Gas Absorption

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 Henrys 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 KxA = 0.9 x 10-4 kg CO2 /kg H2O

Required: PA of CO2 Solution:From A.3-18 for Henrys law constant (Geankoplis p. 884)H = 0.186 x 104 atm/mol frac.PA= HxA

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 systemPAT = 1 atmPA of SO2 = 0.2 atm

REQUIRED:Degrees of freedom, FVariables that can be set

SOLUTION:F = C P + 2F = 3 2 + 2F = 3Variables that can be set:1. total pressure2. temperature3. mole fraction composition xA of SO2

10.3-2 Equilibrium Swtage Contact for Gas-Liquid System. A gas mixture at 2.026 x 105 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 SO2 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. 10.2-1PT = 2.026 x 105 Pa =T = 293 KPA of SO2 = 1.52 x 104 Pa = .15 atmInlet gas = 5.70 kg molInlet H2O = 2.20 total kgmol

REQUIREDXA1, yA1, L1 V1

SOLUTION:xAo = 0amount of entering acetone = yAN+1vAN+1 = 0.01(30) = 0.30 = 29.7 kgmol/air hacetone leaving in Vi = 0.10(0.30) = 0.30 kgmol/hacetone leaving in Ln = 0.9 (0.30) = 0.27 kgmol/hV1 = 29.7 + 0.03 = 29.73 kgmolH2O + acetone/hr

Ln = 108 + 0.27 = 108.27 kgmol H2O + acetone/hr

Using equation:

10.3-3 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/hrYAN + 1 = 0.01VN + 1 = 30.0 kg mol/hT = 300 KPT = 101.3 kPayA = 2.53 xA

REQUIRED:Theoretical stages graphically using analytical Kremser equation.

SOLUTION:xAo = 0amount of enetering acetone = yAN + 1 VN + 1 = 0.01(30) = 0.30entering air = (1-yAN + 1) VN + 1 = (1 0.01)(30) = 29.7 kg mol/air hacetone leaving in V1 = 0.10(0.30) = 0.30 kg mol/hacetone leaving in Ln = 0.9(0.30) = 0.27 kg mol/hV1 = 29.7 + 0.03 = 29.73 kg mol air + acetone /hrLn = 108 + 0.27 = 108.27 kg mol H2O + acetone/hr10.4-1 Interface Concentrations and Overall Mass-Transfer Coefficients. Use the same equilibrium data and film coefficients ky and kx 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 Ky and Ky and flux NA.c. Overall mass transfer coefficient Kx and flux NA.

GIVEN:Equilibrium datayAo = 0.380 mol fractionxAL = 0.10T = 298 KP = 1.013 x 105 Paky = 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:KX, KX

SOLUTION:Trial 1:

y = mx + b0.38 = -0.10 (1.342) + bb = 0.5142xA1 = 0.247yAi = 0.183

Trial 2:

10.4-2 Use the same equilibrium data and film coefficients ky and kx 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 Ky and Ky and flux NA.(c) Overall mass-transfer coefficient Kx and flux NA.

GIVEN:yAG = 0.25xAL = 0.05ky = 1.465 x103 kgmol A / sm3 mol fractionkx = 1.967 x103 kgmol A / sm2 mol fraction

REQUIRED: a.) Interface concentrations yAi & xAi & NA SOLUTION:

Trial 1:

Trial 2:

Trial 3:

or,

b)

c)

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 kya = 0.0739 kgmol.sm3 mol frac and kxa = 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 Kxa to calculate the tower height.(a) Calculate the tower height using kya(b) Calculate the tower height using Kya.

Solution:

(a) La = 68 kgmol/h ya = 0.005 xa = 0

(b)Vb = 57.8 kgmol/h yb = 0.04

height = z = Nx Hx = 0.2250 (8.6692) = 2.2106 m

XyXYmbyi

00.010.0150.020.0290.00500.01700.02290.02870.040000.01010.01520.02040.02990.0050.01730.02340.02950.0417-2.2754-2.2707-2.2685-2.2667-2.26100.0050.03970.05690.07400.01080.00140.00960.01410.01880.0272

xi1/(y yi)1/(x xi)1/(y yi) aveyx

0.00180.01320.01900.02460.0348277.7780135.1351113.6364101.010178.125555.5556312.50250217.3913172.4138206.4565124.3858107.323389.56780.0120.00590.00580.01130.010.0050.0050.009

dy/(y-yi)Avedx/(x-xi)AveKy=kya/(1-y)Kx =kxa/(1-x)

