Mass and Heat Transfer (1st Edition, 2008) – Solutions Manual – Russell

CHAPTERS 2-7 COVERED




SOLUTIONS MANUAL

,PROBLEM SOLUTIONS
MASS & HEAT TRANSFER
ANALYSIS OF MASS AND HEAT EXCHANGERS

Cambridge University Press

CHAPTER 2 SOLUTIONS

Problem 2.1

2A + B ⎯
⎯→ nD

Use a mole balance on each component.

dC A dC B 1 1 dC D n n
= −rA = −kC A C B = − rA = − kC A C B = rA = kC A C B
2 2 2
A: B: D:
dt dt 2 2 dt 2 2


From equilibrium, CA – CAi = 2(CB – CBi)

C A − C Ai
CB = + C Bi
2

-For the condition when CAi = CBi:

C A − C Ai C + C Ai
CB = + C Ai = A
2 2

dC A 2 ⎛ C + C Ai ⎞
Therefore, = − kC A ⎜ A ⎟
dt ⎝ 2 ⎠


-For the condition when CAi ≠ CBi:

C A − C Ai
CB = + C Bi
2

dC A 2 ⎛ C − C Ai ⎞
Therefore, = −kC A ⎜ A + C Bi ⎟
dt ⎝ 2 ⎠


Experiments can be performed to find the concentration of component A as a function of time.
The above equations can be solved, and the data plotted to obtain the value of the rate constant k
for both cases. Note that if the CB(t) or CD(t) are more easily obtainable, the differential


1

,equations for the other components can be used and the equilibrium relationship can be
manipulated to fit these equations.




2

,Problem 2.2

Use the data given to check if r=k’CA is a good model.

dC A
= r = − k ' C A (negative since losing component A)
dt

Integration of the above equation gives:

C A (t )
ln = −k ' t
C Ai

which equals:

ln C A (t ) = −k ' t + ln C Ai

Plot ln CA vs. t. Slope will give k’ if the data are linear.

1.8


1.7


1.6
y = -0.0026x + 1.6872
R2 = 0.992
1.5
ln CA




1.4


1.3


1.2


1.1


1
0 50 100 150 200 250
Tim e (m in)




Therefore, k’=0.0026 min-1. The plot is linear, so this is a good model.




3

,Problem 2.3




CB t
dCB
i.e., ∫
CB 0 K eCB + { K e (C A0 − CB 0 ) + 1} CB − CB 0
2
= − k2 ∫ dt
0
....(2)



4

, a2
dx
Left hand side of (2) is of the form ∫ ax
a1
2
+ bx + c
which is equal to


1 ⎛ 2ax + b − b 2 − 4ac ⎞
ln ⎜ ⎟ if b 2 − 4ac > 0 (which is certainly the case here).
b − 4ac ⎝ 2ax + b + b − 4ac ⎟⎠
2 ⎜ 2




Integration of (2) gives,

CB
⎡ 1 ⎛ 2 K eCB + K e (C A0 − CB 0 ) + 1 − q ⎞ ⎤
⎢ ln ⎜ ⎟ ⎥ = − k2t
⎢⎣ q ⎜⎝ 2 K eCB + K e (C A0 − CB 0 ) + 1 − q ⎟⎠ ⎥⎦ CB 0


( where q = { K e (C A0 − CB 0 ) + 1} + 4 K eCB 0 )
2




⇒
1
ln ⎜
{
⎛ 2 K eCB + K e (C A0 − CB 0 ) + 1 − q } {2 K Ce B0 }
+ K e (C A0 − CB 0 ) + 1 + q ⎞
⎟ = −k t
⎝ {
q ⎜ 2 K eCB + K e (C A0 − CB 0 ) + 1 + q } {2 K Ce B0 + K e (C A0 − C B0 ) + 1 −}q ⎟
⎠
2



....(3)

Since Ke, CA0 and CB0 are known, q can be calculated and comes out to be 6.6487.
Substituting the known values of Ke, CA0, CB0 and q in (3), we get

⎛ 27.74CB − 1.0154 ⎞
0.3878 ln ⎜ 5.7429 × ⎟ = −k2t where CB should be in moles/liter.
⎝ 27.74CB + 4.1416 ⎠

⎛ 27.74CB − 1.0154 ⎞
Plotting 0.3878 ln ⎜ 5.7429 × ⎟ versus t and fitting the data to a straight line will
⎝ 27.74CB + 4.1416 ⎠
give us k2 as the negative of the slope of the fitted line. Figure below shows the plotted data and
corresponding linear fit from which the slope comes out to be
–0.0005687 min-1.


