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Chemical Engineering 140 September 26, 2005
Midterm #1 Solution (total pts: 185)
(40) i. Use transient species mass balance on salt in upper chamber
CV = Upper chamber volume = V
Transient mass balance for species i = salt. No reaction, no outlet streams
( ) , ,
m iin iout i
i m m G dt
dm = − + iin i
i i n M dt
d nM ,
Constant species molecular weight Mi allows us to reduce mass balance to mole balance
i in
i
V
i n dt
dCV C rdV dt
d ,
(Well mixed tank Ci is indep. of position) (*)
But, rate of moles in varies with the concentration difference across the SCL:
, ˆ^ ( c^ C ( t ))
L
PA
n (^) i in = NiA = o − i (Relationship for flux N is given in part (i).) (**)
Combining () and (*), we obtain
c C t L
PA
dt
dC V dt
dV C dt
d CV o i
i i
i = + = − (V const)
Receptor cell
SCL
Donor cell
Conductivity probe V
c (^) o
Nsalt
L
A
Csalt(t)
Given in problem statement:
A ≡ Area of membrane (SCL)
V ≡ Volume of receptor cell, CONSTANT
L ≡ Thickness of membrane (SCL)
c (^) o ≡ Concentration of salt in donor cell,
CONSTANT
Both tanks are WELL MIXED
Csalt(t = 0) = 0 (Upper chamber initially
contains no salt)
( c C ( t ))
L
PA
dt
dC V (^) o i
i = −
Rearranging, we have a first order, linear ODE for Ci(t):
i o
i c VL
PA
C t VL
PA
dt
dC
(20) (ii) Solve the ODE and use BC’s to get expression for concentration
Defining an integrating factor μ and writing the general solution to this linear first order ODE,
where α is a constant of integration (Can also be solved by Separation of Variables)
= exp[ dt ] VL
PA
μ
α μ μ
dt VL
PAc C t
o i
( ) , which simplifies to
= + t VL
PA
C (^) i ( t ) co αexp
Applying the initial condition that Csalt (t = 0) = 0 and simplifying, gives
= − t VL
PA
C (^) i ( t ) co 1 exp
(20) (iii) Express similar to Nicolas et al
− = t VL
PA
c
C t
o
i exp
1 t VL
PA
c
C t
o
i − =
ln 1
Equation 1 given by Nicolas is correct.
(25) 2i If the process is at steady state, what is the feed rate of high fructose corn syrup in
gal/min? Assume corn syrup specific gravity is 1.34, relative to water at 4°C.
Solution: min mout dt
dm = ^ − = 0 ; min mout (^) = (overall mass balance)
F + 750 kg/hr + 5 kg/hr + 10 kg/hr = 1000 kg/hr F = 235 kg/hr
3 3
2
2
4 2 83.^6
ft
lb F
ft
lb H O
lb HO
lb F sp gr
m m
m
C m F =^ HO = =
ρ ρ (any correct conversions accepted)
(25) 2ii Steady state, unsteady state, vs equilibrium state
Steady state – no time dependence, but possible spatial dependence
Unsteady state – process variables change with time
Equilibrium – no time or spatial dependence
(15) 2iv In a steady-state problem, why is it necessary to choose a basis?
Selecting and declaring a basis is necessary so that you and others can later identify the
foundation for the calculations made and appropriately scale the problem. In addition, the basis
can provide an additional piece of information for solving the problem.
Dyes, flavorings,
& preservatives
Soft Drink
High fructose
Corn syrup
Water
CO 2
1000 kg/hr
10 kg/hr
5 kg/hr
750 kg/hr
F
min
60 min
3
3 hr gal
kg
lb
ft
gal
lb F
ft
hr
kgF F
m
m