2.47750.73390.62251.01214.34031.40631.16851.75410.07430.07520.07560.07610.07560.1690.17070.17160.17240.1740

Ny = 4.8549Nx = 8.6692kyAve = 0.0756kxAve = 0.1715

XymmKyKx

0.00720.02240.02920.03600.048400.00720.0110.01480.02240.77780.75000.81580.86960.82760.66670.80430.86270.86840.94120.05900.05550.05590.05620.05690.04500.04550.04570.04590.0463

AVE: 0.8082Ave: 0.8287

1/(y y)1/(x x)1/(y y)ave1/(x x)avedy/(y y)avedx/(x x)ave

200102.040884.033671.942456.8182138.888980.645270.422562.551.5464151.020493.037277.988064.3803109.767175.533966.461357.02031.81220.54810.45230.72751.09770.37770.33230.5132

Noy= 3.5409Nox= 2.3209

10.6-10. Repeat Example 10.6-2 but use transfer units and calculate HL, NL, and tower height.

Given:Acetone-H2O systemA = 0.186 m2T = 293 KP = 101.32 kPay1 = 0.026y = 0.005V = 13.65 kgmol inert air / h L = 45.36 kgmol H2O / h kya = 3.78 x 10-2 kgmol / sm3 mol frackxa = 6.16 x 10-2 kgmol / s m3 mol frac

REQUIRED:HL, NL, z

SOLUTION:

from the graph:xi1 = 0.0136 xi2 = 0.0019

10.7-1Liquid 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 kxa. [Note: the interface values xi have alre4ady been obtained. Usi a graphical integration of Eq. (10.6-15).

Given:A = 0.0929 m2T= 293 KP = 1.013 x 105y1= 0.20y1 = 0.02x1 = 0 V = 6.53 x 10-4 kgmol air/ sL = 4.20 x 10-2 kgmol / skya = 0.0594 Gy0.7Gx0.25kxa = 0.152 Gx 0.82

REQUIRED: Z using kxa

SOLUTION:

yxVLGyGxkya

0.020.040.070.130.2000.0003320.0008550.002010.003556.66 x 10-46.68 x 10-47.02 x 10-47.51 x 10-48.16 x 10-40.0420.04210.042030.042080.042150.21300.22260.23780.27180.31648.1388.1478.1628.1968.2410.033980.035040.036730.040320.04496

kxambxiyimb

0.8480.8490.8500.8530.85724.456723.268021.540318.442615.30340.020.047720.088420.16710.25430.000460.001090.001880.003560.005720.00880.0260.048240.10160.166324.747623.625222.270519.074215.98290.020.047840.089040.16830.2567

xiyimbxdy/(xi x)

0.0004590.001040.001860.003540.005700.00870.024160.04760.10040.164224.750123.668722.081419.099316.0037------0.047620.088880.16840.25680.0003320.0005230.0011550.001540.58970.62470.96870.8679

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 % efficiencyt = 3.11 h

REQUIRED:

(a) DAeff(b) T if 90% efficient with same diameter

SOLUTION:

(a)

(b)

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, x1V2, x2

Lo, No, yo, BL1, N1, y1, B

GIVEN:

V2 = 80 kg solventxA2 = 0.03xC2 = 0.97xA1 = 0.22N = 1.5 kg insoluble solid/kg solution

REQUIRED:(a) amount and composition of overflow, V1(b) amount and composition of the underflow, L1

SOLUTION:

For Overall Material Balance:

For Solute Balance:

To find NM

To solve for the value of exit underflow

For exit Overflow:

20.1. Roasted copper ore containing the copper as CuSO4 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 H2O 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/hr1.2 tons CuSO4 and 0.5 ton H20 is to be treated per hourStrong solution 90% H2O, 10% CuSO4CuSO4 recovery of 98% of that in oreyb = 0after each stage, 1 ton inert solids retains 2 tons of H2O + CuSO4 dissolved