Thus, k2 = 0.0005687 min-1


liter
k1 = K e k 2 = 0.0078879
mole. min

Thus, the constitutive equation for the rate of (CH3)2CCNOH production is


5

,dCD
= k1C ACB − k2CD
dt
with k1 = 0.0078879 liter/(mole.min) and k2 = 0.0005687 min-1.




Figure: Plot to obtain the reverse rate constant, k2




6

,Problem 2.4

First, we write the species mole balances for A and R, respectively, for an isothermal batch
system:

dC A dCR
= − rA = −k1C A and = rA − rs = k1C A − k2CR
dt dt


The species A balance is a separable first order ODE, and can thus be solved independently for
CA(t) with the initial condition CA(t=0)= CA0:

CA t
dC A'
∫ CA' = −k1 ∫0 dt
CA 0

⎛C ⎞
⇒ ln ⎜ A ⎟ = −k1t
⎝ C A0 ⎠
⇒ C A (t ) = C A0 e − k1t

Substituting this result into the species R balance yields

dCR
= k1C A0 e − k1t − k2CR .
dt

This first order ODE is not separable, and thus requires an integrating factor in order to solve.
We have the model equation in the general form

dx
= q (t ) − p(t ) x
dt

whose general solution is

− p ( t ) dt ⎡ ∫ p (t ) dt q(t )dt ⎤ .
x(t ) = e ∫ ⎢⎣ ∫ e
⎦⎥

Substituting the appropriate terms from the R balance yields

CR (t ) = e ∫ ⎡⎢ ∫ e ∫ k1C A0 e − k1t dt ⎤⎥
− k2 dt k2 dt

⎣ ⎦
⇒ CR (t ) = k1C A0 e ⎡⎣ ∫ e
k2t ( k2 − k1 ) t
dt ⎤⎦
k1
⇒ CR (t ) = C A0 e − k2t ⎡⎣ A − e( k2 − k1 )t ⎤⎦
k2 − k1



7

, k1
⇒ CR (t ) = C A0 ⎡⎣ Ae − k2t − e − k1t ⎤⎦ .
k2 − k1

Finally, using the initial condition CR(t=0)= 0, we obtain

k1
CR (t ) = C A0 ⎡⎣ e − k2t − e − k1t ⎤⎦ .
k2 − k1

dCR
The maximum value of CR is obtained when = 0 , such that
dt t =tmax

dCR (t ) k1 d
=0= C A0 ⎡⎣e− k2t − e− k1t ⎤⎦
dt k2 − k1 dt

⇒ 0 = −k2 e − k2tmax + k1e− k1tmax
⎛k ⎞
ln ⎜ 2 ⎟
= ⎝ 1⎠ .
k
⇒ tmax
(k2 − k1 )

Now plug this value into the equation for CR to find CR(tmax).

k1 ⎡ ⎛ − k ln (k 2 / k1 ) ⎞ ⎛ − k ln (k 2 / k1 ) ⎞⎤
C R (t max ) = C A0 ⎢exp⎜⎜ 2 ⎟⎟ − exp⎜⎜ 1 ⎟⎟⎥
k 2 − k1 ⎣ ⎝ (k 2 − k1 ) ⎠ ⎝ (k 2 − k 1 ) ⎠⎦




8

, Problem 2.5

Extension of Problem 2.4…

First, we write the species mole balances for A and R, respectively, for an isothermal batch
system:

dC A dCR
= rA = −k1C A and = −rA − rs = k1C A − k2CR
dt dt

The species A balance is a separable first order ODE, and can thus be solved independently for
CA(t) with the initial condition CA(t=0)= CA0:

CA t
dC A'
∫ CA' = −k1 ∫0 dt
CA 0

⎛C ⎞
⇒ ln ⎜ A ⎟ = −k1t
⎝ C A0 ⎠
⇒ C A (t ) = C A0 e− k1t

Substituting this result into the species R balance yields

dCR
= k1C A0 e − k1t − k2CR .
dt

With the solution as outlined in 2.4,

k1
CR (t ) = C A0 ⎡⎣ e− k2t − e− k1t ⎤⎦ .
k2 − k1


a) We are given reference values of the rate constants at 25°C. Therefore, we can cast the rate
constants in terms of the reference values by dividing the Arrhenius expression at any
temperature by that for the reference temperature,

Ea
−
ki (T ) k e RT ⎡ E ⎛1 1 ⎞⎤
= i ,0 Ea ⇒ ki (T ) = ki (Tref ) exp ⎢ − a ⎜ −
ki (Tref ) −
⎢ R ⎜ T T ⎟⎟ ⎥⎥
RT ⎣ ⎝ ref ⎠ ⎦
ki ,0 e ref

where here the given reference temperature is 298.15K. Now, CR(t), tmax, and CR,max can be
computed using the rate constants calculated from the above expression (given here in SI units):




9