10.2.1
Gas Solubility in Aqueous Solution. At 303 K the concentration of CO2 in water is 0.90 x 10-4 kg CO2/kg water. Using the
Henry’s law constant from Appendix A.3, what partial pressure of CO2 must be kept in the ags to keep the CO2 from
vaporizing from the aqueous solution?
Given:
T = 303 K
xA = 0.9 x 10-4 kg CO2 /kg H2O
Required: PA of CO2
Solution:
From A.3-18 for Henry’s law constant (Geankoplis p. 884)
H = 0.186 x 104 atm/mol frac.
PA= HxA

kgCO2 18 kgmolH2O 
PA = 0.186 x 104 atm x  0.9 x 10-4
x
  0.06848 atm
kgH2O 44 kgmol CO2 

1.01325 x 105
= 6.939 x 103 Pa
1 atm
PA = 0.06848 atm x
10.3.1
Phase Rule for a Gas-Liquid System. For the systen SO2-air-water, then total pressure is set at 1 atm abs and the partial
pressure of SO2 in the vapor is set at 0.20 atm. Calculate the number of degrees of freedom, F. What variables are
unspecified that can be arbitrarily set?
GIVEN:
SO2 – air – H2O system
PAT = 1 atm
PA of SO2 = 0.2 atm
REQUIRED:
Degrees of freedom, F
Variables that can be set
SOLUTION:
F=C–P+2
F=3–2+2
F=3
Variables that can be set:
10.3.2
1.
total pressure
2.
temperature
3.
mole fraction composition xA of SO2
Equilibrium Swtage Contact for Gas-Liquid System. A gas mixture at 2.026 x 10 5 Pa total pressure containing air and SO2 is
contacted in a single-stage equilibrium mixer with pure water at 293 K. The partial pressure of SO 2 in the original gas is 1.52
x 104 Pa. The inlet gas contains 5.70 total kg mol and the inlet water 2.20 total kg mol. The exit gas and liquid leaving are in
equilibrium. Calculate the amounts and compositions of the outlet phases. Use equilibrium data from Fig.10.2-1.
GIVEN:
Use equilibrium data in Fig. 10.2-1
PT = 2.026 x 105 Pa =
T = 293 K
4
PA of SO2 = 1.52 x 10 Pa = .15 atm
Inlet gas = 5.70 kg mol
Inlet H2O = 2.20 total kgmol
REQUIRED
XA1, yA1, L1 V1
SOLUTION:
xAo = 0
amount of entering acetone = yAN+1vAN+1
= 0.01(30) = 0.30
= 29.7 kgmol/air h
acetone leaving in Vi = 0.10(0.30) = 0.30 kgmol/h
acetone leaving in Ln = 0.9 (0.30) = 0.27 kgmol/h
V1 = 29.7 + 0.03 = 29.73 kgmolH2O + acetone/hr
yA1 =
0.030
=0.00101
29.73
Ln = 108 + 0.27 = 108.27 kgmol H2O + acetone/hr
XAN =
0.27
0.002493
108.27
Using equation:
A1 =
L
Lo
108
=

=1.4358
mv mv1
2.53 (29.73)
AN =
LN
108.27
=
=1.4265
mvN+1 2.53 (30)
A= A1AN  1.4358(1.4265)  1.4311
N=
1 1

log  yN+1 - mxo   1   
A A

log A
N=
10.3.3
 0.01 - 2.53(0)  
1
1 
log 
  1  1.4311  1.4311 
0.00101
2.53(0)



3.7662 stages
log 1.4311
Absorption in a Countercurrent Stage Tower. Repeat example 10.3-2 using the same conditions but with the following
change. Use a pure water flow to the tower of 108 kg mol H2O/h, that is, 20% above the 90 used in Example 10.3-2.
Determine the number of stages required graphically. Repeat using the analytical Kremser equation.
GIVEN:
LO = 108 kg mol H2O/hr
YAN + 1 = 0.01
VN + 1 = 30.0 kg mol/h
T = 300 K
PT = 101.3 kPa
yA = 2.53 xA
REQUIRED:
Theoretical stages graphically using analytical Kremser equation.
SOLUTION:
xAo = 0
amount of enetering acetone = yAN + 1 VN + 1
= 0.01(30) = 0.30
entering air = (1-yAN + 1) VN + 1 = (1 – 0.01)(30) = 29.7 kg mol/air h
acetone leaving in V1 = 0.10(0.30) = 0.30 kg mol/h
acetone leaving in Ln = 0.9(0.30) = 0.27 kg mol/h
V1 = 29.7 + 0.03 = 29.73 kg mol air + acetone /hr
Ln = 108 + 0.27 = 108.27 kg mol H2O + acetone/hr
10.4.1
Interface Concentrations and Overall Mass-Transfer Coefficients. Use the same equilibrium data and film coefficients k’ y and
k’x as in Example 10.4-1. However, use bulk concentrations of yAG = 0.25 and xAL = 0.05. Calculate the following:
a.
Interface concentrations yAi and xAi and flux NA.
b.
Overall mass transfer coefficients K’y and Ky and flux NA.
c.
Overall mass transfer coefficient K’x and flux NA.
GIVEN:
Equilibrium data
yAo = 0.380 mol fraction
xAL = 0.10
T = 298 K
P = 1.013 x 105 Pa
ky = 1.465 x 10-3 kg mol/s.m3 mol fraction
kx = 1.967 x 10-3 kg mol A/s.m2 mol fraction
REQUIRED:
K’X, KX
SOLUTION:
Trial 1:
k'x/(1-xA)iM
1.967 x 10

k'y/(1-yA)iM
1.465 x 10
-3
-3
/1.0
 - 1.342
/1.0
y = mx + b
0.38 = -0.10 (1.342) + b
b = 0.5142
xA1 = 0.247
yAi = 0.183
(1-yA)iM =
(1-xA)iM =
(1-yAi) - (1-yAG) (1-0.183) - (1-0.380)
=
= 0.715
  1-yAi  
  1-0.183  
ln 
ln 


1-y
AG 



  1-0.380  
(1-xAL) - (1-xAi) (1-0.1) - (1-0.247)
=
= 0.825
  1-xAL  
  1-0.1 
ln 
ln 


  1-xAi  
  1-0.247  
Trial 2:
k'x/(1-xA)iM
-1.967 x 10 -3 / 0.825

 - 1.163
k'y/(1-yA)iM
1.465 x 10 -3 / 0.715
xAi = 0.254
yAi = 0.194
0.38 = - 1.163 (0.10) + b
b = 0.4963
(1-yA)iM =
(1-xA)iM =
(1-0.194) - (1-0.380)
= 0.7089
  1-0.194  
ln 

  1-0.380  
(1-0.1) - (1-0.254) (1-0.1) - (1-0.247)
=
= 0.8206
  1-0.1 
  1-0.1 
ln 
ln 


  1-.254  
  1-0.247  
k'x/(1-xA)iM
-1.967 x 10 -3 / 0.8206

 - 1.160
k'y/(1-yA)iM
1.465 x 10 -3 / 0.7089
-1.163  -1.60
 xAi = 0.254
yAi = 0.194
m'' =
yAG - yAi
0.38 - 0.194
=
 2.0217
*
xA - xAi
0.346 - 0.254
1
1
1
=
+
K'x m''k'y
k'x
1
1
1
=
+
-3
K'x
2.0217 (1.465 x 10 )
1.967 x 10-3
K'x = 0.00118 kgmol/s m2 frac
K'x =
K'x
(1- xA) * M
(1- xA) * M =
(1- xA) * M =
Kx =
(1- xAL) - (1- xA * )
 (1- xAL) 
ln 
* 
 (1- xA ) 
(1- 0.10) - (1- 0.346)
0.7705
 (1- 0.10) 
ln 

 (1- 0.346) 
0.00118
0.00153
0.7705
 K'x 
*
NA  
 (xA -xAL)
(1-x
A
)
*M


 0.00118 
NA = 
 (0.346-0.10)
 0.7705 
NA = 0.000377 kgmol/sm2
1
1
1


K'x/(1-xA)*M
m'' / (1-yA)*M K'x/(1-xA)*M
(1-yAi) - (1-yAG)
(1-yAi)
ln
(1-yAG)
(1-0.194) - (1-0.38)
(1-yA)iM =
0.7089
(1-0.194)
ln
(1-038)
(1-yA)iM =
(1-xA)iM 
(1-xAL) - (1-xAi) (1-0.1) - (1-0.254)

0.8206
(1-xAL)
(1-0.1)
ln
ln
(1-xAi)
(1-0.254)
1
1
1


-3
K'x/(1-xA)*M
2.0217(1.465 x 10 / 0.7089 1.967 x 10-3 /0.8206
= 656.5320
% resistance in the gas film
239.3425
=
(100) 36.46%
656.5392
% resistance in the liquid film
417.1835
=
(100) 63.54%
656.54
10.4-2 Use the same equilibrium data and film coefficients k’y and k’x as in Example 10.4-1. However, use bulk concentrations of y AG =
0.25 and xAL = 0.05. Calculate the following.
(a) Interface concentrations yAi and xAi and flux NA.
(b) Overall mass-transfer coefficients K’y and Ky and flux NA.
(c) Overall mass-transfer coefficient K’x and flux NA.
GIVEN:
yAG = 0.25
xAL = 0.05
ky = 1.465 x10–3 kgmol A / sm3 mol fraction
kx = 1.967 x10–3 kgmol A / sm2 mol fraction
REQUIRED:
a.) Interface concentrations yAi & xAi & NA
SOLUTION:
Trial 1:
k'x/(1-xA)iM
1.967 x 10

k'y/(1-yA)iM
1.465 x 10
-3
-3
/1.0
 - 1.342
/1.0
y = mx + b
0.25 = -1.342 (0.05) + b
b = 0.3171
xAi = 0.1634
yAi = 0.1010
Trial 2:
(1-yA)iM =
(1-xA)iM =
(1-yAi) - (1-yAG) (1-0.1010) - (1-0.25)
=
= 0.8223
  1-yAi  
  1-0.1010  
ln 
ln 


  1-yAG  
  1-0.25  
(1-xAL) - (1-xAi) (1- 0.05) - (1- 0.1634)
=
= 0.8921
  1-xAL  
  1- 0.05  
ln 
ln 


  1-xAi  
  1- 0.1634  
k'x/(1-xA)iM
1.967 x 10 -3 / 0.8921

0.8223  - 1.2376
k'y/(1-yA)iM
1.465 x 10 -3 /1.0
y = mx + b
0.25 = -1.2376 (0.05) + b
b = 0.3119
xAi = 0.1686
yAi = 0.1054
Trial 3:
(1-yA)iM =
(1-xA)iM =
-
(1-yAi) - (1-yAG) (1-0.1054) - (1-0.25)
=
= 0.8202
  1-yAi  
  1-0.1054  
ln 
ln 


  1-yAG  
  1-0.25  
(1-xAL) - (1-xAi) (1- 0.05) - (1- 0.1686)
=
= 0.889
  1-xAL  
  1- 0.05  
ln 
ln 


  1-xAi  
  1- 0.1686  
k'x/(1-xA)iM
-1.967 x 10 -3 / 0.8894

0.8223  - 1.2382
k'y/(1-yA)iM
1.465 x 10 -3 / 0.8202
-1.2382  1.2376
 xAi = 0.1686
yAi = 0.1054
 k'y 
NA  
 (yAG -yAi)
 (1- yA)iM 
 1.967 x 10 -3 
NA = 
 (0.25 - 0.1054)
0.8202


-4
NA = 2.583 x 10 kgmol/sm2
or,
 k'x 
NA  
 (xAi -xAL)
 (1- xA)iM 
 1.967 x 10 -3 
NA = 
 (0.1686 - 0.05)
 0.8894 
NA = 2.62 x 10-4 kgmol/sm2
b)
m' =
yAi - yA *
0.1054 - 0.0243
=
 0.6838
xAi - xAL
0.1686 - 0.05
(1-yA)*M =
(1-yA * ) - (1-yAG)
 (1-yA * ) 
ln 

 (1-yAG) 
yA * = 0.0243
(1-yA)*M =
(1-0.0243) - (1-0.25
0.8579
 (1- 0.0243) 
ln 

 (1- 0.25) 
1
1
m'


K'y/(1-yA)*M
k'y / (1-yA)iM k'x / (1-xA)iM
1
0.6838
=

-3
1.465 x 10 / (0.8202) 1.967 X 10 -3 / 0.8894
K'y = 9.872 x 10-4
Ky =
K'y
9.872 x 10-4
=
0.00115
(1  yA )* M
0.8579
 K'y
 9.872 x 10-4
NA = 
(yAG-yA * ) 
(0.25 - 0.0243)
0.8579
 (1  yA )* M

NA = 2.5972 x 10 - 4 kgmol/sm2
c)
m" =
yAG - yAi
0.25 - 0.1054
=
 1.1814
*
xA - xAi
0.291 - 0.1686
(1-xA)*M =
(1- xA L ) - (1-xA * ) (1- 0.05) - (1-0.291)

0.8236
 (1-xA L ) 
 (1-0.05) 
ln 
ln 
*


 (1-xA ) 
 (1-0.291) 
1
1
1


K'x/(1-xA)*M
m"k'y/(1-yA)*M k'x / (1-xA)*M
1
1
1


K'x/0.8236
1.1814(1-yA)*M 1.967 x 10 -3 / 0.8894
K'x = 7.8775 x 10-4
10.6-6. - 10.6-7. A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at
293 K and 1.013 x 105. The inlet pure water flow is 68.0 kg mol/h and the total inlet gas flow is 57.8 kgmol/h. The tower diameter is
0.747 m. The film mass-transfer coefficients are k’ya = 0.0739 kgmol.sm3 mol frac and k’xa = 0.169 kgmol/sm3 mol frac. Using the design
methods for dilute gas mixtures, do as follows.
Repeat Example 10.6-2, using the overall liquid mass-transfer coefficient K’xa to calculate the tower height.
(a) Calculate the tower height using k’ya
(b) Calculate the tower height using K’ya.
Solution:
(a)
La = 68 kgmol/h
ya = 0.005
xa = 0
(b)
Vb = 57.8 kgmol/h
yb = 0.04
V' = V (1 - y b ) 57.8(1  0.04) 55.488 kgmol/h
L' = L (1 - xb )  68 (1 - 0) = 68 kgmol/h
V' (y b -y a ) = L'(x b - x a )
55.488 (0.04-0.005) = 68x b
x b 0.029
YN+1 
L'
68
0.005
Xn + Yn 
Xn +
V'
55.488'
0.995
YN+1  1.2255 Xn + 0.0050
if Xn 0,
YN+1 = 0.0050
x
1-x
X=
y=
slope = m = -
Y
1+Y
k'xa/(1-x)
k'ya/(1-y)
y = mx + b
L
kxaS
V'
V=
1- y
Hx =
S = Area =
Nx =
dx
xi - x
d2
(0.747)2
=
= 0.4383 m3
4
4
height = z = Nx Hx = 0.2250 (8.6692) = 2.2106 m
m' =
y A1 -y A *
x A1 -x AL
K'y =
k'y k'x
k'x + m' k'y
K'y =
Ky =
m" =
y AG -y Ai
x A *-x Ai
K'x =
m" k'y k'x
k'x + m''k'y
(0.0739)(0.169)
0.0450
0.169 + 0.8287(0.0739)
K'y
KyaS
Kx =
K'x
(1- X)
55.488
57.8
55.7668
V5 =
1-0.005
1- 0.040
55.7668 + 57.8
V=
56.7834
2
V1 =
Hoy =
V
56.7864 / 3600
=
=0.6438
KyaS
0.0559 / 0.4383
= 0.6438
z = 0.6438 (3.54009) = 2.2796 m
68
68
68
L5 =
70.0309
1- 0
1- 0.029
68 + 70.0309
L=
69.0155
2
L1 =
L
69.0155 / 3600
=
=0.9571
KxaS
0.0457 / 0.4383
= 0.6438
z = 0.9571 (2.3209) = 2.2213 m
Hox =
X
0
y
0.0050
X
0
Y
0.005
m
-2.2754
b
0.005
yi
0.0014
0.01
0.0170
0.0101
0.0173
-2.2707
0.0397
0.0096
0.015
0.0229
0.0152
0.0234
-2.2685
0.0569
0.0141
0.02
0.0287
0.0204
0.0295
-2.2667
0.0740
0.0188
0.029
0.0400
0.0299
0.0417
-2.2610
0.0108
0.0272
xi
1/(y – yi)
1/(x – xi)
0.0018
277.7780
555.5556
0.0132
135.1351
312.50
0.0190
113.6364
250
0.0246
101.0101
217.3913
0.0348
78.125
172.4138
dy/(y-yi)Ave
dx/(x-xi)Ave
2.4775
4.3403
0.7339
1.4063
0.6225
1.1685
1.0121
1.7541
Ny = 4.8549
Nx = 8.6692
1/(y – yi) ave
∆y
∆x
206.4565
0.012
0.01
124.3858
0.0059
0.005
107.3233
0.0058
0.005
89.5678
0.0113
0.009
Ky=k’ya/(1-y)
Kx =k’xa/(1-x)
0.0743
0.169
0.0752
0.1707
0.0756
0.1716
0.0761
0.1724
0.0756
0.1740
kyAve = 0.0756
kxAve = 0.1715
X‫٭‬
0.0072
y‫٭‬
0
m’
0.7778
m”
0.6667
Ky
0.0590
Kx
0.0450
0.0224
0.0072
0.7500
0.8043
0.0555
0.0455
0.0292
0.011
0.8158
0.8627
0.0559
0.0457
0.0360
0.0148
0.8696
0.8684
0.0562
0.0459
0.0484
0.0224
0.8276
0.9412
0.0569
0.0463
AVE: 0.8082
1/(y – y‫)٭‬
200
1/(x‫ – ٭‬x)
138.8889
102.0408
80.6452
84.0336
70.4225
71.9424
62.5
56.8182
51.5464
Ave: 0.8287
1/(y – y‫)٭‬ave
1/(x‫ – ٭‬x)ave
151.0204
109.7671
93.0372
75.5339
77.9880
66.4613
64.3803
57.0203
dy/(y – y‫)٭‬ave
1.8122
dx/(x‫ – ٭‬x)ave
1.0977
0.5481
0.3777
0.4523
0.3323
0.7275
0.5132
Noy= 3.5409
Nox= 2.3209
10.6-10. Repeat Example 10.6-2 but use transfer units and calculate HL, NL, and tower height.
Given:
Acetone-H2O system
A = 0.186 m2
T = 293 K
P = 101.32 kPa
y1 = 0.026
y = 0.005
V = 13.65 kgmol inert air / h
L = 45.36 kgmol H2O / h
k’ya = 3.78 x 10-2 kgmol / sm3 mol frac
k’xa = 6.16 x 10-2 kgmol / s m3 mol frac
REQUIRED:
HL , NL , z
SOLUTION:
 y1 
 y 
 xo 
 x 
L' 
 + V'  1 - y1   L'  1 - x   V'  1 - y 
 1 - xo 






 0 
 0.026 
 x1 
 0.005 
45.36 
 + 13.65 
  45.36 
 +13.65 

1 - 0
 1 - 0.026 
 1 - x1 
 1 - 0.005 
x1 = 0.00648
 1h 
45.36 kgmolH2O / h 

L
 3600 s 
HL =
=
= 1.0264
K'xa S
6.6 x 10 -2kgmol / sm3 mol frac. (0.186 m2 )
from the graph:
xi1 = 0.0136
xi2 = 0.0019
(xi - x)M =
(xi1 - x1) - (xi2 - x2)
(0.0132 - 0.00648) - (0.0019 - 0)
=
=0.0038
 (xi1 - x1) 
 (0.0132 - 0.00648) 
ln 
ln 


(0.0019)
 (xi2 - x2) 


NL =
dx
0.00648

1.7053
xi - x
0.0038
z = NL HL = 1.7053 (1.0264)
z = 1.75 m
10.7-1
Liquid Film Coefficients and Design of SO2 Tower. Using the data of example 10.7-1, calculate the height of the tower using
Eq. (10.6-15), which is based on the liquid film mass-transfer cofficient k’xa. [Note: the interface values xi have alre4ady been
obtained. Usi a graphical integration of Eq. (10.6-15).
Given:
A = 0.0929 m2
T= 293 K
P = 1.013 x 105
y1= 0.20
y1 = 0.02
x1 = 0
V’ = 6.53 x 10-4 kgmol air/ s
L’ = 4.20 x 10-2 kgmol / s
k’ya = 0.0594 Gy0.7Gx0.25
k’xa = 0.152 Gx 0.82
REQUIRED:
Z using kxa
SOLUTION:
 y2 
 y1 
 x2 
 x1 
L' 

  L' 
 + V' 
  V' 
 1 - x2 
 1 - x1 
 1 - y1 
 1 - y2 
0.2 
x1 
0.02 
 0 
-4 
-2 
-4 
4.20 x10  2 
 + 6.53x10 
  4.20x10 
 +6.23x10 

1 - 0 
 1 - 0.2 
 1 - x1 
 1 - 0.02 
x1 = 0.00355
 y 
6.53 x 10-4 (29)kg air / s + 6.53 x 10-4 
64.1
1  y 

Gy =
0.0929
if y = 0.20
 0.2 
6.53 x 10-4 (29)kg air / s + 6.53 x 10-4 
 64.1
1  0.2 

Gy =
0.0929
2
= 0.3164kg / sm
V=
V'
6.53 x 10-4

8.16 x10  4
1-y
1  0.2
L=
L'
4.20 x 10 -2

= 0.0421
1- x
1 - 0.00355
 x 
4.20 x 10-2  18  + 4.20 x 10 -2 
 64.1
 1 x 
Gx =
0.0929
 0.00355 
4.20 x 10-2 + 4.20 x 10 -2 
 64.1
1  0.00355 


8.2410
0.0929
k'xa = 0.152 Gx 0.82 = 0.152 (8.2410)0.82 0.8569
k'ya = 0.0594Gy 0.7Gx 0.25 = 0.0594 (0.2130)0.7 (8.2410)0.25 = 0.0341
L
0.04207
=
0.5319
kxa (1-x)iM
0.85356(0.9975)(0.0929)
z = NLHL  3.051 (0.5319) = 1.6228
HL 
y
x
V
L
Gy
Gx
k’ya
0.02
0
6.66 x 10-4
0.042
0.2130
8.138
0.03398
0.2226
8.147
0.03504
8.162
0.03673
8.196
0.04032
8.241
0.04496
0.04
k’xa
0.07
0.848
0.13
0.849
0.20
0.850
0.000332
6.68 x 10-4
m
b
xi
0.000855
7.02 x 10-4
24.4567
0.02
0.00046
0.00201
7.51 x 10-4
23.2680
0.04772
0.00109
0.00355
8.16 x 10-4
21.5403
0.08842
0.00188
0.0421
yi
0.04203
0.0088
0.04208
0.026
0.04215
0.04824
b
0.2378
24.7476
0.02
0.2718
23.6252
0.04784
0.3164
22.2705
0.08904
0.853
18.4426
0.1671
0.00356
0.1016
19.0742
0.1683
0.857
15.3034
0.2543
0.00572
0.1663
15.9829
0.2567
xi
0.000459
yi
0.0087
m”
24.7501
b
------
0.00104
0.02416
23.6687
0.04762
0.00186
0.0476
22.0814
0.08888
0.00354
0.1004
19.0993
0.1684
0.00570
0.1642
16.0037
0.2568
m’
∆x
dy/(xi – x)
0.000332
0.5897
0.000523
0.6247
0.001155
0.9687
0.00154
0.8679
NL = 3.051
Leaching
SOLVED PROBLEMS:
12.8-1. Effective Diffusivity in Leaching Particles. In Example 12.8-1 a time of leaching of the solid particle of 3.11 h is needed to
remove 80% of the solute. Do the following calculations.
(a)
Using the experimental data, calculate the effective diffusivity, DAeff.
(b)
Predict the time to leach 90% of the solute from the 2.0 mm particle.
GIVEN:
80 % efficiency
t = 3.11 h
REQUIRED:
(a)
DAeff
(b)
T if 90% efficient with same diameter
SOLUTION:
(a)
D Aeff t
0.112
a2
0.112 a 2
D Aeff 
t
2
 2
0.112   mm 2
 1hr 
 2



3.11 hr
 3600 s 
D Aeff 1.0004 x 10-5 mm2 /s
(b)
D Aeff t
0.18
a2
t
0.18 a 2
D Aeff
2
 2
0.18  mm 2
 2
t
1.0004 x 10 -5 mm 2 /s
 1 hr 
t 18000 s 

 3600 s 
t 5 hr
12.9.1.
Leaching of Oil from Soybeans in a Single Stage. Repeat Example 12.9-1 for single stage leaching of oil from soybeans.
The 100 kg of soybeans contains 22 wt % oil and the solvent feed is 80 kg of solvent containing 3 wt % soybean oil.
V1, x1
Lo, No, yo, B
V2, x2
L1, N1, y1, B
GIVEN:
V2 = 80 kg solvent
xA2 = 0.03
xC2 = 0.97
xA1 = 0.22
N = 1.5 kg insoluble solid/kg solution
REQUIRED:
(a) amount and composition of overflow, V1
(b) amount and composition of the underflow, L1
SOLUTION:
B 100 (1 - 0.22) 78 kg insoluble solid
L o 100 - 78 22 kg of A
 B   78 
kg solid
N o     3.5455
kg solution
 L o   22 
y Ao 1.0
For Overall Material Balance:
L o  V2 L1  V1 M
L o  V2 M 22  80
M 102
For Solute Balance:
L o y Ao  V2 xA 2 Mx AM
22(1.0)  80(0.03) 102x AM
x AM 0.2392
To find NM
B N o L o N M M
78 N M (102)
N M 0.7647
To solve for the value of exit underflow
N1L1 N M M
(1.5)L1 (0.7647)(1 02)
L1 51.9996 52 kg
For exit Overflow:
L1  V1 M
52  V1 102
V1 50 kg
20.1. Roasted copper ore containing the copper as CuSO 4 is to be extracted in a countercurrent stage extractor. Each hour a charge
consisting of 10 tons of gangue, 1.2 tons of copper sulfate, and 0.5 ton of water is to be treated. The strong solution
produced is to consist of 90 percent H 2O and 10 percent CuSO4 by weight. The recovery of CuSO4 is to be 98 percent of
that in the ore. Pure water is to be used as the fresh solvent. After each stage, 1 ton of inert gangue retains 2 tons of water
plus the copper sulfate dissolved in that water. Equilibrium is attained in each stage. How many stages are required?
GIVEN:
10 tons of inert solids/hr
1.2 tons CuSO4 and 0.5 ton H20 is to be treated per hour
Strong solution – 90% H2O, 10% CuSO4
CuSO4 recovery of 98% of that in ore
yb = 0
after each stage, 1 ton inert solids retains 2 tons of H2O + CuSO4 dissolved
REQUIRED:
Stages required
SOLUTION:
yb = 0
ya = 0.1
La = 1.2 tons CuSO4 + 0.5 ton H20
= 1.7 tons solution/hr
1.2
x a  0.7059
1.7
CuSO4 recovery = 0.98(1.2)
= 1.176 tons
CuSO4 retained = 1.2 – 1.176
= 0.024 tons
solution retained 

Lb 
2 tons H 2 0  0.024 tons CuSO 4
1 ton inert solid
2.024 tons solution
1 ton inert solid
2.024 tons solution
10tons solid 
1 ton inert solid
L b 20.24 tons solution/h r
Solute Balance:
L a x a  Vb y b L b x b  Va y a
1.7 (0.7059)  Vb (0) 20.24(0.0012)  0.1(Va )
Va 1.1757 tons /hr
By Material Balance:
La  Vb Lb  Va
1.7  Vb 20.24  1.1757
Vb 19.7157 tons H2O
from the graph :
No. of stages 3.70
20.4. Oil is to be extracted from halibut livers by means of ether in a countercurrent extraction battery. The entrainment of solution by the
granulated liver mass was found by experiment to be as shown in Table 20.5. In the extraction battery, the charge per cell is
to be 100 lb, based on completely exhausted livers. The unextracted livers contain 0.043 gal of oil per pound of exhausted
material. A 95 percent` recovery of oil is desired. The final extract is to contain 0.65 gal of oil per gallon of extract.
Solution retained by 1
Solution
Solution retained by
Solution
lb exhausted livers, gal
concentration, gal
1 lb exhausted livers,
concentration, gal
0.035
oil/gal solution
0
gal
0.068
oil/gal solution
0.4
0.042
0.1
0.081
0.5
0.050
0.2
0.099
0.6
0.058
0.3
0.120
0.68
The ether fed to the system is oil free. (a) How many gallons of ether are needed per charge of livers? (b) How many
extractors are needed?
GIVEN:
Charge per cell = 100 lbs
95% oil recovery
yA = 0.65
yB = 0
assuming xA = 1, LA = 0.043 gal oil/lb exhaust liver(100)
LA = 4.3 gal
REQUIRED:
(a)
gal of ether needed per charge of liver
(b)
number of extractors needed
SOLUTION:
For XB:
Oil retained = (0.043 gal oil/ lb exhausted liver) (0.05) (100) = 0.215 gal oil
For the solution in the spent solids: (by trial & error)
Trial 1:
Let XB = 0.1
Solution retained = 0.042
LB = 0.042 x 100 = 4.2 gal
xB 
0.251
0.051
4.2
Trial 2:
XB = 0.051, by linear regression:
Solution retained = 0.0386
LB = 0.0386 x 100 = 3.86 gal
xB 
0.251
0.065
3.86
Trial 3
XB = 0.065, by linear regression
Solution retained = 0.0396
LB = 0.0396 x 100 = 3.96
xB 
0.251
0.0634
3.96
Trial 4
xB = 0.0634, by linear regression
solution retained = 0.0394
LB = 0.0394 x 100 = 3.94
xB 
0.251
0.0637
3.94
0.0637 (close to 0.0634)
assuming XA = 1, LA = 0.043 gal oil/lb exhaust liver x 100
LA = 4.3
By solute balance:
LaxA +VBYB = LBXB +VAYA
4.3(1) + VB(0) = 3.94(0.0637) +Va(0.65)
VA = 6.2293 gal
By OMB:
LA + VB = LB +VA
4.3 + VB = 3.94 + 6.2293
VB = 5.8693 gal
If X1 = ya = 0.65, solution retained = 0.1121
L1 = 0.1121 x 100
L1 = 11.21 gal
By OMB:
La + V2 = L1 + Va
V2 = 11.21 +6.2293-4.3
V2 = 13.1393 gal
By Solute Balance:
Lax + V2y2 = L1x1 + Vaya
4.3(1)+ 13.1393y2 = 11.21(0.65) + 6.2293(0.65)
y2 = 0.5355
If xN = 0.4, solution retained = 0.068
LN = 0.068 x 100
LN = 6.8 gal
By OMB:
La + VN+1 = LN +Va
4.3 + VN+1 = 6.8 + 6.2293
VN+1 = 8.7293
By Solute Balance:
Laxa +VN+1 yN+1 = LNxN + Vaya
4.3(1) + 8.8293yN+1 = (6.8) (0.4) + (6.2293) (0.65)
yN+1 = 0.2828
From the Graph:
Number of stages = 6.7813
Geankoplis
11.2-1 Single-Stage Contact of Vapor-Liquid System. A mixture of 100 mol containing 60 mol% n-pentane and 40 mol% n-heptane is
vaporized at 101.32 kPa abs pressure until 40 mol of vapor and 60 mol of liquid in equilibrium with each other is produced. This occurs
in a single-stage system and the vapor and liquid are kept in contace with other until the vaporization is complete. The equilibrium data
are given in Example 11.3-2. Calculate the composition of the vapor and the liquid.
Given:
x
1.000
0.867
0.594
0.398
y
1.000
0.984
0.925
0.836
x
0.254
0.145
0.159
0
y
0.701
0.521
0.271
0
Solution:
C5H12 Balance:
0.60(100)  YA (40)  X A (60)
YA 1.5  1.5X A
when XA = 0.6 , YA = 0.6……….pt 1
when XA = 0.4, YA = 0.9………...pt 2
1) Create a graph of the equilibrium curve for n-pentane.
2) Plot points at pt.2.
3) Get the value of XA and YA, on the point of intersection between the equilibrium curve and the segment joining the two points.
4) Get YB and XB.
Answers:
From the graph:
XA = 0.43 and YA= 0.855
XB = 1 - 0.43 = 0.57 and YB = 1 – 0.855 = 0.14
11.3-2 Comparison of Differential and Flash Distillation. A mixture of 100 kg mol which contains 60 mol % n-pentane (A) and 40 mol
% n-heptane (B) is vaporized at 101.32 kPa pressure under differential conditions until 40 kg mol are distilled. Used equilibrium data
from Example 11.3-2.
(a)
What is average composition of the total vapor distilled and the composition of the liquid left.
(b) If this same vaporization is done in an equilibrium or flash distillation and 40 kg mol are distilled, what is the composition of the
vapor distilled and of the liquid left?
Given:
x
1.000
0.867
0.594
0.398
0.254
0.145
0.059
0
y
1.000
0.984
0.925
0.836
0.701
0.521
0.271
0
1/ ( y-x )
8.547
3.021
2.283
2.237
2.660
4.717
Solution:
X
ln
ln
2
L1
dx