REQUIRED:Stages required

SOLUTION:

yb = 0ya = 0.1La = 1.2 tons CuSO4 + 0.5 ton H20 = 1.7 tons solution/hr

CuSO4 recovery = 0.98(1.2) = 1.176 tonsCuSO4 retained = 1.2 1.176 = 0.024 tons

Solute Balance:

By Material Balance:

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 lb exhausted livers, galSolution concentration, gal oil/gal solutionSolution retained by 1 lb exhausted livers, galSolution concentration, gal oil/gal solution

0.03500.0680.4

0.0420.10.0810.5

0.0500.20.0990.6

0.0580.30.1200.68

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 lbs95% oil recoveryyA = 0.65yB = 0assuming 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 oilFor the solution in the spent solids: (by trial & error)

Trial 1:Let XB = 0.1Solution retained = 0.042LB = 0.042 x 100 = 4.2 gal

Trial 2:XB = 0.051, by linear regression:Solution retained = 0.0386LB = 0.0386 x 100 = 3.86 gal

Trial 3

XB = 0.065, by linear regressionSolution retained = 0.0396 LB = 0.0396 x 100 = 3.96

Trial 4

xB = 0.0634, by linear regressionsolution retained = 0.0394LB = 0.0394 x 100 = 3.94

0.0637 (close to 0.0634)

assuming XA = 1, LA = 0.043 gal oil/lb exhaust liver x 100LA = 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 100L1 = 11.21 gal

By OMB:

La + V2 = L1 + VaV2 = 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 +Va4.3 + VN+1 = 6.8 + 6.2293

VN+1 = 8.7293

By Solute Balance:

Laxa +VN+1 yN+1 = LNxN + Vaya4.3(1) + 8.8293yN+1 = (6.8) (0.4) + (6.2293) (0.65)yN+1 = 0.2828

From the Graph:

Number of stages = 6.7813

Geankoplis11.2-1 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:xyxy

1.0001.0000.2540.701

0.8670.9840.1450.521

0.5940.9250.1590.271

0.3980.83600

Solution:

C5H12 Balance:

when XA = 0.6 , YA = 0.6.pt 1when 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= 0.855XB = 1 - 0.43 = 0.57 and YB = 1 0.855 = 0.1411.3-2 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:xy1/ ( y-x )

1.0001.000

0.8670.9848.547

0.5940.9253.021

0.3980.8362.283

0.2540.7012.237

0.1450.5212.660

0.0590.2714.717

00

Solution:

from the graph:

____________ AT = 0.5048therefore X2 = 0.4075

Answers:XA = 0.4075yA = 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

from the graph:

______________ AT = 0.4163Therefore x2 = 0.6325Answer:xA = 0.6325 and yA = 0.7975xB = 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/h0.5 benzene0.5 tolueneR = Ln/D = 4.52XD = 0.9XB = 0.10

Solution:F = D + B100 = D + BXF F = DXD + BXB100 ( 0.5 ) = 0.9D + 0.1 ( 100 D )50 = 0.9D 0.1D + 1040 = 0.8DD = 50 kg mol/hB = 50 kg mol/hR= 4.52

if x = 0 ,Yn+1 = 0.16( 0,0.16)..........plotted on the graphAnswer: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 xH and yH.

Temperature TemperatureKCxHyHKCxHyH

409.3136.200383.8110.60.4850.730

402.6129.40.080.230376.0102.80.7900.904

392.6119.40.2500.154371.598.31.0001.000

Given:F = 200 mol/h0.42 heptane0.58 ethyl benzeneXD = 0.97 heptaneXB = 0.011 heptaneR = Ln/D = 2.5

Solution:F = D + B200 = D + BXF F = DXD + BXB200( 0.42 ) = 0.97D + 0.011 ( 200 D )84 = 0.97D - 0.011 + 2.281.8 = 0.959DD = 85.2972 mol/hB = 114.7028 mol/h

R = 2.5

= 0.277Answer: 10.77 theoretical stages or 9.77 trays + a reboilerfeed 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 XD = 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

(a)Rm 0.4687512Rm = 0.46875Rm = 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 = 0.68x = 0.42

Solution:

Rm = 0.5273 ( Rm + 1 )0.4227Rm = 0.5273Rm = 1.12

Minimum number of trays: 7.64 theoretical stages or 6.64 trays + a reboiler

Foust7.27. 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 toluenexD = 0.95 benzenexW = 0.04 benzeneq = 0

Solution:

(a) Rm = 0.8176Rm + 0.8176 Rm = 4.4827(b) Minimum number of equilibrium stages @ Total Reflux = 6.89 stages(c) R = 8

@ Xn = 0 , Yn+1 = 0.1056@ Xn = 0. 95 , Yn+1 = 0.95Number of equilibrium stages @ reflux = 8 : 8.79 stages(d) R = 8, feed is liquid at boiling pointNumber of equilibrium stages = 8.75 stages

7.35. An equimolar mixture of ethanol and water is to be fractionally distilled to produce a distillate of composition 0.80 mole fraction ethanol and bottoms of 0.05 mole fraction of ethanol. The feed is saturated liquid and there is a total condenser and a total reboiler.(a) At areflux ratio of 2.0, how many equilibrium stages are required?(b) How many stages are required at total reflux?(c) What is the minimum refux ratio?(d) What percentage of the feed ethanol is recovered in the distillate?

SOLVED PROBLEMS:1. An air stream at C having a humidity of 0.03 is contacted in adiabatic humidifier. It is cooled & humidified at 90 % RH. Determine the temperature of the humidified air & the make up H2O for every of inlet air. GIVEN:

COOLER

T1 = 90 OCH=

RH2 = 90 %

REQUIRED: L in every m3 of airSOLUTION: Step 1: From Psychrometric Chart:

Step 2: Compute for humid volume.

balance: L = W = T K = = 1.0769 Basis: 1 W = 1

W = 0.9285 kg da

L = W = 0.9285 kg da ( 0.51-.0.03) 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: qHOLUTION:

Step 1: From Psychrometric ChartStep 2: Compute for the Humid Volume

VH = ( 2.83 x 103 + 4.56 x 103 H)(T+273) = [2.83 x 103 + (4.56 x 103 x 0.022)] (30+273) VH = 0.8879 m3/ kg dry air

Step 3: Calculate Mass flowrate of air

W = W = 225.25 kg dry air/hr

Step 4: Calculate the specific heat

cs = (1.005 + 1.88 H)cs = [1.005 + 1.88 ( 0.022) ]= 1.046

Step 5: Calculate the heat load of the heater=

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 75m3, the air in the room is changed every 15 min.

GIVEN:

AC

ROOM25 0C90%RH

qH

COOLER

qc

25 0C 15 0C

L

REQUIRED: qc

SOLUTION:

Step 1: Compute for h1 & h2 , then vH

+ + 2.501 + +

+ + 2.501 + +

(25 + 273)

Step 2: Compute for W.

W=345.5027

Step 3: Solve for L, and finally qc.

L=3.0059

TW = 115 0F

- (15-25) K

x

hp

5. 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:

PRE-HEATER H1=H2 TW2=TW3 H3=H4

REHEATERADIABATICHUMIDIFIER 90% RH

TD = 200 0F 70 0F

REQUIRED: T of air leaving the preheater and humidifierSOLUTION:

Step 1: From Psychrometric Chart

Tw = 115 oF

H3=H4

0.048 lb H2O/lb da

Td = 200 oF

90 % RHT3= 100 oFTw2=Tw3=103.8 oF

H3=0.048lbH2O/lb da

Step 2: Solve for H2.

H2O Balance at Humidifier:

WH2 = WH1 + LL = W( H3 H2)L = W (0.048 H2) eqn. 1

eqn. 2

0.0008= =

Step 3: Use H2 to read T2 from Psychrometric Chart.

Tw2 =103.8 0FT2 =152 oFH2 = 0.0367LbH2O/lb da

6. In a plant lab having a floor area of 100 m2 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:

ROOM23.9 oC80%RHAC

A=100 m2H=3m80% void10 min

23.9 oC80 % RH18.3 oCqHCOOLER

L

REQUIRED: L

SOLUTION:

Step 1: From Psychrometric Chart.

80% RH

H=0.0151

H=0.0106

23.9 oC18.3 oC

Step 2: Compute for vh. (23.9 + 273) =

Step 3: Solve for W and then L, the condensate.

