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# Week 5 [compatibility mode]

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### Week 5 [compatibility mode]

1. 1. KNF1023 Engineering Mathematics II Application of First Prepared By Order ODEs Annie ak JosephPrepared ByAnnie ak Joseph Session 2008/2009
2. 2. Learning Objectives Apply the first order ODEs in mixing problems Apply the Torricelli’s law in the real life application
3. 3. Example of application: MixingproblemsThe tank in the figure below contains 1000 galof water in which initially 100lb of salt is dissolved.Brine runs in at a rate of 10 gal/min, and eachGallon contains 5 lb of dissolved salt. The mixturein the tank is kept uniform by stirring. Brine runsout at 10 gal/min. Find the amount of salt in thetank at any time t.
4. 4. Solution: Step1. Setting up a model.Let y(t)denote the amount of salt in the tankat time t (unit is lb/1000gal). Its time rate ofchange is dy = Salt inflow rate - Salt outflow rate dt “Balance Law”The amount of salt entering the tank is (5 × 10)lb/min which is 50lb/min.
5. 5. Continue…The amount of salt flow out the tank is  y (t )  lb/min.  ×10  1000 Now, the outflow is 10 gal of brine. This is 10/1000= 0.01 (=1%) of the total brine content in thetank, hence 0.01 of the salt content y(t), that is,0.01y(t). Thus the model is the ODE dy = 50 − 0.01 y (t ) = −0.01( y (t ) − 5000) dt
6. 6. Step 2. Solution of the model.The ODE is separable. Separation, integration,And taking exponents on both sides gives dy = −0.01dt y − 5000 ln y − 5000 = −0.01t + c * −0.01t + c * y − 5000 = e −0.01t y − 5000 = ce
7. 7. Continue…Initially the tank contains 100 lb of salt. Hencey(0)=100 is the initial condition that will give theunique solution. Substituting y=100 and t=0 inThe last equation gives .Hence c=-4900. HenceThe amount of salt in the tank at time t is −0.01t y (t ) = 5000 − 4900e
8. 8. Example 2: Mixing problems inCascade TankTwo tanks are cascaded as in Figure 1. Tank1 initially contains 30 kilograms of saltdissolved in 100 liters of brine, while tank 2contains 150 liters of brine in which 70 kilogramsOf salt are dissolved. At time zero, a brine solutioncontaining ½ kilogram of salt per liter is added totank 1 at the rate of 10 liters per minute. Tank 1has an output that discharges brine into tank 2 atthe rate of 5 liters per minute, and tank 2 has anoutput of 15 liters per minute. Determine theamount of salt in each tank at any time t (y1(t) andy2(t)).
9. 9. Figure 1
10. 10. Solution:Let y1(t) denotes the amount ofsalt in tank 1 at time t. dy salt inflow salt outflowRate of change : = − dt rate rateSalt entering tank 1 :  l   1 kg  10 ×  = 5 kg/min  min   2 l 
11. 11. Continue…Salt out of tank 1 :  y1 (t ) kg   l   ×5  = 0.05 y1 kg/min  100 l   min  dy1 = 5 − 0.05 y1 dt = −0.05 ( y1 − 100 ) dy1 ∫ y1 − 100 = ∫ −0.05dt ln y1 − 100 = −0.05t + c1 y1 − 100 = exp ( −0.05t + c1 ) = ce −0.05t ( where c = ec1 )
12. 12. Continue…Initially, the tank contains 30kg of salt. Hence, y1 ( 0 ) = 30 −0.05( 0 ) 30 − 100 = ce c = −70 −0.05t ∴ y1 (t ) = 100 − 70e − t / 20 = 100 − 70e
13. 13. Continue…Let y2(t) denotes the amount of salt in tank2 at time t.Rate of change : Salt entering tank 2-Salt out of tank 2Salt entering tank 2: = salt out tank 1 = 0.05 y1 = 0.05 (100 − 70e − t / 20 ) kg/min − t / 20 = 5 − 3.5e