L 2 X1 y  x
L1
100kmol
 ln
 0.5108
L2
60kmol
from the graph:
 3.04  2.78 
A1  
 0.05  0.1455
2


 2.78  2.58 
A 2 
 0.05  0.1340
2


 2.58  2.42 
A 3 
 0.05  0.1250
2


 2.42  2.3 
A 4 
 0.0425  0.1003
2


____________
AT = 0.5048
therefore X2 = 0.4075
Answers:
XA = 0.4075
yA = 0.845
11.3-3 Difeerential Distillation of Benzene-Toluene. A mixture containing 70 mol % benzene and 30 mol % toluene is distilled under
differential conditions at 101.32 kPa (1 atm). A total of one third of the moles in the feed is vaporized. Calculate the average composition
of the distillate and the composition of the remaining liquid. Use equilibrium data Table 11.1-1.
Given:
F = 0.7 benzene and 0.3 toluene
Solution:
OMB:
B = F – D = F – 1/3F = 2/3F
X
2
L1
dx
ln

L 2 X1 y  x
ln
L1
F
 ln
 0.4055
2
L2
F
3
from the graph:
 6.68  5.92 
A1  
  0.05  0.3150
2

 5.92  5.66 
A2 
  0.0175  0.10
2

______________
AT = 0.4163
Therefore x2 = 0.6325
Answer:
xA = 0.6325 and yA = 0.7975
xB = 1 – 0.6325 = 0.3675 and yB = 1 – 0.7975 = 0.2025
11.4-1 Distillation Using McCabe-Thiele Method. A rectification column is fed 100 kg mol/h of a mixture of 50 mol % benzene and 50
mol % toluene at 101.32 kPa abs.pressure. The feed is liquid at the boling point. The distillate is to contain 90 mol % benzene and the
bottoms 10 mol % benzene. The reflux ratio is 4.52:1. calculate the kg mol/h distillate, kg mol/ h bottoms and the theoretical number of
trays needed using the McCabe –Thiele method.
Given:
F = 100 kg mol/h
0.5 benzene
0.5 toluene
R = Ln/D = 4.52
XD = 0.9
XB = 0.10
Solution:
F=D+B
100 = D + B
XF F = DXD + BXB
100 ( 0.5 ) = 0.9D + 0.1 ( 100 – D )
50 = 0.9D – 0.1D + 10
40 = 0.8D
D = 50 kg mol/h
B = 50 kg mol/h
R= 4.52
R
Ln
D
 R 
 1 
Yn 1  
Xn  

XD
1  R 
 R  1
 4.52 
 1 
Yn  1  
Xn  

  0.9
 5.52 
 5.52 
if x = 0 ,Yn+1 = 0.16
( 0,0.16)..........plotted on the graph
Answer:
4.9 trays + a reboiler
( see preceeding graph )
11.4-2. Rectification of a Heptane-Ethyl Benzene Mixture. A saturated liquid feed of 200 mol/h at the boiling point containing 42 mol
% heptane and 58 mol% ethyl benzene is to be fractionated at 101.32 kPa abs to give a distillate containing 97 mol% heptane and a
bottoms containing 1.1 mol% heptane. The reflux ratio used is 2.5:1. Calculate the mol/h distillate. Mol/h bottoms, theoretical number of
trays, and the feed tray number. Equilibrium data are given below at 101.32 kPa abs pressure for the mole fraction n-heptane x H and yH.
Temperature
K
409.3
402.6
392.6
Temperature
ºC
136.2
129.4
119.4
xH
0
0.08
0.250
Given:
F = 200 mol/h
0.42 heptane
0.58 ethyl benzene
XD = 0.97 heptane
XB = 0.011 heptane
R = Ln/D = 2.5
Solution:
F=D+B
200 = D + B
XF F = DXD + BXB
200( 0.42 ) = 0.97D + 0.011 ( 200 – D )
84 = 0.97D - 0.011 + 2.2
81.8 = 0.959D
D = 85.2972 mol/h
B = 114.7028 mol/h
R = 2.5
yH
0
0.230
0.154
K
383.8
376.0
371.5
ºC
110.6
102.8
98.3
xH
0.485
0.790
1.000
yH
0.730
0.904
1.000
 2.5 
 X n    1   X D 
Yn 1  

 3.3 
 3.5 
= 0.277
Answer:
10.77 theoretical stages or 9.77 trays + a reboiler
feed enters at 6yh tray from the top
11.4-3. Graphical Solution for Minimum Reflux Ratio and Total Reflux. For the rectification in problem 11.4-1, where an equimolar
liquid feed of benzene and toluene is being distilled to give a distillate of composition X D = 0.90 and a bottoms of composition XW = 0.10,
calculate the following using graphical methods.
(a) Minimum reflux ratio Rm.
(b) Minimum number of theoretical plates at total reflux.
Solution:
From the graph x’ = 0.5 and y’ = 0.7125
x  y'
Rm
0.96  0.5
 D

Rm  1 x D  x' 0.96  0.7125
(a)
Rm – 0.4687512Rm = 0.46875
Rm = 0.88
(b)
4 trays + a reboiler ( see preceeding graph )
11.4-4. Minimum number of Theoretical Plates and Minimum Reflux Ratio. Determine the minimum reflux ratio Rm and the
minimum number of theoretical plates at total reflux for the rectification of a mixture of heptane and ethyl benzene as given in Problem
11.4-2. Do this graphical methods of Mc Cabe-Thiele.
Given:
y’ = 0.68
x’ = 0.42
Solution:
Rm
0.97  0.68

Rm  1 0.97  0.42
Rm = 0.5273 ( Rm + 1 )
0.4227Rm = 0.5273
Rm = 1.12
Minimum number of trays: 7.64 theoretical stages or 6.64 trays + a reboiler
Foust
7.27. A mixture containing 30 mol% benzene and 70 mole% toluene is to be fractionated at normal atmospheric pressure in a column
with a total condenser and a still from which the bottoms are withdrawn. The distillate is to contain 95 mole percent benzene and the
bottoms 4 mole percent benzene. The feed is at its dew point.
(a) What is the minimum reflux ratio ( LO / D )
(b) What is the minimum number of equilibrium stages in the column required at total reflux?
(c) How many equilibrium stages are required at a reflux ratio of 8?
(d) How many equilibrium stages would be required at a reflux ratio of 8 if the feed were a liquid at its bubble point?
Given:
F = 0.3 benzene and 0.7 toluene
xD = 0.95 benzene
xW = 0.04 benzene
q=0
Solution:
(a)
x  y'
Rm
0.95  0.3
 D

Rm  1 x D  x' 0.95  0.155
Rm = 0.8176Rm + 0.8176
Rm = 4.4827
(b) Minimum number of equilibrium stages @ Total Reflux = 6.89 stages
(c) R = 8
 R 
Yn 1  
 X n    1   X D 

1  R 
 R  1
8
1
Yn 1   (X n )     0.95
9
 9
@ Xn = 0 , Yn+1 = 0.1056
@ Xn = 0. 95 , Yn+1 = 0.95
Number of equilibrium stages @ reflux = 8 :
8.79 stages
(d) R = 8, feed is liquid at boiling point
Number of equilibrium stages = 8.75 stages
7.35. An equimolar mixture of ethanol and water is to be fractionally distilled to produce a distillate of composition 0.80 mole fraction
ethanol and bottoms of 0.05 mole fraction of ethanol. The feed is saturated liquid and there is a total condenser and a total reboiler.
(a) At areflux ratio of 2.0, how many equilibrium stages are required?
(b) How many stages are required at total reflux?
(c) What is the minimum refux ratio?
(d) What percentage of the feed ethanol is recovered in the distillate?
SOLVED PROBLEMS:
1.
kg H2O

An air stream at 90 C having a humidity of 0.03 kg dry air is contacted in adiabatic humidifier. It is cooled & humidified at 90 %
3
RH. Determine the temperature of the humidified air & the make up H O for every m
2
of inlet air.
GIVEN:
COOLER
T1 = 90 OC
H=
kg H2O
0.03
kg dry air
RH2 = 90 %
REQUIRED:
T2
L in every m3 of air
SOLUTION:
Step 1: From Psychrometric Chart:
Step 2: Compute for humid volume.
H2O
balance:


L=W 
VH


= 


=
H2 
2.83 x 10
3
2.83 x 10
H1



 4.56 x 10
3
3
 4.56 x 10
3
H


 TK



0.03
3
VH
= 1.0769
m
kg da
3
Basis: 1
m

kg da 

3 
m
1.0679
m


W=1
3
W = 0.9285 kg da






90  273






L=W



H2 
H1



kg H2O
= 0.9285 kg da ( 0.51-.0.03) kg da
H2O
L = 0.0195 kg
2.
3. 200 m3/h of air at 300C and 80% relative humidity is to be heated to 500C using a heater. Calculate the heat load of the heater in KW.
GIVEN:
qH
T1 = 300C
T2 = 50 0C
HEATER
RH = 80%
Vf =200 m3/h
REQUIRED:
qH
OLUTION:
Step 1: From Psychrometric Chart
Step 2: Compute for the Humid Volume
VH = ( 2.83 x 10–3 + 4.56 x 10–3 H)(T+273)
= [2.83 x 10–3 + (4.56 x 10–3 x 0.022)] (30+273)
VH = 0.8879 m3/ kg dry air
Step 3: Calculate Mass flowrate of air
3


m
 200
 1kg dry air
hr   0.8879m3 

W=
W = 225.25 kg dry air/hr
Step 4: Calculate the specific heat
cs = (1.005 + 1.88 H)
cs = [1.005 + 1.88 ( 0.022) ]
cs
kJ
= 1.046 kg k
Step 5: Calculate the heat load of the heater

q H 225.25
=
kg
kJ 
  1.0464

hr 
kg K
kJ
q H 4714.032 hr
4.
An air conditioning unit must keep the air inside the room at 25 0C and 90% relative humidity. The air coming from the air
conditioning unit has a T of 15 0C. Calculate the hp rating of the air conditioning unit if the room space is 75m 3, the air in the room is
changed every 15 min.
GIVEN:
AC
qH
ROOM
25 0C
90%RH
qc
25 0C
COOLER
15 0C
L
REQUIRED: qc
SOLUTION:
Step 1: Compute for h1 & h2 , then vH
h1  1.005



h1  1.005








+




 1.88H
1
+
3
x10  t1  0
3
6
+ 2.501
x10 H1
6

1.880.0184 x10 25  0 + 2.501 x10 0.0189




J
h1 72,008.2 kg
h2  1.005



h2  1.005








+




 1.88H
2
+
3
x10  t1  0
3


1.88 0.0097 x10 15  0









6
+ 2.501


+
x10 H2
6
2.501 x10  0.0097


J
h2 39,608.79 kg


vH  2.83x10  3  4.56x10  3H1 t1  273




vH 2.83x10  3  4.56x10  30.0184 
3
m

vH 0.8683 kg dry air
Step 2: Compute for W.
3


75 m  kg dry air360 min hr
W

15 min  0.8683 m  1
kg dry air
hr
W=345.5027
Step 3: Solve for L, and finally qc.
L WH1  H2
(25 + 273)




L 86.3757

 kg H2O 
kg dry air


 0.184  0.0097
hr
 kg dry air

kg H2O
hr
L=3.0059
TW = 115 0F


q C Wh1  h2 Lc pLt2  t1
-

q C  345.5027


H O
kg dry air 72,008.2  39,608.79
 3.0059 kg 2 

kJ
hr
J
hr 



4.187x10  3 kg K
kg


(15-25) K
1 hr
7 J
q C 1.132x10 hr 3600 s
x
q c 4.2168
5.
hp
A chemical plant in Baguio is going to build a system that will contain ambient air to 200 0F and 115 0F wet bulb. The air in
Baguio is foggy and at an average T of 70 0F, the analysis shows that 0.0008 lb H2O per cubic feet of air is entrained. The system shall
consist of a heater, then an adiabatic humidifier and a reheater system. The exhaust of adiabatic humidifier has a humidity of 90% RH.
What is the T of air leaving the preheater and humidifier.
GIVEN:
H1=H2
TW2=TW3
H3=H4
90% RH
PREHEATER
ADIABATIC
HUMIDIFIER
70 0F
REHEATER
TD = 200 0F
0.0008 lbH2O
3
ft air
REQUIRED: T of air leaving the preheater and humidifier
SOLUTION:
Step 1: From Psychrometric Chart
Tw = 115 oF
H3=H4
0.048 lb H2O/lb da
Td = 200 oF
90 % RH
Tw2=Tw3=1
03.8 oF
H3=0.048
lbH2O/lb da
Step 2: Solve for H2.
T3= 100 oF
H2O Balance at Humidifier:
WH2 = WH1 + L
L = W( H3 – H2)
L = W (0.048 – H2)
eqn. 1
1
W
vH
eqn. 2
0.048  H2 70  460
0.0252  0.0405H2
0.0008=






0.0367 lb H2O
H2
lb dry air
=
Step 3: Use H2 to read T2 from Psychrometric Chart.



Tw2 =103.8 0F
H2 = 0.0367
LbH2O/lb da
6.
T2 =152 oF
In a plant lab having a floor area of 100 m 2 and a ceiling height of 3 m, the T and RH are kept at 23.9 0F and 80 %
respectively. The closed loop air conditioning unit installed for the purpose has an air capacity to change the air in the room, which 80 %
is void space, every 10 minutes. The air leaving the condenser f the aircon unit has a T of 18.3 0C. Calculate the quantity of condensate
which has to be dashed from the aircon unit in kg/hr
GIVEN:
AC
ROOM
2
A=100 m
H=3m
23.9 oC
80%RH
80% void
10 min
qH
23.9 oC
80 % RH
18.3 oC
COOLER
L
REQUIRED: L
SOLUTION:
Step 1: From Psychrometric Chart.
80% RH
H=0.0151
H=0.0106
18.3 oC
o
23.9
C
Step 2: Compute for vh.




vH 2.83x10  3  4.56x10  30.0151 
(23.9 + 273)
3
vH
=
0.8607 m
kg dry air
Step 3: Solve for W and then L, the condensate.
 kg
 1 
2
dry air
 100m x3m

 0.8607 
 0.8
W 
 435.71
kg dry air60 min


10 min
  1hr 
W 

W=2614026 kg dry air/ hr
L = W (H1-H2)
 2614.26 kg

hr
L =
da0.0151  0.0106kg H2O 


kg da



11.7642 kg H2O
hr
L=


GEANKOPLIS:
9.3-1. Humidity from Vapor Pressure
The air in a room is at 37.8 oC and a total pressure of 101.3 kPa abs containing water vapor with a partial pressure pA = 3.59
kPa. Calculate:
a. Humidity
b. Saturation humidity and percentage humidity
c. Percentage relative humidity
GIVEN:
ROOM
T= 37.8  C
PT = 101.3 KPa
PA = 3.59 Pa
REQUIRED:
Humidity, Saturation humidity & Percentage Humidity,
Percentage Relative Humidity
SOLUTION:
Step 1: Calculate H.
PA
MA
P  PA x MB
H= T
KPa
18


H = 3.59 101.3  3.59 KPa X 29



kg H2O
H = 0.0228 kg dry air
Step 2: Compute Hs & Hp.
From steam table, by interpolation:
PAs of H2O = 6.59 KPa
Hs
=
p As
MA
P T  P As MB
x
Hs
=



6.59 KPa
18

101.3  6.59 KPa X 29
kg H2O
HS
= 0.0432 kg dry air
Hp
=
H
Hs
x 100
kg H2O
kg dry air
kg H2O
0.0432
kg dry air
0.0228
Hp
=
Hp
x 100
= 57.78 %
Step 3: Solve for HR.
HR
=
PA
P As
x 100
3.59 KPa
HR 6.59 KPa
=
x 100
HR
= 54.48 %
9.3-2. Percentage and Relative Humidity
The air in a room has a humidity H of 0.021 kg H2O/ kg dry air at 32.2 oC and 101.3 kPa abs. Pressure. Calculate:
a.
Percentage humidity Hp
b.
Percentage relative humidity HR
GIVEN:
ROOM
T= 32.2 C
PT = 101.3 KPa
H = 0.021 kg H2O/ kg da
REQUIRED:
Percentage Humidity & Percentage Relative Humidity
SOLUTION:
Step 1: Compute for PA.
PA
MA
P  PA x MB
H= T
PA
18
101.3  P A 29
0.021 =
x
61.69 - 0.609
61.69
PA
PA
PA
= 18
PA
= 18.609
= 3.32 KPa
Step 2: Compute for Hs to solve HP & HR.
From steam table, by interpolation:
PAs of H2O = 4.82 KPa
p As
MA
Hs P T  P As MB
=
x
4.82 KPa
18
Hs 101.3  4.82 KPa 29
=
X
HS
kg H2O
= 0.031 kg dry air
Hp
=
H
Hs
x 100
kg H2O
kg dry air
kg H2O
0.031
kg dry air
0.021
Hp
=
x 100
Hp
= 67.74 %
HR
=
PA
P As
x 100
3.32 KPa
HR 4.82 KPa
=
x 100
HR
= 68.88 %
9.3-3. Use of the Humidity Chart
The air entering a dryer has a temperature of 65.6 oC (150 oF) and a dew point of 15.6oC (60 oF). Using the humidity chart,
determine the actual humidity and percentage humidity. Calculate the humid volume of this mixture and also calculate CS using SI and
English units.
GIVEN:
Td = 65.6 C
Dp = 15.6 C
DRYER
REQUIRED:
Actual humidity and % humidity
Humid volume & humid heat in SI & English units.
SOLUTION:
Step 1: From Psychrometric Chart.
Dp = 60 F
5.3 % RH
H = 0.011
kg H2O/ kg da
Td = 150 F
Step 2: Compute for humid heat.
CS
= 1.005 +
1.88H
= 1.005 + 1.88 (0.011)
CS
KJ
= 1.026 kg dry air K
CS
= 0.24 +
( SI )
0.45H
= 0.24 + 0.45 (0.011)
CS
= 0.245
lb m
Btu
dry air  F
( English )
Step 3: Solve for VH.
vH


vH = 


=
2.83 x 10
2.83 x 10
3
3
 4.56 x 10
 4.56 x 10
3



3
0.011



H





 TK
( 65.6 + 273 )
3
m
vH
= 0.975 kg dry air
vH


=
( SI )
0.0204  0.0405 H



T

F
vH = ( 0.0204 + 0.0405(0.011) ) ( 150 + 460 )
3
vH
= 15.64
lbm
ft
dry air
( English )
9.3-4. Properties of Air to a Dryer.
An air-water vapor mixture going to a drying process has a dry bulb temperature of 57.2 oC and a humidity of 0.030 kg H 2O/kg
dry air. Using the humidity chart and appropriate equations, determine the percentage humidity, saturation humidity at 57.2 oC, dew
point, humid heat and humid volume.
GIVEN:
DRYER
T= 52.7 C
H = 0.030 kg H2O/kg da
REQUIRED:
Percentage humidity, saturation humidity at 57.2 oC, dew point,
humid heat & humid volume
SOLUTION:
Step 1: From Psychrometric Chart.
Step 2: Look for PAs at 57.2 oC and solve for HS & HP.
From steam table, by interpolation:
PAs of H2O at 57.2 oC = 17.60 Kpa
p As
MA
Hs P T  P As MB
=
x
=



17.60 KPa
18



101.3 17.60 KPa x 29
kg H2O
HS
= 0.1305 kg dry air
HP
=
H
HS
=
0.030
0.1305 x 100
HP 
23 %
Step 3: Compute for humid heat in SI and English.
CS
= 1.005 +
1.88H
= 1.005 + 1.88 (0.030)
CS
KJ
= 1.0614 kg dry air K
CS
= 0.24 +
( SI )
0.45H
= 0.24 + 0.45 (0.030)
CS
= 0.2535
lb m
Btu
dry air  F
( English )
Step 4: Solve for VH.
vH


vH = 


=
2.83 x 10
2.83 x 10
3
3
 4.56 x 10
 4.56 x 10
3



3
H
0.030



 TK


 ( 57.2 + 273 )
3
vH
vH
= 0.9796


=
m
kg dry air
( SI )
0.0204  0.0405 H



T

F
vH = ( 0.0204 + 0.0405(0.030) ) ( 134.96 + 460 )
3
vH
= 15.72
lb m
ft
dry air
( English )
9.3-5. Adiabatic Saturation Temperature
Air at 82.2 oC and having a humidity H=0.0655 kg H2O/kg dry air is contacted in an adiabatic saturator with water. It leaves at
80% saturation.
a.
What are the final values of H and T oC?
b.
For 100% saturation, what would be the values of H and T?
GIVEN:
T= 82.2 oC
H = 0.0655
kg H2O/kg da
REQUIRED:
Final values of H and T oC
ADIABATIC
SATURATOR
80% RH
Values of H and T at 100 % saturation
SOLUTION:
Step 1: From Psychrometric Chart.
80 % RH
Tw = 120 oF
H2 = 0.079
kg H2O/kg da
H1 = 0.0655
kg H2O/kg da
T2 = 52.78 oC
or 127 oF
T1 = 82.2 oC
or 180 oF
Step 2: For 100 % saturation, take readings in the Psychrometric Chart.
H2 = 0.0802
kg H2O/kg da
H1 = 0.0655
kg H2O/kg da
T2 = 49 oC
or 120 oF
T1 = 82.2 oC
or 180 oF
9.3-6 Adiabatic Saturation of Air
Air enters an adiabatic saturator having a temperature of 76.7 oC and a dew-point temperature of 40.6 oC. It leaves the
saturator 90 % saturated. What are the final values of H and T oC?
GIVEN:
90% RH
Td = 76.7 oC
ADIABATIC
SATURATOR
REQUIRED:
Dp = 40.6 oC
Final values for H and T oC
4
SOLUTION:
Step 1: From Psychrometric Chart.
Tw = 113 oF
90 % RH
Dp = 105 oF
or 40.6 oC
H2 = 0.0075
kg H2O/ kg da
T2 = 116 oF
or 46.7 oC
T1 = 140 oF
or 76.7 oC
9.3-7 Humidity from Wet and Dry Bulb Temperature
An air-water vapor mixture has a dry bulb temperature of 65.6 oC and a wet bulb temperature of 32.2 oC. What is the
humidity of the mixture?
GIVEN:
Td = 65.6 oC & Tw = 32.2 oC
REQUIRED:
Humidity
SOLUTION:
Step 1: From Psychrometric Chart.
Tw = 32.2 C
or 90 F
Td = 65.6 oC
or 149.9 F
H = 0.0175
kg H2O/kg da
9.3.8
Humidity and Wet Bulb Temperature
The humidity of an air-water vapor mixture is H = 0.030 kg H2O/kg dry air. The wet bulb temperature of the mixture of 60 oC.
What is the wet bulb temperature?
GIVEN:
H = 0.030 kg H2O/kg da & Td = 60 C
REQUIRED:
Wet bulb temperature, Tw
SOLUTION:
Step 1: From Psychrometric Chart.
Tw = 98 F
or 36.67 C
H = 0.030
kg H2O/kg da
Td = 140 F
or 60 C
9.3.10
Cooling and Dehumidifying Air
Air is entering an adiabatic cooling chamber has a temperature of 32.2 oC and a percentage humidity of 65%. It is cooled by a
cold water spray and saturated with water vapor in the chamber. After leaving, it is heated to 23.9 oC. The final air has a percentage
humidity of 40%.
(a)
What is the initial humidity of the air?
(b)
What is the final humidity after heating?
GIVEN:
ADIABATIC
COOLING
CHAMBER
T1 = 32.2 C
RH1 = 65 %
REQUIRED:
Initial and final humidity
SOLUTION:
HEATER
T3 = 23.9 C
RH1 = 40 %
Step 1: Obtain H1 from Psychrometric Chart.
65 % RH
H1 = 0.02
kg H2O/ kg da
T1 = 32.2 C
Step 2: Obtain H3 from Psychrometric Chart.
40 % RH
H3 = 0.0075
kg H2O/kg da
T3 = 23.9 C
PROBLEMS
(GEANKOPLIS)
9.6-1) Time for Drying in Constant-Rate Period. A batch of wet solid was dried on a tray dryer using constant drying conditions and a
thickness of material on the tray of 25.4 mm. Only the top surface was exposed. The drying rate during the constant rate period was R =
2.05 kg H2O/ h – m2 ( 0.42 lb H2O/ h – ft2 ). The ratio Ls/A used was 24.4 kg dry solid/ m2 exposed surface ( 5 lbm dry solid/ft2 ). The
initial free moisture was X1 = 0.55 and the critical moisture content Xc = 0.22 kg free moisture/ kg dry solid.
Calculate the time to dry a batch of this material from X1 = 0.45 to X2 = 0.3 using the same drying conditions but a thickness of
50.8 mm, with drying from the top and bottom surface.
Given:
R = 2.05 kg H2O/h-m2
Ls/A = 24.4 kg d.s/m2
X1= 0.45 kg free moisture/kg d.s
X1 = 0.55 kg free moisture/kg d.s
Xc = 0.22 kg free moisture/kg d.s
X2 = 0.3 kg free moisture/kg d.s
Required:
Time (t)
Solution:
t
Ls 
XC 
 X1  X2   XC ln

A 
X2 
For Ls/A of 2nd condition:
Since
Ls VA  V(m) V(m)



A
A
vol
bA
 Ls   Ls 
   
 A 1  A  2
V (m)
V(m)

25.4 A 1 (50.8)A
2
V (m)
V (m )

25.4 A 25.4 A
So,
Therefore,
kg d.s
 Ls   Ls 
      24.4
m2
 A 1  A  2
t
Ls
 X1  X2 
AR C
kg d.s
m2  0.45  0.3  kg H2O
t
kg H2O
kg d.s
2.05
2
h m
24.4
t = 1.7854 hrs
9.6-2) Prediction of Effect of Process Variables on Drying Rate. Using the conditions in example 9.6-3 for the constant rate
drying period, do as follows:
a. Predict the effect on Rc if the air velocity is only 3.05 m/s
b.
Predict the effect if the gas temperature is raised tom 76.7ºC and it remains the same.
Given:
From example 9.6-3
A = 0.457 x 0.457 m
b = 25.4 mm
v = 6.1 m/s
T db = 65.6ºC
H = 0.010 kg H2O/kg d.a
Required:
a.
Rc if v = 3.05 m/s
b.
Rc with T = 76.6ºC
Solution:
a)
From ex. 9.6-3
 1.037
kg
m3
G  v
3.05
m
s 
kg 
 3600  1.037 3 
s
hr 
m 
G = 11386.26 kg m2/h
h  0.0204G 0.8
0.020411386.26
W
h  35.8703 2
m  K
0.8
h
 T  TW 
W
35.8703
 65.6  28.9 3600

24331000
Rc 
Rc = 1.9479 kg/ m2-h
b)
@ T = 76.7ºC
H = 0.010 kg H2O/kg d.a


 2.83x10   4.56 x10  0.010   273  76.7 
VH  2.83 x10  3  4.56 x10  3 H T
-3
3
VH = 1.0056 m3/kg da

1.0  0.010
kg
1.0044 3
1.0056
m
G  v
6.1
m
s 
kg 
 3600  1.0044 3 
s
hr 
m 
G = 22056.1447 kg/ h-m2
h  0.0204 G0.8
h
 0. 8
Rc 0.0204
 T  T22056.1447
W
W
W
h 60
60
.8771 2
.8771

 76.7K 28.9  3600 
m
2433 1000 
Rc = 4.21 kg/h-m2
9.6-3) Prediction in Constant-Rate Drying Region. A granular insoluble solid material wet with water is being dried in the constant rate
period in a pan 0.61 m x 0.61 m and the depth of material is 25.4 mm. The sides and bottom are insulated. Air flows parallel to the top
drying surface at a velocity of 3.05 m/s and has fry bulb temperature of 60 C and wet bulb temperature of 29.4 C . The pan contains
11.34 kg of dry solid having a free moisture content of 0.35 kg H2O per kg dry solid and the material is to be dried in the constant-rate
period to 0.22 kg H2O / kg dry solid.
a. Predict the drying rate and time in hours needed.
b. Predict the time needed if the depth of mat’l is increased to 44.5 mm
Given:
A = 0.61m x 0.61m
b = 25.4 mm
v = 3.05 m/s
Tdb = 60C
Twb = 29.4C
X1 = 0.35 kg H2O/ kg d.s
X2 = 0.22 kg H2O/ kg d.s
Ls = 11.34 kg d.s
Required:
t, Rc
Solution:
a.) @ Tdb = 60C, H = 0.0141 kg H2O/ Kg d.a

