ChE 4253 Design I
Mid-Term Exam, page 1 of 6
Open Book – Open Notes Exam
Problem 1___________/30
Problem 2___________/40
Problem 3___________/30
Total _______________/100
Name:_______________________
ChE 4253 Design I
Mid-Term Exam, page 2 of 6
Name:_______________________
1.) 2 short answer problems – 15 points each plus Bonus question
a.) Assuming an annual compound interest rate of 5.8%, which costs more, a 8 year long
$1,250 annual maintenance contract with an additional payment of $100 each year after
year one or a new $15,000 pump after 4 years of service? (Ignore salvage values.)
i := 0.058
n := 8
A1 := 1250
G := 100
n
A := A1 + G
( 1 + i) - i n - 1
3
= 1.571  10
This is the average annuity for one with a gradient
n
i ( 1 + i) - i
F := A
( 1 + i) n - 1
4
= 1.543  10
F
P1 :=
3
( 1 + i)
C1 := 15000
P2 :=
This is the future value of the average annuity
i
n
= 9.83  10
n := 4
C1
4
( 1 + i)
Present Value of Service Contract
of this future value
n
= 1.197  10
Present value of pump to be purchased at yr 4
The pump costs more!!
b.) For a $30,230,000 (=CTDC) plant, there are sales of $1,620,000 and a cost of
manufacturing of $750,000 (excludes depreciation which is 15 class life MARCS).
Determine the payback period (PBP) given a tax rate of 40.2%.
CTDC := 30230000
S := 1620000 
t := 0.402
1
yr
C := 750000
1
yr
1
D := CTDC 0.05 
See MARCS 15 yr Table - use year 1 %
yr
PBP :=
CTDC
( 1 - t )  ( S - C) + D
= 14.879  yr
c.) Bonus Question: (extra 5 points) For the plant in b) above, what is the annual cost of
maintenance, wages and benefits? Assume this is a solids handling process.
Maintenance, Wages & Benefits See Cost Table using solids handling process for % of C.TDC
6
MWB := 0.05  CTDC = 1.512  10
This problem is somewhat decieving so equal credit is given if S&B were added @ 25% of MW&B
6
MWB2 := 0.05  CTDC + 0.25  MWB = 1.889  10
ChE 4253 Design I
Mid-Term Exam, page 3 of 6
Name:_______________________
2) (40 points) Methyl methacrylate (MMA) CH2=C(CH3)COOCH3 is produced by
two overall reactions:
a) the reaction of methyl propionate with formaldehyde producing MMA and
methanol
CH3(C3H5COO) + CH2O  CH2=C(CH3)COOCH3 +CH3OH
b) And the oxidation of tert-butanol (C4H10O) followed by reaction with methanol
(CH3OH) producing MMA, water and hydrogen gas
C4H10O + O2 + CH3OH CH2=C(CH3)COOCH3 + 2H2O + 2 H2
Determine which reaction is the best economically using gross profit analysis.
Data
MW gm/mole
MMA
tert-butanol
$/kg
$ 2.689
$2.50
$0.22
$0.427
$3.00
$0.01
$2.20
100.12
74.12
31.9988
32.04
O2
methanol
H2
H2O
2.00
18.01
88.11
30.03
methyl propionate
formaldehyde
$0.266
Solution
Problem 2
CH3(C3H5COO) + CH2O <----> CH2=C(CH3)COOCH3 +CH3OH
Profit1 := 1  mole 100.12 
gm
mole
 2.689
1
kg
+ 1  mole 32.04
gm
mole
 0.427
1
kg
-  1  mole 88.11

gm
mole
 2.20 
1
kg
+ 1  mole 30.03
gm
mole
 0.266
Profit1
mole
 100.12  gm 


mole 

= 0.81
1
$/kg
kg
C4H10O + O2 + CH3OH --> CH2=C(CH3)COOCH3 + 2H2O + 2 H2
gm
1
gm
1
gm
1
 2.689
+ 2  mole 18.01
 0.01 
+ 2  mole 2 
 3.0
...
mole
kg
mole
kg
mole
kg
gm
1
gm
1
gm
1
+ - 1  mole 74.12
 2.50 
+ 1  mole 31.998 
 0.22 
+ 1  mole 32.04
 0.427 
mole
kg
mole
kg
mole
kg 

Profit2 := 1  mole 100.12 
Profit2
mole
 100.12  gm 


mole 

= 0.755
1
kg
$/kg
= 0.076 $ per mole
1 
 = 0.081
kg 
$ per mole
ChE 4253 Design I
Mid-Term Exam, page 4 of 6
Name:_______________________
3) (30 Points) Using property estimation methods determine the heat of mixing
between phenol and benzene.
Properties Phenol
Molecular formula C6H6O
Molar mass 94.11 g mol−1
Appearance transparent crystalline solid
Density 1.07 g/cm3
Melting point 40.5°C, 314 K, 105°F
Boiling point 181.7°C, 455 K, 359°F
Properties Benzene
Molecular formula C6H6
Molar mass 78.11 g mol−1
Appearance Colorless liquid
Density 0.8765(20) g/cm3
Melting point 5.5 °C, 278.7 K
Boiling point 80.1 °C, 353.3 K
Problem 3
Drago E and C
Data for Phenol
EA := 4.58  
kcal
CA := 0.30  
kcal
Data for Benzene


 mole 
0.5


EB := 0.40  


 mole 
0.5
CB := 0.8 
 mole 
(
kcal
kcal


 mole 
)
kcal
Δ HAB := - EA EB + CA CB = -2.072
mole
Heat of Mixing
0.5
0.5
ChE 4253 Design I
Mid-Term Exam, page 5 of 6
EXTRA PAGE
Name:_______________________
ChE 4253 Design I
Mid-Term Exam, page 6 of 6
EXTRA PAGE
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