t = e a t Cc : b = c $eK : at b K > Tcoke t = 20Kc1$e 0.0005088877241 at Tcoke t = 20 Kc1 eK Initial conditions Tcoke(0)=3 (8.4) a K $0 b > 3 = 20Kc1$e 3 = 20 K1.0 c1 (8.5) c1 := 17 (8.6) > c1 d 17 The model now is a$t b K > Tcoke d t/20 K c1$e at b K Tcoke := t/20 Kc1 e (8.7) Verify > e d 2.71828; e := 2.71828 (8.8) 9.80495532 (8.9) > Tcoke 900 So, this is also an exam question, now you can do it easily :)