W=2614026 kg dry air/ hr

L = W (H1-H2)

L =

L =

GEANKOPLIS:

9.3-1. Humidity from Vapor PressureThe 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. Humidityb. Saturation humidity and percentage humidityc. Percentage relative humidity

GIVEN:

ROOMT= 37.8 CPT = 101.3 KPa PA = 3.59 Pa

REQUIRED:Humidity, Saturation humidity & Percentage Humidity, Percentage Relative Humidity

SOLUTION:

Step 1: Calculate H.H = x

H = 3.59 X

H = 0.0228

Step 2: Compute Hs & Hp.

From steam table, by interpolation:

PAs of H2O = 6.59 KPa

= x

= X

= 0.0432

= x 100

= x 100

= 57.78 %Step 3: Solve for HR.

= x 100

= x 100

= 54.48 %

9.3-2. Percentage and Relative HumidityThe 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 Hpb. Percentage relative humidity HR

GIVEN:

ROOMT= 32.2 CPT = 101.3 KPa H = 0.021 kg H2O/ kg da

REQUIRED:Percentage Humidity & Percentage Relative Humidity

SOLUTION:

Step 1: Compute for PA.

H = x

0.021 = x 61.69 - 0.609 = 18 61.69 = 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

= x

= X

= 0.031

= x 100

= x 100

= 67.74 %

= x 100 = x 100

= 68.88 %

9.3-3. Use of the Humidity ChartThe air entering a dryer has a temperature of 65.6oC (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:

DRYER

Td = 65.6 CDp = 15.6 C

REQUIRED:Actual humidity and % humidityHumid volume & humid heat in SI & English units.

SOLUTION:Step 1: From Psychrometric Chart.

Dp = 60 F

H = 0.011kg H2O/ kg da

5.3 % RH

Td = 150 F

Step 2: Compute for humid heat.= 1.005 +

= 1.005 + 1.88 (0.011)

= 1.026 ( SI )

= 0.24 +

= 0.24 + 0.45 (0.011)

= 0.245 ( English )

Step 3: Solve for VH.

= T K

vH = ( 65.6 + 273 )

= 0.975 ( SI )

= T F

vH = ( 0.0204 + 0.0405(0.011) ) ( 150 + 460 )

= 15.64 ( 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 H2O/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:

DRYERT= 52.7 CH = 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

= x

= x = 0.1305

= = x 100 23 %

Step 3: Compute for humid heat in SI and English.

= 1.005 + = 1.005 + 1.88 (0.030)

= 1.0614 ( SI )

= 0.24 + = 0.24 + 0.45 (0.030)

= 0.2535 ( English )

Step 4: Solve for VH.

= T K

vH = ( 57.2 + 273 )

= 0.9796 ( SI )

= T F

vH = ( 0.0204 + 0.0405(0.030) ) ( 134.96 + 460 ) = 15.72 ( English )

9.3-5. Adiabatic Saturation TemperatureAir 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 oCADIABATIC SATURATOR

80% RHH = 0.0655 kg H2O/kg da

REQUIRED:Final values of H and T oCValues of H and T at 100 % saturation

SOLUTION:Step 1: From Psychrometric Chart.

80 % RH

H2 = 0.079kg H2O/kg da

Tw = 120 oF

H1 = 0.0655kg H2O/kg da

T1 = 82.2 oCor 180 oFT2 = 52.78 oCor 127 oF

Step 2: For 100 % saturation, take readings in the Psychrometric Chart.

H2 = 0.0802kg H2O/kg da

H1 = 0.0655kg H2O/kg da

T1 = 82.2 oCor 180 oFT2 = 49 oCor 120 oF

9.3-6 Adiabatic Saturation of AirAir 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 oCADIABATIC SATURATOR

Dp = 40.6 oC

REQUIRED: Final values for H and T oC

4SOLUTION:

Step 1: From Psychrometric Chart.

Tw = 113 oF 90 % RH

Dp = 105 oFor 40.6 oCH2 = 0.0075kg H2O/ kg da

T1 = 140 oFor 76.7 oCT2 = 116 oFor 46.7 oC

9.3-7 Humidity from Wet and Dry Bulb TemperatureAn 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 oCREQUIRED:HumiditySOLUTION:Step 1: From Psychrometric Chart.

