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51_Example_1_mw

# 51_Example_1_mw - KaRE TExT Unit 51 Example 1 O restart...

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O O O O O O O O O O O O O O O O O O O O O O O KaRE TExT Unit 51, Example 1 restart : Steady state solution in mol, m, min cal and K R d 1.987 : pi d 3.1417 : diam d 0.1 : L d 5.0 : T0 d 30.0 C 273.15 : Tinitial d 25.0 C 273.15 : cAin d 1.0 \$ 1000 : cBin d 1.2 \$ 1000 : vFlow d 75 1000 : nAin d cAin \$ vFlow : nBin d cBin \$ vFlow : dH d K 10700 : cP d 1.0 : dens d 10 6 : k0 d 8.72 \$ 10 5 1000. : E d 7200 : Enter the rate expression and the design equations (a lower case t preceeds each of the variables to denote it as the transient solution) r d k0 \$ exp K E R \$ tT z , t \$ tnA z , t vFlow \$ tnB z , t vFlow : tbalA d v v z tnA z , t = K pi \$ diam 2 4 \$ r K pi \$ diam 2 4 \$ vFlow \$ v v t tnA z , t : tbalB d v v z tnB z , t = K pi \$ diam 2 4 \$ r K pi \$ diam 2 4 \$ vFlow \$ v v t tnB z , t : tbalY d v v z tnY z , t = pi \$ diam 2 4 \$ r K pi \$ diam 2 4 \$ vFlow \$ v v t tnY z , t : tbalZ d v v z tnZ z , t = pi \$ diam 2 4 \$ r K pi \$ diam 2 4 \$ vFlow \$ v v t tnZ z , t : tbalT d v v z tT z , t = K pi \$ diam 2 \$ dH 4 \$ dens \$ cP \$ vFlow \$ r K pi \$ diam 2 4 \$ vFlow v v t tT z , t :

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O O O (1) O O O O O O O Then set up the initial and boundary conditions ibcA d tnA z , 0 = 0.0, tnA 0, t = nAin : ibcB d tnB z , 0 = 0.0, tnB 0, t = nBin : ibcY d tnY z , 0 = 0.0, tnY 0, t = 0.0 : ibcZ d tnZ z , 0 = 0.0, tnZ 0, t = 0.0 : ibcT d tT z , 0 = Tinitial, tT 0, t = T0 : pdes d tbalA , tbalB , tbalY , tbalZ , tbalT : ibcs d ibcA union ibcB union ibcY union ibcZ union ibcT : pdsoln d pdsolve pdes , ibcs , numeric , time = t , range = 0 .. L , spacestep = 0.01,
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