CERTIFICADO DE ORIGEN



 +

Substitute the initial conditions [y(0)=10, y(1)=6] in the above equation we get: C₁ = −0.0693 and K = −0.02877

(b) The time to dissolve 80% of A means that the undissolved amount is 20%, then y 2= kg

0693 . 0 02877 .

10 0 10

1  = − −

 





+ t

y Ln y Q

0693 . 0 02877 .

10 0 2

2 10

1  = − −



 





+ t

Q Ln

hours t =3.82

∴

4.4.4 Water Tanks

Example 24 A circular cylinder of radius 10 m and height 25 m whose axis is vertical as shown in Fig.2, is filled with water. How long will it take for all the water to escape through an orifice with 50 cm radius at the bottom of the tank?

Assuming the velocity of escape v in terms of instantaneous height h is given by v =0.6 2gh.

Chapter Four 129 Solution: Assume incremental volume dV will take time dt to escape from the tank. So from the geometry of the tank the incremental volume is: dV = −

π

*r² dh ₍₅₇₎ Where, dh is the height of the incremental volume dV .

The minus sign has to be in the above equation because the height of the water decreases with time.

The incremental volume dV will take time dt to escape from the orifice of the tank at speed v. So, the incremental Volume dV can be calculated also from the following equation:

dt v A

dV = . . (58)

Where A is the area of the orifice. So if the orifice is circle with radius of r_o the A=

π

r_o². And assume v =0.6 2gh. Then the incremental volume is given by the following equation:

dt gh r

dV =

π

_o²*0.6 2 (59)

by equating (57) and (59) we get the following equation:

dt gh r

dh

r _o *0.6 2

* ²

π

²

π

=

−

The above differential equation can be solved by separation of

variables. dt

r gh r h

dh _o

2 2

2 6 .

−0

=

∴

By integrating both sides of the above equation we get the

following: dt c

r gh r h

dh =

∫

− _o +

∫

^{0.6 2} ₂²

Where c is the constant of integration, c

r t g r

h =− ^o +

∴2 0.6 2 ₂² (60)

Ordinary Differential Equations 130

R R r h

R-h

dh

Fig.3

So, at time t =0 h= H_o. Substitute this condition in the above equation we get the following values for c. 2 H_o = 0+c

Ho

c 2=

∴ Substitute this value in (60) we get the final

results: ^o t H_o

r g r

h 0.6 2 2

2 =− ₂² +

To get the time required for tank to get empty we can substitute in the above equation for h =0,

2 2/ 2

6 . 0

2

r r g t H

o e = o

∴ (61)

So, we can substitute the data in our example in the above equation to obtain the time for the tank to get empty, t_e.

. min 08 . 25 . sec 10 1505

/ 5 . 0

* 81 . 9

* 2 6 . 0

25 2

2

2 = =

e = t

Example 25 A spherical tank of radius R=100 cm which contains water and has outlet of radius

cm

r_o

= 5

at the bottom of the tank as

shown in Fig.3. At time t

= 0

the outlet is

opened and the water flows out. Determine the time when the tank will be empty, assuming the initial height of water

( )

R cm

h 0 + =100 . The velocity with which liquid issues from an orifice is v

= 0 . 6 2

gh, where (g = 9.81m/s²) is the acceleration of the gravity.

Chapter Four 131 Solution: Let the origin be chosen at the lowest point of the tank and let h be the instantaneous depth, V the instantaneous volume, and r the instantaneous radius of the free surface of the water as shown in Fig.3. Then in an infinitesimal time dt, the water level will fall by the amount dh, and the resultant decrease in volume of the water in the tank will be:

dh r

dV =−

π

² (62)

Now by Torricelli’s law the velocity with which a liquid issues from an orifice is: v= 0.6 2gh where g is the acceleration of gravity and h is the instantaneous height, or head, of the liquid above the orifice. In the interval dt, then, a stream of water of length

dt gh

vdt =0.6 2 and of cross-sectional area

π

r_o² will emerge from the outlet. The volume of this amount of water is:

gh r

length

area* =

π

_o²*0.6 2 (63)

So from the above equation and (62) we can get the following equation: −

π

r²dh =

π

r_o²*0.6 2gh dt (64) Before this equation be solved, r must be expressed in terms of h.