 

 2.83 x10   4.56 x10  0.0141 60  273 
VH  2.83 x10  3  4.56 x10  3 H T
VH
3
3
VH 0.9638 m3 / kg d.a
m 1.0 kg d.a  0.0141 kg H 2 O
m
m 1.0141  
VH
1.0141
0.9638
 1.0522 kg/m 3

G v(3600 )
G  3.05 1.0522 (3600 )
G 11,553 .031 kg/h - m 2
h 0.0204 G0.08
h 0.0204(11,553 .031)0.08
h 36.29 W/m2  K
From steam table: @ Twb = 29.4C , hw = 2432.14 x 103
RC 
h
 T  Tw  3600 
hw
RC 
36.29
 60  24.9  3600 
2432 .14 x103
Rc = 1.6437 kg/ h- m2
t
Ls
 X1  X 2 
AR c
t
11.34
0.35  0.22 
 0.61x0.611.6437 
t = 2.4103 hrs
b.) by ratio & Proportion:
t1 b1

t 2 b2
t2 
t1b2  2.4103hr  44.5mm

b1
25.4mm
t2 = 4.2228 hrs
9.6-4) Drying a Filter Cake in the constant-rate region. A wet filter cake in a pan 1 ft x 1 ft square and 1 in. thick is dried on the top
surface with air at wet bulb temperature of 80F and a dry bulb of 120F flowing parallel to the surface at a velocity of 2.5 ft/s. The dry
density of the cake is 120 lbm/ ft3 and the critical free moisture content is 0.09 lb H2O / lb dry solid. How long will it take to dry the
material from a free moisture content of 0.20 lb H2O / lb dry material to the critical moisture content?
Given
A = 1 ft2
t = 1in
Twb = 80F
Tdb = 120F
V = 2.5 ft/s
 = 120 lbm/ft3
Xc = 0.90 lbm H2O/ lbm d.s
X1 = 0.20 lbm H2O/ lbm d.s
Required:
t
Solution:
@ Twb = 80F Tdb = 120F : H = 0.013 kg H2O/ kg da
VH  0.0252  0.0405  T R
VH   0.0252   0.0405  0.013    580 
VH 14.92
ft 3
lbmda

1 
VH

1  0.013
lbm
0.0679 3
14.92
ft
G 
G  2.5  3600  0.0679  1,080,000
lbm
h  ft 2
Rc 
h
 T  Tw  3600
w
h 0.0128 G 0.8
h 0.0128  656.1 2.1681
Btu
h  ft 2   F
@ Twb = 80F ; w = 1048.4 (from steam table)
2.2950
120  80  3600 
1048 .4
lbm
Rc 0.0827
h  ft 3
Rc 
tc = 13.30 hrs
tc 
s X  Xc 
Rc
1
120  0.20  0.09 
t c  12
0.0827
9.7-1) Graphical Integration for Drying in Falling-Rate Region. A wet solid is to be dried in a tray under steady state conditions from a
free moisture content of X1 = 0.4 kg H2O/ kg d.s to X2 = 0.02 kg H2O/ kg d.s. The dry solid weight is 99.8 kg dry solid and the top surface
area for drying is 4.645 m2. The drying rate curve can be represented by Fig. 9-.5-1b.
Calculate the time for drying, but use a straight line through the origin for the drying rate in the falling rate period.
Given:
X1 = 0.4
X2 = 0.02
Ms = 99.8 kg d.s
A = 4.645 m2
Required:
T using straight line through the origin
Solution:
Using Fig 9-5-1b;
Rc = 1.51 kg H2O/h-m2
Xc = 0.1950
mS
 X1  XC 
ARc
99.8
tC 
 0.4  0.1950 
4.645 1.51
tC 
tC = 2.9169 hrs
tF = 6.3185 hrs
tF 
mS
X
XC ln C
AR C
X2
tF 
99.8
 0.1950  ln 0.1950
4.6451.51
0.02
t T  t C  tF
t T  2.9169  6.3185
tT = 9.2354 hrs
9.7-2) Drying Tests with a Foodstuff. In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer
with air flow over the top exposed surface having an area of 0.186m2. The bone-dry sample weight was 3.765 kg dry solid. At
equilibrium after a long period, the wet sample weight was 3.955kg H2O + solid. Hence, 3.955-3.765, or 0.190, kg of equilibrium
moisture was present. The following sample weights versus time were obtained in the drying test.
______________________________________________________________________
Time(h)
Weight(kg)
Time(hr) Weight(kg)
Time(hr) Weight(kg)__
0
4.944
2.2
4.554
7.0
4.019
0.4
4.885
3.0
4.404
9.0
3.978
0.8
4.808
4.2
4.241
12.0
3.955
1.4
4.699
5.0
4.150
______________________________________________________________________
Calculate the free moisture content X kg H2O/kg d.s. for each data point and plot X versus time. (Hint: For 0h, 4.944-0.1903.765=0.989) kg free moisture in 3.765kg dry solid, Hence, X=0.989/3.765)
SOLUTION:
Getting the free moisture content: (at the given time and weight)
Free moisture content wet sample weight- equilibrium moisture content-bone dry
Free moisture content:
.267
0.2470
0.2266
0.1976
0.1591
0.1193
0.07596
0.0518
0.01699
601088x10-3
0
sample weight
9.8-2) Drying when radiation, Conduction, and Convection are present. A material is granular and wet with water and is being dried in a
layer 25.4 mm deep in a batch-tray dryer pan. The pan has a metal bottom having a thermal conductivity of km=43.3 W/m. K and a
thickness of 1.59mm. The thermal conductivity of the solid is Ks= 1.125 W/m.K. The air flows parallel to the top exposed surface and the
bottom metal at a velocity of 3.05m/s and a temperature of 60oC and humidity H=0.010kgH2O/kg dry solid. Direct radiation heat from
steam pipes having a surface temperature of 104.4oC falls on to exposed top surface, whose emissivity is 0.94. Estimate the surface
temperature and the drying rate for the constant-rate period.
GIVEN:
Z s 0.0254m
K m 43.3 W / mK
Z m 0.00159 m
K s 1.125 W / mK
V 3.05m / s
T 60 o C
H 0.010kgH 2 O / kgd.s.
TR 104 .4 0 C ~
377 .4K
 0.94
Required:
Ts and Rc
Solution:
1  0.010
3
0.8
1.0547
kgG
/m
h
0
.
0204
c
0.9576
G sv
 3600
G  3.05
(1.0547 )
 h 1  H
kgVH
11,580.606
VHh.m
(22 .83 x10  3  4.56 x10  3 )(T  273 )
hc  0.0204(11580
.606 )0.8 -3  4.56 x10  3 (0.010 )](60  273 )
[(2.83x10

3
W
o
306K
hc  36.3593 VH2 0.9576 m / kgda Tsshould be  Tw   Ts 33 C ~
mK
4
4
 TR 
 T 

   s 
100 
 100 
from psychrometric chart
hR (5.676 ) 
TR  Ts
@ Td  60 o C;H  0.01kgH 2O / kgd.a.
4
4
 377 .4 
 306 
Tw  28.8889 o C

  

100 
 100 
hR 0.94(5.676 ) 
377 .4  306
W
m2K
at Ts 33o Cfrom steam table :
hR 8.6076
 s 2423 .37KJ / kg
hc / k yMB ~
Cs
Cs [1.005  1.88H]103
[1.005  1.88(0.010 )]103
1023.8 J/kgK
1
UK  W
19.9530 12 Zm Zs
mh K K  K
c
m
s
1
 chart : @ T  33 o C,
by psychrometric
s0.00159
1
0.0254


kgH 236
O .3593
43.3
1.125
Hs  0.0328
kgd.a.
for trial 1
Hs - H s
 U 
h
  1  k  T  Ts   R  TR  Ts 
hc / kyHB
hc 
he

 0.0328  0.01 2423 .37 1000    1  19.9530  60  T   8.6076 (104.4  T )


s
s
1023 .8
36.3593 
36.3593

Ts  31.8355 o C
33 o C  31.8355 o C
TRIAL 2 :
Ts  32.3  305 .3K
4
 377 .4 

 
100 
hR  0.94(5.676 ) 
377 .4 
hR  8.5832 W / m2K
 305 .3 


 100 
305 .3
4
@ Ts  32.3o C
 s  2425 .036KJ / Kg
@ Ts  32.3o C;Hs  0.0312 kgH 2O / Kgd.a.
(0.0312  0.01)2425 .036
8.5832
104 .4  Ts 
1.5488( 60  Ts ) 
1023 .8
36 .3593
50.2156 1.5488 (60  Ts )  0.2361(104 .4  Ts )
TS  33 .68 O C
32.3o C  33 .68 o C
Trial 3 :
Ts 32.6o C 305 .6K
4
 377 .4 

 
100 

hR 0.94(5.676 )
377.4 
 305 .6 


 100 
305 .6
4
hR  8.3936 W / m2K
@ Ts  32.6o C,  s  2424 .322KJ / Kg
(0.0318  0.01)2424 .322(1000 )
8.5936
1.5488 (60  Ts ) 
(104 .4  Ts )
1023 .8
36.3593
Ts  32.9908
TRIAL 4 :
@Ts  32 .7o C  305 .7O K
Hs  0.032;  s  2424 .084
4
 377 .4 

 
100 

hR  0.94(5.676 )
377.4 
hR  8.5971
 305.7 


 100 
305 .7
4
(0.032  0.01)2424 .084(1000 )
8.5971
1.5488( 60  Ts ) 
(104.4  Ts )
1023 .28
36 .3593
Ts  32.75 o C
Rc 
(hc  Uk )(T  Ts )  hR ( TR  Ts )
(3600 )
s
(36.3593  19.9538)(6 0 - 32.75)  [8.5991(1. 4.4 - 32.75)]
(3600 )
2424 .084 x10 3
3.1937

Rc = 3.1937 kg/h-m2
9.9-1) Diffusion Drying of Wood. Repeat example 9.9-1 using the physical properties given but the following changes.
a.
Calculate the time needed to dry the wood from a total moisture of 0.22 to 0.13. Use Fig. 5.3-13.
b.
Calculate the time needed to dry planks of wood 12.7 mm thick from Xt1 = 0.29 to Xt = 0.09. Compare with the time needed for 25.4
mm thickness.
Given:
From Ex.9.9-1
DL = 2.97 x 10-6
X* = 0.04 kg H2O/kg dry wood
x1 = 0.0127 m
a. Xt1 = 0.22
Xt = 0.13
b. Xt1 = 0.29
Xt = 0.09
x1 = 12.7 mm
Required:
a.
time using fig. 5.3-13
b.
time
Solution:
a.)
X = Xt - X*
= 0.13 – 0.04
= 0.09
X1 = Xt1 - X*
= 0.22 – 0.04
= 0.18
Ea = X/ X1 = 0.09/0.18 = 0.50
From Fig.5.3-13
DLt/x12 = 0.20
2
t

x1 (0.20 )
DL
 0.0127 2  0.20 
2.97 x10  6
t= 10.86 hrs
b.)
X1 = Xt1 –X* = 0.29-0.04 = 0.25
X = Xt –X* = 0.09-0.04 = 0.05
x1 = 12.7/ 2 (1000) = 6.35 x 10-3 m
Equation 9.9-6
2
t
4 x1
8X
ln 2 1
2
 DL  X


2
4 6.35x10 -3
8 0.25 
 2
ln 2
6
 2.97 x10
  0.05 


t = 7.70 hrs
9.10-1) Drying a Bed of Solids by Through Circulation. Repeat example 9.10-1 for drying of a packed bed of wet cylinders by through
circulation of the drying air. Use the same conditions except that the air velocity is 0.381m/s
Given:
V 0.381m / s
*
X  1 1.0kgH 2 O / kgdrysolid X1 X t1 X
1  0.01
X 0.01
 0.99kgH 2O / kgd.s.
 1602kg / m 3
ds
Xc  X tc  X*
bed thickness 50.8mm 0.0508m
 0.50 - 0.01
 s 641kg / m 3
 0.49kgH 2O / kgd.a.
H 0.04kgH O / kgd.s.
1
2
for solid cylinder :
D c 0.00635 m
h 0.0254m
X  Xt  X*
 0.10 - 0.01
 0.09kgH 2O / kgd.a.
Required:
total time
Solution:
by psychrometric chart :
if radiation
are neglected,
@T1 and
121 .conduction
1o C; H 0.04
solid temp
at.2T;w.H 0.074
T is47
w
w
VH ( 2.83 x10  3  4.56 x10  3 H)(T oK )
VH [2.83 x10  3  4.56 x10  3 (0.04 )](121 .1  273 )
VH 1.1872m3 / kgd.a.

1  0.04
 0.8760kgd.a.  H2O / m3
1.1872
1


G    
  0.381m / s(0.8760kg / m3 ) (3600 s / h)(1/ 1.04 )
 1.0  0.04 
G 1155 .3092kgd.a. / hm 2
for
heat
t :0.074
sin
ce H-1transfer
 0.040coefficien
and Hw 
assume H2  0.05
assuming Tair  93.3o C
GT 1155.3092
 1155 .3092
(0.05 )3600s
5
air  2.15 x10 kg / m.s x
hr
kgair  H2O
GT 1161 .0747
-22
 7.74x10
h.m kg / m.h
Dsolid
GT with
0.0135
(1161
.0747and
)
for dry
641kg
present
Nre 


2
 kg / m3 7.74 x10
ds 1602
 202.5130
641kg
volume 
 0.04001m3 .
using eq' n 9.10
1602- 14;
kg / m3
0.49
0.49
 1  0.G
4  0.6 0.214(1161 .0747 )
h  0.214 T0.51 
(0.0135 )0.51
D
4(1   )(h  0.5Dc )
a
Dc h :
from steam table
Tw  47.2o C,  w  2389
KJ
 2.389 x10 6 J / kg
kg
Cs 14.(005
H
1  0.61)[.88
0.0254
 0.5(0.00635 )]
a
Cs 1.005  10..88
(0.05
0254
(0.)00635 )
a 283 .4646m2
Cs 1.099KJ / kgd.a.K1 or 1.099x10 3 J / kgd.a.K
2
D (Dch  0.5Dc ) 2
D 
[0.00635(0
0.0135m .0254)  0.5(0.0063 5)2 ]
1155 .3092kgd.a. / h.m2
3600 s / h
G  0.3209kgd.a. / s.m2
G
for time of drying at constant rate period
t
t
t
s w x1( X1  Xc )
GC s ( T1  Tw )(1  e  hax / GC s )
641( 2.389 x10 6 )(0.0508 )(0.99  0.49 )
   61.0555 x 283 .4646 x 0.30508  

3

0.3209(1.099 x10 )(121 .1  47.2)1   e  0.3209 x1.099 x10




1626 s  h 

  0.4519h
0.9230  3600 s 
@ falling - rate period :
X 
s w x1Xc ln c 
 X 
t
 h a X1



GC s ( T1  Tw ) 1  e GC s 




T=1.2024 hrs
9.10-3) Through Circulation Drying in the Constant-Rate Period. Spherical wet catalyst pellets having a diameter of 12.7mm are being
dried in a through circulation dryer. The pellets are in a bed 63.5mm thick on a screen. The solids are being dried by air entering with a
superficial velocity of 0.914m/s at 82.2oC and having a humidity H=0.01 kgH2O/kg dry air. The dry solid density is determined as
1522kg/m3, and the void fraction in the bed is 0.35. The initial free moisture content is 0.90 kgH 2O/kg solid and the solids are to be dried
to a free moisture content of 0.45, which is above the critical-moisture content. Calculate the time for drying in this constant-rate period.
Given:
DP=12.7
Z=63.5
V=0.914
T=821.2C
H=0.01kgH2O/kgd.a.
P=1522kg/m3
E=0.35
X1=0.90
X2-0.45
Required:
tCRP
Solution:
tCRP =Ps (X1-X2)
ah (T-Tw)
 0.49 
641( 2.389 x10 6 )(0.0508 )0.49 ln

0.09 

t

  61.055 x 283 .4646 x 0.0508

3

0.3209 x1.099 x10 3
0.3209(1.099 x10 )(121 .1  47.2)1  e






h
t  2701 .83735 x
 0.7505h
3600 s
total time t (0.4519  0.7505)h
1.2024hrs
setting X2=Xc ; z=X1m
@T =82.2 C & H= 0.01
from P.C. Tw=32.2 C
@82.2C u=0.019x10-3
Basis:1 kg d.a.
 10.01
V
H
VH=[2.83x10-3+4.56x10-3H]TK
=[2.86x10-3+4.56x10-3(0.1)](82.2+273)
=1.02m3/kg d.a.
So,
  1.01
1.02
=0.99 kg/m3
NRE=DPVP/u


NRe  
12.7 / 1000   0.914   0.99 
0.019 x 10 3
=604.83 > 350
Gt0.59
h 0.151
DP0.49
Gt=(0.914m/s)0.99
=0.905kg/m2s

0.905 0.59
h 0.151
12.7 / 10000.49
=0.853W/m2K
 1 

S

 1   
S
1522 (1 - 0.35 )
989.3 kg/m 3
a=6(1-E)/DP
a
6(1   ) 61  0.35 

12.7 / 1000
Dp
=307.09
@Tw=32.2 C ;
t CRP 
w =2425.27kJ/kg(by interpolation)
G LS 1
H  HC
ln W
L S A K YMB
H W  H1
t CRP 
989.3 2425.27 0.9  0.451000
307.09 0.853 82.2  32.2 3600
tCRP
=7.75 h
9.10-4) Material and Heat Balances on a Continuous Dryer. Repeat Example 9.10-2 making heat and material balances, but with the
following changes. The solid enters at 15.6 oC and leaves at 60oC. The gas enters at 87.8 oC and leaves at 32.2oC. Heat losses from the
dryer are estimated as 293
Given:
Q=2931W
TG1,H1
Gas
solid
G,TG2,H2
Ls,Ts1,X1
Ts2,X2
Ls=453.6
Ts1=15.6C
X1=0.04
Ts2=60C
X2=0.002
TG2=87.8C
H2=0.01
TG1=32.2C
CpA=4.187 kJ/kgK
CpB=1.465 kJ/kgK
Required:
a.)G
b.)H1
Solution:
Material Balance:
GH2 + LsX1 = GH1 + LsX2
G(0.01) +453.6(0.04) =GH1 + 453.6(0.002)
G(0.01-H1)=-17.24
(1)
Heat Balance :
@TG2=87.8C & To=0C
H’G2= CS(TG2-To) + H2
=[1.005 + 1.88(0.01)](87.8-0) + 0.01(2501)
=114.9 kJ/kgd.a.
For the exit gas’
H’G1=CS(HG1-To) + H1
=[1.005 + 1.88H1](32.2-0) +H1(2501)
=32.36 +2561.54H1
For the entering solid,
H’S1=Cps(Ts1-To) + X1Cpa(Ts1-To)
=1.456(15.6-) + 0.04(4.187)(15.6-0)
=25.33kJ/kgd.a.
H’S2=Cps(Ts2-To) + X2Cpa(TS2-To)
=1.456(60-0) + 0.002(4.187)(60-0)
=87.86 kJ/kgd.s.
Heat Balance on dryer,
GH’G2+LsH’S1=GH’G1+LsH’S2+Q
G(114.9)+453.6(25.33)=G(32.36+2561.54H1)+39853.3+10551.6
G(82.54-2561.54H1)=38915.21
(2)
From Eq 1,
G=-17.24/(0.01-H1)
From Eq 2,
-17.24/(0.01-H1) (82.54-2561.54H1)=38915.21
-1422.99+444160.95H1=389.15-
38915.21H1
83076.16H1=1812.14
So,
H1=0.0218kgH2O/kgd.a.
Subs. To Eq1,
G=-17.24/(0.01-0.0218kgH2O/kgd.a.)
G=1461.02kgd.a./h
9.10-5) Drying in a Continuous Tunnel Dryer. A rate of feed of 700lbm dry solid/h containing a free moisture content of X 1=0.4133lb
H2O/lb dry solid is to be dried to X=0.0374 lbH 2O/lb d.s. in a continuous counterflow tunnel dryer. A flow of 13280lbm/dry air/h enters at
203oF with an H2=0.0562lbH2O/lb dry air. The stock enters at the wet bulb temperature of 119 oF and remains essentially constant in
temperature in the dryer. The saturation humidity at 119 oF from the humidity chart is Hw=0.0786 lb H 2O/lbd.a. The surface area
available for drying is (A/Ls)=0.30ft2/lbmd.s..
A small batch experiment was performed using constant drying conditions, air velocity, and temperature of the solid
approximately the same as in the continuous dryer. The equilibrium critical moisture content was found to be X c=0.0959lbH2O/lb d.s.,
and the experimental value of kyMB was found as 30.15lbm air/hr.ft2. In the falling-rate was directly proportional to X.
For the continuous dryer, calculate the time in the dryer in the constant-rate zone and in the falling-rate zone.
Given:
Ls=700lbmds/h
Tw=119F
X1=0.4133
w=0.0786lbH2O/lbda
H
X2=0.0374
A/Ls=0.30ft2/lbds
G=13280lbda/h
Xc=0.0959
TG2=203F
KyMB=30.15lbmair/hft2
H2=0.0562lbH2O/lbda
Required
a.)tCRP
b.)tFRP
Solution:
Material Balance:
Ls(Xc-X2)=G(Hc-H2)
700(0.0959-0.0374)=13280(Hc-0.0562)
Hc=0.0593lbH2O/lbda
t CRP 
G LS 1
H  HC
ln W
LS A K YMB HW  H1
GH2 + LsX1 = GH1 + LsX2
13280(0.0562)+700(0.4133)=13280H1+700(0.0374)
H1=0.076 lbH2O/lbda
So,
t CRP 
13280 1
1
0.0786  0.0593
ln
700 0.3 30.15
0.0786  0.076
tCRP=4.204h
For Falling rate zone,
G L S XC
1
X H  H2 
ln C W
L S A K YMB HW  H2 G / L S  X 2 X2 HW  HC 
13280 1 0.0959
1
0.059 0.0786  0.0562 

ln
700 0.3 30.15  0.086  0.0562 13281 / 700  0.0374 0.0374  0.0786  0.0593 
t FRP 
t FRP
tFRP=0.47h
9.10-6) Air Recirculation in a Continuous Dryer. The wet feed material to a continuous dryer contains 50 wt% water on a wet basis and
is dried to 27wt% by countercurrent air flow. The dried product leaves at the rate of 907.2 kg/h. Fresh air to the system is at 25.6ºC and
has a humidity of H = 0.007 kg H2O/kg d.a. The moist leaves the dryer at 37.8ºC and H = 0.020 and part of it is recirculated and mixed
air enters the dryer at 65.6ºC and H = 0.01. Calculate the fresh air flow, the percent air leaving the dryer that is recycled, the heat added
in the heater, and the heat loss from the dryer.
Given:
Required:
a.
fresh air
b.
% air leaving
c.
heat added (heater)
d.
heat loss (dryer)
Solution:
H2O Bal. On the heater
G1H1 + G6H2 = ( G1 +G6 ) H4
G1 (0.007) + G6 (0.020) = G1 (0.010) +0.01G6
1.1
G6 = 0.003 G1 ………….(1)
H2O Bal. On the dryer
(G1 +G6)H4 + LsX1 = (G1 + G6)H2 +LsX2
for Ls
Ls = L ( 1 – X )
= 907.2 (1 – 0.27)
= 662.256 kg d.s
dry basis:
since G3 = G4 = G1+G6 = 41728.75 d.a/h
0.5
1.0
1  0.5
0.27
X2 
0.37
1  027
X1 
at junction A
G1H1 + G6H6 = G4H4
G1  0.007    41728 .75  G1  0.02  41728 .75 0.01
G1  0.007   834.57  0.02G 1 417 .29
 G 1  0.013   417 .289
G1 = 32,099.15 kg fresh d.a /h
From eqn 1:
G1  G6 G3
G6 41728 .75  32099 .15
9629.6 kg/h
% recycled 
G6
9629.6

0.2306
G4 41728.75
% recycled = 23.06% recycled air
Heat Balance on Heater
G6H'G6  G1H'G1  G 4H'G 4   Q
H' G  Cs( TG  TD )  H o
H' G 4  Cs4 ( TG 4  TD )  H4 o
Cs 1.005  1.88(0.01) 1.0238
H' G 4 1.0238( 65.6  0)  0.01 2501
92.17 kJ/kg d.a
H' G1  Cs1(TG1  T0 )  H1 o
1.0182(25. 6 - 0)  0.007(2501 )
43.57 KJ/kg d.a
H' G 6  Cs6 (TG6  T0 )  H6  o
1.0426(37. 8 - 0)  0.02(2501)
89.43 KJ/kg d.a
subs to equation
- Q G6H'G6 G1H'G1  G4H'G 4
9629.60(89 .43)  32099.15(4 3.57) - 41722.38(9 2.17)
- 1586178.59
KJ  hr 


hr  3600 s 
Q = 440.605 KW
Amount of Heat loss from the dryer
Heat bal.
Q = 32.06 KW
MC CABE
G4H'G4  L sHS1 G2H'G2 LsHS2  Q
G2 G4 41722 .38 kg d.a/h
H'G2 Cs 2 (TG2  T0 )  H2 o
Whe re
CS2 1.0426 KJ/kg d.a
H' G2 1.0426 (37.8  0)  0.02 2501
89.43 KJ/kg d.a
H' S2 1.63781 H'S1
subs to the eqn :
41722 .38(92.17 )  662 .26(H' S2 1.63781)  41722 .38(89.43 )  662 .26H'S2 Q
Q 115403 .98
KJ  hr 


hr  3600 s 
24.1) Fluorspar (CaF2) is to be dried from 6 to 0.4 percent moisture (dry basis) in a countercurrent adiabatic rotary dryer at a rate of
18,000 lb/hr
of bone-dry solids. The heating air enters at 1000 oF with a humidity of 0.03 and wet bulb temperature of 150 oF. The solids have a
specific heat of 0.48 Btu/lb oF; they enter the dryer at 70 oF and leave at 200oF. The maximum allowable mass velocity of the air is 2,000
lb/ft2.h.
a.)
Assuming Eq.(24.8) applies, what would be the diameter and length of the dryer if Nt= 2.2? Is this reasonable design?
b.)
Repeat part (a) with Nt=1.8.
Given:
Nt ln
Thb  Twb
Tha  Twb
Twb inlet Tw
Twa outlet Tw
Thb outlet Tb
Tha outlet Tb
for Tha
2.2 ln
1000  150
850
ln
Tha  150
Thc  150
Solution:
850
also, 2.2
e 
x a  0.06 Thc  150
2. 2
x b  4 xe10 (3Tha  150 )  850
Tha  244 .18 o F
Ms 18000
needed
so, the other
rate ofquantities
mass transfer
: are :
so,
Mv Ms@150
( x a  xobF) 1008 .1Btu ( App 7)
lb
Btu
18000 (0.06  4 x10  3 ) o
qT 123.7111(18000 )  2.2268 x10 6
specific heats in Btu/lb F are,
h
Mv 1008lb / h
Cps  0.48
Cpv  0.45 (App14) CpL 1 The flow rate of entering air is found a heat balance and the humid heat Cs ,
B
The heat duty is found from subs
the eq(24.1)
qT
 Cps( Tsb  Tsa )  x acpL ( Tv  Tsa )  ( x a  xb )  x bcpL ( Tsb  Tv )  ( x a  xb )cpv ( Tv a  Tv )
M' s
where :
Tsa  feedT
Tv  vapor ' nT
Tsb  final solids T
Tv a  final vapor T
 heat of vapor
Cps , CpL , Cpv  sp. heat of solid. liquid and vapor
-3
00.48(200
- 70)
70)
CsB 
.24  0.45
 00.06(1)(15
.24  0.45(00.-03
)  (0.06 - 4x10)1008. 1  4x10 1
 150 )  (0.06  4 x10  3 )(0.45 )(244 .18  150 )
(0200
.2535
qt
Btu
123 .7111
Ms
lb
L = 35.842 ft
mg(1  Hb ) 
qt
Csb (Thb  Tha )
2.2268 x10 6
(1000  244 .18)0.2535
11622 .10699

since, Hb  0.03;mg 11622 .10699 / 1  0.03
mg 11283 .59902
lb
h
the outlet humidity
Ha Hb  mv / mg
 0.03 
1008
lb
 0.1193
11283 .59902
lb
1/ 2
 4A 
for D  

  
mg (1  Hb ) 11622 .10699
A

 5.8105 ft 2
G
2000
so,D  2.72ft
The dryer length is given by eq' n (24.23) :
qT
L
0.125 DG0.67 T
T  TL 
Thb  Twb  (Tha  Twa )
ln[( Thb  Twb ) /( Tha  Twa )]
since, Twb  Twa
1000  150  (244 .18  180 )
ln[(1000  150 ) /( 244.18  150 )]
785 .82

ln 9.0253
357.1863 o F

so,
L
2.226 x10 6
0.125 ( 2.72)(2000 )0.67 (357 .1863 )
24.2)A porous solid is dried in a batch dryer under constant drying conditions. Seven hours are required to reduce the moisture content
from 35 to 10 percent. The critical moisture content was found to be 20 percent and the equilibrium moisture content is 4 percent. All
moisture contents are on the dry basis. Assuming that the rate of drying during the falling rate period is proportional to the free moisture
content, how long should it take to dry a sample of the same solid from 35 to 5 percent under the same conditions?
Given
tT = 7 hrs
1st condition
2nd condition
x1= 35
x1 = 35
x2 = 10
x2 = 5
xc = 20
x* = 4
Required:
tT @ second condition
Solution:
Basis: 100 total moisture
Total moisture – equilibrium moisture = Free moisture
1st condition
X1: 35 – 4 = 31
X2: 10 – 4 =6
Xc: 20 – 4 = 16
2nd condition
x2 = 5 – 4 =1
at:
31 dx 31  16 15
t 


c
Rc
Rc
Rc
16
X
X
16 16
t  c ln c  ln
f Rc X
Rc Rc
2
t
T
t  t
c f
Rc = 4.3848 kg H2O/ h – m2
At 2nd condition:
X1 =31 ; X2 = 1
31 dx 31  1
30
t 


