This document provides an overview of applying first order ordinary differential equations (ODEs) to mixing problems and real-life applications. It includes two examples of using first order ODEs to model mixing problems, such as determining the amount of salt in a tank over time. It also applies Torricelli's law to model the outflow of water from a leaking tank and calculates the time until the tank empties. The document demonstrates setting up the mathematical models as first order ODEs, finding the general and particular solutions, and interpreting the solutions in the context of the problems.

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