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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 -3 /1.0   - 1.342 k'y/(1-yA)iM 1.465 x 10 -3 /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 -3 /1.0   - 1.342 k'y/(1-yA)iM 1.465 x 10 -3 /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

y=

slope = m = -

k'xa/(1-x) k'ya/(1-y)

X=

Y 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

y AG -y Ai x A *-x Ai

m' =

y A1 -y A * x A1 -x AL

K'y =

k'y k'x k'x + m' k'y

K'y =

(0.0739)(0.169) 0.0450 0.169 + 0.8287(0.0739)

Ky =

K'y KyaS

m" =

K'x =

m" k'y k'x k'x + m''k'y

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:

kg H2O  1. 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 of inlet air.

2

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  3

  = 

2.83 x 10

  =

2.83 x 10

H1   4.56 x 10

3

3

 4.56 x 10

3



H  T K   

0.03

3

VH

= 1.0769

m kg da

3

Basis: 1

m



3 

kg da



3 

W = 1 m  1.0679 m  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

W=

3   m  200  1kg dry air hr   0.8879m3  

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

 kg kJ      225.25 1.0464 qH  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  

  

+ 2.501

     +

1.88 0.0097 x10 15  0   

6

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

0.8607 m vH kg dry air = Step 3: Solve for W and then L, the condensate.

 1   kg dry air  2  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 da 0.0151  0.0106kg H2O     hr kg da   L =

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 Hp kg dry air = 0.0228

Hp

x 100

= 57.78 %

Step 3: Solve for HR.

PA HR P As = x 100

3.59 KPa HR 6.59 KPa =

HR

x 100

= 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

0.021 =

PA 18 101.3  P A 29 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

x 100

kg H2O kg dry air kg H2O 0.031 kg dry air x 100 = 0.021

Hp

=

H Hs

Hp

= 67.74 %

PA HR P As = x 100

3.32 KPa HR 4.82 KPa =

HR

x 100

= 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)

Btu lb m dry air  F

CS

= 0.245

  =

2.83 x 10

( English )

Step 3: Solve for VH.

vH  

vH = 

2.83 x 10

3

3

 4.56 x 10

 4.56 x 10

3

  

3

0.011

  



H  T K   

( 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

ft lbm 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

Btu lb m 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  T K 

0.030  ( 57.2 + 273 ) 

3

m vH = 0.9796 kg dry air vH

 

=

( SI )

0.0204  0.0405 H  T  F 

vH = ( 0.0204 + 0.0405(0.030) ) ( 134.96 + 460 )

3

vH

= 15.72

ft lb m 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

    V  2.83 x10   4.56 x10  0.0141 60  273  VH  2.83 x10  3  4.56 x10  3 H T 3

H

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 tb  2.4103hr  44.5mm t2  1 2  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   1  Uk  T  T   hR  T  T 

s R s  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 

hc / kyHB

Ts  31.8355 o C 33 o C  31.8355 o C TRIAL 2 : Ts  32.3  305 .3K

4

 377 .4   305 .3       100   100  hR  0.94(5.676 )  377 .4  305 .3 hR  8.5832 W / m2K

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   305 .6      100   100   hR 0.94(5.676 ) 377.4  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   305.7      100   100   hR  0.94(5.676 ) 377.4  305 .7 hR  8.5971

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

x (0.20 ) t 1 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

4x 8X t  2 1 ln 2 1  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

H

w=0.0786lbH2O/lbda

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  t FRP  ln 700 0.3 30.15  0.086  0.0562 13281 / 700  0.0374 0.0374  0.0786  0.0593  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.

25. In _____cyclic operations, product is delivered continuously while the unit is in operation, but the rate of delivery decreases with time. dy. a. batch dz. b. continuous ea. c. semi-batch eb. d. semi-continuous ec.

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 ed. a. batch ee. b. continuous ef. c. semi-batch eg. d. semi-continuous eh.

