Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2008/2009

KNF1023Engineering

Mathematics II

Application of First Order ODEs

Learning Objectives

Apply the Torricelli’s law in the real life application

Apply the first order ODEs in mixing problems

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.

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

= Salt inflow rate - Salt outflow rate

“Balance Law”

The amount of salt entering the tank is

lb/min which is 50lb/min.

dt

dy

)105( ×

Continue…

The amount of salt flow out the tank is

lb/min.

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

×10

1000

)(ty

)5000)((01.0)(01.050 −−=−= tytydt

dy

Step 2. Solution of the model.

The ODE is separable. Separation, integration,

And taking exponents on both sides gives

0.015000

dydt

y= −

−

ln 5000 0.01 *y t c− = − +

*01.05000 ctey

+−=−

0.015000 ty ce

−− =

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

0.01( ) 5000 4900 ty t e

−= −

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)).

Figure 1

Solution:

Let y1(t) denotes the amount of

salt in tank 1 at time t.

Rate of change :

Salt entering tank 1 :

salt inflow salt outflow

rate rate

dy

dt= −

1 kg10 5 kg/min

min 2

l

l

× =

Continue…

Salt out of tank 1 :

11

( ) kg5 0.05 kg/min

100 min

y t ly

l

× =

( )

( )

( )1

11

1

1

1

1 1

1 1

0.05

5 0.05

0.05 100

0.05100

ln 100 0.05

100 exp 0.05

where ct

dyy

dt

y

dydt

y

y t c

y t c

ce c e−

= −

= − −

= −−

− = − +

− = − +

= =

∫ ∫

Continue…

Initially, the tank contains 30kg of salt.

( )( )

1

0.05 0

0.05

1

/ 20

Hence, 0 30

30 100

70

( ) 100 70

100 70

t

t

y

ce

c

y t e

e

−

−

−

=

− =

= −

∴ = −

= −

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:

( )1

/ 20

/ 20

salt out tank 1

0.05

0.05 100 70 kg/min

5 3.5

t

t

y

e

e

−

−

=

=

= −

= −

Continue…

Salt out of tank 2 :

This is a non-homogeneous 1st order ODE

22

( ) kg 115 kg/min

150 min 10

y t ly

l

× =

/ 2022

/ 2022

15 3.5

10

15 3.5

10

t

t

dye y

dt

dyy e

dt

−

−

= − −

+ = −

Continue…

Solve the homogeneous part :

22

2

2

2

/10

2

10

10

1 1

10

1 ln

10

t

dyy

dt

dy dty

y t

y e−

+ =

= −

= −

=

∫ ∫

Continue…

Substitute into the ODE :-

/10

2

/10 /10

2

1' '

10

t

t t

y e v

y e v v e

−

− −

= ⋅

= ⋅ + −

( )/10 /10 /10 /20

/10 /20

1 1' 5 3.5

10 10

5 3.5

t t t t

t t

e v v e e v e

dve e

dt

− − − −

− −

⋅ − + ⋅ = −

= −

Continue…

( )

/10 / 20

/10 / 20

2

/10 /10 / 20

2 2

/ 20 /10

2

5 3.5

50 70

50 70

50 70

t t

t t

t t t

t t

dv e e dt

v e e c

y e e e c

e c e

−

− −

= −

= − +

∴ = − +

= − +

∫ ∫

Continue…

Initially, the tank contains 70kg of salt.

( )( ) ( )

2

0 0

2

2

/ 20 /10

2

Hence, 0 70

70 50 70

90

50 70 90t t

y

e c e

c

y e e− −

=

= − +

=

∴ = − +

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?

Physical information

Under the influence of gravity the out-flowing

Water has velocity

(Torricelli’s Law)

Where h(t) is the height of the water above the

hole at time t, and is

the acceleration of gravity at the surface of the

earth.

( ) 0.600 2 ( )v t gh t=

22 /17.32/980 sftscmg ==

Solution. Step 1. Setting up the model.

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

(A=Area of hole)

∆V must equal the change ∆V* of the volume ofthe water in the tank. Now

(B=Cross-sectional area of tank)

V Av t∆ = ∆

*V B h∆ = − ∆

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− ∆ = ∆

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

And by letting ∆t→0 we obtain the ODE

0.600 2 ( )h A A

v gh tt B B

∆= − = −

∆

26.56dh A

hdt B

= −

Continue…

Where

This is our model, a first-order ODE.

9802600.056.26 ×=

Step 2. General solution.

Our ODE is separable. A/B is constant. Separation

and integration gives

26.56dh A

dtBh

= −

1

2 26.56A

h dh dtB

−

= −∫ ∫1

2

26.561

2

h At c

B= − +

2 26.56A

h t cB

= − +

Continue…

Dividing by 2 and squaring gives

Here

Inserting yields

The general solution

2 26.56A

h t cB

= − +

2

cc =

000332.0100

5.028.1328.13

2

2

=×=π

π

B

A

2( 13.28 )2

A ch t

B= − +

2( ) ( 0.000332 )h t c t= −

Step 3. Particular solution.

The initial height (the initial condition) is

h(0)=225 cm. Substitution of t=0 and h=225

Gives from the general solution

and thus the particular solution

( ) 00.15,2252

== cc

2( ) (15.00 0.000332 )ph t t= −

Step 4. Tank empty.

if t=15.00/0.000332 = 45181 s

= 12.6 hours

0)( =th p

Prepared By

Annie ak Joseph

Prepared ByAnnie ak Joseph Session 2007/2008