c
Rc
4.3848
1 Rc
6.84 hrs
X
X
16
16
t  c ln c 
ln
f Rc X
4.3848 1
2
10.12 hrs
t
T
 6.84  10.12
tT = 16.36 hrs
FOUST
18.10
A rotary countercurrent dryer is fed with wet sand containing 50 percent moisture and is discharging sand containing 3 percent
moisture. The entering air is at 120Cand has an absolute humidity of 0.007g H2O/g air. The wet sand enters at 25C and leaves at
40C.The air leaves at 45C.The wet sand input is 10 kg/s. Radiation amounts to 5 cal/g dry air.
Calculate the pounds of dry air passing through the dryer and the humidity of the air leaving the dryer.
Latent heat of H2O at 40C = 575
Cp for dry sand
= 0.21 cal/gC
Cp for dry air
= 0.238 cal/gC
Cp for H2O vapor
= o.48cal/gC
Given:
TG1=45C,H1=?
G1=?,TG2=120C,H2=0.007
Ls = 10 kg/s
Ts2 = 49C
Ts1 = 25C
X2 = 0.03
X1 = 0.5
Required:
G, H1
Solution:
GH2 + LsX1 = H1 + LsX2
G(0.007) + 10 x3(0.5) = GH1+10 x 3(0.003)
0.007G + 4.7 x103 = GH1
(1)
H’G2 = Cs(TG2- To) + H2
= [1.005 + 1.88(0.007)](120-0) + 0.007(2501)
H’G2 = 139.6862 kJ/kg d.a.
For the exit gas,
H’G1 = Cs(TG1- To) + H1
= [(1.005 + 1.88H1)(45-0) + H1 (2501)
H’G1 = 45.225 + 2585.6 H1
For the entering solid,
H’s1 = Cps(Ts1-To) + X1 Cpa(Ts1-To)
= 0.221(25-0) + 0.5(0.48)(25-0)
H’s1 = 11.25 cal/g
H’s2 = Cps(Ts2-To) + X2Cpa(Ts2-To)
= 0.21(400-0) + 0.03(0.48)(40-0)
H’s2 = 8.976 cal/g
Substituting into Eq,
G H’G2 +Ls H’s1 = GH’G1+Ls H’s2 + Q
G(139.6862) +10(11.25) = G(45.225 + 2585.6 H1) + 10(8.976) +5
94.4612G –17.74 = 2585.6GH1
(2)
Using (1) &(2),
G=159140.7017g/h~159.1407 kg/h
H1=0.03653 gH2O/g air
EVAPORATION
1.
2.
3.
4.
5.
6.
It is a unit operation which means to concentrate a solution considering a non-volatile solute and volatile solvent.
a.
condensation
b.
evaporation
c.
precipitation
d.
extraction
It is conducted by vaporization of a portion of the solvent to produce a concentrated solution
a.
condensation
b.
evaporation
c.
precipitation
d.
extraction
Which of the following is not a processing factor in evaporation?
a.
solubility
b.
temperature
c.
frothing
d.
specific heat
Which of the following is not a proper assumption in evaporator calculations?
a.
The standard temperature difference is based on the temp of the boiling liquid at the liquid vapor interface
b.
The standard temperature difference is based on the temp of the liquid at the liquid interface
c.
The log mean temperature difference is based on the temp of the boiling liquid at the liquid vapor interface
d.
The log mean temperature difference is based on the temp of the liquid at the liquid interface
The heat required to vaporize 1lb of the solvent is taken as the ____ at the exposed surface temperature of the solution
a.
latent heat of vaporization
b.
specific heat
c.
heat transfer
d.
specific heat of condensation
What is the usual solvent in an evaporation process?
a.
water
b.
alcohol
c.
benzene
d.
ether
7.
8.
9.
For feed of organic salts in water, the heat capacity may be assumed ___ to that of water alone
a.
greater
b.
lesser
c.
equal
d.
none of the above
The effect of boiling point rise due to _____ is commonly neglected in evaporation calculations
a.
heat transfer
b.
temperature change
c.
hydrostatic head
d.
change in velocity
It is the difference in temperature of the liquid at the heat source and the liquid at the top evaporating liquid
a.
temperature change
b.
heat transfer
c.
boiling point rise
d.
log mean temperature
10. When two or more evaporators are connected, it is called
a.
single effect evaporator
b.
double effect evaporator
c.
multiple effect evaporator
d.
multiple effect forward evaporator
11. ___ is equivalent to the number of effects but it will never reach it unless the temp of the feed is higher than the boiling point
of the liquid in the evaporator body
a.
economy
b.
capacity
c.
percent recovery
d.
efficiency
12. ___ of the evaporator is the total amount of evaporation it is capable of producing per unit time
a.
economy
b.
capacity
c.
percent recovery
d.
efficiency
13. This can be determined by the measurement of the total heat transferred to the evaporating mixture
a.
economy
b.
capacity
c.
percent recovery
d.
efficiency
14. It is the most important single factor in evaporator design, since the heating surface represents the largest part of the
evaporator cost.
a.
temperature
b.
temperature difference
c.
heat transfer
d.
pressure
15. The greatest increase in steam economy is achieved by reusing the vaporized solvent, which is done in the___
a.
single effect evaporator
b.
double effect evaporator
c.
multiple effect evaporator
d.
multiple effect backward evaporator
16. In the ___, the desirability of producing crystals of a definite uniform size usually limits the choice to evaporators having a
positive means of circulation.
a.
multiple effect evaporator
b.
single effect evaporator
c.
forced circulation evaporator
d.
crystallizing evaporator
17. ___ which is the growth on body and heating surface walls of a material having a solubility that increases with increase in
temperature
a.
salting
b.
scaling
c.
crystals
d.
fouling
18. It is the deposition and growth on body walls, and especially on heating surfaces, of a material undergoing an irreversible
chemical reaction in the evaporator
a.
salting
b.
scaling
c.
crystals
d.
fouling
19. It is the formation of deposits other than salt or scale and maybe due to corrosion
a.
salting
b.
scaling
c.
crystals
d.
foulingThis evaporator is suitable for the widest variety of evaporator applications
a.
multiple effect evaporator
b.
Short-tube vertical evaporators
c.
forced circulation evaporator
d.
crystallizing evaporator
20. This is one of the earliest types still in widespread commercial uses, and its principal use is in the cane industry
a.
multiple effect evaporator
b.
Short-tube vertical evaporators
c.
forced circulation evaporator
d.
crystallizing evaporator
21. When the ratio of feed to evaporation is low, it is desirable to provide for
a.
recirculation of product
b.
steam recycling
c.
used of multiple effect evaporator
d.
high powered evaporator
22. This is normally the cheapest per unit capacity type of evaporator
a.
long-tube evaporator
b.
single effect evaporator
c.
short-tube evaporator
d.
falling film long tube evaporator
23. The principal problem in the falling film long tube evaporator is the
a.
feed distribution
b.
temperature control
c.
high pressure
d.
heat loss
24. The falling film long tube evaporator is widely used for concentrating ___
a.
heat sensitive materials
b.
highly soluble materials
c.
high boiling point rise
d.
corrosive materials
25. this usually results from the presence in the evaporating liquid of colloids or of surface-tension depressants
a.
splashing loss
b.
entrainment losses
c.
heat losses
d.
foaming losses
26. ___ are usually insignificant if a reasonable height has been provided between the liquid level and the top of the vapor head
a.
splashing losses
b.
entrainment losses
c.
heat losses
d.
foaming losses
27. These are frequently encountered in an evaporator if feed is above the boiling point
a.
splashing losses
b.
entrainment losses
c.
heat losses
d.
foaming losses
28. These are frequently used to reduce product losses
a.
knitted wire mesh
b.
mechanical thermo compressors
c.
steam jacket
d.
entrainment separators
29. This employ reciprocating, rotary positive-displacement , centrifugal or axial flow direction
a.
knitted wire mesh
b.
mechanical thermo compressors
c.
steam jacket
d.
entrainment separators
30. the requirement of low temperature for fruit-juice concentration has led to the development of an evaporator employing a ___
a.
steam economy
b.
multiple effect
c.
temperature distribution
d.
secondary fluid
31. The approximate ____ in a multiple effect evaporator is under the control of the designer, but once built, the evaporator
establishes its own equilibrium.
a.
recirculation of feed
b.
multiple effect
c.
temperature distribution
d.
secondary fluid
32. The ___ of a multiple effect evaporator will increase in proportion to the number of effects
a.
steam economy
b.
multiple effect
c.
temperature distribution
d.
secondary fluid
33. It may influence the evaporator selection, since the advantages of evaporators having high heat transfer coefficients are more
apparent when expensive materials of constructions are indicated
a.
heat sensitivity
b.
corrosion
c.
fouling factor
d.
scaling
34. ____ usually range from a minimum of about 1.2 m/s
a.
horizontal tube velocity
b.
tube velocity
c.
tube rate
d.
none of the above
35. s solutions are heated and concentration of the solute increases, ___ limit of the material may be exceeded and crystals may
form.
a.
crystallization
b.
seeding
c.
solubility
d.
frothing
36. The simplest form of evaporator consists of an open pan in which the liquid is boiled
a.
open kettle
b.
vertical type natural circulation
c.
falling film type evaporator
d.
agitated film evaporator
37. A variation of the long-tube evaporator
a.
forced circulation type evaporator
b.
vertical type natural circulation
c.
falling film type evaporator
d.
agitated film evaporator
38. This is done in a modified falling film evaporator
a.
forced circulation type evaporation
b.
vertical type natural circulation
c.
falling film type evaporation
d.
agitated film evaporation
39. In this operation, the fresh feed is added to the first effect and flows to the next in the same direction as the vapor flow
a.
Feed-forward
b.
Feed-backward
c.
Parallel-feed
d.
None of the above
40. Involves the adding of fresh feed and the withdrawal of concentrated product from each effect
a.
Feed-forward
b.
Feed-backward
c.
Parallel-feed
d.
None of the above
41. The fresh feed enters at the coldest effect
a.
Feed-forward
b.
Feed-backward
c.
Parallel-feed
d.
None of the above
42. For strong solutions of dissolved solutes the boiling point rise due to the solutes in the solution cannot be predicted but the
empirical law known as ___ is used
a.
Enthalpy concentration chart
b.
Duhring chart
c.
Equilibrium chart
d.
Mollier chart
43. When a multiple effect evaporator is at steady state operation, the flow rates and rate of evaporation in each effect are ___
a.
greater
b.
lesser
c.
constant
d.
equal
44. In commercial practice, the areas in all effects for a multiple effect evaporator are
a.
greater
b.
lesser
c.
constant
d.
equal
45. The over-all material balance made over the whole system of a multiple effect evaporator must be satisfied, and if the final
solution is concentrated, the feed rate will be ______.
a.
increased
b.
decreased
c.
held constant
d.
neglected
46. ___ of the vapor is recovered and reuse in a multiple effect evaporator
a.
heat transfer
b.
latent heat
c.
specific heat
d.
sensible heat
47. 48. The increase in steam economy obtained by using multiple-effect evaporator is at the expense of ___.
a.
reduced capacity
b.
product quality
c.
evaporator’s efficiency
d.
percent yield
48. 49. Evaporation is also sometimes called as___
a.
Thermal recompression
b.
Liquid compression
c.
Water distillation
d.
Vapor compression
49. 50. In a ____ system, the vapor is compressed by acting on it with high-pressure steam in a jet ejector.
a.
Thermal recompression
b.
Liquid compression
c.
Water distillation
d.
Vapor compression
e.
f.
g.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
Answer key
Evaporation (objective questions)
b
b
d
a
a
a
c
c
c
c
a
b
b
c
c
d
a
b
d
c
b
a
a
a
a
d
a
b
d
b
d
c
a
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
h.
b
b
c
a
c
d
a
c
b
b
c
d
a
b
a
c
a
i.
j.
Unit Operations Economics
k.
l.
m. 1. ____ is the money returned to the owners of capital for use of their capital.
a.
Interest
b.
Cash
c. Loan
d. credit
n. 2. Any profit obtained through the uses of capital
o.
a. Interest
p.
b. Cash
q.
c. Loan
r.
d. credit
s.
3. In economic terminology, the amount of capital on which interest is paid is designated as the principal, and the rate of
interest is defined as the amount of interest earned by a unit of principal in a unit of time
t.
a. simple interest
u.
b. compound interest
v.
c. nominal interest
w.
d. interest on loan
X.
4. When an interest period of less than 1 year is involved
y.
a. simple interest
z.
b. ordinary interest
aa.
c. nominal interest
ab.
d .compound interest
ac. 5. This interest stipulates that interest is due regularly at the end of each interest period.
ad.
a. simple interest
ae.
b. ordinary interest
af.
c. nominal interest
ag.
d.compoundinterest
ah. 6. In ______interest rate, other time units are employed butthe interest rate is still expressed on an annual basis.
ai.
a. simple interest
aj.
b. ordinary interest
ak.
c. nominal interest
al.
d.compoundinterest
am. 7. The concept of ______is that the cost or income due to interest flows regularly, and this is just as reasonable an
assumption for most cases as the concept of interest accumulating only at discrete intervals.
an.
a. continuous Interest
ao.
b. Cash flow
ap.
c. Loan
aq.
d. credit
ar.
8. The present worth of a future amount is the present ___which must be deposited at a given interest yield tom yield the
desired amount at some future date.
as.
a. principal
at.
b. Cash flow
au.
c. capital
av.
d. credit
aw. 9. An ____is a series of equal payments occurring at equal time intervals.
ax.
a. principal
ay.
b. interest
az.
c. capital
ba.
d. annuity
bb.
10. The amount of an ___is the sum of all the payments plus interest if allowed to accumulate at a definite rate of interest
from the time of initial payment to the end of the term.
a. principal
bc.
b. interest
bd.
c. capital
be.
d. annuity
bf.
11. Common type of ____which involves payments, which occur at the end of each interest period
a. Perpetuity
bh.
b. Present worth of annuity
bi.
c. capital
bj.
d. annuity
bk.
12. Defined as the principal which would have to be invested at the present time at compound interest rate to yield a total
bl.
a. Perpetuity
bm.
b. Present worth of annuity
bn.
c. capital
bo.
d. annuity
bp.
13. An annuity in which the first payment is due after a definite number of years.
bq.
a. Perpetuity
br.
b. Present worth of annuity
bs.
c. deferred interest
bt.
d. deferred annuity
bu. 14. It is an annuity in which the periodic payments continue indifferently.
bv.
a. Perpetuity
bw.
b. Present worth of annuity
bx.
c. deferred interest
by.
d.deferredannuity
bz. 15. Defined as the original cost of the equipment plus the present value of the renewable perpetuity.
ca.
a. Perpetuity
cb.
b. Present worth of annuity
cc.
c. deferred interest
cd.
d.capitalizedcost
ce. 16. An ____design could be based on conditions giving the least cost per unit of time or the maximum profit per unit of
production.
cf.
a. optimum economic
cg.
b. plant
ch.
c. simple economic
ci.
d.plant economic
cj.
17. A determination of an “____condition” serves as a base point for a cost or design analyses, and it can often be
quantitized in specific mathematical form.
ck.
a. optimum
cl.
b. plant design
cm.
c. simple
cn.
d.plant economic
co.
18. The ___has one distinct advantage over the analytical method.
cp.
a. optimum
cq.
b. graphical
cr.
c. simple
cs.
d.economical
ct.
19. The point where total product cost equals total income represents the_____, and the optimum production schedule
must be at a production rate higher than that.
cu.
a. optimum
cv.
b. break-even point
cw.
c. annuity
cx.
d. balanced
cy.
20. The total product cost per unit of time may be divided into the two classifications of operating costs and ____costs.
cz.
a. optimum operating costs
da.
b. Organization costs
db.
c. annuity costs
dc.
d. superproduction costs
dd.
21. This depends on the rate of production and include expenses for direct labor, raw materials, power, heat, supplies,
and similar items which are function of the amount of material produced.
de.
a. operating costs
df.
b. Organization costs
dg.
c. annuity costs
dh.
d. superproduction costs
di.
22. ______are due to expenses for directive personnel, physical equipment, and other services or facilities which must
be maintained irrespective of the amount of material produced. These are independent of the rate of production.
dj.
a. operating costs
dk.
b. Organization costs
dl.
c. annuity costs
dm.
d. superproduction costs
dn.
23. Extra expenses due to increasing the rate of production are known as ____.
do.
a. operating costs
dp.
b. Organization costs
dq.
c. annuity costs
dr.
d. superproduction costs
ds.
24. In a true ___operation, no product is obtained until the unit is shut down for discharging.
dt.
a. batch
du.
b. continuous
dv.
c. semi-batch
dw.
d. semi-continuous
bg.
dx.
dy.
dz.
ea.
eb.
ec.
ed.
ee.
ef.
eg.
eh.
ei.
ej.
ek.
el.
em.
en.
eo.
ep.
eq.
er.
es.
et.
eu.
ev.
ew.
ex.
ey.
ez.
fa.
fb.
fc.
fd.
fe.
ff.
fg.
fh.
fi.
fj.
fk.
fl.
fm.
fn.
fo.
fp.
fq.
fr.
fs.
ft.
fu.
fv.
fw.
fx.
fy.
fz.
ga.
gb.
gc.
gd.
ge.
gf.
gg.
gh.
gi.
25. In _____cyclic operations, product is delivered continuously while the unit is in operation, but the rate of delivery
decreases with time.
a. batch
b. continuous
c. semi-batch
d. semi-continuous
26. _____cyclic operations are often encountered in the chemical industry, and the design engineer should understand
the methods for determining optimum cycle times in this type of operation
a. batch
b. continuous
c. semi-batch
d. semi-continuous
27. It is a mathematical technique, for determining optimum conditions for allocation of resources and operating
capabilities to attain a definite objective.
a. economics
b. programming
c. linear programming
d. dynamic programming
28. The concept of this is based on converting an overall decision situation involving many variables into a series of
simpler individual problems with each of these involving a small number of total variables.
a. economics
b. programming
c. linear programming
d. dynamic programming
29. When quality constraints or restrictions on certain variables exist in an optimization situation, a powerful analytical
technique is the use of :
a. Lagrange Multipliers
b. Geometric Programming
c. steepest Ascent or Descent
d. dynamic programming
30. For the optimization situation in which two or more independent variables are involved, response surface can often
be prepared to show the relationship among the variables one of the early methods proposed for establishing optimum conditions
from response surfaces is known as the method:
a. Lagrange Multipliers
b. Geometric Programming
c. linear programming
d. steepest Ascent or Descent
31. A technique for optimization, based on the inequality relating the arithmetic mean to geometric mean, with this
method the basic idea is to start by finding the optimum way to distribute the total cost
a. Lagrange Multipliers
b. Geometric Programming
c. linear programming
d. steepest Ascent or Descent
32. ____(CPM) and program evaluation and review technique) have received particular attention and have shown the
desirability of applying mathematical and graphical analysis
a. critical path method
b. critical planning method
c. careful planning and management
d. none of the above
33. The word ____is used as a general term for the measure of the amount of profit that can be obtained from a given
situation.
a. profitability
b. income
c. net income
d. sales
34. It is the common denominator for all business activities before capital is invested to a project or enterprise
a. profitability
b. income
c. net income
d. sales
35. ____on investments is ordinarily expressed on an annual percentage basis. The yearly profit divided the by the total
initial investment necessary represents the fractional return:
a. rate of return
b. interest income
c. profitability
d. annuity
36. ___is defined as the difference between income and expense.
a. profit
b. gross income
c. operating expenses
d. sales
37. This is a function of the quantity of goods and services produced and the selling price.
a. sales
b. gross income
c. operating expenses
d. profit
gj.
gk.
gl.
gm.
gn.
go.
gp.
gq.
gr.
gs.
gt.
gu.
gv.
gw.
gx.
gy.
gz.
ha.
hb.
hc.
hd.
he.
hf.
hg.
hh.
hi.
hj.
hk.
hl.
hm.
hn.
ho.
hp.
hq.
hr.
hs.
ht.
hu.
hv.
hw.
hx.
hy.
hz.
ia.
ib.
ic.
id.
ie.
if.
ig.
ih.
ii.
ij.
ik.
il.
im.
in.
io.
ip.
iq.
ir.
is.
it.
iu.
iv.
iw.
ix.
38. The method of approach for a profitability evaluation is by ____, which takes into account the time value of the
money and is based on the amount of the investment that is unreturned at the end of each year during the estimated life of the
project.
a. marketing analysis
b. gross income
c. discounted cash flow
d. law of demand
39. In the net present worth study, the procedure has involved the determination of an index or interest rate which
discounts the annual ____to a zero present value when properly compared to the initial investment.
a. marketing analysis
b. gross income
c. cash flow
d. selling price
40. The ____concept is useful for comparing alternatives which exist as possible investment choices within a single
overall project.
a. marketing analysis
b. sales inventory
c. discounted cash flow
d. capitalized cost profitability
41. ____ is defined as the minimum length of time theoretically necessary to recover the original capital investment in the
form of cash flow to the project based on total income minus all cost except depreciation.
a. deadline
b. due date
c. payout period
d. maturity date
42. The term ___ refers to a special type of alternative in which facilities are currently in existence and it may be
desirable to replace these facilities with different ones.
a. reserves
b. replacement
c. substitute
d. none of the above
43. In the formula, P = F(1+I )-N , where I = interest rate per period and N the number of interest periods is known as
a. sinking fund factor
b. single payment present worth factor
c. uniform series
d. capital recovery factor
44. Which of the following relationships between compound interest factors is not correct?
a. Single payment compound amount factor and single payment present worth factor are
reciprocals
b. Capital recovery factor and uniform series present worth are reciprocals
c. Capital recovery factor equals the sinking fund factor plus the interest rate
d. capital recovery factor and sinking fund factor are reciprocals
45. An “annuity” is defined as
a. Earned interest due at the end of each interest period
b. Cost of producing a product or rendering a service
c. Amount of interest earned by a unit principal in a unit of time
d. A series of equal payments made at equal interval of time
46. Which one of the following method is not a method of depreciating plant equipment for accounting and engineering
economic analysis purposes?
a. double entry method
b. fixed percentage method
c. sum-of-years digit method
d. sinking fund method
47. ___ is a science which deals with the attainment of the maximum fulfillment of society’s unlimited demands for goods
and services.
a. Political science
b. Economics
c. Engineering Economy
d. Social Science
48. It is the branch of economics which deals with the application of economics laws and theories involving engineering
and technical projects or equipments.
a. Political science
b. Economics
c. Engineering Economy
d. Social Science
49. This refers to the products or services that are directly used by people to satisfy their wants.
a. Producer goods and services
b. Consumer goods and services
c. Luxury Products
d. Supply
50. Those that are used to produce the goods and services for the consumer are called
a. Producer goods and services
b. Consumer goods and services
c. Machineries and Technology
d. Supply
51. This refers to the satisfaction or pleasure derived from the consumer goods and services.
a. Product Satisfaction
iy.
iz.
ja.
jb.
jc.
jd.
je.
jf.
jg.
jh.
ji.
jj.
jk.
jl.
jm.
jn.
jo.
jp.
jq.
jr.
js.
jt.
ju.
jv.
jw.
jx.
jy.
jz.
b. Consumer’s contentment
c. Luxury Products
d. Utility
52. These are products that have an income-elasticity of demand greater than one.
a. Basic Commodities
b. Consumer goods
c. Luxury Products
d.Supply
53. If the income elasticity of demand is greater than one, this means that
a. as income decreases, more income will be spent on the basic commodities
b. as income increases, more expenses will be made on the basic commodity products
c. as income increases, more income will be spent on luxury items
d. as income decreases, less will be spent on the basic commodity products
54. It is the amount of goods and products that are available for sale by the suppliers
a. Producer goods and services
b. Consumer goods and services
c. Machineries and Technology
d. Supply
55. The want or desire or need for a product using money to purchase it.
a. Supply
b. Consumer goods and services
c. Luxury Products
d. Demand
56. The law of supply and demand, states that
a. “When free competition exists, the price of the product will be that value where supply is equal to the demand”
b. “When perfect competition exists, the price of the product will be that value where supply is equal to the demand”
c. “When free competition exists, the price of the product will be that value where supply is greater to the demand”
d. “When perfect competition exists, the price of the product will be that value where supply is greater to the
demand”
ka.
kb.
kc.
kd.
ke.
kf.
kg.
kh.
ki.
kj.
kk.
kl.
km.
kn.
ko.
kp.
kq.
kr.
ks.
kt.
ku.
kv.
kw.
kx.
ky.
kz.
la.
lb.
lc.
ld.
le.
lf.
lg.
lh.
li.
lj.
lk.
ll.
lm.
ln.
lo.
lp.
lq.
lr.
ls.
57. This is a form of market structure where the number of suppliers is used to determine the type of market.
a. Market Structures
b. Competition
c. Free competition market
d. Perfect Competition
58. This is a market situation wherein a given product is supplied by a very large number of vendors and there is no
restriction of any additional vendor from entering the market.
a. Market Structures
b. Competition
c. Free competition market
d. Perfect Competition
59. A place where the vendors or the sellers and vendees or the buyers come together is called
a. Market Structures
b. Market
c. Free competition market
d. Perfect Competition market
60. This is defined as the interest on a loan or principal that is based on the original amount of the loan or principal.
a. Ordinary Simple Interest
b. Simple Interest
c. Exact Simple Interest
d. Compound Interest
61. This is based on one banker’s year.
a. Ordinary Simple Interest
b. Simple Interest
c. Exact Simple Interest
d. Compound Interest
62. __ is based on the exact number of days in a given year.
a. Ordinary Simple Interest
b. Simple Interest
c. Exact Simple Interest
d. Compound Interest
63. One banker’s year is equivalent to
a. 365 days
b. 350 days
c. 12 months and 30 days
d. 12 months and 15 days
64. This is defined as the interest of loan or principal which is based on the amount of the original loan and the previous
accumulated interest
a. Ordinary Simple Interest
b. Simple Interest
c. Exact Simple Interest
d. Compound Interest
65. It is the amount of money or payment for the use of a borrowed money or capital
a. Annuity
b. Interest
c. Simple Interest
d. Simple Annuity
lt.
lu.
lv.
lw.
lx.
ly.
lz.
ma.
mb.
mc.
md.
me.
mf.
mg.
mh.
mi.
mj.
mk.
ml.
mm.
mn.
mo.
mp.
mq.
mr.
ms.
mt.
mu.
mv.
mw.
mx.
my.
mz.
na.
nb.
nc.
nd.
ne.
nf.
ng.
nh.
ni.
nj.
nk.
nl.
nm.
nn.
no.
np.
nq.
nr.
ns.
nt.
nu.
nv.
nw.
nx.
ny.
nz.
oa.
ob.
oc.
od.
oe.
of.
og.
oh.
oi.
oj.
ok.
ol.
om.
on.
oo.
op.
66. This is defined as the basic annual rate of interest
a. effective rate of annuity
b. effective rate of interest
c. nominal rate or annuity
d. nominal rate of interest
67. It is defined as the actual or exact rate of interest earned on a principal during1 year period
a. effective rate of annuity
b. effective rate of interest
c. nominal rate or annuity
d. exact simple interest
68. This refers to the difference between the future worth of a negotiable paper and its present worth.
a. Depreciation
b. Market value
c. Discount
d. Fair value
69. This refers to the sales of stock or share at reduced price
a. Depreciation
b. Market value
c. Discount
d. Fair value
70. This is the deduction from the published price of services or goods.
a. Depreciation
b. Salvage value
c. Discount
d. Fair value
71. “Annuity” is simply defined as
a. series of equal payments occurring at equal time interval of time
b. series of payments occurring at equal time interval
c. series of equal payments occurring at certain time interval
d. series of divided payments at pay periods
72. An annuity of a fixed time span is also called as
a. Ordinary annuity
b. Annuity Certain
c. Annuity Due
d. Deferred Annuity
73. __ is the type of annuity where the payments are made at the beginning of each period starting from the first period
a. Ordinary annuity
b. Annuity Certain
c. Annuity Due
d. Deferred Annuity
74. An annuity where the payments are made at the end of each period beginning on the first period
a. Ordinary annuity
b. Annuity Certain
c. Annuity Due
d. Deferred Annuity
75. It is when the first payment does not begin until some later date in the cash flow
a. Ordinary annuity
b. Perpetuity
c. Annuity Due
d. Deferred Annuity
76. When an annuity does not have a fixed time span but continues indefinitely, then it is referred to as
a. Ordinary annuity
b. Perpetuity
c. Annuity Due
d. Deferred Annuity
77. This refers to the rate of interest that is quoted in the bond.
a. bond
b. bond rate
c. bond value
d. interest rate of bond value
78. It is the present worth of the future payments that will be received
a. book value
b. bond value
c. fair value
d. salvage value
79. It is the amount a willing buyer will pay to a willing seller for a property where each has equal advantage and neither
one of them is under compulsion to buy or sell
a. Book value
b. Market value
c. Use value
d. Fair value
80. The amount of the property which the owner believed to be its worth as an operating unit is called
a. value
b. Market value
c. Use value
d. Fair value
oq.
or.
os.
ot.
ou.
ov.
ow.
ox.
oy.
oz.
pa.
pb.
pc.
pd.
pe.
pf.
pg.
ph.
pi.
pj.
pk.
pl.
pm.
pn.
po.
pp.
pq.
pr.
ps.
pt.
pu.
pv.
pw.
px.
py.
pz.
qa.
qb.
qc.
qd.
qe.
qf.
qg.
qh.
qi.
qj.
qk.
ql.
qm.
qn.
qo.
qp.
qq.
qr.
qs.
qt.
qu.
qv.
qw.
qx.
qy.
qz.
ra.
rb.
rc.
rd.
re.
rf.
rg.
rh.
ri.
rj.
81. ___ is the amount obtained from the sale of the property.
a. book value
b. bond value
c. market value
d. salvage value
82. It is the worth of the property determined by a disinterested person
a. Book value
b. Market value
c. Use value
d. Fair value
83. It is the reduction or fall in the value of an asset or physical property during the course of its working life due to the
passage of time
a. Depreciation
b. Depleting value
c. Discounted value
d. Fair value
84. __ is the money worth of an asset or product
a. Price
b. Market value
c. value
d. Fair value
85. This refers to the present worth of all profits that are to be received through ownership of a particular property
a. Price
b. Market value
c. value
d. capitalized cost
86. Refers to the sum of its first cost and cost of perpetual maintenance of a property
a. Price
b. Market value
c. perpetual value
d. capitalized cost
87. __ is a long term note or a financial security issued by businesses or corporation and guaranteed on certain assets of
the corporation or its subsidiaries.
a. Contract
b. Receipt
c. bond value
d. bond
88. This value implies that the property will still be use for the purpose it is intended
a. salvage value
b. book value
c. use value
d. depreciation
89. A depreciation computation method which is also known as Diminishing Balance Method
a. Sinking Fund Method
b. Straight-line Method
c. Declining Balance method
d. Break-even analysis
90. Also called as Constant- Percentage Method in depreciation method
a. Sinking Fund Method
b. Straight-line Method
c. Declining Balance method
d. Break-even analysis
91. This is also known as “Resale value”
a. salvage value
b. book value
c. use value
d. depreciation
92. ___ is considered as the simplest type of business organization wherein the firm is controlled and owned by a single
person
a. sole-proprietorship
b. Partnership
c. Corporation
d. Entrepreneurship
93. __ is a firm owned and controlled by two or more persons who are bind to an agreement
a. sole-proprietorship
b. Partnership
c. Corporation
d. Entrepreneurship
94. It is a firm owned by a group of ordinary shareholders and the capital of which is divided up to the number of shares.
a. sole-proprietorship
b. Partnership
c. Corporation
d. Entrepreneurship
95. It is defined as a distinct legal entity separate from the individuals who owns it and can engage in any business
transaction which a real person could do.
a. sole-proprietorship
rk.
rl.
rm.
rn.
ro.
rp.
rq.
rr.
rs.
rt.
ru.
rv.
rw.
rx.
ry.
rz.
sa.
sb.
sc.
sd.
se.
sf.
sg.
sh.
si.
sj.
sk.
sl.
sm.
sn.
so.
sp.
sq.
sr.
ss.
st.
su.
sv.
sw.
sx.
sy.
sz.
ta.
tb.
tc.
td.
te.
tf.
tg.
th.
ti.
tj.
tk.
tl.
tm.
tn.
to.
tp.
tq.
tr.
ts.
tt.
tu.
tv.
tw.
tx.
ty.
tz.
ua.
ub.
uc.
ud.
b. Partnership
c. Corporation
d. Entrepreneurship
96. Also known as joint-stock company or a cooperative
a. Incorporation
b. Partnership
c. Corporation
d. Entrepreneurship
97. This refers to the situation where the sales generated are just enough to cover the fixed and variable cost.
a. break-even
b. Break even chart
c. break even point
d. none of the above
98. The break even chart shows
a. the relationship between income and fixed cost
b. the relationship between the volume and fixed cost, variable costs, and income
c. the relationship between the assets and fixed cost, variable costs, and income
d. the relationship between the liabilities and fixed cost, variable costs, and income
99. The level of production where the total income is equal to the total expenses is known as
a. break-even
b. Break even chart
c. break even point
d. none of the above
100. The sum of perpetuity is a/an
a. book value
b. zero value
c. infinite value
d. bond value
Board Problems:
1. Design based on conditions giving the least cost per unit time and maximum profit per unit production
a. battery limit
b. break-even point
c. optimum economic design
d. plant design
2. A flow diagram, indicating the flow of materials, unit operations involved, equipment necessary and special information
on operating temperature and pressure is a
a. schematic diagram
b. qualitative flow diagram
c. process flow chart
d. quantitative flow diagram
3. In plant design implementation, soil testing is done to determine
a. pH
b. load bearing capacity
c. porosity
d, viscosity
4. This includes all engineering aspects involved in the development of either a new, modified or expanded industrial
plant is called
a. plant design
b. optimum design
c. process design
d. engineering design
5. A chemical engineering specializing in the economic aspects of design is called
a. plant engineer
b. cost engineer
c. design engineer
d. process engineer
6. This refers to the actual design of the equipment and facilities for carrying out the process
a. process engineering
b. plant design
c. process design
d. optimum design
7. The final step before construction plans for plant and includes complete specifications for all components of the plant
and accurate costs based on quoted prices are obtained
a. preliminary design
b. quick estimate design
c. firm process design
d. detailed design estimate
8. A thorough and systematic analysis of all factors that affect the possibility of success of a proposed undertaking
usually dealing with the market, technical, financial, socio-economic and management aspects is called
a. pr5oject feasibility study
b. plant design
c. project development and research
ue.
uf.
ug.
uh.
ui.
uj.
uk.
ul.
um.
un.
uo.
up.
uq.
ur.
us.
ut.
uu.
uv.
uw.
ux.
uy.
d. product development
9. Discusses the nature of the product line, the technology necessary for production, its availability, the proper mix of
production resources and the optimum production volume
a. market feasibility
b. socio-economic feasibility
c. technical feasibility
d. management feasibility
10. Discusses the nature of the unsatisfied demand which the project seeks to meet growth and the manner in which it is
to be met
a. financial feasibility
b. management feasibility
c. market feasibility
d. technical feasibility
11. A multiple effect evaporator produces 10, 000 kg of salt from a 20% brine solution per day. One kg of steam
evaporates 0.7 N kg of water in N effects at a cost of P25 per 1000 kg of steam. The cost of the first effect is P450, 000 and the
additional effects at P300, 000 each. The life of the evaporator is 10 years with no salvage value. The annual average cost of
repair and maintenance is 10% and taxes and insurance is 5%. The optimum number of effects for minimum annual cost is
a. 3 effect
b. 5 effects
c. 4 effects
d. 2 effects
12. A process requires 20, 000 lb/hr of saturated steam at 115 psig. This is purchased from a neighboring plant at P18.
00 per short ton and the total energy content rate (mechanical) in the steam may be valued at P7. 5 x 10 -6 per Btu. Hours of
operation per year are 7200. The friction loss in the line is given by the following equation
F = 187.5 Lq 1.8 mc 0.20 in ft-lbs/lbm
d 0.20 Di 4.8
uz.
Cf = 1.44 Di 1.5 L , in p/yr
va.
vb.
Where: L = length of straight pipe, ft.
vc.
q = steam flow rate, cu. ft. per sec
vd.
mc = steam viscosity cp.
ve.
d = steam density, lb per ft3
vf.
Di = inside diameter of pipe, inches
vg. The optimum pipe diameter that should be used for transporting the above steam is
vh.
a. 6 in
b. 4 in
c. 3 in
d. 5 in
vi.
13. A smelting furnace operating at 2, 400 oF is to be insulated on the outside to reduce heat losses and same on energy.
The furnace wall consists of a ½ inch steel plate and 4 inch thick refractory inner lining. During operation without other insulation,
the outer surface of the steel plate exposed to air has a temperature of 300 oF. Ambient air temperature is at 90oF. Operation is 300
days per year. Thermal conductivities in Btu per hr-ftoF are: steel plate = 26; refractory =0.1; insulation to be installed = 0.025. The
combined radiation and convection loss to air irrespective of material exposed is 3 Btu per hr-ft 2-oF, annual fixed charge is 20 % of
the initial insulation cost. If heat energy is P5.0 per 10, 000 Btu and installed cost of insulation is P100 per inch ft 2 of area, the
optimum thickness of the outer insulation that should be installed is
vj.
a. 8 in
b. 12 in
c. 10 in
d. 6 in
vk.
14. An organic chemical is produced by a batch process. In this process chemical X and Y react to form Z. Since the
reaction rate is very high, the total time required per batch has been found to be independent of the amount of materials and each
batch requires 2 hr, including time for charging, heating, and dumping. The following equation shows the relation between the
pound of Z produced (lbz) and the pound of X (lbx) and Y (lby) supplied:
vl.
vm.
lbz = 1.5(1.1lbxlbz +1.3lbylbz – lbxlby)0.5
vn.
vo.
Chemical X costs P0.09 per pound, chemical Y costs P0.04 per pound and chemical Z sells for P0.8 per pound. If
half of the selling price for chemical Z is due to cost other than raw materials, the maximum profit obtainable per pound of chemical
Z is
vp.
a. P0.3 per lbz
vq.
b. P0.5 per lbz
vr.
c. P0.12 per lbz
vs.
d. P0.25 per lbz
vt.
15. One hundred gram moles of R are to be produced hourly from a feed consisting of a saturated solution of A. (Cao =
0.1 gmole per liter). The reaction is A
R with rate ГR = (0.2/hr) Ca.
vu.
Cost of reactant at Cao = 0.1 gram mol per liter is P3.75/g-mole A; cost of backmix reactor, installed complete with
auxiliary equipment, instrumentation, overhead, labor depreciation, etc. is P0.075 per hr-liter. The conversion that should be used
for optimum operation is
vv.
a. 45%
b. 60%
c. 50%
d. 40%
vw.
16. One hundred lb moles of reactant A at a concentration of 0.01 lbmole per cuft is to be reacted with reactant B to
produce R and S. The reaction follows the aqueous phase elementary reaction:
vx.
vy.
A + B ------- R +S where k = 500 cuft/(lbmole-hr).
vz.
wa.
The amount of R required is 95 lbmoles/hr. Data:
wb. * in extracting R from the reacted mixture, A and B are destroyed
wc. * B costs P15 per lbmole in crystalline form, and is very soluble in aqueous solutions such that even when present in
large amount does not affect A in solution
wd. * capital and operating cost for backmix reactor is P0.10 per(cuft-hr).
we.
The optimum backmix reactor size is:
wf.
wg.
wh.
wi.
wj.
wk.
wl.
wm.
wn.
wo.
wp.
wq.
wr.
ws.
wt.
wu.
wv.
ww.
wx.
wy.
wz.
xa.
xb.
xc.
xd.
xe.
xf.
xg.
xh.
xi.
xj.
xk.
xl.
xm.
xn.
xo.
xp.
xq.
xr.
xs.
xt.
xu.
xv.
xw.
xx.
xy.
xz.
ya.
yb.
yc.
a. 23, 900 ft3
b. 24, 200 ft3
c. 25, 900 ft3
d. 23, 500 ft3
17. April 1992.
A unit of welding machine cost P45, 000 with an estimated life of 5 years. Its salvage value id P2, 500. Find its
depreciation rate by straight-line method.
a. 17.75%
b. 19.88%
c. 18.89%
d.15.56%
18. April 1997.
A machine has an initial cost of P50, 000 and a salvage value of P10, 000 after 10 years. Find the book value after
5 years of using straight-line depreciation.
a. P12, 500
b.P30, 000
c. P16, 400
d.P22, 300
19. October 1992.
The initial cost of a paint sand mill, including its installation, is P800, 000. The BIR approved life of this machine is
10 years for depreciation. The estimated salvage value of the mill is P50, 000 and the cost of dismantling is estimated is estimated
to be P15, 000. Using straight-line depreciation, what is the annual depreciation charge and what is the book value of the machine
at the end of six years.
a. P74, 500 ; P340, 250
b. P76, 500 ; P341, 000
c. P76, 500 ; P 342, 500
d. P77, 500 ; P343, 250
20. November 1997.
The cost of the equipment is P500, 000.00 and the cost of installation is P30, 000. If the salvage value is 10% of the
cost of equipment at the end of five years, determine the book value at the end of the fourth year. Use straight-line method.
a. P155, 000
b.P140, 000
c. P146, 000
d. P132,600
21. April 1998.
An equipment costs P10, 000 with a salvage value of P500 at the end of 10 years. Calculate the annual
depreciation cost by sinking fund method at 4% interest.
a. P791.26
b.P950.00
c. P971.12
d. P845.32
22. November 1995.
A machine costing P720, 000 is estimated to have a book value of P40, 545.73 when retired at the end of 10 years.
Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in
%?
a.28
b. 25
c. 16
d.30
23. November 1998.
AMD Corporation makes it a policy that for any new equipment purchased; the annual depreciation cost should not
exceed 20 % of the first cost at any time with no salvage value. Determine the length of service life is necessary if the depreciation
used is the SYD method.
a.9 years
b. 10 years
c. 12 years
d.19 years
24. November 1996.
At 6%, find the capitalized cost of a bridged whose cost is P250M and the life is 20 years, if the bridge must be
partially rebuilt at a cost of P100M at the end of each 20 years.
a. P275.3M
b.P265.5M
c. P295.3M
d. P282.1M
25. October 1995
A company must relocate one of its factories in three years. Equipment for the loading dock is being considered for
purchase. The original cost is P20, 000; the salvage value of the equipment after three years is P8, 000. The company’s rate return
on the money is 10% . Determine the capital recovery per years.
a. P5, 115
b.P4, 946
c. P5, 625
d. P4, 805
26. October 1998.
The annual maintenance cost of a machine shop is P69, 994. if the cost of making a forging is P56 per unit and its
selling price is P135 per forged unit, find the number of units to be forged to break-even.
a.886 units
b. 885 units
c. 688 units
d.668 units
27. October 1990.
Compute for the number of locks that an ice plant must be able to sell per month to break even based on the
following data:
Cost of electricity
P20.00
Tax to be paid per block
2.00
Real Estate Tax
3, 500.00
per month
Salaries and wages
25, 000.00 per month
Others
12, 000.00 per month
Selling price of Ice
55.00 per block
a. 1228
b. 1285
c. 1373
d.1312
28. April 1998
XYX Corporation manufactures bookcases that sell for P65.00 each. It costs XYZ Corporation P35, 000 per year to
operate its plant. This sum includes rent, depreciation charges on equipment, and salary payments. If the cost to produce one
bookcase is P50.00, how many cases must be sold each year for XYZ to avoid taking a loss?
a. 2334
b. 539
c. 750
d.2333
yd.
ye. Answer Key
yf. Board Problems
yg.
1. c
2. b
3. b
4. a
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
b
c
c
a
c
c
d
a
c
c
c
a
c
b
b
c
a
b
a
c
c
a
a
a
yh.
yi.
yj.
yk.
a
a
a
b
d
c
a
a
d
d
d
b
d
a
d
a
a
a
b
b
a
b
d
a
d
d
c
a
d
b
a
a
a
a
a
d
c
c
d
c
a
a
b
d
d
a
yl.
ym.
Answer Key
Objective Problems
47. b
48. c
yn. 49. b
yo. 50. a
yp. 51. d
yq. 52. c
yr. 53. c
ys. 54. d
yt. 55. d
yu. 56. a
yv. 57. b
yw. 58. d
yx. 59. b
yy. 60. b
yz. 61. a
za. 62. c
zb. 63. c
zc. 64. d
zd. 65. b
ze. 66. d
zf. 67. b
zg. 68. c
zh. 69. c
zi. 70. c
zj. 71. a
zk. 72. b
zl. 73. c
zm. 74. a
zn. 75. d
zo. 76. b
zp. 77. b
zq. 78. b
zr. 79. b
zs. 80. c
zt. 81. d
zu. 82. d
zv. 83. a
zw. 84. c
zx. 85. c
zy. 86. d
zz. 87. d
aaa. 88. a
aab. 89. c
aac. 90. c
aad. 91. a
aae. 92. a
aaf. 93. b
aag. 94. c
aah. 95. c
aai. 96. c
aaj. 97. a
aak. 98. b
aal. 99. c
aam.100. c
aan.
aao.
aap.
Problem Solving
aaq.
1. The purchased cost of a shell-and-tube heat exchanger (floating head and carbon-steel tubes) with 100 ft 2 of heating surface
was $3,000 in 1980. What will be the purchased cost of a similar heat exchanger with 200 ft 2 of heating surface in 1980 if
purchased-cost-capacity exponent is 0.60 for surface area ranging from 100-400 ft2?
a. $4 547.15
b. $4 127.25
c. $4 567.10
d. $4523.00
2. efer to problem no. 1. If the purchased-cost-capacity exponent for this type of exchanger is 0.81 for surface areas ranging
from 400-2,000 ft2, what will be the purchased cost of a heat exchanger with1, 000 ft2 of heating surface in 1985?
a. $20 423.38
b. $18 527.12
c. $14 547.00
d. $24 542.05
3. If the purchased cost of a shell-and-tube heat exchanger (floating head and carbon-steel tubes) with 100 ft 2 of heating surface
was $3,000 in 1980. Note that the purchase-cost-capacity exponent is not constant over range surface area requested. What
will be the purchased cost of a heat exchanger with 700 ft2 of heating surface in 1985?
a. $15 398.79
b. $15 298.79
c. $16 498.79
d. $16 598.79
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
From the preceding question, what will be the purchased cost of the heat exchanger with 800 ft2 of heating surface in 1985?
a. $18 046.32
b. $18 146.32
c. $17 046.32
d. $17 146.32
Refer to problem no. 3. What will be the purchased cost of the heat exchanger with 1500 ft2 of heating surface in 1985?
a. $38 363.62
b. $48 363.62
c. $18 363.62
d. $28363.62
aar.
The purchase and installation cost of some pieces of equipment are given as a function of weight rather than capacity. An
example of this is the installed cost of large tanks. The 1980 cost for an installed aluminum tank weighing 100,000 lb was
$390,000. For a size range from 200,000 to 1,000,000 ld, the installed cost-weight exponent for aluminum tanks is 0.93. If an
aluminum tank weighing 700,000 lb is required, what is the present capital investment needed?
a. $3 160 100
b. $3 060 000
c. $6 160 100
d. $6 060 000
The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000. The plant
is to be an addition to an existing formaldehyde plant. The major part of the building cost will be for indoor construction, and
the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values found for typical
chemical plants. On the basis of this information, estimate the total direct plant cost.
a. $879 000
b. $879 253
c. $890 560
d. $825 020
The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000. The plant
is to be an addition to an existing formaldehyde plant. The major part of the building cost will be for indoor construction, and
the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values found for typical
chemical plants. On the basis of this information, estimate the fixed capital investment.
a. $ 1 246 530
b. $ 2 336 230
c. $ 2 546 531
d. $ 1 246 830
Refer to the previous problem and on the basis of this information, what is the total capital investment?
a. $1 542 125
b. $1 478 570
c. $1 532 120
d. $1 468 530
The total capital investment for a chemical plant is $1,500,000 and the plant process 3M kg of product annually. The selling
price of the product is $0.82/kg. Working capital amounts to 15% of the total capital investment. The investment is from
company funds, and no interest is charged. Raw-material costs for the product are $0.09/kg, labor $0.08/kg, utilities $0.05/kg,
and packaging $0.008/kg. Distribution costs are 5% of the total product cost. Determine the percent change in total cost.
a. 36.16%
b. 45.32%
c. 12.00%
d. 35.61%
It is desired to have a $ 9000 available from 12 years from now. If $ 5000 is available for investment at the present time, what
is discrete annual rate of compound interest on the investment would be necessary to give the desired amount?
a. 6.02%
b. 5.02%
c. 4.02%
d. 3.05%
An original loan of $2000 was made at 6 percent simple interest per year for 4 years. At the end of this time, no interest had
been paid and the loan was extended for 6 more years at a new, effective, compound-interest rate of 8 percent per year.
What is the total amount owned at the end of ten years if no intermediate payments are made?
a. $4 935.45
b. $3 945.45
c. $4 945.45
d. $3935.45 The original cost for a distillation tower is $24,000 and the useful life of the tower is estimated to be 8 years.
The sinking fund method for determining the arte of depreciation is used, and the effective annual interest for the
depreciation fund is 6 percent. If the scrap value of the distillation tower is $ 4000, determine the asset value at the end
of 5 years.
a. $ 3,547.928
b. $2 235.623
c. $2 547.928
d. $3 235.623
An annuity is due to being used to accumulate money. Interest is compounded at an effective annual rate of 8 %, and $1000
is deposited at the beginning of each year. What will be the total amount of annuity due be after 5 years?
a. $1233.24
B. $1173.32
C. $3573.2
D. $183.23
For the total year payments of $5000 for ten years, what will be the compound amount accumulated at the end of ten years if
the payment is at the end of the year? The effective (annual) interest is 20% and payments are uniform.
15.
16.
17.
18.
19.
20.
21.
22.
23.
a. $154 793.41
b. $456 793.41
c. $129 793.41
d. $209 793.41
Referring to the previous problem, estimate the compound amount accumulated at the end of ten years, if the payment is
made weekly?
a. $143 951.49
b. $243 951.49
c. $443 951.49
d. $643 951.49
A multiple – effect evaporator is to be used for evaporating 400,000 lb of water per day from a salt solution. The total initial
cost for the 1st effect is $18,000 and each additional effect costs $15,000. The life period is estimated to be 10 years, and the
scrap value at the end of the life period may be assumed to be zero. The straight-line depreciation method is used. Fixed
charges minus depreciation are 15% yearly based on the first cost of the equipment. Steam cost $1.50 per 1000 lb. Annual
maintenance charges are 5% of the initial equipment cost. All other cost is independent of the number of effects. The unit will
operate 300 days per year. If the lb of water evaporated per pound of steam equals 0.85 x numbers of effects, determine the
optimum number of effects for minimum annual cost.
a. 2 effects
b. 3 effects
c. 5 effects
d. 4 effects
Determine the optimum economic thickness of insulation that should be used under the following conditions: Standard steam
is being passed continuously through a steel pipe with an outside diameter of 10.75 in. The temperature of the steam is
400F, and the steam is valued at $1.80 per 1000 lb. The pipe is to be insulated with material that has a thermal conductivity
of 0.03 Btu/h-ft2-F/ft. The cost of installed insulation per foot of pipe length is $4.5xI t, where It is the thickness of the insulation
in inches. Annual fixed charges including maintenance amount to 20% of the initial installed cost. The total length of the pipe
is 1000 ft, and the average temperature of the surrounding may be taken as 70F. Heat transfer resistance due to the steam
film, scale and pipe wall are negligible. The air –film coefficient at the outside of the insulation may be assumed constant at
2.0 Btu/h-ft2-F for all insulation thickness.
a. $3257
b. $4057
c. $3057
d. $4257
A proposed chemical plant will require a fixed capital investment of $10 million. It is estimated that the working capital will
amount to 25% of the total investment and annual depreciation costs are estimated to be 10 percent of the fixed capital
investment. If the annual profit will be $3 million, determine the standard percent return on the total investment
a. 22.5%
b. 11.25%
c. 34.5%
d. 15.98%
If a plant will require a fixed capital investment of $10 million and the working capital will amount to 25% of the total
investment and annual depreciation costs are estimated to be 10 percent of the fixed capital investment. If the annual profit
will be $3 million, what is the minimum payout period (POP)?
a. 2.5 years
b. 4 years
c. 3.5 years
d. 2 years
An annual investigation of a proposed investment has been made. The following result has been presented to management.
The minimum payout period based on capital recovery using a minimum annual return of 10 percent as a fictitious expense is
10 years; annual depreciation costs amount top 8 percent of the total investment. Using this information, determine the
standard rate of return on the investment.
a. 21%
b. 10.5%
c. 56.93%
d. 11.5%
The information given in the previous problem, applies to conditions before income taxes. If 34% percent of all profits must be
paid out for income taxes, determine the standard rate of return after taxes using the figures given in the previous problem.
a. 7.93%
b. 6.93%
c. 4.93%
d. 5.93%
A capitalized cost for a piece of equipment has been found to be $55,000. This cost is based on the original cost plus the
present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the capitalized
cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under these conditions,
what would be the original cost of the equipment?
a. $57 391.47
b. $27 291.47
c. $47 391.47
d. $37 291.47
aas. For problems 24- 26.On Aug. 1, a concern had 10,000 lb of raw material on hand, which was purchased at a cost of
$0.030 per pound. In order to build up the reserve, 8000 lb of additional raw material was purchased on Aug. 15 at a cost
of $0.028 per pound. If none of the raw material was used until after the last purchase.
Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by
current average method.
a. $0.00297/lb
b. $0.0297/lb
c. $0.2297/lb
d. $0.2970/lb
24. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by
“FiFo” method.
a. $0.281/lb
b. $0.028/lb
c. $0.038/lb
d. $0.128/lb
25. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of August by
lifo method.
a. $0.031/lb
b. $0.041/lb
c. $2.031/lb
d. $1.041/lb
e.
For question nos. 27 – 31.
f.
The following are the data gathered from the AMD Food Corporation:
g.
h.
i.
j.
k.
l.
m.
n.
o.
p.
q.
r.
s.
t.
u.
Cash
Accounts payable:
B Company
C Company
v.
$20
w.
x.
Accounts receivable
Inventories
Mortgage payable
Common stock sold
Machinery and equipment (at present value)
Furniture and fixtures (at present value)
Government bonds
Surplus
y.
2,0
z.
8,0
aa.
ab.
6,0
ac.
15,
ad.
5,0
ae.
50,
af.
18,
ag.
5,0
ah.
3,0
ai.
2,0
aj.
26. From the data given above, determine the total asset of the AMD Food Corporation.
a. $ 47, 050
b. $ 67, 000
c. $ 87, 050
d. $ 57, 000
27. Determine the total current assets of the AMD Food Corporation.
a. $ 11,000
b. $ 44,000
c. $ 22,000
d. $ 33,000
28. What is the total amount of the Current liabilities of the AMD Corporation?
a. $ 45,097
b. $ 85,097
c. $ 25,000
d. $ 15,000
29. What is the total amount of the Fixed Assets of the AMD Food Corporation?
a. $ 23,000
b.
c.
d.
ak.
al.
$ 33,050
$ 32,000
$ 23,050
During the month of October, the following information was obtained in the AC antifreeze retailing company:
am.
an.
ao.
ap.
aq.
ar.
as.
at.
Salaries
Delivery expenses
Rent
Sales
Antifreeze available for sale during October (at cost)
Antifreeze inventory on Oct. 31 (at cost)
Other expenses
Earned surplus before income taxes as of Sept.30
au.
$
av.
700
aw.
400
ax.
15,
ay.
20,
az.
11,
ba.
1,2
bb.
800
bc.
30. Prepare an income statement for the month of October to determine the net income is for the month of October.
a. $ 29,100
b. $ 19,200
c. $ 29,200
d. $ 19,100
31. From the data above, determine also the total gross income of the AC Antifreeze Company.
a. $26,900
b. $36,100
c. $26,100
d. $46,900
bd. The following information applies to MADSteel Company on a given date:
be.
bf.
bg.
bh.
bi.
bj.
bk.
bl.
bm.
bn.
bo.
bp.
bq.
Long-term debts
Debts due within 1 year
Accounts payable
Machinery and equipment (at cost)
Cash in bank
Prepaid rent
Government bonds
Social security taxes payable
Reserve for depreciation
Reserve for expansion
Inventory
Accounts receivable
br.
bs.
bt.
bu.
bv.
bw.
bx.
$
1
,
6
0
0
1
,
0
0
0
2
,
3
0
0
1
0
,
0
0
0
3
,
1
0
0
3
0
0
3
,
0
0
0
by. 2
4
0
bz. 6
0
0
ca. 1
,
2
0
0
cb. 1
,
6
0
0
cc. 1
,
7
0
0
cd.
32. Determine the cash asset for the MADSteel Company at the given date.
a. $6,100
b. $7,100
c. $7,900
d. $6,900
33. Determine the current asset for MADSteel Company
a. $8, 700
b. $9, 500
c. $8, 500
d. $9, 700
34. Determine the current liabilities of MADSteel Company
a. $5, 540
b. $1, 540
c. $3, 540
d. $2, 540
35. Determine the Quick ratio for MADSteel Company
a. 2.29
b. 9.87
c. 2.56
d. 9.56
36. Determine the current ratio of MADSteel Company
a. 2.74
b. 2.56
c. 2.64
d. 1.98
37. A reactor of special design is the major item of equipment in a small chemical plant. The initial cost of a completely installed
reactor is $60,000, and the salvage value at the end of the useful life is estimated to be $10,000. Excluding depreciation costs
for the reactor, the total annual expenses for the plant are $100,000. How many years of useful life should be estimated for
the reactor if 12 % of the total annual expenses for the plant are due to the cost for the reactor depreciation? The straight-line
method for determining depreciation should be used.
a. 4 years
b. 2 years
c. 8years
d. 1year
38. The initial installed cost for a new piece of equipment is $10,000, and its scrap value at the end of its useful life is estimated to
be $2,000. The useful life is estimated to be 10 years. After the equipment has been in use for 4 years, it is sold for $7,000.
The company which originally owned the equipment employs the straight-line method for determining depreciation costs. If
the company had used an alternative method for determining depreciation cost, the asset (or book) value for the piece of
equipment at the end of 4 years would have been $5240. The total income-tax rate for the company is 34% of all gross
earnings. Capital-gains taxes amount to 34% of the gain. How much net saving after taxes would the company have achieved
by using the alternative (in this case, reducing-balance) depreciation method instead of the straight-line depreciation method?
a. $2161.60
b. $1261.07
c. $1171.70
d. $1161.60
39. A piece of equipment is originally costing $40,000 was put into use 12 years ago. At the time the equipment was put into use,
the service life was estimated to be 20 years and the salvage and scrap value at the end of the service life were assumed to
be zero. On this basis, the straight-line depreciation fund was set up. The equipment can now be sold for $10,000, and a
40.
41.
42.
43.
44.
45.
46.
47.
50.
more advanced model can be installed for $55,000. Assuming the depreciation fund is available for use, how much new
capital must be supplied to make the purchase?
a. $22,000
b. $21,000
c. $12,000
d. $11,000
The original investment for an asset was $10,000, and the asset was assumed to have a service life of 12 years with $2,000
salvage value at the end of the service life. After the asset has been in use for 5 years, the remaining service life and the final
salvage value are reestimated at 10 years and $ 1,000, respectively. Under these conditions, what is the depreciation cost
during the sixth year of the total life is straight-line depreciation is used?
a. $596.65/yr
b. $561.66/yr
c. $678.65/yr
d. $566.66/yr
e. A piece of equipment having a negligible salvage and scrap value is estimated to have a service life of 10 years.
The original cost of the equipment was $40,000. Determine the following:
Based on the above data, determine the depreciation charge for the fifth year if double-declining balance depreciation is used.
a. $ 3,567.2
b. $ 3,276.80
c. $ 4, 245.80
d. $ 4,252.80
The depreciation charge for the fifth year if sum-of-the-years-digits depreciation is used.
a. $ 4,363.64
b. $ 5,572.14
c. $ 8,265.52
d. $ 9,356.45
The percent of the original investment paid off in the first half of the service life using the double-declining balance method.
a. 32.8%
b. 67.4%
c. 23.4%
d. 54.2%
The percent of the original investment paid off in the first half of the service life using the sum-of-the-years-digits method.
a. 45.45%
b. 54.54%
c. 23.23%
d. 35.53%
The original cost of the property is $30,000, and it is depreciated by a 6 percent sinking-fund method. What is the annual
depreciation charge if the book value of the property after 10 years is the same as if it had depreciated at $2,500/year by the
straight-line method?
a. $ 167.67/yr
b. $ 267.67/yr
c. $ 289.67/yr
d. $ 189.67/yr
A concern has a total income of $1 million/year, and all expenses except depreciation amount to $600,000/year. At the start of
the first year of the concern’s operation, a composite account of all depreciable items show a value of $850,000, and the
overall service life is estimated to be 20 years. The total salvage value at the end of the service life is estimated to be
$50,000. Thirty percent of all profits before taxed must be paid out as income taxes. What would be the reduction in incometax charges for the first year of operation if the sum-of-the-years-digits methods were used for depreciation accounting instead
of the straight-line method?
a.
a. $16,356.72
b.
b. $26,285.72
c. c. $26,356.72
d. d. $16,285.72
The total value of anew plant is $2 million. A certificate of necessity has been obtained permitting a write-off of 60 percent of
the initial value ay 5 years. The balance of the plant requires a write-off period of 15 years. Using the straight-line method and
assuming negligible salvage and scrap value, determine the total depreciation cost during the first year.
a. $ 35, 555. 33/ yr
b. $ 55, 333. 33/ yr
c. $ 35, 333. 33/ yr
d. $ 53, 333. 33/ yr
A profit-producing property has an initial value of $50,000, a service life of 10 years, and zero salvage and scraps value. By
how much would annual profits before taxes be increased if a 5 percent sinking-fund method were used to determine
depreciation costs instead of straight-line method?
e.
a. $ 3 556.24
f.
b. $ 3 354.35
g.
c. $ 3 484.31
h.
d. $ 3.034.35
51. In order to make it worthwhile to purchase a new piece of equipment, the annual depreciation costs for the
equipment cannot exceed $3,000 at any time. The original cost of the equipment is $30,000, and it has a zero salvage and
scrap value. Determine the length of service life necessary if the equipment is depreciated by the sum=of-the-years-digits
method by the straight-line method.
j.
a. 19 years
k.
b. 23 years
l.
c. 17 years
m.
d. 18 years
n. 52. Referring to the previous number, determine the length of service life necessary if the equipment is depreciated by the straightline method.
o.
a. 8 years
p.
b.10 years
q.
c. 12 years
r.
d. 9 years
s.
t.
A materials-testing machine was purchased for $20,000 and was to be used for 5 years with an expected residual
salvage value of $5,000. Graph the annual depreciation charges and year-end book values obtained by using:
u.
53. By using Straight-line depreciation
v.
a. $ 6,000
w.
b. $ 5,000
x. c. $ 9,000
y.
d. $ 7,000
z.
54.By using Sum-of-digits depreciation
aa.
a. $ 6,000
ab.
b. $ 5,000
ac.
c. $ 9,000
ad.
d. $ 7,000
ae.
af.
55. By using Double-declining balance depreciation
a. $ 625
b. $ 675
c. $278
d. $955
ag.
56. An asset with an original cost of $10,000 and no salvage value has a depreciation charge of $2381 during its
second year of service when depreciated by the sum-of-digits method. What is its expected useful life?
a. 6 years
b. 4 years
c. 9 years
d. 2 years
ah.
57. An electronic balance costs P90, 000 and has an estimated salvage value of P8, 000 at the end of its 10 years
lifetime. What would be the book value after 3 years, using straight-line method in solving for the depreciation?
a. P 65,200
b. P 76,466
c. P 24, 674
d. P 21,758
ai. A broadcasting corporation purchased equipment for P53, 000 and paid P1, 5000 for freight and delivery charges
top the job sites. The equipment has a normal life of 10 year with a trade-in value of P5, 000 against the purchase
of anew equipment at the end of the life.
aj.
58. Referring to the problem above, determine the annual depreciation by straight-line method.
a. P4, 950
b. P5, 950
c. P7, 950
d. P3, 950
ak. 59. From the preceding number, determine annual depreciation by sinking fund method. Assuming interest 6 ½%
compounded annually.
al.
a. P 3,668
am.
b. P 1,575
an.
c. P 4,245
ao.
d. P6,258
ap.
aq. A firm brought equipment for P56, 000. Other expenses including installation amounted to P4, 000. The equipment
is expected to have a life of 16 years with a salvage value of 10% of the original cost.
ar. 60. Determine the book value at the end of 12 years by SLM:
as.
a. P19, 500
at.
b. P20,555
au. c. P15,582
av. d. P18, 562
i.
aw. 61. Determine the book value at the end of 12 years by SFM:
P29, 520
P29, 520
P29, 520
P29, 520
ax.
ay.
az. For numbers 62 – 64. A certain type of machine losses 10% of its value each year. The machine cost P2, 000
originally. Make cut a schedule showing the following:
By yearly depreciation
ba.
a.131.32
bb.
b.132.12
bc. c.123.23
bd. d. 213.20
The total depreciation after 5 years.
a. P819.12
b. P562.23
c. P456.26
d. P895.23
Estimate the book value at the end year for 5 years.
a. P1, 180.98
b. P5, 125.25
c. P1, 256.32
d. P5, 235.00
Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8 years for an
asset that costs P15,000 new and has an estimated scrap value of P2,000 at the end of 10 years by DBM.
a. P 12,108
b. P12, 008
c. P11,355
d. P10,245
Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8 years for an
asset that costs P15, 000 new and has an estimated scrap value of P2, 000 at the end of 10 years by DDBM.
a. P 12,583
b. P 10,483
c. P 12,483
d. P 10,583
Mr. Dim bought a calciner for P220, 000 and used it for 10 years, the life span of the equipment. What is the book value of the
calciner after 5 years of use? Assume a scrap value of P20, 000 for SLM;
a. P 120,566
b. P 120,452
c. P 120,000
d. P 130,000
Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of P22, 000
for textbook for DBM.
a. P 96, 570.00
b. P 69, 235.00
c. P 59, 235.00
d. P 69, 570.00
Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of P20, 000
for textbook for DDBM.
a. P72 091
b. P72 356
c. P71 190
d. P72 090
A structure costs P12, 000 new. It is estimated to have a life of 5 years with a salvage value at the end of life of P1, 000.
Determine the book value at the end of three years.
a. 733
b. 562
c. 252
d. 377
Operator A produces 120 spindle/hr on a lathe. His hourly rate is $1.80. Operator B, using an identical lathe, is able to
produce 150 identically units/hr. The overhead charge for a lathe is fixed at $2.50/hr. Determine operator B’s hourly rate so
that his cost per piece is identical to A’s.
a. $2.88
b. $3.88
c. 4.88
d. 1.88
a.
b.
c.
d.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
72. In a type 1 warehouse, initial cost will be $24, 000.This warehouse has adequate capacity for the near future, but 12 years
from now an addition will be required that costs $15, 000. A type 2 warehouse costs $34, 000. This type has the same
capacity as the type 1 warehouse with its addition. What will be the present cost of the type 1 warehouse? Which of these
should be built, assuming that depreciation is negligible and that the interest rate is 7%?
a. $30, 540
b. $30,660
c. $23, 548
d. $15,235
73. Determine the equal (year-end) payments that will be available for the next four years if we invest $4, 000 at 6%.
a. $2533.23
b. $2322.25
c. $1123.25
d. $1154.36
74. A new snow removal machine costs $50, 000. The new machine will operate at a reputed savings of $400 per day over the
present equipment in terms of time and efficiency. If interest is at 5% and the machine’s life is assumed to be 10 years with
zero salvage, how many days per year must the machine be used to make the investment economical?
a. 15 days
b. 16 days
c. 14 days
d. 12 days
75. Based on the sinking fund method and using the data in the previous problem, what number of days must the machine be
used if the amount to be accumulated in 10 years is
a. $50, 000
a.
b. $60,000
b.
c. $55,000
c.
d. $45, 000
76. Based on the sinking fund method and using the data in the previous problem, what number of days must the machine be
used if the amount to be accumulated in 10 years is $50, 000?
a. 12 days
b. 10 days
c. 14 days
d. 13 days
77. A consulting engineer decides to set up an educational fund for his son that will provide $3000 per year for 6 years starting in
16 years. The best interest rate he can expect to get is 5% compounded quarterly. He wants to accumulate the necessary
capital by making quarterly deposits until his son starts college. What will be his needed quarterly deposit?
a. $15, 210
b. $16,213
c. $15, 213
d. $16, 210
78. A low carbon steel machine part, costing $350 installed, lasts 6 years when operating in a corrosive atmosphere. An
identically shaped part, but treated for corrosion resistance, would cost $650 installed. How long would the corrosion
resistance part have to last to be at least as good investment as the untreated part? Assume money is worth 7%
a. 11 years
b. 12 years
c. 14 years
d. 13 years
79. The total cost of a cast product consists of (1) the raw material cost that is directly proportional to the weight, of the casting,
(2) the machining cost that varies inversely as the weight, and (3) overhead cost that remains constant per unit produced
regardless of weight. Find the weight giving the minimum cost per casting.
a. W = (k1/k2)1/4 lb
b. W = (k2/k1)1/2 lb
c. W = (k2/k1)1/4 lb
d. W = (k1/k2)1/2 lb
80. Based on the previous problem, what is the minimum total cost?
a. Ct = [k2(k2 / k1)1/2 = k2 (k1 / k2) 1/4+ Co]
b. Ct = [k1(k2 / k1)1/2 = k2 (k1 / k2) ½ + Co]
c. Ct = [k1(k2 / k1)1/4 = k2 (k2 / k1) ½ + Co]
d. Ct = [k2(k1 / k1)1/4 = k2 (k1 / k2) ½ + Co]
81. Methyl alcohol condensed at 148 F is to be cooled to 100 F for storage at a rate of 10, 000gal/hr by water available at 75 F in
a countercurrent heat exchanger. The over-all heat transfer coefficient is constant and estimated at 200 Btu/ft 2-hr-F. Heat
exchanger annual costs including operation are estimated at $2 per ft 2 including depreciation. The cooler is to operate 5, 000
hr/year, and the value of heat utilized is estimated at $5x10 -7 per Btu. What is the estimated optimum cost of the heat
exchanger if the cost for surface is $9 per ft2?
a. $4, 014
b. $4,123
c. $5,545
d. $5,123
82. What is the most economical number of effects to use in the recovery of black liquor in a paper plant if the following cost data
are available? The annual fixed costs increase essentially linearly with each effect (except for condensing, feeding, and other
equipment costs for multiple units which may be considered to balance each other). If a fixed amount of evaporation is to be
obtained and each units to have 1, 000 ft 2 of heating surface with a service life of five years, the annual fixed costs Cf would
be (using cost data of $25, 000 for a single evaporator of 5, 000 ft2, employing the 0.6 factor, and neglecting the interest).
a.
Cf =( 1, 000)0.6 25, 000 N
…….
dollars per year
b.
5, 000
5
c. where N is the number of effects.
d.
Because of the steam economy in multiple-effect operation, the direct costs for steam will decrease and the total of
all annual direct costs, CD, has been established for this type of operation as
83.
84.
85.
86.
87.
88.
e.
CD = 65, 000 N -0.95 dollars
f.
4 effects
g. 2 effects
h. 6 effects
i.
5 effects
A capitalized cost for a piece of equipment has been found to be $55, 000. This cost is based on the original cost plus the
present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the capitalized
cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under these conditions,
what would be the original cost of the equipment?
a. $36, 861.47
b. $32, 251.47
c. $27, 291.47
d. $37, 291.47
A heat treating furnace is used to preheat small steel parts. The furnace uses fuel oil consisting $0.04 per gallon, with a
heating value of 142, 000Btu/gal. The furnace has a firebrick lining, the outside temperature of which is 1210 F, this is to be
covered with insulation costing $300 per 1000 board feet. The air temperature is 110 F. Operations is 7200 hr/yr. Conductivity
is 0.028 for insulation in Btu/hr-ft 2-F. Calculate the most economical thickness of insulation. Furnace life is 8 years. Assume
negligible temperature drop from insulation to air.
a. 5.52 in
b. 5.24 in
c. 5.48 in
d. 5.45 in
A batch inorganic chemical operations gives product C from two chemicals A and B according to the following empirical
relation:
a.
C = 2.8 (AB – AC – 1.2 BC + 0.5C2 )0.5
b. where A, B and C are pounds of respective components. The reaction rate is sufficiently high to be neglected, and the
time to make any batch is essentially the charging and discharging time, including heating up, which totals 1 hr. If A costs
$0.10 per lb and B costs $0.05 per lb, what is the ratio of B to A to give the minimum costs of raw materials per lb of
product . what is the cost per lb of C?
c. $4.08
d. $3.23
e. $4.23
f.
$3.08
Based from the previous problem, what is the cost per lb of C?
a. $4.08
b. $3.23
c. $4.23
d. $3.08
Seven million pounds of water per year is to be obtained from 8 percent solids slurry to be filtered on a leaf filter to produce a
cake containing 40 percent solids. The area of the filter is 200 ft 2. Tests show a value of 2 x 104 for k in pound units. The cake
is not washed. The dumping and cleaning time is 3 hr and costs $39 each cycle. Filtration costs are $14 per hr, and inventory
charges maybe neglected. What is the cycle time for minimum costs?
a. 0.014 hrs
b. 0.015 hrs
c. 1 hr
d. 0.25 hrs
Referring to the previous problem, calculate the cycle time for maximum production.
a. 3.04 hrs.
b. 2.04 hrs
c. 3.52 hrs
d. 1.02 hrs
e.
f.
For nos. 89- 91.
g. In processing 500 ton/day of ore assaying 50% mineral, 300 tons of concentrate containing 66.7 % are obtained at a
cost of sales ( all fixed operating cost are excluded) of $15 per ton concentrate. An investment of $200, 000 of
concentrate that will assay 71% mineral. If the plant operates 200 days/year, equipment must pay out in 5 years with
interest at 15% and no salvage value and no additional labor or repair costs need to be considered.
h.
89. Based on the stated problem above, calculate the additional cost per ton of concentrate for capital recovery on the new
equipment.
a. $74.6
b. $56.2
c. $15.2
d. $14.0
90. Determine the selling price in dollars per ton (100% mineral basis) required for which the cost of the n new equipment is
justified.
a. $ 123.12
b. $ 123.50
c. $ 263.25
d. $ 263.13
91. What is the % increase in recovery and rejection for the new process based on mineral and gauged?
a. 42.567%
b. 22.224%
c. 52.677%
d. 12.235%
92. A ties on a plant railroad sliding are to be replaced. Untreated ties consisting $ 2.50 installed have a life of 7 years. If created
ties have a life of 10 years, what is the maximum installed cost that should be paid for treated ties if money is worth 8
percent?
a. $3.15
b. $3.10
c. $3.25
d. $3.20
93. Powdered coal having a heating value of 13, 500 Btu/ lb is to be compared with fuel oil worth $2.00 per bbl (42 gal) having a
heating value of 130,000 Btu/gal as a source of fuel in the processing plant. If the efficiency of the conversion of the fuel is
64% for coal and 72 % for oil, with all other costs being equal, what is the maximum allowable selling price for coal per ton?
a. $15.13 / ton
b. $11.13 / ton
c. $21.13 / ton
d. $25.13 / ton
94. A steam boiler is purchased on the basis of guaranteed performance. However, initial tests indicate that the opening (income)
cost will be P400 more per year than guaranteed. If the expected life is 25 years and money is worth 10 %, what deduction
from the purchase price would compensate the buyer for the additional operating cost?
a. 3, 635.82 pesos
b. 3, 660.82 pesos
c. 3,560.82 pesos
d. 3, 630.82 pesos
95. If the sum of P12, 000 is deposited in an account earning interest at the rate of 9% compounded quarterly, what will it become
at the end of 8 years?
a. 14, 857.24
b. 34, 457.24
c. 14, 527.24
d. 24, 457.24
96. At a certain interest rate compounded quarterly, P1, 000 will amount to P4, 500 in 15 years. What is the amount at the end of
10 years?
a. 2, 125.17 pesos
b. 2, 725.17 pesos
c. 2, 625.17 pesos
d. 2, 845.17 pesos
97. A one bagger concrete mixer can be purchased with a down payment of P8, 000 and equal installments of P600 each paid at
the end of every month for the next 12 months. If the money is worth 12% compounded monthly, determine the equivalent
cash prize of the mixer.
a. 4, 753.05 pesos
b. 34, 753.05 pesos
c. 24, 753.05 pesos
d. 14, 753.05 pesos
98. A certain company makes it the policy that for any new piece of equipment, the annual depreciation cost should not exceed
10% of the original cost at any time with no salvage or scrap value. Determine the length of service life necessary if the
depreciation method use is straight line formula.
a. 10 years
b. 12 years
c. 18 years
d. 8 years
99. Solve the previous problem with the sinking fund formula at 8%
a. 8 years
b. 10 years
c. 6 years
d. 12 years
100. Determine the ordinary simple interest on $10, 000 for 9 months and 10 days if the rate of interest is 12%.
a. $433.33
b. $633.33
c. $333.33
d. $933.33
e.
f.
Answers:
g.
h. 1. The purchased cost of a shell-and-tube heat exchanger (floating head and carbon-steel tubes) with 100 ft 2 of heating
surface was $3,000 in 1980. What will be the purchased cost of a similar heat exchanger with 200 ft 2 of heating surface
in 1980 if purchased-cost-capacity exponent is 0.60 for surface area ranging from 100-400 ft2?
i.
GIVEN:
j.
A1 = 100 ft2
i100-400 = 0.60
k.
P1980 = $3, 000
i400-1,000 = 0.81
l.
A2 = 200 ft2
m. REQUIRED:
n. $ of 200 ft2 in 1980
o. SOLUTION:
p.
0.6
 200 ft2 
c ost200 ft2 =$3,000 
2 
 100 ft 
cost200 ft2 =$4,547.15
q.
2. Refer to problem no. 1. If the purchased-cost-capacity exponent for this type of exchanger is 0.81 for surface areas
ranging from 400-2,000 ft2, what will be the purchased cost of a heat exchanger with1, 000 ft2 of heating surface in 1985?
s. GIVEN:
t.
A1 = 100 ft2
i100-400 = 0.60
u.
P1980 = S3,000
i400-1,000 = 0.81
v.
A2 = 200 ft2
w. REQUIRED:
x. $ of 1,000 ft2 in 1985
y.
SOLUTION:
z.
using table 3, page 163 (Cost indexes as annual average)
aa.
where: 1980 = 560, and 1985 = 790
ab.
r.
0.6
 400 ft2 
cost =$3,000 
2 
 100 ft 
cost =$6,892.19
0.81
 1,000 ft2 
cost1985 =$6,892.19 
2 
 400ft 
 790 