27. It is a mathematical technique, for determining optimum conditions for allocation of resources and operating capabilities to attain a definite objective. ei. a. economics ej. b. programming ek. c. linear programming el. d. dynamic programming em.

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. en. a. economics eo. b. programming ep. c. linear programming eq. d. dynamic programming er.

29. When quality constraints or restrictions on certain variables exist in an optimization situation, a powerful analytical technique is the use of : es. a. Lagrange Multipliers et. b. Geometric Programming eu. c. steepest Ascent or Descent ev. d. dynamic programming ew.

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: ex. a. Lagrange Multipliers ey. b. Geometric Programming ez. c. linear programming fa. d. steepest Ascent or Descent fb.

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 fc. a. Lagrange Multipliers fd. b. Geometric Programming fe. c. linear programming ff. d. steepest Ascent or Descent fg.

32. ____(CPM) and program evaluation and review technique) have received particular attention and have shown the desirability of applying mathematical and graphical analysis fh. a. critical path method fi. b. critical planning method fj. c. careful planning and management fk. d. none of the above fl.

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. fm. a. profitability fn. b. income fo. c. net income fp. d. sales fq.

34. It is the common denominator for all business activities before capital is invested to a project or enterprise fr. a. profitability fs. b. income ft. c. net income fu. d. sales fv.

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: fw. a. rate of return fx. b. interest income fy. c. profitability fz. d. annuity ga.

36. ___is defined as the difference between income and expense. gb. a. profit gc. b. gross income gd. c. operating expenses ge. d. sales gf.

37. This is a function of the quantity of goods and services produced and the selling price. a. sales gg. b. gross income gh. c. operating expenses gi. d. profit

gj.

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. gk. a. marketing analysis gl. b. gross income gm. c. discounted cash flow gn. d. law of demand go.

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. gp. a. marketing analysis gq. b. gross income gr. c. cash flow gs. d. selling price gt.

40. The ____concept is useful for comparing alternatives which exist as possible investment choices within a single overall project. gu. a. marketing analysis gv. b. sales inventory gw. c. discounted cash flow gx. d. capitalized cost profitability gy.

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. gz. a. deadline ha. b. due date hb. c. payout period hc. d. maturity date hd.

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. he. a. reserves hf. b. replacement hg. c. substitute hh. d. none of the above hi.

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 hj. a. sinking fund factor hk. b. single payment present worth factor hl. c. uniform series hm. d. capital recovery factor hn.

44. Which of the following relationships between compound interest factors is not correct? ho. a. Single payment compound amount factor and single payment present worth factor are reciprocals hp. b. Capital recovery factor and uniform series present worth are reciprocals hq. c. Capital recovery factor equals the sinking fund factor plus the interest rate hr. d. capital recovery factor and sinking fund factor are reciprocals hs.

45. An “annuity” is defined as ht. a. Earned interest due at the end of each interest period hu. b. Cost of producing a product or rendering a service hv. c. Amount of interest earned by a unit principal in a unit of time hw. d. A series of equal payments made at equal interval of time hx.

46. Which one of the following method is not a method of depreciating plant equipment for accounting and engineering economic analysis purposes? hy. a. double entry method hz. b. fixed percentage method ia. c. sum-of-years digit method ib. d. sinking fund method ic.

47. ___ is a science which deals with the attainment of the maximum fulfillment of society’s unlimited demands for goods and services. id. a. Political science ie. b. Economics if. c. Engineering Economy ig. d. Social Science ih.

48. It is the branch of economics which deals with the application of economics laws and theories involving engineering and technical projects or equipments. ii. a. Political science ij. b. Economics ik. c. Engineering Economy il. d. Social Science im.

49. This refers to the products or services that are directly used by people to satisfy their wants. in. a. Producer goods and services io. b. Consumer goods and services ip. c. Luxury Products iq. d. Supply ir.

50. Those that are used to produce the goods and services for the consumer are called is. a. Producer goods and services it. b. Consumer goods and services iu. c. Machineries and Technology iv. d. Supply iw.