This is easily done through the use of equation of the circle which describes the vertical cross section of the tank:

( )

^{2 2 2 2}

2 h R R or r 2hR h

r + − = = −

∴ (65)

With this, the differential equation (65) can be written as:

(

2^{hR h2}

)

^dh

π

^ro²*0.6 2^{gh dt}

π

− =−

∴ (66)

Ordinary Differential Equations 132

This is a simple separable equation.

(

2^Rh1/2 −^h3/2

)

^dh= −^ro²*0.6 2^gdt

To fiend how long it will take the tank to empty, we must determine

the value of t when h=0. ^{2 5/2} Substitute these values in the above equation we get:

.

Example 26 The tank shown in Fig.4 is consists of two portions, the top one is cylinder with 20 m height and 10m radius of its base and the other portion is half sphere with 10 m radius and has 50 cm² outlet area at the bottom. If this tank filled with water and at time t=0 the outlet is opened and the water flows out. Determine the time when the tank will be empty, assuming the tank was initially filled with water. The velocity with which liquid issues from an orifice is

gh

v =0.6 2 , where (g =9.81m/s²) is the acceleration of the gravity.

Chapter Four 133 Solution: Let the origin be chosen at the lowest point of the tank and let h be the instantaneous depth, V the instantaneous volume, and r the instantaneous radius of the free surface of the water as shown in Fig.4.

Let us start first with finding the time required for the cylindrical portion to get empty. Then in an infinitesimal time dt, the water level will fall by the amount dh, and the resultant decrease in volume of the water in the tank

will be:

dh dV =−

π

10²

Now by torricelli’s law the velocity with which a liquid issues from an orifice is:

gh

v

= 0 . 6 2

where g is the acceleration of gravity and h is the instantaneous height, or head, of the liquid above the orifice. In the interval dt.

The unit volume escape from the orifice at time dt is:

dt gh

dV =0.6 2 *50*10⁻⁴

So, by equating both volumes we get the following equation:

dh dt

gh

* 50 * 10

⁴

10

²

2

6 .

0

⁻

= − π

, dt

h

dh ₅

10

* 23 .

4 ⁻

−

=

∴

10m

20m

gh v=0.6 2

50 cm2

Area=

Fig.4

Ordinary Differential Equations 134

It is clear that the above equation is first order differential equation in separable form, so it is easy to solve this equation by integrating both sides.

c h dt

dh =

∫

− +

∫

^{4.23*10−5 ∴ 2 h = −4.23*10−5t +c}

It is clear from the initial condition that at t =0, h = 20+10=30m Substitute this initial condition in the above equation we get:

+c

−

= 4.23*10⁻ *0 30

2 ⁵ ∴ c= 2 30 =10.954

So, the solution of differential equation becomes, 954

. 10 10

* 23 . 4

2 h =− ⁻⁵t +

At the bottom of the cylinder portion h 10

=

m so we can obtain the time required for the cylindrical portion to get empty if we substitute in the solution of differential equation for h 10

=

m. Then,

954 . 10 10

* 23 . 4 10

2 = − ⁻⁵t + hours t_cylinder =30.4

∴

In the same way we can obtain the time required for spherical portion to get empty. We use also unit element in spherical portion

dh r

dV =−

π

² . But r = 10² −

(

10−h

)

² = 20h−h²

(

^{h h}

)

^dh

dV =− 20 − ²

∴

π

This volume will take time dt to escape from the orifice of the tank.