 560 
cost1985 =$20,423.38
ac.
ad. 3. If the purchased cost of a shell-and-tube heat exchanger (floating head and carbon-steel tubes) with 100 ft 2 of heating
surface was $3,000 in 1980. Note that the purchase-cost-capacity exponent is not constant over range surface area
requested. What will be the purchased cost of a heat exchanger with 700 ft2 of heating surface in 1985?
ae. GIVEN:
af.
A = 100 to 2,000 ft2
ag.
Present year = 1980
ah.
P1980 = $3,000
ai. REQUIRED:
aj.
1985 purchase cost with 700 ft2
ak. SOLUTION:
al.
Using table 3, page 163 (Cost indexes as annual average)
am.
Where: 1980 = 560, and 1985 = 790
an.
For 200 ft2
ao.
0.60
 200 
C 1985 =$3,000 

 100 
C 1985 =$6,414.73
ap.
 790 


 560 
For 300 ft2
aq.
0.60
 300 
C 1985 =$6,414.73 

 200 
C 1985 =$8,181.50
ar.
For 400 ft2
as.
0.60
 400 
C 1985 =$8,181.50 

 300 
C 1985 =$9,722.91
at.
For 500 ft2
au.
0.81
 500 
C 1985 =$9,722.91

 400 
C 1985 =$11,649.13
av.
For 600 ft2
aw.
0.81
 600 
C 1985 =$11,649.13 

 500 
C 1985 =$13,503.00
ax.
For 700 ft2
ay.
0.81
 700 
C 1985 =$13,503.00 

 600 
C 1985 =$15,298.79
az.
ba. 4. From the preceding question, what will be the purchased cost of the heat exchanger with 800 ft 2 of heating surface in
1985?
bb. GIVEN:
bc.
A = 100 to 2,000 ft2
bd.
Present year = 1980
be.
P1980 = $3,000
bf. REQUIRED:
bg.
1985 purchase cost with 800 ft2
bh. SOLUTION:
bi.
For 700 ft2
bj.
0.81
 700 
C 1985 =$13,503.00 

 600 
C 1985 =$15,298.79
bk.
bl.
bm.
For 800 ft2
bn.
0.81
 800 
C 1985 =$15,298.79 

 700 
C 1985 =$17,046.32
bo.
bp. 5. Refer to problem no. 3. What will be the purchased cost of the heat exchanger with 1500 ft 2 of heating surface in
1985?
bq. GIVEN:
br.
A = 100 to 2,000 ft2
bs.
Present year = 1980
bt.
P1980 = $3,000
bu. REQUIRED:
bv.
1985 purchase cost with 1500 ft2
bw. SOLUTION:
bx.
For 800 ft2
by.
0.81
 800 
C 1985 =$15,298.79 

 700 
C 1985 =$17,046.32
bz.
For 1,000 ft2
ca.
0.81
 1,000 
C 1985 =$17,046.32 

 800 
C 1985 =$20,423.39
cb.
For 1,500 ft2
cc.
0.81
 1,500 
C 1985 =$20,423.39 

 1,000 
C 1985 =$28,363.62
cd.
ce.
cf. 6. The purchase and installation cost of some pieces of equipment are given as a function of weight rather than
capacity. An example of this is the installed cost of large tanks. The 1980 cost for an installed aluminum tank weighing
100,000 lb was $390,000. For a size range from 200,000 to 1,000,000 ld, the installed cost-weight exponent for
aluminum tanks is 0.93. If an aluminum tank weighing 700,000 lb is required, what is the present capital investment
needed?
cg. GIVEN:
ch.
W1980 = 100,000 lb
i200, 000-1,000,000 = 0.93
ci.
C1980 = $390,000
cj. REQUIRED:
ck.
C in $ for 700,000 lb
cl.
cm. SOLUTION:
cn.
0.60
 200,000 
C =$390,000 

 1,000,000 
0.93
 700,000 


 200,000 
C =$1,900,000
co.
cp.
cq.
Present Cost
using table 3, page 163 (Cost indexes as annual average)
where: 1980 = 560, and 1985 = 904
cr.
 904 
C 1990 =$1,900,000 

 560 
C 1990 =$3,060,000
cs.
7. The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000.
The plant is to be an addition to an existing formaldehyde plant. The major part of the building cost will be for indoor
construction, and the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values
found for typical chemical plants. On the basis of this information, estimate the total direct plant cost.
cu. GIVEN:
cv.
Equipment Cost = $300,000
cw.
Contractor’s Fee = 7% of the direct plant cost
cx.
cy. REQUIRED:
cz. Total direct plant cost
da. SOLUTION:
db. Purchased Equipment
100% ($300,000)= $300,000
dc. Purchased Installation
39% ($300,000)
=
$117,000
dd. Instrumentation
13% ($300,000)
=
$39,000
de. Piping
31% ($300,000)
=
$93,000
df. Electrical
10% ($300,000)
=
$30,000
dg. Building
29% ($300,000)
=
$87,000
dh. Yard Improvements
10% ($300,000)
=
$30,000
di. Service Facilities
55% ($300,000)
=
$165,000
dj. Land
6% ($300,000)
=
$18,000
ct.
dk.
Total DirectPlant Cost =$879,000
dl.
dm. 8. The purchased equipment cost for a plant which produces pentaerythritol (solid-fuel processing plant) is $300,000.
The plant is to be an addition to an existing formaldehyde plant. The major part of the building cost will be for indoor
construction, and the contractor’s fee will be 7% of the direct plant cost. All other costs are close to the average values
found for typical chemical plants. On the basis of this information, estimate the fixed capital investment.
dn. GIVEN:
do.
Equipment Cost = $300,000
dp.
Contractor’s Fee = 7% of the direct plant cost
dq. REQUIRED:
dr. Fixed capital investment
ds.
dt.
du. SOLUTION:
dv.
dw.
dx.
dy.
dz.
ea.
eb.
ec.
ed.
ee.
ef.
eg.
eh.
ei.
ej.
ek.
Purchased Equipment
Purchased Installation
Instrumentation
Piping
Electrical
Building
Yard Improvements
Service Facilities
Land
100% ($300,000)= $300,000
39% ($300,000)
=
13% ($300,000)
31% ($300,000)
10% ($300,000)
=
29% ($300,000)
=
10% ($300,000)
=
55% ($300,000)
=
6% ($300,000)
=
Engineering and Supervision
Construction Expense
32% ($300,000)
34% ($300,000)
=
=
Constructor’s Fee
Contingencies
7% ($879,000)
36% ($879,000)
=
=
$117,000
=
$39,000
=
$93,000
$30,000
$87,000
$30,000
$165,000
$18,000
$96,000
$102,000
$198,000
$61,530
$108,000
$169,530
el.
Fixed Capital Investment =$1,246,530
em.
en.
eo.
ep.
eq.
er.
es.
et.
eu.
ev.
ew.
9. Refer to the previous problem and on the basis of this information, what is the total capital investment?
GIVEN:
Equipment Cost = $300,000
Contractor’s Fee = 7% of the direct plant cost
Fixed capital investment = $1, 246, 530
REQUIRED:
Total capital investment
SOLUTION:
Fixed Capital Investment
$1, 246, 530
Working Capital
74% ($300,000)
=
$222,000
Total Capital Investment =$1,468,530
ex.
ey. 10. The total capital investment for a chemical plant is $1,500,000 and the plant process 3M kg of product annually. The
selling price of the product is $0.82/kg. Working capital amounts to 15% of the total capital investment. The investment is
from company funds, and no interest is charged. Raw-material costs for the product are $0.09/kg, labor $0.08/kg, utilities
$0.05/kg, and packaging $0.008/kg. Distribution costs are 5% of the total product cost. Determine the percent change in
total cost.
ez. GIVEN:
fa.
Total Capital Investment
$1,500,000
fb.
Product
3,000,000 kg annually
fc.
Selling Price
$0.82/kg
fd.
Working Capital
15% of the total capital
fe.
Raw Materials Cost
$0.09/kg
ff.
Labor
$0.08/kg
fg.
Utilities
$0.05/kg
fh.
Packaging
$0.008/kg
fi.
Distribution Cost
5% of total capital cost
fj. REQUIRED: % change in total cost
fk.
fl. SOLUTION:
fm.
TPC = FC + WC +VO
fn.
TPC = 0.35 + 0.040 + 0.25
fo.
TPC =1or100%
fp.
by ratio and proportion
1.5
x  500,000 
=

25  1,000,000 
x =8.84%
fq.
fr.
after producing only 500,000 kg of product
fs.
TPC = 0.35 + 0.20 + 0.084 = 0.6384
ft.
fu.
% change =100 - 63.84
% change =36.16%
fv.
fw.
11. It is desired to have a $ 9000 available from 12 years from now. If $ 5000 is available for investment at the present
time, what is discrete annual rate of compound interest on the investment would be necessary to give the desired
amount?
FX.
fy.
fz.
ga.
gb.
GC.
gd.
ge.
gf.
gg.
gh.
GIVEN:
S = $ 9000
P = $ 5000
n = 12
REQUIRED: Interest, i
SOLUTION:
S = P (1 + i )n
$9000 = $5000(1 + i) 12
i = 0.05 or 5.02 %
12. An original loan of $2000 was made at 6 percent simple interest per year for 4 years. At the end of this time, no
interest had been paid and the loan was extended for 6 more years at a new, effective, compound-interest rate of 8
percent per year. What is the total amount owned at the end of ten years if no intermediate payments are made?
GIVEN:
P = $2000
i = 6%
n = 4 years
REQUIRED: SC
SOLUTION:
SS = P (1 + in)
= 2000 [1 + 0.06(4)]
SS = $2480
Extended for 6 years
SC = $2480 (1 + 0.08)6
gi.
gj.
gk.
gl.
gm.
gn.
go.
gp.
gq.
gr.
gs.
gt.
SC = $ 3,935.45
gu.
gv.
gw. 13. The original cost for a distillation tower is $24,000 and the useful life of the tower is estimated to be 8 years. The
sinking fund method for determining the arte of depreciation is used, and the effective annual interest for the depreciation
fund is 6 percent. If the scrap value of the distillation tower is $ 4000, determine the asset value at the end of 5 years.
gx. GIVEN: n = 8 years
gy.
V = $ 24,000
gz.
VS = $4000
ha. REQUIRED: R
HB.
HC.
HD.
HE. SOLUTION:
hf.
R (V  VS )
i
(1  i )n  1
R (24000  4000)
0.06
(1  0.6)5  1
hg.
R = $ 3,547.928
hh.
hi.
hj. 14. An annuity is due to being used to accumulate money. Interest is compounded at an effective annual rate of 8 %, and
$1000 is deposited at the beginning of each year. What will be the total amount of annuity due be after 5 years?
HK. GIVEN:
HL.
R = $1000
hm.
N = 5 yrs.
I=8%
hn. REQUIRED: S5
ho. SOLUTION:
hp.
If no interest:
S5 = R [ ( 1+ i) n – 1]/i
hq.
S5= $5000
hr.
But with interest:
hs.
S5 = 1000[(1 + 0.08) 5-1]/0.08
ht.
hu.
S5 = $ 5866.6009
Per year:
S = $5866.6009/5 = $ 1173.32
hv.
hw. 15. For the total year payments of $5000 for ten years, what will be the compound amount accumulated at the end of ten
years if the payment is at the end of the year? The effective (annual) interest is 20% and payments are uniform.
HX. GIVEN:
hy.
R = $5000
hz.
20% annual interest
ia. REQUIRED:
ib.
S at the end of the year
IC. SOLUTION:
id.
ie.
if.
S = R [(1+ i) n – 1]/i
S = $5000[(1 +0.2) 10-1]/0.2
S = $ 129793.41
ig.
ih.
ii.
16. Referring to the previous problem, estimate the compound amount accumulated at the end of ten years, if the
payment is made weekly?
IJ. GIVEN:
ik.
R = $5000
il.
20% annual interest
im. REQUIRED:
in.
S at the end of the year
IO.
IP.
IQ. SOLUTION:
ir.
is.
For weekly i ≠ 0.2
it.
For new ieff = [1+ r/m) m-1
iu.
Ieff = [1 + 0.2/52) -1 = 0.2209
iv.
Solving for S,
iw.
S = R[1+ ieff) – 1]/ieff
ix.
S = $5000 [1.02209) 10-1]/0.2209
iy.
S = $ 143951.4873
iz.
ja.
jb.
jc.
jd.
je.
jf.
jg.
jh.
ji.
jj.
jk.
jl.
JM.
jn.
jo.
JP.
jq.
jr.
js.
jt.
ju.
jv.
jw.
17. A multiple – effect evaporator is to be used for evaporating 400,000 lb of water per day from a salt solution. The total
initial cost for the 1st effect is $18,000 and each additional effect costs $15,000. The life period is estimated to be 10
years, and the scrap value at the end of the life period may be assumed to be zero. The straight-line depreciation
method is used. Fixed charges minus depreciation are 15% yearly based on the first cost of the equipment. Steam cost
$1.50 per 1000 lb. Annual maintenance charges are 5% of the initial equipment cost. All other cost is independent of the
number of effects. The unit will operate 300 days per year. If the lb of water evaporated per pound of steam equals 0.85
x numbers of effects, determine the optimum number of effects for minimum annual cost.
GIVEN:
Initial cost = $18000
Lb of water = 400000 lb
Cost of additional effect = $15000
Life period = 10
Salvage value = 0
Fixed charges = 15% of first cost
Steam cost = $1.50/1000 lb
Maintenance charges = 5% of first cost
Operation = 300 days/year
Lb of steam/lb water = 0.85 x no. of effects
REQUIRED:
Optimum number of effects for minimum annual cost
SOLUTION:
Let x = additional no. of effects
Cost of equipment = 18000 + 15000x
FC = 0.15 (18000 +15000x)
= 2700 + 2250 x
Annual maintenance = 0.05(18000+15000x)
= 900 + 750x
jx.
jy.
jz.
ka.
kb.
kc.
kd.
Depreciation/yr = (18000+15000x)/10
d = 1800 + 1500x
kf.
lb of steam = 400000lbH2O/day (300 days/yr)
0.85(x+1)
= 1.41176 x 108
ke.
(x+1)
Steam cost / yr =
1.50  1.41176x108 