51. This refers to the satisfaction or pleasure derived from the consumer goods and services. ix. 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.

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. kg. a. Market Structures kh. b. Competition ki. c. Free competition market kj. d. Perfect Competition kk.

59. A place where the vendors or the sellers and vendees or the buyers come together is called kl. a. Market Structures km. b. Market kn. c. Free competition market ko. d. Perfect Competition market kp.

60. This is defined as the interest on a loan or principal that is based on the original amount of the loan or principal. kq. a. Ordinary Simple Interest kr. b. Simple Interest ks. c. Exact Simple Interest kt. d. Compound Interest ku.

61. This is based on one banker’s year. kv. a. Ordinary Simple Interest kw. b. Simple Interest kx. c. Exact Simple Interest ky. d. Compound Interest kz.

62. __ is based on the exact number of days in a given year. la. a. Ordinary Simple Interest lb. b. Simple Interest lc. c. Exact Simple Interest ld. d. Compound Interest le.

63. One banker’s year is equivalent to lf. a. 365 days lg. b. 350 days lh. c. 12 months and 30 days li. d. 12 months and 15 days lj.

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 lk. a. Ordinary Simple Interest ll. b. Simple Interest lm. c. Exact Simple Interest ln. d. Compound Interest lo.

65. It is the amount of money or payment for the use of a borrowed money or capital lp. a. Annuity lq. b. Interest lr. c. Simple Interest ls. 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.

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 oh. a. Book value oi. b. Market value oj. c. Use value ok. d. Fair value ol.

80. The amount of the property which the owner believed to be its worth as an operating unit is called om. a. value on. b. Market value oo. c. Use value op. d. Fair value

oq. or. os. ot. ou. ov. ow. ox. oy. oz. pa.

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 pb. a. Depreciation pc. b. Depleting value pd. c. Discounted value pe. d. Fair value pf.

84. __ is the money worth of an asset or product pg. a. Price ph. b. Market value pi. c. value pj. d. Fair value pk.

85. This refers to the present worth of all profits that are to be received through ownership of a particular property pl. a. Price pm. b. Market value pn. c. value po. d. capitalized cost pp.

86. Refers to the sum of its first cost and cost of perpetual maintenance of a property pq. a. Price pr. b. Market value ps. c. perpetual value pt. d. capitalized cost pu.

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. pv. a. Contract pw. b. Receipt px. c. bond value py. d. bond pz.

88. This value implies that the property will still be use for the purpose it is intended qa. a. salvage value qb. b. book value qc. c. use value qd. d. depreciation qe.

89. A depreciation computation method which is also known as Diminishing Balance Method qf. a. Sinking Fund Method qg. b. Straight-line Method qh. c. Declining Balance method qi. d. Break-even analysis qj.

90. Also called as Constant- Percentage Method in depreciation method qk. a. Sinking Fund Method ql. b. Straight-line Method qm. c. Declining Balance method qn. d. Break-even analysis qo.

91. This is also known as “Resale value” qp. a. salvage value qq. b. book value qr. c. use value qs. d. depreciation qt.

92. ___ is considered as the simplest type of business organization wherein the firm is controlled and owned by a single person qu. a. sole-proprietorship qv. b. Partnership qw. c. Corporation qx. d. Entrepreneurship qy.

93. __ is a firm owned and controlled by two or more persons who are bind to an agreement qz. a. sole-proprietorship ra. b. Partnership rb. c. Corporation rc. d. Entrepreneurship rd.

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. re. a. sole-proprietorship rf. b. Partnership rg. c. Corporation rh. d. Entrepreneurship ri.

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. rj. 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.

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 ug. a. market feasibility uh. b. socio-economic feasibility ui. c. technical feasibility uj. d. management feasibility uk.

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 ul. a. financial feasibility um. b. management feasibility un. c. market feasibility uo. d. technical feasibility up.

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 uq. a. 3 effect ur. b. 5 effects us. c. 4 effects ut. d. 2 effects uu.

