Week 5 [compatibility mode]
- 1. KNF1023 Engineering Mathematics II Application of First Prepared By Order ODEs Annie ak Joseph Prepared By Annie ak Joseph Session 2008/2009
- 2. Learning Objectives Apply the first order ODEs in mixing problems Apply the Torricelli’s law in the real life application
- 3. Example of application: Mixing problems The tank in the figure below contains 1000 gal of water in which initially 100lb of salt is dissolved. Brine runs in at a rate of 10 gal/min, and each Gallon contains 5 lb of dissolved salt. The mixture in the tank is kept uniform by stirring. Brine runs out at 10 gal/min. Find the amount of salt in the tank at any time t.
- 4. Solution: Step1. Setting up a model. Let y(t)denote the amount of salt in the tank at time t (unit is lb/1000gal). Its time rate of change 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. 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 the tank, 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. 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. Continue… Initially the tank contains 100 lb of salt. Hence y(0)=100 is the initial condition that will give the unique solution. Substituting y=100 and t=0 in The last equation gives .Hence c=-4900. Hence The amount of salt in the tank at time t is −0.01t y (t ) = 5000 − 4900e
- 8. Example 2: Mixing problems in Cascade Tank Two tanks are cascaded as in Figure 1. Tank 1 initially contains 30 kilograms of salt dissolved in 100 liters of brine, while tank 2 contains 150 liters of brine in which 70 kilograms Of salt are dissolved. At time zero, a brine solution containing ½ kilogram of salt per liter is added to tank 1 at the rate of 10 liters per minute. Tank 1 has an output that discharges brine into tank 2 at the rate of 5 liters per minute, and tank 2 has an output of 15 liters per minute. Determine the amount of salt in each tank at any time t (y1(t) and y2(t)).
- 9. Figure 1
- 10. Solution: Let y1(t) denotes the amount of salt in tank 1 at time t. dy salt inflow salt outflow Rate of change : = − dt rate rate Salt entering tank 1 : l 1 kg 10 × = 5 kg/min min 2 l
- 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. 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. Continue… Let y2(t) denotes the amount of salt in tank 2 at time t. Rate of change : Salt entering tank 2-Salt out of tank 2 Salt entering tank 2: = salt out tank 1 = 0.05 y1 = 0.05 (100 − 70e − t / 20 ) kg/min − t / 20 = 5 − 3.5e
- 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 10 This is a non-homogeneous 1st order ODE
- 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. 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. 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. 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. Leaking Tank. Outflow of Water Through a Hole (Torricelli’s Law) The law concerns the outflow of water from a cylindrical tank with a hole at the bottom. You are asked to find the height of the water in the tank at any time if the tank has diameter 2 m, the hole has diameter 1 cm, and the initial height of the water when the hole is opened is 2.25 m. When will the tank be empty?
- 20. Physical information Under the influence of gravity the out-flowing Water has velocity v(t ) = 0.600 2 gh(t ) (Torricelli’s Law) Where h(t) is the height of the water above the hole at time t, g = 980cm / s 2 = 32.17 ft / s 2 and is the acceleration of gravity at the surface of the earth.
- 21. Solution. Step 1. Setting up the model. To get an equation, we relate the decrease in Water level h(t) to the outflow. The volume ∆V of the outflow during a short time ∆t is ∆V = Av∆t (A=Area of hole) ∆V must equal the change ∆V* of the volume of the water in the tank. Now (B=Cross-sectional ∆V * = − B∆h area of tank)
- 22. Continue… Where ∆h(>0) is the decrease of the height h(t) of the water. The minus sign appears because the volume of the water in the tank decreases. Equating ∆V and ∆V * gives − B∆h = Av∆t
- 23. Continue… Now we express v according to Torricelli’s Law And then let ∆t (the length of the time interval considered) approach 0 – this is a standard way Of obtaining an ODE as a model. That is, we Have ∆h A A = − v = − 0.600 2 gh(t ) ∆t B B And by letting ∆t→0 we obtain the ODE dh A = −26.56 h dt B
- 24. Continue… Where 26.56 = 0.600 2 × 980 This is our model, a first-order ODE.
- 25. Step 2. General solution. Our ODE is separable. A/B is constant. Separation and 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. Continue… A 2 h = −26.56 t + c B A c Dividing by 2 and squaring givesh = ( −13.28 t + ) 2 B 2 c Here c = 2 A Inserting 13 .28 = 13 .28 × 0 .5 2 π 2 = 0 .000332 yields B 100 π The general solution 2 h(t ) = (c − 0.000332t )
- 27. Step 3. Particular solution. The initial height (the initial condition) is h(0)=225 cm. Substitution of t=0 and h=225 2 Gives from the general solution (c ) = 225, c = 15.00 and thus the particular solution 2 hp (t ) = (15.00 − 0.000332t )
- 28. Step 4. Tank empty. h p (t ) = 0 if t=15.00/0.000332 = 45181 s = 12.6 hours
- 29. Prepared By Annie ak Joseph Prepared By Annie ak Joseph Session 2007/2008