dt gh

dV =0.6 2 *50*10⁻⁴

∴

Chapter Four 135

(

^20h−h2

)

^{dh= 0.6 2gh*50*10−4dt}

−

∴

π

∴

_∫ (

^{20h0.5 −h1.5}

)

^{dh= −0.0042298dt +k}

∴ h t k

h − =−0.0042298 + 5

. 2 5

. 1

20 _1.5 ^2.5

At t =30.4*3600sec, h =10m

∴ − = −0.0042298*30.4*3600+k 5

. 2 10 10

5 . 1

20 _1.5 ^2.5

Then, k = 758.055

∴ 0.0042298*30.4*3600 758.055 5

. 2 10 10

5 . 1

20 _1.5 − ^2.5 = − +

The total time required for the total tank to get empty can be obtained by applying h =0 in the above equation. Then,

∴ t =179218 sec= 49.7827hours Another Solution

For cylindrical portion we can use (60) to get the time required for this portion to get empty

c r t

g r

h =− ^o +

∴2 0.6 2 ₂² (60)

So, at time t

= 0

h 30

=

m. Substitute this condition in the above equation we get the following values for c.

+c

=

∴2 30 0 ∴c= 2 30 30 2 2

6 . 0

2 ₂

2 +

−

=

∴ t

r g r

h ^o

Ordinary Differential Equations 136

To get the time required for cylindrical portion of the tank to get empty we can substitute in the above equation for h 10

=

m. Then,

In the same way we can obtain the time required for spherical portion to get empty. We use (67) as explained before.

c

Chapter Four 137 Example 27 A right circular

cylinder of radius 8m and Length of 16m as shown in Fig.5, whose horizontal axis, is filled with water. How long will it take for all the water to escape through circular orifice has 10 cm radius at the bottom of the tank? Assume v= 0.6 2gh Solution: In this case we will

calculate the time required for the upper half to get empty. Then we can do the same for the lower half.

First, For upper half:

The horizontal incremental volume can be taken in the upper half of the tank. It is clear from geometry of the tank, the width of the incremental volume,dV is

2

r and the length L of the cylinder.

L

h-R R r

h

dh

Fig.6 16m

16m

Fig.5

Ordinary Differential Equations 138

It is clear from Fig.6 that ^r = ^R2 −

(

^h−^R

)

² = ^2Rh−^h2

So the incremental volume dV is given by the following equation:

dh L h Rh

dV =−2* 2 − ² * (69)

Also, this volume dV can be calculated at the orifice of the tank as following: dV =

π

r_o²*0.6 2gh dt (70) By equating (69) and (70) we get the following result:

dt gh r

dh L h

Rh * _o *0.6 2

2

*

2 − ² =

π

²

−

∴

dt g L r

dh h

R _o * 2

2 − = −2

π

²

∴

By integrating both sides of the above equation we can get the following equation:

( )

r g t c

dh L h

R− = − _o +

∴ * 2

2 2 3

2 _3/2

π

₂

(71) Where c is the constant of integration.

At time t =0, h=2R Substitute this condition in the above equation we get the value of the constant c,

= 0

∴ c (72)

Substitute from (72) into (71).

( )

r g t

h L

R _o * 2

2 2 3

2 − _3/2 = −

π

₂

∴ (73)

To obtain the time required for the upper half to get empty we have to put h

=

R in the above equation, (73).

g r

t LR

o

u 3 2

4

2 2 / 3

=

π

∴ (74)

Chapter Four 139 For the lower half

The horizontal incremental volume dV can be taken in the upper half of the tank. It is clear from geometry of the tank, the width of the incremental volume,dV is

2

r and the length L of the cylinder.