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 uv.

F = 187.5 Lq 1.8 mc 0.20 in ft-lbs/lbm uw. ux. d 0.20 Di 4.8 uy. 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.

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. wm. a. 17.75% b. 19.88% c. 18.89% d.15.56% wn.

18. April 1997. wo.

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. wp. a. P12, 500 b.P30, 000 c. P16, 400 d.P22, 300 wq.

19. October 1992. wr.

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. ws. a. P74, 500 ; P340, 250 wt. b. P76, 500 ; P341, 000 wu. c. P76, 500 ; P 342, 500 wv. d. P77, 500 ; P343, 250 ww.

20. November 1997. wx.

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. wy. a. P155, 000 b.P140, 000 c. P146, 000 d. P132,600 wz.

21. April 1998. xa.

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. xb. a. P791.26 b.P950.00 c. P971.12 d. P845.32 xc.

22. November 1995. xd.

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 %? xe. a.28 b. 25 c. 16 d.30 xf.

23. November 1998. xg.

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. xh. a.9 years b. 10 years c. 12 years d.19 years xi.

24. November 1996. xj.

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. xk. a. P275.3M b.P265.5M c. P295.3M d. P282.1M xl.

25. October 1995 xm.

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. xn. a. P5, 115 b.P4, 946 c. P5, 625 d. P4, 805 xo.

26. October 1998. xp.

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. xq. a.886 units b. 885 units c. 688 units d.668 units xr.

27. October 1990. xs.

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: xt.

Cost of electricity P20.00 xu.

Tax to be paid per block 2.00 xv.

Real Estate Tax 3, 500.00 per month xw.

Salaries and wages 25, 000.00 per month xx.

Others 12, 000.00 per month xy.

Selling price of Ice 55.00 per block xz. a. 1228 b. 1285 c. 1373 d.1312 ya.

28. April 1998 yb.

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? yc. a. 2334 b. 539 c. 750 d.2333 yd. ye. Answer Key yf. Board Problems yg.

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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. Answer Key yj. Objective Problems 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. 47. b ym. 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

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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.

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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. Salaries an. Delivery expenses ao. Rent ap. Sales aq. Antifreeze available for sale during October (at cost) ar. Antifreeze inventory on Oct. 31 (at cost) as. Other expenses at. 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. Long-term debts bg. Debts due within 1 year bh. Accounts payable bi. Machinery and equipment (at cost) bj. Cash in bank bk. Prepaid rent bl. Government bonds bm. Social security taxes payable bn. Reserve for depreciation bo. Reserve for expansion bp. Inventory bq. 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

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44.

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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.

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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) = $117,000 13% ($300,000) = $39,000 31% ($300,000) = $93,000 10% ($300,000) = $30,000 29% ($300,000) = $87,000 10% ($300,000) = $30,000 55% ($300,000) = $165,000 6% ($300,000) = $18,000

Engineering and Supervision Construction Expense

32% ($300,000) 34% ($300,000)

= =

Constructor’s Fee Contingencies

7% ($879,000) 36% ($879,000)

= =

$96,000 $102,000 $198,000 $61,530 $108,000 $169,530

el.

Fixed Capital Investment =$1,246,530 em. 9. Refer to the previous problem and on the basis of this information, what is the total capital investment? en. GIVEN: eo.

Equipment Cost = $300,000 ep.

Contractor’s Fee = 7% of the direct plant cost eq.

Fixed capital investment = $1, 246, 530 er. REQUIRED: es.

Total capital investment et. SOLUTION: eu.

Fixed Capital Investment $1, 246, 530 ev.

Working Capital 74% ($300,000) = $222,000 ew.

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. GIVEN: fy.

S = $ 9000 fz.

P = $ 5000 ga. n = 12 gb. REQUIRED: Interest, i GC. SOLUTION: gd.