14. 14. Continue…Salt out of tank 2 :  y2 (t ) kg   l  1   ×  15 = y2 kg/min  150 l   min  10 dy2 − t / 20 1 = 5 − 3.5e − y2 dt 10 dy2 1 − t / 20 + y2 = 5 − 3.5e dt 10This is a non-homogeneous 1st order ODE
15. 15. Continue…Solve the homogeneous part : dy2 1 + y2 = 0 dt 10 1 1 ∫ y2 dy2 = − 10 ∫ dt 1 ln y2 = − t 10 − t /10 y2 = e
16. 16. Continue… −t /10 y2 = e ⋅v −t /10  1  −t /10 y2 = e ⋅ v+ v − e  10 Substitute into the ODE :-  1  − t /10 1 − t /10 e− t /10 ⋅ v − v   e + (e ⋅ v ) = 5 − 3.5e − t /20  10  10 − t /10 dv − t /20 e = 5 − 3.5e dt
17. 17. Continue… ∫ ∫  5 e − 3.5 e dt t /10 t / 20 dv =  t /10 t / 20 v = 50 e − 70 e + c2∴ y2 = e − t /10 ( 50 e t /10 − 70 e t / 20 + c2 ) − t / 20 − t /10 = 50 − 70 e + c2 e
18. 18. Continue…Initially, the tank contains 70kg of salt. Hence, y2 ( 0 ) = 70 ( 0) ( 0) 70 = 50 − 70e + c2 e c2 = 90 − t / 20 − t /10 ∴ y2 = 50 − 70e + 90e
19. 19. Leaking Tank. Outflow of WaterThrough a Hole (Torricelli’s Law)The law concerns the outflow of water from acylindrical tank with a hole at the bottom. You areasked to find the height of the water in the tank atany time if the tank has diameter 2 m, the holehas diameter 1 cm, and the initial height of thewater when the hole is opened is 2.25 m. When willthe tank be empty?
20. 20. Physical informationUnder the influence of gravity the out-flowingWater has velocity v(t ) = 0.600 2 gh(t ) (Torricelli’s Law)Where h(t) is the height of the water above thehole at time t, g = 980cm / s 2 = 32.17 ft / s 2 and isthe acceleration of gravity at the surface of theearth.
21. 21. Solution. Step 1. Setting up themodel.To get an equation, we relate the decrease inWater level h(t) to the outflow. The volume ∆Vof the outflow during a short time ∆t is ∆V = Av∆t (A=Area of hole)∆V must equal the change ∆V* of the volume ofthe water in the tank. Now (B=Cross-sectional ∆V * = − B∆h area of tank)
22. 22. Continue…Where ∆h(>0) is the decrease of the heighth(t) of the water. The minus sign appearsbecause the volume of the water in the tankdecreases.Equating ∆V and ∆V * gives − B∆h = Av∆t
23. 23. Continue…Now we express v according to Torricelli’s LawAnd then let ∆t (the length of the time intervalconsidered) approach 0 – this is a standard wayOf obtaining an ODE as a model. That is, weHave ∆h A A = − v = − 0.600 2 gh(t ) ∆t B BAnd by letting ∆t→0 we obtain the ODE dh A = −26.56 h dt B
24. 24. Continue…Where 26.56 = 0.600 2 × 980This is our model, a first-order ODE.
25. 25. Step 2. General solution.Our ODE is separable. A/B is constant. Separationand integration gives dh A = −26.56 dt h B 1 − A ∫ h dh = − ∫ 26.56 B dt 2 1 2 h A = −26.56 t + c 1 B 2 A 2 h = −26.56 t + c B
26. 26. Continue… A 2 h = −26.56 t + c B A cDividing by 2 and squaring givesh = ( −13.28 t + ) 2 B 2 cHere c = 2 AInserting 13 .28 = 13 .28 × 0 .5 2 π 2 = 0 .000332 yields B 100 πThe general solution 2 h(t ) = (c − 0.000332t )
27. 27. Step 3. Particular solution.The initial height (the initial condition) ish(0)=225 cm. Substitution of t=0 and h=225 2Gives from the general solution (c ) = 225, c = 15.00and thus the particular solution 2 hp (t ) = (15.00 − 0.000332t )
28. 28. Step 4. Tank empty.h p (t ) = 0 if t=15.00/0.000332 = 45181 s= 12.6 hours
29. 29. Prepared By Annie ak JosephPrepared ByAnnie ak Joseph Session 2007/2008