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Prof.Dr. Nadeem FerozeSana Zahid
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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.
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0),(),( dyyxNdxyxM
dFdyy
Fdx
x
F
General solution: F (x,y) = C
x
N
y
M
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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).
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(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
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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
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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
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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
Homogeneous Equations
Example
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)(....................
).........(
).........(
).....(..........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
Homogeneous Equations
Example
Diving equation (ii) by (i) to eliminate d
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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
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=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
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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
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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.
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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
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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
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))......(()()( 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
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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
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