1000
x 1

kg.
= 70,588.23 / (x+1)
kh.
ki. Total Cost per year = CT = FC + Steam cost + depreciation + maintenance cost
kj.
kk. CT = (2700 + 2250x) + (1800+1500x) + (900+750x) + (70588.23/x+1)
kl. CT = 5400 + 4500x + 70588/(x+1)2
km.
dCT
70588
 4500
dx
(x  1) 2
kn. 0 = 4500(x2+2x+1) – 70588
ko. 0 = 4500x2 + 9000x – 66088
kp.
kq.
Using quadratic equation:
kr.
x=
 9000
 9000 2  4 4500 
66088
2(4500)
ks.
x = 2.96
kt.
ku.
Use total effect = 3
kv.
kw. 18. Determine the optimum economic thickness of insulation that should be used under the following conditions:
Standard steam is being passed continuously through a steel pipe with an outside diameter of 10.75 in. The temperature
of the steam is 400F, and the steam is valued at $1.80 per 1000 lb. The pipe is to be insulated with material that has a
thermal conductivity of 0.03 Btu/h-ft2-F/ft. The cost of installed insulation per foot of pipe length is $4.5xI t, where It is the
thickness of the insulation in inches. Annual fixed charges including maintenance amount to 20% of the initial installed
cost. The total length of the pipe is 1000 ft, and the average temperature of the surrounding may be taken as 70F. Heat
transfer resistance due to the steam film, scale and pipe wall are negligible. The air –film coefficient at the outside of the
insulation may be assumed constant at 2.0
Btu/h-ft2-F for all insulation thickness.
KX.
GIVEN:
ky.
Steam
kz.
400F
do = 10.75 in
la.
lb.
Do (with insulation) = 10.75 + 2 It
lc.
Cost of steam = $1.80/1000lb
ld.
Thermal conductivity = 0.03 Btu/h-ft2-F/ft
le.
Cost of insulation/ft = $4.5 It
lf.
Annual fixed charges + maintenance = 20% of Initial cost
lg.
Length = 1000 ft
lh.
T surrounding = 70F
li.
h = 2 btu/h-ft2F
lj.
REQUIRED:
lk.
Optimum economic thickness insulation
LL.
LM. SOLUTION:
ln.
Cost of insulation = (4.5 It) (1000 ft ) = 4500 It
lo.
lp.
FC = 0.20 (4500 It)
lq.
FC = 500 It
lr.
ls.
cost of steam = m = UAo∆T
lt.
∆T = 400 – 70 =330F
lu.
 = 826 btu/lb
lv.
A = DoL =
10.75  2I t (1000)
12
lw.
Uo 
1
1  xwDo
ho
kDln
Do  do
2

10.75  2I t   (10.75)
Dln 
10.75  I t
2
1
Uo 
I
(
10
.
75

2
I
)
t
1  t
2
0.03(10.75  I t )(12)
Dln 
lx.
Cost of steam=
lz.
dTAC
 900I t 458.8(10.75  2It)
dI t
ma.
mb.




1


 1  I t (10.75  2I t )

0.03(10.75  I t )(12) 
 2
mc.
md.
TAC = 900
It
0 = 900
2.31
It
( 10.75 I t )(10.75  2I t )
(10.75 I t )(2I t ) 2  0.935
2.31
( 10.75 I t )(10.75  2I t )
(10.75 I t )(2I t ) 2  0.935
me. It = 1 TAC = $3228
mf.
= 1.5 TAC = $3054
mg.
mh.
It = 1.4 TAC = $3057
MI.
mj. 19. A proposed chemical plant will require a fixed capital investment of $10 million. It is estimated that the working capital
will amount to 25% of the total investment and annual depreciation costs are estimated to be 10 percent of the fixed
capital investment. If the annual profit will be $3 million, determine the standard percent return on the total investment
MK.
ML.
MM. GIVEN:
mn.
FCI = $10 M
mo.
Working capital = 25% TCI
mp.
Annual depreciation costs = 10% FCI
mq.
Annual profit = $3 M
MR. REQUIRED:
ms.
% return on investment (ROR)
MT. SOLUTION:
mu.
ROR =
annualprof it
TCI
mv.
mw.
mx.
my.
mz.
na.
nb.
But TCI = FCI + WC
Let x = TCI
x = 10 M + 0.25x
x = TCI = $13.33M
ROR =
$3M
x100
$13.33M
ROR = 22.5%
nc.
nd. 20. If a plant will require a fixed capital investment of $10 million and the working capital will amount to 25% of the total
investment and annual depreciation costs are estimated to be 10 percent of the fixed capital investment. If the annual
profit will be $3 million, what is the minimum payout period (POP)?
NE. GIVEN:
nf.
FCI = $10 M
ng.
Working capital = 25% TCI
nh.
Annual depreciation costs = 10% FCI
ni.
Annual profit = $3 M
NJ. REQUIRED:
nk.
minimum payout period
NL. SOLUTION:
nm.
POP =
depreciabl eFCI
profit dep' n

yr
yr
nn.
no.
ns.
nt.
nu.
nw.
ny.
=
10M
$3M
 0.1($10M)
yr
POP = 2.5 years
np.
nq. 21. An annual investigation of a proposed investment has been made. The following result has been presented to
management. The minimum payout period based on capital recovery using a minimum annual return of 10 percent as a
fictitious expense is 10 years; annual depreciation costs amount top 8 percent of the total investment. Using this
information, determine the standard rate of return on the investment.
NR. GIVEN
Payout period = 10 years
Minimum annual return = 0.10 of fictitious expense
Annual depreciation cost = 8% TCI
NV. REQUIRED
Standard rate of return
NX. SOLUTION
Payout period =
FCI
averagepro fit avedep' n

yr
yr
nz.
oa.
ob.
oc.
Ave. profit = annual profit – expenses
= x – 0.10TCI
10 =
od.
oe.
0.85TCI
(x  0.10TCI)  0.8TCI
x = 0.105 TCI
%rate of return =
of.
annualprof it
x10 0
TCI
ROR
= 10.50%
og.
oh. 22. The information given in the previous problem, applies to conditions before income taxes. If 34% percent of all profits
must be paid out for income taxes, determine the standard rate of return after taxes using the figures given in the
previous problem.
OI. GIVEN:
oj.
Payout period = 10 years
ok.
Minimum annual return = 0.10 of fictitious expense
ol.
Annual depreciation cost = 8% TCI
om.
34% of all profits must be paid out for income taxes
ON. REQUIRED
oo.
Standard rate of return after taxes
OP. SOLUTION:
oq.
Profit = 0.105TCI (before taxes)
or.
Profit = 0.105TCI – 0.34(0.105TCI)
(after taxes)
os.
Profit = 0.0693TCI
ot.
ROR =
annualprof it
x10 0
TCI
ou.
ov.
=
ROR
0.0693TCI
x100
TCI
= 6.93%
ow.
ox. 23. A capitalized cost for a piece of equipment has been found to be $55,000. This cost is based on the original cost plus
the present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the
capitalized cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under
these conditions, what would be the original cost of the equipment?
OY. GIVEN:
oz.
k = $55,000
i = 12%
pa.
Vs = 0
n = 10 yrs.
PB. REQUIRED: Cv
PC. SOLUTION
pd.
k=
C R (1  i) n
 Vs
(1  i) n  1
pe.
Vs = 0
pf.
pg.
ph.
55 000 =
CR = $37 291.47
k=
Cv 
pi.
C R (1  0.12)10
(1  0.12)10 1
CR
(1  i) n  1
Cv = 55 000-
37291.47
(1  0.12)10  1
pj.
Cv =$37 291.47
pk.
pl.
pm. For problems 24- 26.On Aug. 1, a concern had 10,000 lb of raw material on hand, which was purchased at a cost of
$0.030 per pound. In order to build up the reserve, 8000 lb of additional raw material was purchased on Aug. 15 at a cost
of $0.028 per pound. If none of the raw material was used until after the last purchase.
pn.
po. 24. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of
August by current average method.
pp. SOLUTION:
pq.
Ave = $ (0.030 + 0.028 + 0.031) / 3
pr.
Ave = $ 0.0297/ lb
ps.
pt. 25. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of
August by “FiFo” method.
pu. SOLUTION:
pv.
10,000 lb = $ 0.030
pw.
other 2000 lb cost = $ 0.028/lb
px. 26. Determine the total cost of 12,000 lb of the raw material on an inventory or cost-of-sales account for the month of
August by lifo method.
py. ANSWER:
recent price = $ 0.031/lb
pz.
qa.
qb.
qc.
For question nos. 27 – 31.
The following are the data gathered from the AMD Food Corporation:
qd.
qe.
qf.
qg.
qh.
qi.
qj.
qk.
ql.
qm.
qn.
qo.
qp.
qq.
Cash
Accounts payable:
B Company
C Company
qr.
$20
qs.
qt.
Accounts receivable
Inventories
Mortgage payable
Common stock sold
Machinery and equipment (at present value)
Furniture and fixtures (at present value)
Government bonds
Surplus
qu.
2,0
qv.
8,0
qw.
qx.
6,0
qy.
15,
qz.
5,0
ra.
50,
rb.
18,
rc.
5,0
rd.
3,0
re.
2,0
rf. 27. From the data given above, determine the total asset of the AMD Food Corporation.
rg. ANSWER:
RH. BALANCE SHEET
ri.
ASSETS
rj.
rk.
EQUITIES
rl.
rm. Current Assets
rn.
ro.
Current
Liabilities
rp.
rq.
rr.
rs.
Accounts
Payable
rt.
Cash
$
ru.
Accounts
Receivable
rv.
6,0
rw.
MC
Company
rx.
$
ry.
Inventories
rz.
15,
sa.
MD
Company
sb.
8,0
sd.
3,0
se. Mortgage
Payable
sf.
5,0
sh.
$
si.
sj.
sc. Government
Bonds
sg. Total
$
sk. Fixed Assets
sl.
sm. Stockholde
r’s Equity
sn.
so. Machinery &
Equipment
sp.
18,
sq. Common
Stocks
Sold
sr.
50,
ss.
st.
5,0
Furniture &
Fixtures
sw. Total
ta.
te.
Total Assets
sx.
$
su. Total
sy.
tn.
to.
tp.
Surplus
Total
td.
$
tf.
tg. Total
Equities
th.
$
tl.
tm. 28. Determine the total current assets of the AMD Food Corporation.
ANSWER: From the balance sheet;
Total Current Asset = $ 44,000
tq.
tt.
tu.
tv.
tw.
tx.
sz.
2,0
tc.
Total Assets: $ 67, 000
tr.
ts.
sv.
$
tb.
$
ti.
tj.
tk.
Total
29. What is the total amount of the Current liabilities of the AMD Corporation?
ANSWER: From the balance sheet;
Total Current Liabilities = $ 15,000
30. What is the total amount of the Fixed Assets of the AMD Food Corporation?
ANSWER: From the balance sheet;
Total Fixed Assets = $ 23,000
ty.
tz.
During the month of October, the following information was obtained in the AC antifreeze retailing company:
ua.
ub.
uc.
ud.
ue.
uf.
ug.
uh.
Salaries
Delivery expenses
Rent
Sales
Antifreeze available for sale during October (at cost)
Antifreeze inventory on Oct. 31 (at cost)
Other expenses
Earned surplus before income taxes as of Sept.30
ui.
$
uj.
700
uk.
400
ul.
15,
um.
20,
un.
11,
uo.
1,2
up.
800
uq.
ur. 31. Prepare an income statement for the month of October to determine the net income is for the month of October.
us. ANSWER:
ut. AC Antifreeze Retailing Company
uu. Income Statement
uv. As of October
uw. Income
ux.
Sales
$ 15,100
uy.
Antifreeze available for sale
20,200
uz.
Earned Surplus before income
800
va.
Total Gross Income 36,100
vb. Deductions
vc.
Antifreeze inventory on Oct.31
11,600
vd.
Salaries
3,000
ve.
Delivery
700
vf.
Rent
400
vg.
Other Expenses
1,200
vh.
Net Income
$ 19,200
vi.
vj.
vk. 32. From the data above, determine also the total gross income of the AC Antifreeze Company.
vl. ANSWER:
vm.
From the income statement;
vn.
Gross Income = $36,100
vo.
vp. The following information applies to MADSteel Company on a given date:
vq.
vr.
vs.
vt.
vu.
vv.
vw.
vx.
vy.
vz.
wa.
wb.
wc.
Long-term debts
Debts due within 1 year
Accounts payable
Machinery and equipment (at cost)
Cash in bank
Prepaid rent
Government bonds
Social security taxes payable
Reserve for depreciation
Reserve for expansion
Inventory
Accounts receivable
wd. $
1
,
6
0
0
we. 1
,
0
0
0
wf. 2
,
3
0
0
wg. 1
0
,
0
0
0
wh. 3
,
wi.
wj.
wk.
wl.
wm.
wn.
wo.
1
0
0
3
0
0
3
,
0
0
0
2
4
0
6
0
0
1
,
2
0
0
1
,
6
0
0
1
,
7
0
0
wp.
wq.
wr. 33. Determine the cash asset for the MADSteel Company at the given date.
ws.
ANSWER: Cash Asset = $6,100
wt.
wu.
wv.
ww.
wx.
wy.
wz.
xa.
xb.
xc.
xd.
xe.
xf.
xg.
xh.
xi.
xj.
xk.
xl.
Cash Assets
Cash in Bank
Government Bonds
$ 3,100
3,000
Total
Current Assets
Accounts Receivable
Cash in Bank
Government Bonds
Inventory
Prepaid Rent
Total
6,100
1,700
3,100
3,000
1,600
300
9,700
Long-term Debts
Current Liabilities
Departments due within 1 year
Accounts Payable
Social Security Taxes payable
Total
1,600
1,000
2,300
240
5,140
xm.
xn. 34. Determine the current asset for MADSteel Company
xo.
ANSWER: Current Asset = $9, 700
xp. 35. Determine the current liabilities of MADSteel Company
xq.
ANSWER: Current Liabilities= $3, 540
xr.
xs. 36. Determine the Quick ratio for MADSteel Company
xt.
xu.
SOLUTION:
Quick Ratio = Current assets – Inventory
Current liabilities
xv.
xw.
= $9,700– $1, 600
xx.
$3, 540
xy.
xz.
Quick Ratio = 2.29
ya.
yb. 37. Determine the current ratio of MADSteel Company
yc. SOLUTION:
Current ratio =
yd.
current assets
Current liabilities
ye.
= $9, 700
yf.
yg.
yh.
$3, 540
yi.
yj.
yk.
yl.
ym.
yn.
yo.
yp.
YQ.
yr.
ys.
yt.
yu.
yv.
Current ratio = 2.74
38. A reactor of special design is the major item of equipment in a small chemical plant. The initial cost of a completely
installed reactor is $60,000, and the salvage value at the end of the useful life is estimated to be $10,000. Excluding
depreciation costs for the reactor, the total annual expenses for the plant are $100,000. How many years of useful life
should be estimated for the reactor if 12 % of the total annual expenses for the plant are due to the cost for the reactor
depreciation? The straight-line method for determining depreciation should be used.
GIVEN:
Vo = $60,000
Vs = $10,000
Total expenses annually exc. Depreciation = $ 100,000
i = 0.12
REQUIRED: n using SLM
SOLUTION:
d=
Vo  Vs
n
x = total annual expenses inc dep’n/yr
x = 100,000 + 0.12x
x = $ 13,636.36 = d
60,000  10,000
n
yw.
13,636.36 =
yx.
n = 3.67
yy.
n = 4 years
yz. 39. The initial installed cost for a new piece of equipment is $10,000, and its scrap value at the end of its useful life is
estimated to be $2,000. The useful life is estimated to be 10 years. After the equipment has been in use for 4 years, it is
sold for $7,000. The company which originally owned the equipment employs the straight-line method for determining
depreciation costs. If the company had used an alternative method for determining depreciation cost, the asset (or book)
value for the piece of equipment at the end of 4 years would have been $5240. The total income-tax rate for the
company is 34% of all gross earnings. Capital-gains taxes amount to 34% of the gain. How much net saving after taxes
would the company have achieved by using the alternative (in this case, reducing-balance) depreciation method instead
of the straight-line depreciation method? GIVEN:
za.
Vo = $10,000
zb.
Vs = $2,000
zc.
V4 DBM = $5,240
zd.
n = 10 years
ze.
V4 SLM = $7,000
zf.
Tax = 0.34
ZG. REQUIRED: Net saving after taxes
ZH. SOLUTION:
zi.
SLM:
zj.
$7,000 – 7,000 (0.34) = $4,620
zk.
DDB:
zl.
$5,240 – 5,240 (0.34) = $3,458.40
zm.
Net saving after taxes = $4,620 - $3,458.40 = $1161.60
zn.
zo. 40. A piece of equipment is originally costing $40,000 was put into use 12 years ago. At the time the equipment was put
into use, the service life was estimated to be 20 years and the salvage and scrap value at the end of the service life were
zp.
zq.
zr.
zs.
zt.
ZU.
ZV.
assumed to be zero. On this basis, the straight-line depreciation fund was set up. The equipment can now be sold for
$10,000, and a more advanced model can be installed for $55,000. Assuming the depreciation fund is available for use,
how much new capital must be supplied to make the purchase?
GIVEN:
Vo = $40,000
Vs = 0
a = 12
n = 20
REQUIRED: New Capital needed
SOLUTION:
zw.
zx.
zy.
Vo  Vs
n
d=
d=
d=
40,000 
20
0
$2,000
yr
zz. Va = Vo – ad
aaa.
aab. Va = 40,000 - 8 ( 2000 )
aac. Va = $24,000
aad. New Capital = 55,000 – (24,000 + 10,000)
aae. New Capital = $21,000
aaf.
aag. 41. The original investment for an asset was $10,000, and the asset was assumed to have a service life of 12 years with
$2,000 salvage value at the end of the service life. After the asset has been in use for 5 years, the remaining service life
and the final salvage value are reestimated at 10 years and $ 1,000, respectively. Under these conditions, what is the
depreciation cost during the sixth year of the total life is straight-line depreciation is used?
aah. GIVEN:
aai.
Vo = $10,000
aaj.
Vs = $2,000
aak.
a=5
aal.
n = 12
AAM.
REQUIRED: d during the sixth day
AAN.
SOLUTION:
aao.
d=
aap.
aaq.
aar.
aas.
aat.
Vo  Vs
n
d=
d=
V5
10,000  2,000
12
$666 .67
yr
Va = Vo – ad
= 10,000 - 5 (666.67) = $6666.65
aau. After 5 years of use:
6666 .65  1,000
10
aav.
da =
aaw.
da = $566.65/yr
aax.
aay. A piece of equipment having a negligible salvage and scrap value is estimated to have a service life of 10 years. The
original cost of the equipment was $40,000. Determine the following:
aaz. 42. Based on the above data, determine the depreciation charge for the fifth year if double-declining balance
depreciation is used.
aba. GIVEN:
abb.
Vo = $40,000
abc.
Vs = 0
abd. n = 10
ABE.
REQUIRED: da & % paid off
ABF.Solution:
abg.
da = 40,000 ( 1-2/10)5-1 2/10
abh.
da = $ 3,276.80
abi. 43. The depreciation charge for the fifth year if sum-of-the-years-digits depreciation is used.
abj. GIVEN:
abk.
Vo = $40,000
abl.
Vs = 0
abm. n = 10
ABN.
REQUIRED: da & % paid off
ABO.
SOLUTION:
abp.
SYDM



10  5  1 
10  10  1 
abq.
da = 2
( 40,000)
abr.
da = $ 4,363.64
ABS.
abt. 44. The percent of the original investment paid off in the first half of the service life using the double-declining balance
method.
abu. GIVEN:
abv.
Vo = $40,000
abw.
Vs = 0
abx. n = 10
ABY.
REQUIRED: da & % paid off
ABZ.
SOLUTION:
5
ACA.
acb.
acc.

2
 1 

V5
10 
= 40,000 
= $13,107.20
13,107 .2
40,000
% paid off =
= 32.768%
% = 32.8
acd.
ace. 45. The percent of the original investment paid off in the first half of the service life using the sum-of-the-years-digits
method.
acf. GIVEN:
acg.
Vo = $40,000
ach.
Vs = 0
aci. n = 10
ACJ.
REQUIRED: da & % paid off
ACK.
ACL.
SOLUTION:
acm.
SYDM
acn.
V1 = 40,000 – 4,363.64 = $ 35,636.36
aco.
V2 = 35,636.36 - 4,363.64 = $ 31,272.72
acp.
V3 = 31,272.72 - 4,363.64 = $ 26,909.08
acq.
V4 = 26,909.08 - 4,363.64 = $ 22,545.94
acr.
V5 = 22,545.94 - 4,363.64 = $ 18,181.8
acs.
act.
acu.
% paid off =
18,181 .80
40,000
= 45.45%
% = 45.45
acv.
acw. 46. The original cost of the property is $30,000, and it is depreciated by a 6 percent sinking-fund method. What is the
annual depreciation charge if the book value of the property after 10 years is the same as if it had depreciated at
$2,500/year by the straight-line method?
acx. GIVEN:
acy.
Vo = $30,000
acz.
i = 0.06
ada.
n = 10 years
ADB.
REQUIRED:
adc.
da = $2,500/yr
add.
da using SFM if Va SLM = Va SFM
ADE.
SOLUTION:
adf.
Using SLM:
adg.
Va = 30,000 – 10 (2,500)
adh.
Va = $ 5,000
adi.
adj. Using SFM:
adk.
30,000 –5,000 = R (1+ 0.06)10 - 1 / (0.06)
adl.
R = $ 1,896.70 / 10 yr
adm.
da = $ 189.67/yr
adn.
ado. 47. A concern has a total income of $1 million/year, and all expenses except depreciation amount to $600,000/year. At
the start of the first year of the concern’s operation, a composite account of all depreciable items show a value of
$850,000, and the overall service life is estimated to be 20 years. The total salvage value at the end of the service life is
estimated to be $50,000. Thirty percent of all profits before taxed must be paid out as income taxes. What would be the
reduction in income-tax charges for the first year of operation if the sum-of-the-years-digits methods were used for
depreciation accounting instead of the straight-line method?
adp. GIVEN:
adq.
Total income = $ 1 million/yr
adr.
Annual expenses exc dep’n = $600,000
ads.
For the 1st yr of operation:
adt.
y = $ 850,000
adu.
n = 20 yr
adv.
Vs = $ 50,000
adw.
Income tax rate = 0.45
ADX.
ADY.
REQUIRED:
adz.
Reduction in income taxes if SYDM is used instead SLM
AEA.
SOLUTION:
aeb.
Using SYDM:
aec.
da = 2 ( 850,000 –50,000 ) = $ 76,190.48
aed.
Gross Earnings = Total income – total expense
aee.
GE = $ 1,000,000 – ( $600,000 + 76,190.48 ) = $ 323,809.52
Net Profit = $ 323,809.52 (1 - 0.45) = $ 178,095.24
aef.
aeg.
aeh.
Income tax = 0.45 (323,809.52) = $ 145,714.28
Using SLM:
aei.
d=
aej.
d=
aek.
ael.
aem.
aen.
aeo.
aep.
aeq.
aer.
Vo  Vs
n
850 ,000  50,000
20
d=
$40 ,000
yr
Basis: 1 yr
GE = $ 1,000,000 – (600,000 + 40,000) = $ 360,000
Net Profit = $ 360,000 (1 - 0.45) = $ 198,000
Income Tax = 0.45 ($ 360,000) = $ 162,000
Reduction in net profit = $198,000 - $178,095.24 = $19,904.76
Reduction in income tax = $162,000 - $ 145.714.28 = $16,285.72
Reduction in income tax = $16,285.72
aes.
aet. 48. The total value of anew plant is $2 million. A certificate of necessity has been obtained permitting a write-off of 60
percent of the initial value ay 5 years. The balance of the plant requires a write-off period of 15 years. Using the straightline method and assuming negligible salvage and scrap value, determine the total depreciation cost during the first year.
aeu. GIVEN:
aev.
Vo = $2,000,000
aew.
Vs = 0
aex.
n = 15
aey.
a=5
aez.
Vs = negligible
AFA.
REQUIRED: d after 1 year
AFB.
SOLUTION:
afc.
Vo new = 2,000,000 – 2,000,000 ( 0.6) = $ 800,000
afd.
d=
Vo  Vs
n
80,000 
15
0
afe.
d=
aff.
d = $ 53, 333. 33/ yr
afg.
afh.
afi. 50. A profit-producing property has an initial value of $50,000, a service life of 10 years, and zero salvage and scraps
value. By how much would annual profits before taxes be increased if a 5 percent sinking-fund method were used to
determine depreciation costs instead of straight-line method?
afj. GIVEN:
afk.
Vo = $50,000,000
afl.
Vs = 0
afm. n = 10
AFN.
REQUIRED: Annual profits increase
afo. SOLUTION:
afp.
afq.
afr.
Using SFM:
 
 1
 

 1

5

 0.05    1
 0.05 
Va = 50,000 - (50,000-0)
Va = $28034.35

!0


1
Using SLM
Vo  Vs
n
afs.
d=
aft.
d = $ 25,000
afu.
inc = $28034.3 - $25,000
afv.
inc = $ 3.034.35
afw.
afx. 51. In order to make it worthwhile to purchase a new piece of equipment, the annual depreciation costs for the
equipment cannot exceed $3,000 at any time. The original cost of the equipment is $30,000, and it has a zero salvage
and scrap value. Determine the length of service life necessary if the equipment is depreciated by the sum=of-the-yearsdigits method by the straight-line method.
afy. GIVEN:
afz.
da = $ 3,000
Vo = $30,000
Vs = 0
aga. REQUIRED: n
AGB.
SOLUTION:
agc.
Using SYDM:
a=1
agd.
da = 2 ( n-1+1)/ n (n+1) x ( Vo –Vs )
age.
3000n2 + 3000 n = 60000n
agf.
n = 19 years
agg.
agh. 52. Referring to the previous number, determine the length of service life necessary if the equipment is depreciated by
the straight-line method.
agi. GIVEN:
agj.
da = $ 3,000
Vo = $30,000
Vs = 0
agk. REQUIRED: n
AGL.
AGM.
SOLUTION:
agn.
Using SLM:
a=1
Vo  Vs
n
ago.
d=
agp.
3000 = 30000n
n= 10 years
agq.
agr.
ags. A materials-testing machine was purchased for $20,000 and was to be used for 5 years with an expected residual
salvage value of $5,000. Graph the annual depreciation charges and year-end book values obtained by using:
agt.
agu. 53. By using Straight-line depreciation
agv. GIVEN:
agw.
Vo = $20,000
agx.
Vs = $5,000
agy.
AGZ.
AHA.
n=5
REQUIRED: da & Va
SOLUTION:
ahb.
SLM
ahc.
d=
Vo  Vs
n
20,000  5,000
5
ahd.
d=
= $ 3,000
ahe.
ahf.
Va1 = Vo – ad = $17,000
ahg.
Va2 = $14,000
ahh.
Va3 = $11,000
ahi.
Va4 = $ 8,000
ahj.
Va5 = $ 5,000
ahk.
ahl. 54. By using Sum-of-digits depreciation
ahm.GIVEN:
ahn.
Vo = $20,000
aho.
Vs = $5,000
ahp.
n=5
AHQ.
REQUIRED: da & Va
AHR.
SOLUTION:
n
nt
ahs.
aht.
d=
(Vo-Vs)
nt = 15
5
 
15 
=
(20,000-5,000) = $ 5.000
ahu.
d1
ahv.
d2 = $ 4,000
ahw.
d3 = $ 3,000
ahx.
d4 = $ 2,000
ahy.
d5 = $ 1,000
ahz.
aia.
Va1 = Vo – d = $ 15,000
aib.
Va2 = $ 11,000
aic.
Va3 = $ 8,000
aid.
Va4 = $ 6,000
aie.
Va5 = $ 5,000
aif.
aig. 55. By using Double-declining balance depreciation
aih. GIVEN:
aii.
Vo = $20,000
aij.
Vs = $5,000
aik.
n=5
AIL. REQUIRED: da & Va
AIM. SOLUTION:
ain.
DDBM
 Vs 
 5,000 
2 