It is clear from Fig.7 that: r = R² −

(

R−h

)

² = 2Rh−h²

So the incremental volume dV is given by the following equation:

dh L h Rh

dV = −2* 2 − ² * (75)

Also, this volume dV can be calculated at the orifice of the tank as following: dV =

π

r_o²*0.6 2gh dt (76) By equating (75) and (76) we get the following result:

dt gh r

dh L h

Rh * _o *0.6 2

2

*

2 − ² =

π

²

−

dt g L r

dh h

R _o * 2

2 − = −2

π

²

∴

L

R-h h r

R dh

Fig.7

By integrating both sides of the above equation we can get the

following equation:

( )

r g t c

h L

R− = − _o +

∴ * 2

2 2 3

2 _3/2

π

₂

(77)

Ordinary Differential Equations 140

Where c is the constant of integration. At time R

π

substitute this condition in the above equation we get the value of the constant c,

( )

c

The total time required for the whole tank to get empty can be obtained by Substituting in the above equation for h

= 0

.

( )

So the time required for the lower portion only to get empty is given by the following equation: t_L

=

t_total

−

t_u (80) The data collected from our example is r_o

= 0 . 1

m, R 8

=

m, and

m

L 16

=

. Apply these data to equation (74), (79), and, (80) respectively we get the following results:

., 81 3468 . 81 9811 .

Chapter Four 141 Example 28 A conical tank shown in Fig.8, filled with water and has outlet of radius r_o = 20cm

at the bottom. At time t=0 the outlet is opened and the water flows out. Determine the time required for the tank to get empty assuming the tank was initially completely filled with water. The velocity with which liquid issues from an orifice is

gh

v =0.6 2 , where (g =9.81m/s²) is the acceleration of the gravity.

Solution: Assume the radius of the base of the tank is R and its height is L. Assume a horizontal incremental volume dV at height h from the bottom of the tank and radius r. The thickness of the incremental volume dV is dh. It is clear from the geometry of the tank that the incremental volume looks like a cylinder with radius r and height dh. So, the incremental volume can be obtained as

following: dV

= − π

r²dh (81)

But L R

hr = h L r = R

∴ (82)

Substitute the value of r in (82) into (81) we get the following result:

dh Lh

dV R

2

* 



 



− 

=

π

(83)

20 m

gh v=0.6 2

cm Radius 20=

10 m

θ

Fig.8

h dh

Ordinary Differential Equations 142

In the same way dV can be obtained at the bottom of the tank as following: dV

=

πr_o²

* 0 . 6 2

gh dt (84)

By equating (83) and (84) we get the following result:

dt

By integrating both sides of the above equation we get the

following: t c

where c is the constant of integration

At time t = 0, h= . Substitute this initial condition into (85): L

Substitute the value of c in (86) into (85) we get the following:

2

So the time required for the tank to get empty is given by applying for h

= 0

in the above equation. So,

Chapter Four 143 4.4.5 Half Life Of Nuclear Materials

Example 29 Experiment shows that radium disintegrates at a rate proportional to the amount of radium instantly present. Its half life (that is the time in which 50% of a given amount will disappear), is 1590 years. What percent will disappear in one year?

Solution: Assume the instantaneous amount and starting amount of radium exists are y and y^o respectively.

dt ky dy =

∴ (88)

Where k is the proportional constant. Equation (88) can be easily solved by variable separation. ∴y =Ce^kt (89) Where C is integration constant. C and k can be determined from initial conditions. The first initial condition is y

=

y₀ at t

= 0

substitute this condition into (89).

ekt

y y = ₀

∴ (90)

The second initial condition is y

=

y₀

/ 2

at t

= 1590

years. Substitute this condition into (90)

1590 0 *

0

2

ek

y = y

∴ ∴ k =−4.36 *10⁻⁴

e t

y

y = ₀ ^{−4.36*10−4}

∴ (91)

To obtain the percent disappear after one year we can do the following:

% 0436 . 0 100

* 1

100 ) *

1 1 (

0

1

* 10

* 36 . 0 4

0

4

 =











 −

 =



 



 −

− −

y e y y

y

Ordinary Differential Equations 144

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