ODE Examples Lecture Continue

7/29/2019 ODE Examples Lecture Continue

1/19

Prof.Dr. Nadeem FerozeSana Zahid


2/19

The equation M(x, y) dx + N(x, y) dy = 0 isexact if and only if M/y =N/x.

In this case there exists a function w = f(x, y) such that f /x = M, f /y = N,

and f(x, y) = C is the required solution.


3/19

0),(),( dyyxNdxyxM

dFdyy

Fdx

x

F

General solution: F (x,y) = C

x

N

y

M


4/19

f(x, y) is found as follows: evaluate M(x, y) dx treating y constant.

evaluate N(x, y) dy treating x constant. The sum of all unlike terms in these two

integrals (including no repetitions) is f(x, y).


5/19

(2xy cos x) dx + (x2 1) dy = 0 is exact asM/y = 2x , N/x =2x.

Now integrate M considering y constant. M dx = (2xy cos x) dx = x2y sin x,

Integrate N considering x constant N dy = (x2 1) dy = x2y y.

The solution isx2y sin x y = C


6/19

A differential equation of the type,

Such an equation can be solved by making the

substitution u = y/xand thereafter integrating the

transformed equation as

x

yf

dx

dy

is termed a homogeneous differential equationofthe first order.

)(vfdx

dvxv

dx

dy

Cvvf

dvx

)(ln


7/19

C6H6+Cl2C6H5Cl +HClC6H5Cl+Cl2C6H4Cl2 + HClC6H4Cl2 + Cl2C6H3Cl3 + HCl

Homogeneous Equations

ExampleLiquid benzene is to be chlorinated batch-wise by spargingchlorine gas into a reaction kettle containing the benzene. Ifthe reactor contains such an efficient agitator that all thechlorine which enters the reactor undergoes chemical

reaction, and only the hydrogen chloride gas liberatedescapes from the vessel, estimate how much chlorine mustbe added to give the maximum yield of monochlorobenzene.

The reaction is assumed to take place isothermally at 55Cwhen the ratios of the specific reaction rateconstants are: k1 = 8k2 ; k2 = 30 k3


8/19

Take a basis of 1 mole of benzene fed to the reactor andintroducethe following variables to represent the stage of system at

time ,p = moles of chlorine presentq = moles of benzene presentr = moles of monochlorbenzene present

s = moles of dichlorbenzene presentt = moles of trichlorbenzene present

Then q + r + s + t = 1and the total amount of chlorine consumed is : y = r + 2s + 3t

From the mass balances : in - out = accumulation


Example


9/19

)(....................

).........(

).........(

).....(..........0

3

32

21

1

ivddtVpsk

iiid

dsVpskprk

iiddrVprkpqk

id

dqVpqk

q

qr

q

r

k

k

dq

dr

8

81)(

1

2

v= r/q


Example

Diving equation (ii) by (i) to eliminate d


10/19

Hence the equation is homogeneous 1st orderdifferential equation

By substitution

1)(1

2 q

r

k

k

dq

dr

dqdvqv

dqdr 1

81 v

dqdvqv )87ln(7/8lnln vKq

7/8

87

q

rKq

v= r/q


11/19

=0, q=1, r=0 7/88K ).......(7

)(8 8/1v

qqr

).........(

)....(..........0

32

1

iiiddsVpskprk

id

dqVpqk

q

r

q

s

dq

ds

8240

Solution:

Diving equation (iii) by (i) to eliminate d


12/19

1st order linear differential equation can besolved by integrating factor method.

))........(21023929(

239297

240 240/18/1 viqqqs

y = r + 2s + 3t

q + r + s + t = 1 t=?

y=?

For any value of q , thevalue of r and s can be foundfrom equation (v) and (vi)

qr

qs

dqds

8240


13/19


14/19

An elevated horizontal cylindrical tank 1 mdiameter and 2 m long is insulated with asbestos

lagging of thickness l= 4 cm, and is employed as a

maturing vessel for a batch chemical process.Liquid at 95C is charged into the tank and allowedto mature over 5 days. If the data below applies,

calculated the final temperature of the liquid andgive a plot of the liquid temperature as a functionof time.


15/19

Liquid film coefficient of heat transfer (h1) = 150 W/m2C

Thermal conductivity of asbestos (k) = 0.2 W/mC Surface coefficient of heat transfer by convection and

radiation (h2) = 10 W/m2

C Density of liquid () = 103 kg/m3 Heat capacity of liquid (s) = 2500 J/kgC Atmospheric temperature at time of charging = 20C Atmospheric temperature (t) t = 10 + 10 cos (/12) Time in hours () Heat loss through supports is negligible. The thermal

capacity of the lagging can be ignored


16/19

T represents the bulk liquid temperatureTw represents the inside wall temperature of the tankTs represents the outside surface temperature of the lagging

TTw

Tst

Rate of heat loss by liquid = h1A (T - Tw)Rate of heat loss through lagging = kA/l(Tw - Ts)Rate of heat loss from the exposed surface of

the lagging = h2A (Ts - t)

Area of tank (A) = ( x 1 x 2) + 2 ( 1 / 4 x 12 ) = 2.5 m2


17/19

))......(()()( 21 itTAhTTl

kATTAh ssww

tTTs

674.0326.0

sw Tl

kTh

l

khT

11

)(2121

1 tTkhkhlhh

khtTs

At steady state, the three rates are equal:

Equating 1st two rates

Substituting the value of Tw

in the last part of the equation

Simplifying the equation


18/19


19/19

0235.092.84262.0sin89.0262.0cos08.010 eT

0 25 50 75 100 125

0

20

40

60

80

100

Temperature(oC)

5days

15C

The 2nd term can only

contribute to +/- 0.08 C and3rd term can only contributeto +/- so these terms can beneglected and the equationcan be simplified to

0235.08510 eT

ODE Examples Lecture Continue

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