20,000 
Vo 
f= 
=2
= 0.5
aio.
aip.
aiq.
Va 1


=
Vo  1  f



a


= 20,000 
1
0.5 
1
= $ 10,000
air.
Va2 = $ 5,000
ais.
Va3 = $ 2,500
ait.
Va4 = $ 1,250
aiu.
Va5 = $ 625
aiv.
aiw. 56. An asset with an original cost of $10,000 and no salvage value has a depreciation charge of $2381 during its second
year of service when depreciated by the sum-of-digits method. What is its expected useful life?
aix. GIVEN:
aiy.
Vo = 120,000
aiz.
Vs = 0
aja.
a=2
ajb.
da = $2381/yr
AJC.
REQUIRED: n
AJD.
SOLUTION:
n = 2(n-a+1)/(n (n+1) x ( Vo-Vs)
aje.
2381n2 + 2381n = 2n –2 (10,000)
ajf.
2381 n2 - 17619n + 20,000 = 0
ajg. By quadratic formula:
n = 5.9998
ajh.
n = 6 years
aji. 57. An electronic balance costs P90, 000 and has an estimated salvage value of P8, 000 at the end of its 10 years
lifetime. What would be the book value after 3 years, using straight-line method in solving for the depreciation?
ajj. GIVEN:
ajk.
Co = P90, 000
ajl.
CL = P8, 000
ajm.
L = 10
ajn.
n=3
ajo. REQUIRED: d using SLM
AJP. SOLUTION:
CO  C
ajq.
d=
ajr.
d=
90,000  8,000
10
ajs.
D3
ajt.
C3 CO D3
= P 8,200
= n(d) = 3(8,200) = P24,600
=
-
C3
aju.
L
L
= 90,000 - 24,000
= P 65,200
ajv.
ajw. A broadcasting corporation purchased equipment for P53, 000 and paid P1, 5000 for freight and delivery charges top the
job sites. The equipment has a normal life of 10 year with a trade-in value of P5, 000 against the purchase of anew
equipment at the end of the life.
ajx. 58. referring to the problem above, determine the annual depreciation by straight-line method.
ajy. GIVEN:
ajz.
Co = P53, 000 + 1,500 = 54,500
aka. CL = P5, 000
L = 10
akb. REQUIRED: da using SLM
AKC.
SOLUTION:
CO  C
akd.
d=
L
L
54,500  5,000
10
ake.
d=
akf.
d = P4, 950
akg. 59. From the preceding number, determine annual depreciation by sinking fund method. Assuming interest 6 ½%
compounded annually.
akh. GIVEN:
aki.
Co = P53,000 + 1,500 = 54,500
akj.
CL = P5, 000
L = 10
akk. REQUIRED: da using SFM
AKL.
SOLUTION:
CO 
CL
F
, 6.5% , 10
A
54,500  5,000
F
A,6.5%,10
P49 ,500
13.3846
akm.
d=
=
=
akn.
d= P 3,668
ako. A firm brought equipment for P56, 000. Other expenses including installation amounted to P4, 000. The equipment is
expected to have a life of 16 years with a salvage value of 10% of the original cost.
akp. 60. Determine the book value at the end of 12 years by SLM:
akq. GIVEN:
akr.
Co = P56,000 + 4,000 = P60,000
aks.
CL = 0.1 Co
akt.
L = 10
aku.
i = 0.12
akv. REQUIRED: C12 using SLM
AKW.
SOLUTION:
CO  C
akx.
d=
aky.
akz.
L
L
d12 = P3, 375 (12) = P 40,500
C12 = Co - d12 = 60,000 – 40,500
ALA.
C12=
alb.
alc. 61. Determine the book value at the end of 12 years by SFM:
ald. GIVEN:
ale.
Co = P56, 000 + 4,000 = P60, 000 CL = 0.1 Co
alf.
L = 10
alg.
i = 0.12
alh. REQUIRED: C12 using SFM
ALI. SOLUTION:
P19, 500
CO  CL
alj.
60,000  6,000
F
F
P54 ,000
, 12% , 16
A,6.5%,10
A
42.7533
d=
=
=
alk.
= P 1,263
 F

D 12  A, 12%, 16  
 1263 (24.1331)
=d
d12 = P30, 480
alm. C12 = Co - d12 = 60,000 – 30,480
aln. C12= P29, 520
all.
alo.
alp. For numbers 62 – 64. A certain type of machine losses 10% of its value each year. The machine cost P2, 000 originally.
Make cut a schedule showing the following:
alq. 62. By yearly depreciation
ALR.
als. GIVEN:
alt.
Co = P2, 000
alu.
dep’n = 10%
alv. REQUIRED: C5
alw. SOLUTION:
alx.
Year
Book value at
Dep’n (10%)
Total dep’n
Book value
aly.
the beginning
at the end
alz.
ama.
1
P 2,000
P 200
P 200
P 1,800
amb.
2
1,800
180
380
1,620
amc.
3
1,620
162
542
1,458
amd.
4
1,458
145.8
687.8
1,312
ame.
5
1,312.20
131.22
819.12
1,180.98
amf.
amg.63. The total depreciation after 5 years.
amh.GIVEN:
ami.
Co = P2, 000
dep’n=10%REQUIRED: C5
amj. ANSWER: P819.12
amk.
aml.
amm.
64. Estimate the book value at the end year for 5 years.
amn.
Answer from the above table: P1, 180.98
amo.65. Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8
years for an asset that costs P15,000 new and has an estimated scrap value of P2,000 at the end of 10 years by DBM.
amp.GIVEN:
amq.
Co = P15, 000
amr. CL = P2, 000
ams.REQUIRED: d8 & V8
AMT.
SOLUTION:
L
 CL 
C 
 O =
10
 2,000 


15,000  = 0.1825 or 18.25%
amu.
k=1amv.
C9 = Co (1 – k) 8 = P 2,992
amw.
D8 = Co - C8 = 15,000 – 2,992
amx.
D8 = P 12,008
amy. 66. Determine the rate of depreciation, the total depreciation up to the end of 8 th year and book value at the end of 8
years for an asset that costs P15, 000 new and has an estimated scrap value of P2, 000 at the end of 10 years by
DDBM.
amz.GIVEN:
ana.
Co = P15, 000
anb. CL = P2, 000
anc. REQUIRED: d8 & V8
AND.
SOLUTION:
2 2
d R L 10
ane.
=
=
8
= 0.2 or 20 %
8

2

2
 1 
 1

L
10
 = 15,000 
 = P2, 517
C8 = co 
anf.
ang.
D9 = Co - C8 = 15,000 – 2,517
anh.
D9 = P 12,483
ani.
anj. 67. Mr. Dim bought a calciner for P220, 000 and used it for 10 years, the life span of the equipment. What is the book
value of the calciner after 5 years of use? Assume a scrap value of P20, 000 for SLM;
ank. GIVEN:
anl.
Co = P220, 000
anm. L = 10
ann.
n=5
ANO.
REQUIRED: D5 & C5
ANP.
SOLUTION:
anq. SLM
D5
anr.
n Co  C L

=
L

=
5 220 ,000  20,000 
10
= P100, 000
ans.
C5 = Co – D5 = 220,000 – 100,000 = P 120,000
ant.
C5 = P 120,000
anu.
anv. 68. Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of
P22, 000 for textbook for DBM.
anw. GIVEN:
anx.
any.
Co = P220, 000
anz.
L = 10
aoa.
n=5
AOB.
REQUIRED: D5 & C5
AOC.
SOLUTION:
aod.

 CL  
 Co   
 Co 
Cs=
n
L
=
(220 ,000 ( 22,000 / 220 ,000 )^(5/ 10)= P 69,570
aoe.
Cs = P 69, 570.00
aof.
aog. 69. Refer to the previous problem. What is the book value of the calciner after 5 years of use? Assume a scrap value of
P20, 000 for textbook for DDBM.
aoh. GIVEN:
aoi. Co = P220, 000
aoj.
L = 10
aok.
n=5
AOL.
AOM.
REQUIRED: D5 &
SOLUTION:
C5
5
aon.
aoo.


2 
2 
 1   P220 ,000  1 

C5
L 
10 

= Co 
=
5
Cs = P 72, 090
aop.
aoq. 70. A structure costs P12, 000 new. It is estimated to have a life of 5 years with a salvage value at the end of life of P1,
000. Determine the book value at the end of three years.
aor. GIVEN:
aos. Co= P12, 000
CL =P1, 000
L=5
aot. REQUIRED: Va
AOU.
SOLUTION:
aov.
Co - Cl = 12,000 - 1,000 = P 11,000
aow.
aox. Year Year in reverse order
Dep’n during the year
Book value
aoy.
aoz. 1
5
5/15 (11,000) = 3,664
P 8,333
apa. 2
4
4/15 (11,000) = 2,933
5,400
apb. 3
3
15 (11,000) = 2,200
3,200
apc. 4
2
2/15 (11,000) = 1,467
1,733
apd. 5
1
1/15 (11,000) = 733
1,000
ape.
apf. 71. Operator A produces 120 spindle/hr on a lathe. His hourly rate is $1.80. Operator B, using an identical lathe, is able
to produce 150 identically units/hr. The overhead charge for a lathe is fixed at $2.50/hr. Determine operator B’s hourly
rate so that his cost per piece is identical to A’s.
apg. SOLUTION:
aph. Cost per unit for operator A = 1.80 + 2.50 = $0.0358
api.
120
apj. Let x = hourly rate of operator B. Then;
apk.
x + 2.50 = $0.0358
apl.
150
apm.
x = $2.88
apn.
apo. 72. In a type 1 warehouse, initial cost will be $24, 000.This warehouse has adequate capacity for the near future, but 12
years from now an addition will be required that costs $15, 000. A type 2 warehouse costs $34, 000. This type has the
same capacity as the type 1 warehouse with its addition. What will be the present cost of the type 1 warehouse? Which
of these should be built, assuming that depreciation is negligible and that the interest rate is 7%?
app.
apq. GIVEN:
apr.
Present cost = $15, 000
aps.
Present worth factor, i = 7%
apt.
n = 12
apu. REQUIRED: present cost of warehouse 1
apv. SOLUTION:
apw.
P = S/ (1+i) n
apx.
= 15, 000 x 0.444 = $6660
apy.
apz. the present cost of type 1 is seen to be 24, 000 + 6660 = $30, 660. Since this is smaller than the type 2 building, it is
more economical to build
aqa.
aqb. 73. Determine the equal (year-end) payments that will be available for the next four years if we invest $4, 000 at 6%.
aqc. SOLUTION:
aqd.
R = $4, 000 0.06 (1 + 0.06) 4
aqe.
= $4, 000 (0.28859)
aqf.
(1 + 0.06) 4 - 1
aqg.
aqh.
R = $1154.36
aqi.
aqj. 74. A new snow removal machine costs $50, 000. The new machine will operate at a reputed savings of $400 per day
over the present equipment in terms of time and efficiency. If interest is at 5% and the machine’s life is assumed to be 10
years with zero salvage, how many days per year must the machine be used to make the investment economical?
aqk. SOLUTION:
aql.
Assume straight-line depreciation and no salvage value.
aqm.
Annual depreciation = 50, 000 – 0 = $5, 000
aqn.
aqo.
aqp.
aqq.
aqr.
10
Average annual unearned interest = ½ [50, 000 x 0.05 + 50, 000 (0.05)] = $1375
10
Annual cost = $6375
aqs. To invest in the machine, the yearly savings must at least be equal to $6375. The number of days m the machine must
be used is therefore
aqt.
aqu.
m = 6375 / 400 = 15.9 or 16 days
aqv.
aqw. 75. Based on the sinking fund method and using the data in the previous problem, what number of days must the
machine be used if the amount to be accumulated in 10 years is $50, 000?
aqx. SOLUTION:
aqy.
R = S i / [(1+i)n – 1 ]
aqz. For i = 5% and n = 10
ara.
arb.
i/[(1+i)n – 1 ] = 0.0795
arc.
ard. therefore
R = 50, 000 x 0.0795
are.
R = $3975
arf.
arg.
m = 3975/400 = 9.9 or 10 days
arh. 76. A consulting engineer decides to set up an educational fund for his son that will provide $3000 per year for 6 years
starting in 16 years. The best interest rate he can expect to get is 5% compounded quarterly. He wants to accumulate
the necessary capital by making quarterly deposits until his son starts college. What will be his needed quarterly deposit?
ari. SOLUTION:
arj.
i = (1+ 0.05 )4 - 1
ark.
4
arl.
i = 0.051
arm.
arn.
aro. Six equal payments of $3, 000 are required; their worth at the beginning of the 16th year is
arp.
arq.
P = R [ ( 1+ i )n – 1] / i (1 + i ) n
arr.
= 3000 [ (1+0.051)6 – 1] / 0.051 ( 1+0.051)6
ars.
P = $15, 210
art.
aru. 77. The engineer has 15 years to accumulate this fund. His quarterly deposits are determined by u sing the sinking
factor. There are 15 x 4 or 60 interest payments to be made at a quarterly interest rate of 0.0125
arv. SOLUTION:
arw.
R = Si/[(1+i)n – 1 ]
arx. Using n = 60
ary.
R = $15, 210 ( 1/88.5745)
arz.
R = $172.00 where 88.5745 represents the amount of annuity.
asa.
asb. 78. A low carbon steel machine part, costing $350 installed, lasts 6 years when operating in a corrosive atmosphere. An
identically shaped part, but treated for corrosion resistance, would cost $650 installed. How long would the corrosion
resistance part have to last to be at least as good investment as the untreated part? Assume money is worth 7%.
asc. SOLUTION:
asd.
R=P
[ i(1+i)n ]
ase.
[ (1+i)n - 1]
asf.
= 350 x 0.2098
asg.
R = $73.50 per year required
ash.
asi. $73.50 = $650 x c.r.f.
asj. c.r.f. = 73.50/ 650.00 = 0.113
ask.
asl.
for
n = 14
crf = 0.114
asm.
n=x
crf = 0.113
asn.
n = 15
crf = 0.109
crf = 0.109
aso.
asp. Difference = 15 -14 = 1.00
0.005
0.004
asq.
asr.
0.004/0.005 = (15 – x)/ 1.00
ass.
ast.
x = 14.20 years
asu.
asv. 79. The total cost of a cast product consists of (1) the raw material cost that is directly proportional to the weight, of the
casting, (2) the machining cost that varies inversely as the weight, and (3) overhead cost that remains constant per unit
produced regardless of weight. Find the weight giving the minimum cost per casting.
asw.
asx. SOLUTION:
asy.
Let Ct = total cost
asz.
Cw = cost based on weight
ata.
Cm = machining cost
atb.
Co = overhead unit cost
atc.
W = weight
atd.
Cw = k1W (direct proportion of raw material cost to weight)
ate.
Cm = k2/W (inverse proportion of machining cost to weight)
atf.
Co = Co
(constant value)
atg.
ath. The minimum total cost can be determined by differentiating cost with respect to weight.
ati.
dCt/dW = k1 – k2/W
atj.
atk. the minimum cost can be ascertained by equating the right hand side of the above eqn to 0.
atl.
atm.
W = ($ x lb)1/2 = lb
atn.
( $/lb)1/2
ato.
atp. Therefore, minimum cost occurs when:
atq.
W = (k2/k1)1/2 lb
atr.
ats.
att. 80. Based on the previous problem, what is the minimum total cost?
atu. SOLUTION:
atv.
Ct = [k1(k2 / k1)1/2 = k2 (k1 / k2) ½ + Co]
atw.
atx. 81. Methyl alcohol condensed at 148 F is to be cooled to 100 F for storage at a rate of 10, 000gal/hr by water available
at 75 F in a countercurrent heat exchanger. The over-all heat transfer coefficient is constant and estimated at 200 Btu/ft 2hr-F. Heat exchanger annual costs including operation are estimated at $2 per ft 2 including depreciation. The cooler is to
operate 5, 000 hr/year, and the value of heat utilized is estimated at $5x10-7 per Btu. What is the estimated optimum cost
of the heat exchanger if the cost for surface is $9 per ft2?
aty.
atz. SOLUTION:
aua.
Since the outlet temperature is fixed;
aub.
∆t is fixed at 100 – 75 = 25 F
auc.
aud. Hourly cost:
2
= $4x10-4 per (ft2-hr)
aue.
5, 000
auf.
aug.
∆t1 = (M/R) (T1 – t1) = 4/5 x 103 (148 – 75) = 11.7 F
auh.
U x 25
200 x 25
aui.
auj. Outlet temp. for water :
148 – 11.7 = 136 F
auk. cp = 0.5
aul. ρ methanol = 0.79g/cc
aum.
q = 10, 000 x 8.33 x 0.79 (0.5)(148 -100) = 1, 580, 000 Btu/ hr
aun.
auo.
∆tm = 25 – 11.7 = 17.7
aup.
ln (25/11.7)
auq.
aur.
q = UA∆t
aus.
A=
1, 580, 000
aut.
200 x 17.7
auu.
A = 446 ft2
auv. Estimated optimum cost:
446 x 9 = $4, 014
auw.
aux.
auy. 82. What is the most economical number of effects to use in the recovery of black liquor in a paper plant if the following
cost data are available? The annual fixed costs increase essentially linearly with each effect (except for condensing,
feeding, and other equipment costs for multiple units which may be considered to balance each other). If a fixed amount
of evaporation is to be obtained and each units to have 1, 000 ft 2 of heating surface with a service life of five years, the
annual fixed costs Cf would be (using cost data of $25, 000 for a single evaporator of 5, 000 ft 2, employing the 0.6 factor,
and neglecting the interest).
auz.
ava.
Cf =( 1, 000)0.6 25, 000 N
…….
dollars per year
avb.
5, 000
5
avc.
avd. where N is the number of effects.
ave.
Because of the steam economy in multiple-effect operation, the direct costs for steam will decrease and the total of
all annual direct costs, CD, has been established for this type of operation as
avf.
CD = 65, 000 N -0.95 dollars
avg.
avh. SOLUTION:
avi.
Ct = 1, 900 N = 65, 000 N-0.95
avj. Differentiating: dCt = 1, 900 – 61, 800 N -0.95 = 0
avk.
dN
avl.
N = 5.95 or 6
avm.
avn. 83. A capitalized cost for a piece of equipment has been found to be $55, 000. This cost is based on the original cost
plus the present value of an indefinite number of renewals. An annual interest rate of 12% was used in determining the
capitalized cost. The salvage value of the equipment at the end of the service life was estimated to be 10 years. Under
these conditions, what would be the original cost of the equipment?
avo.
avp. GIVEN:
k = $55, 000
avq.
i = 12%
avr.
Vs= 0
avs.
N = 10 yrs
avt. REQUIRED : Cv
avu. SOLUTION:
avv.
k = CR ( 1 - i)n + Vs
avw.
(1 + i )n – 1
avx.
55000 = CR ( 1 – 0.12)10
avy.
(1 + 0.12 )10 – 1
avz.
CR = $37, 291.47
awa.
awb.
k = Cv +
CR
awc.
(1 + 0.12 )10 – 1
awd.
Cv = $37, 291.47
awe.
awf. 84. A heat treating furnace is used to preheat small steel parts. The furnace uses fuel oil consisting $0.04 per gallon, with
a heating value of 142, 000Btu/gal. The furnace has a firebrick lining, the outside temperature of which is 1210 F, this is
to be covered with insulation costing $300 per 1000 board feet. The air temperature is 110 F. Operations is 7200 hr/yr.
Conductivity is 0.028 for insulation in Btu/hr-ft 2-F. Calculate the most economical thickness of insulation. Furnace life is 8
years. Assume negligible temperature drop from insulation to air.
awg. GIVEN:
awh.
Heat treating furnace
awi.
T1 = 1210 F
awj.
T2 = 110 F
awk.
Cost of fuel oil $ = 0.04 / gal
awl.
Heating value = 142000 Btu/gal
awm.
Cost of insulation = $300/1000bdft
awn.
Operation = 7200 hrs/yr
awo.
k = 0.028 Btu
awp. REQUIRED: Most economical thickness of insulation
awq.SOLUTION:
awr.
Basis: A = 1 ft1
aws.
q = - kA∆T
awt.
= - (0.028)(1)(110-1210)
awu.
x
awv.
q = 30.8
aww.
x
awx.
cost of fuel = 30.8 x 1gal/142000 Btu ($0.04/gal)(7200hr/yr
awy.
x
awz.
axa.
cost of fuel = 0.06246705
axb.
x
axc.
axd.
cost of insulation = 0.06246705 + 0.3 x
axe.
x
axf.
dTc = -0.06246705 + 0.3 = 0
axg.
x2
axh.
axi.
x = 5.48 in
axj. 85. A batch inorganic chemical operations gives product C from two chemicals A and B according to the following
empirical relation:
axk.
axl.
C = 2.8 (AB – AC – 1.2 BC + 0.5C2 )0.5
axm.
axn. where A, B and C are pounds of respective components. The reaction rate is sufficiently high to be neglected, and the
time to make any batch is essentially the charging and discharging time, including heating up, which totals 1 hr. If A costs
$0.10 per lb and B costs $0.05 per lb, what is the ratio of B to A to give the minimum costs of raw materials per lb of
product . what is the cost per lb of C?
axo.
axp.
axq. GIVEN:
axr.
Reaction:
axs.
A+B
C
axt.
1 hr – time per batch
axu.
A = $0.10/lb
axv.
B = $ 0.05/ lb
axw. REQUIRED: minimum cost of raw materials / lb of product
axx. SOLUTION:
axy.
Ratio of B: A
= 0.5
axz.
Assume: 1 LB OF a AND 1LB OF b are used in the making of C
aya.
ayb.
C = 2.8 ( 1 – C – 1.2C + 0.5C2 ) 0.5
ayc.
dC = 1.4 ( 1 – C – 1.2C + 0.5 C2) 0.5
ayd.
aye.
equating to 0
ayf.
ayg.
0 = 1.4 (-2.2 +C)
ayh.
0 = -3.08 +C
ayi.
C = $3.08
ayj.
ayk. 86. Based from the previous problem, what is the cost per lb of C?
ayl. GIVEN:
aym.
Reaction:
ayn.
A+B
C
ayo.
1 hr – time per batch
ayp.
A = $0.10/lb
ayq.
B = $ 0.05/ lb
ayr. REQUIRED: cost per lb of C
ays. SOLUTION:
ayt.
By ratio and proportion:
ayu.
If A = $0.1 /lb and B = $0.05/lb
ayv.
C = $3.08/ lb
ayw.
ayx. 87. Seven million pounds of water per year is to be obtained from 8 percent solids slurry to be filtered on a leaf filter to
produce a cake containing 40 percent solids. The area of the filter is 200 ft 2. Tests show a value of 2 x 104 for k in pound
units. The cake is not washed. The dumping and cleaning time is 3 hr and costs $39 each cycle. Filtration costs are $14
per hr, and inventory charges maybe neglected. What is the cycle time for minimum costs?
ayy. GIVEN:
ayz.
7000000 lbwater/yr – can obtained from 8% slurry
aza.
A filter = 200 ft2
azb.
Dumping and cleaning time = 3 hr
azc.
Filtration costs = $14/hr
azd. REQUIRED: cycle time for minimum cost
aze. SOLUTION:
azf.
Q = A ( k θf) 0.5
azg.
= 200 ( 2x104)(θf)0.5
azh.
azi.
7, 000, 000 lb/ yr is also equal to 799.08 lb/hr
azj.
799.08 = 200 ( 2x104)(θf)0.5
azk.
θf = 0.014 hrs
azl. 88. Referring to the previous problem, calculate the cycle time for maximum production.
azm.GIVEN:
azn.
7000000 lbwater/yr – can obtained from 8% slurry
azo.
A filter = 200 ft2
azp.
Dumping and cleaning time = 3 hr
azq.
Filtration costs = $14/hr
azr.
θf = 0.014 hrs
azs. REQUIRED: cycle time for maximum production
azt.
azu. SOLUTION:
azv.
θtotal = θf + θw + θd
azw.
θw = 0 ;
since the cake is not washed
azx.
θf = 0.014 hrs
azy.
θd = 3 hrs
azz.
baa.
θtotal = ( 0+0.014+3)
bab.
θtotal = 3.04 hrs.
bac.
bad.
bae. For nos. 89- 91.
baf. In processing 500 ton/day of ore assaying 50% mineral, 300 tons of concentrate containing 66.7 % are obtained at a
cost of sales ( all fixed operating cost are excluded) of $15 per ton concentrate. An investment of $200, 000 of
concentrate that will assay 71% mineral. If the plant operates 200 days/year, equipment must pay out in 5 years with
interest at 15% and no salvage value and no additional labor or repair costs need to be considered.
bag.
bah. 89. Based on the stated problem above, calculate the additional cost per ton of concentrate for capital recovery on the
new equipment.
bai. GIVEN:
baj.
500 tons/day of ore with assay of 50 % mineral
bak.
300 tons/day of ore with assay of 66.7% mineral
bal.
fixed cost = $15/ton
bam.
Operation = 200 days/year
ban.
Pay out period = 5 years
bao.
Interests = 15%
bap.
No salvage value
baq. REQUIRED: additional cost per ton concentrate
bar. SOLUTION:
bas.
Let x additional cost of equipment
bat.
FCI = $15 + x
bau.
Annual share with 15% interest
bav.
= 0.15 (15+x)
baw.
= 2.25 + 0.15x
bax.
with 5 year payment period
bay.
15 + 0.15x
baz.
5
bba.
Cost of equipment invested:
bbb.
= $200, 000 x / 300 tons/day (200 days/yr)
bbc.
=3.33 x
bbd.
tons of concentrate per year
bbe.
= [500 (0.5) + 300 (0.667) + 300 (0.71) ]tons/day ( 200d/yr)
bbf.
= 19893 / x +1
bbg.
Cost total/yr = FCI + Cost concentrate + Annual Charges + Investments
bbh.
bbi.
Ct = 15 +3.825x +(19893/x+1) 9+ 2.25 +0.15x +3.33x
bbj.
dCt/dx = 3.48x - 19893/ (x+1)2 = 0
bbk.
3.48 ( x2 +2x +1) – 19893 = 0
bbl.
by quadratic equation:
bbm.
x = $74.6
bbn.
bbo. 90. Determine the selling price in dollars per ton (100% mineral basis) required for which the cost of the n new
equipment is justified.
bbp. GIVEN:
bbq.
500 tons/day of ore with assay of 50 % mineral
bbr.
300 tons/day of ore with assay of 66.7% mineral
bbs.
Fixed cost = $15/ton
bbt.
Operation = 200 days/year
bbu.
Pay out period = 5 years
bbv.
Interests = 15%
bbw.
No salvage value
bbx. REQUIRED: selling price
bby. SOLUTION:
bbz.
Since x = $74.6
bca.
19893 / (74.6 + 1) = $ 263.13
bcb.
bcc. 91. What is the % increase in recovery and rejection for the new process based on mineral and gauged?
bcd. GIVEN:
bce.
500 tons/day of ore with assay of 50 % mineral
bcf.
300 tons/day of ore with assay of 66.7% mineral
bcg.
Fixed cost = $15/ton
bch.
Operation = 200 days/year
bci.
Pay out period = 5 years
bcj.
Interests = 15%
bck.
No salvage value
bcl.
Selling price = $ 263.13
bcm.REQUIRED: efficiency
bcn. SOLUTION:
bco.
E = 500(0.5) + 300(0.6667) – 300(0.71)
bcp.
500(0.5) + 300(0.6667)
bcq.
E = 52.677%
bcr.
bcs. 92. A ties on a plant railroad sliding are to be replaced. Untreated ties consisting $ 2.50 installed have a life of 7 years. If
created ties have a life of 10 years, what is the maximum installed cost that should be paid for treated ties if money is
worth 8 percent?
bct. GIVEN:
bcu.
Untreated ties = $2.50
bcv.
n=7
bcw.
Created ties = X
bcx.
n = 10
bcy. REQUIRED: maximum installed cost
bcz.
bda. SOLUTION:
bdb.
Getting the annual depreciation cost
bdc.
bdd.
0.08 =
(2.50/7) – ( x /10)
bde.
X – 2.50
bdf.
bdg.
X = $3.10
bdh.
bdi. 93. Powdered coal having a heating value of 13, 500 Btu/ lb is to be compared with fuel oil worth $2.00 per bbl (42 gal)
having a heating value of 130,000 Btu/gal as a source of fuel in the processing plant. If the efficiency of the conversion of
the fuel is 64% for coal and 72 % for oil, with all other costs being equal, what is the maximum allowable selling price for
coal per ton?
bdj.
bdk. SOLUTION:
bdl.
Powdered coal
bdm.
Let x be the selling price per ton
bdn.
bdo.
(X / ton)(ton/ 2000lb) (lb / 13500Btu) (0.64)
bdp.
= 2.37 x 10 -8- /Btu
bdq. Fuel Oil
bdr.
(2.00/bbl) (bbl/42gal) (gal/130000Btu)(0.72)
bds.
= 2.67x 10 -7 / Btu
bdt. Equating both prices:
bdu.
X = $11.13 / ton
bdv.
bdw. 94. A steam boiler is purchased on the basis of guaranteed performance. However, initial tests indicate that the opening
(income) cost will be P400 more per year than guaranteed. If the expected life is 25 years and money is worth 10 %,
what deduction from the purchase price would compensate the buyer for the additional operating cost?
bdx. SOLUTION
bdy.
bdz.
A = 400
bea.
N = 25
i = 0.10
beb.
bec.
P = 400 ( 1 – ( 1+0.10)-25)
bed.
0.10
bee.
P = 3, 630.82 pesos
bef.
beg. 95. If the sum of P12, 000 is deposited in an account earning interest at the rate of 9% compounded quarterly, what will it
become at the end of 8 years?
beh.
bei.
bej. SOLUTION:
bek.
bel.
P = 12, 000
bem.
i = 9% / 4 = 2.25%
ben.
n = 8 (4) = 32
beo.
bep.
F = P(1+i)n
beq.
= 12, 000 (1 + 0.0225)32
ber.
F = 24, 457.24
bes. 96. At a certain interest rate compounded quarterly, P1, 000 will amount to P4, 500 in 15 years. What is the amount at
the end of 10 years?
bet. SOLUTION:
beu.
bev.
For 15 years, P = 1000
bew.
n = 4 (15);
6o periods
bex.
F = 4, 500
bey.
bez.
F= P (1+i)n
bfa.
4, 500 = 1, 000 (1+ i)60
bfb.
i = 0.02538
bfc.
bfd.
For 10 years
bfe.
n = 4 (10);
40 periods
bff.
F= 1, 000 (1 + 0.02538)40
bfg.
F = 2, 725.17 pesos
bfh.
bfi. 97. A one bagger concrete mixer can be purchased with a down payment of P8, 000 and equal installments of P600
each paid at the end of every month for the next 12 months. If the money is worth 12% compounded monthly, determine
the equivalent cash prize of the mixer.
bfj. SOLUTION
bfk.
P = 8, 000 = 600 ( 1 – (1+ 0.01)-12)\
bfl.
0.01
bfm.
P = 14, 753.05 pesos
bfn.
bfo. 98. A certain company makes it the policy that for any new piece of equipment, the annual depreciation cost should not
exceed 10% of the original cost at any time with no salvage or scrap value. Determine the length of service life
necessary if the depreciation method use is straight line formula.
bfp. SOLUTION
bfq.
bfr.
Vs = 0
bfs.
d = 0.10 V
bft.
d = V –Vs
bfu.
n
bfv.
0.10V = V – 0
bfw.
n
bfx.
n = 10 years
bfy.
bfz.
bga.
bgb. 99. Solve the previous problem with the sinking fund formula at 8%
bgc. SOLUTION
bgd.
0.10V = (V-0) ( 0.08)
bge.
(1 + 0.08)n – 1
bgf.
bgg.
n = ln 1.08
bgh.
= 7.64
bgi.
n = 8 years
bgj.
bgk. 100. Determine the ordinary simple interest on $10, 000 for 9 months and 10 days if the rate of interest is 12%.
bgl. SOLUTION
bgm.
9 months and 10 days = 9 (30) + 10 = 280 days
bgn.
P = $10, 000
bgo.
I = Pi (d/360)
bgp.
I = 10, 000 (0.12) (280/360)
bgq.
I = $933.33
bgr.