S = P (1 + i )n ge. $9000 = $5000(1 + i) 12 gf. i = 0.05 or 5.02 % gg. gh. 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? gi. GIVEN: P = $2000 gj. i = 6% gk. n = 4 years gl. REQUIRED: SC gm. SOLUTION: gn.

SS = P (1 + in) go. = 2000 [1 + 0.06(4)] gp.

SS = $2480 gq. Extended for 6 years gr.

SC = $2480 (1 + 0.08)6 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.

TAC = 900

md.

0 = 900

It

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. Cash qe. qf. Accounts payable: qg.

B Company qh.

C Company qi. qj. Accounts receivable qk. Inventories ql. Mortgage payable qm. Common stock sold qn. Machinery and equipment (at present value) qo. Furniture and fixtures (at present value) qp. Government bonds qq. Surplus

qr. $20 qs. qt. 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

sc. Government Bonds

sd.

3,0

se. Mortgage Payable

sf.

5,0

sh. $

si.

sj.

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

sg. Total

Furniture & Fixtures sw. Total

ta.

te.

Total Assets

$

su. Total

sv. $

Surplus

sz.

2,0

tb.

tc.

Total

td. $

tf.

tg. Total Equities

th. $

sx. $

sy.

$ ti. tj. tk. tn. to. tp.

Total Assets: $ 67, 000 tl. tm. 28. Determine the total current assets of the AMD Food Corporation.

ANSWER: From the balance sheet; Total Current Asset = $ 44,000 tq.

tr. ts. tt. tu. tv. tw. tx.

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. Long-term debts vs. Debts due within 1 year vt. Accounts payable vu. Machinery and equipment (at cost) vv. Cash in bank vw. Prepaid rent vx. Government bonds vy. Social security taxes payable vz. Reserve for depreciation wa. Reserve for expansion wb. Inventory wc. 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 ,

1 0 0 wi. 3 0 0 wj. 3 , 0 0 0 wk. 2 4 0 wl. 6 0 0 wm. 1 , 2 0 0 wn. 1 , 6 0 0 wo. 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 =

current assets

yd.

Current liabilities

ye.

= $9, 700

yf. yg. yh.

$3, 540 Current ratio = 2.74

yi. yj.

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. yk. GIVEN: yl.

Vo = $60,000 ym.

Vs = $10,000 yn.

Total expenses annually exc. Depreciation = $ 100,000 yo. i = 0.12 yp. REQUIRED: n using SLM YQ. SOLUTION: yr. ys. yt. yu. yv.

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

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? zp. GIVEN: zq.

Vo = $40,000 zr.

Vs = 0 zs. a = 12 zt. n = 20 ZU. REQUIRED: New Capital needed ZV. 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.

Vo  Vs n

850 ,000  50,000 20

d=

$40 ,000 yr

ael. Basis: 1 yr aem. aen. aeo. aep. aeq. aer.

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:   5   1 0.05        !0  1 0.05     Va = 50,000 - (50,000-0) 



afp. afq. afr.

Using SFM:



1 1

Va = $28034.35 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  d = (20,000-5,000) = $ 5.000

ahu.

1 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 L ajq.

L

d=

ajr.

d=

90,000  8,000 10

ajs.

D3

ajt.

C3 CO D3

= n(d) = 3(8,200) = P24,600

=

-

C3

aju.

= P 8,200

= 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 L akd.

d=

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 L akx.

d= aky. akz.

L

d12 = P3, 375 (12) = P 40,500 C12 = Co - d12 = 60,000 – 40,500

ALA. C12= P19, 500 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:

CO  CL

alj.

60,000  6,000 F F P54 ,000 , 12% , 16 A,6.5%,10 A 42.7533 d= = = = P 1,263 alk.

D 12

 F  12%, 16    A,  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

 2,000   CL   C  10  15,000  = 0.1825 or 18.25%  O =

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: n L

aod.

  CL    Co     Co  = (220 ,000 ( 22,000 / 220 ,000 )^(5/ 10)= P 69,570 Cs=

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.