Tw = 32.2 Cor 90 F

H = 0.0175kg H2O/kg da

Td = 65.6 oC or 149.9 F

9.3-8 Humidity and Wet Bulb TemperatureThe 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, TwSOLUTION:

Step 1: From Psychrometric Chart.

Tw = 98 F or 36.67 C

H = 0.030kg H2O/kg da

Td = 140 For 60 C

9.3-10 Cooling and Dehumidifying AirAir 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:

T1 = 32.2 CRH1 = 65 %HEATERADIABATICCOOLINGCHAMBER

T3 = 23.9 CRH1 = 40 %

REQUIRED:Initial and final humidity

SOLUTION:

Step 1: Obtain H1 from Psychrometric Chart.

65 % RH

H1 = 0.02kg H2O/ kg da

T1 = 32.2 C

Step 2: Obtain H3 from Psychrometric Chart.

T3 = 23.9 CH3 = 0.0075kg H2O/kg da40 % RH

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-m2Ls/A = 24.4 kg d.s/m2X1= 0.45 kg free moisture/kg d.sX1 = 0.55 kg free moisture/kg d.sXc = 0.22 kg free moisture/kg d.sX2 = 0.3 kg free moisture/kg d.s

Required:

Time (t)

Solution:

For Ls/A of 2nd condition:

Since

So,

Therefore,

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/sb. Predict the effect if the gas temperature is raised tom 76.7C and it remains the same.

Given:From example 9.6-3A = 0.457 x 0.457 mb = 25.4 mmv = 6.1 m/sT db = 65.6CH = 0.010 kg H2O/kg d.a

Required:a. Rc if v = 3.05 m/sb. Rc with T = 76.6C

Solution:a)From ex. 9.6-3

G = 11386.26 kg m2/h

Rc = 1.9479 kg/ m2-h

b)@ T = 76.7C H = 0.010 kg H2O/kg d.a

VH = 1.0056 m3/kg da

G = 22056.1447 kg/ h-m2

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 matl is increased to 44.5 mm

Given:A = 0.61m x 0.61mb = 25.4 mmv = 3.05 m/sTdb = 60CTwb = 29.4CX1 = 0.35 kg H2O/ kg d.sX2 = 0.22 kg H2O/ kg d.sLs = 11.34 kg d.s

Required:t, Rc

Solution:a.) @ Tdb = 60C, H = 0.0141 kg H2O/ Kg d.a

From steam table: @ Twb = 29.4C , hw = 2432.14 x 103

Rc = 1.6437 kg/ h- m2

t = 2.4103 hrs

b.) by ratio & Proportion:

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 80F and a dry bulb of 120F 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?

GivenA = 1 ft2t = 1inTwb = 80FTdb = 120FV = 2.5 ft/s = 120 lbm/ft3Xc = 0.90 lbm H2O/ lbm d.sX1 = 0.20 lbm H2O/ lbm d.s

Required:t

Solution:@ Twb = 80F Tdb = 120F : H = 0.013 kg H2O/ kg da

@ Twb = 80F ; w = 1048.4 (from steam table)

tc = 13.30 hrs

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.sA = 4.645 m2Required:T using straight line through the originSolution:Using Fig 9-5-1b;Rc = 1.51 kg H2O/h-m2Xc = 0.1950

tC = 2.9169 hrs

tF = 6.3185 hrs

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)__04.9442.24.5547.04.0190.44.8853.04.4049.03.9780.84.8084.24.24112.03.9551.44.6995.04.150______________________________________________________________________

Calculate the free moisture content X kg H2O/kg d.s. for each data point and plot X versus time. (Hint: For 0h, 4.944-0.190-3.765=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 sample weight

Free moisture content:

.2670.24700.22660.1976

0.15910.11930.075960.0518

0.01699601088x10-30

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:

Required: Ts and Rc

Solution:

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.9.9-1DL = 2.97 x 10-6X* = 0.04 kg H2O/kg dry woodx1 = 0.0127 m

a. Xt1 = 0.22b. Xt1 = 0.29 Xt = 0.13 Xt = 0.09 x1 = 12.7 mm

Required:a. time using fig. 5.3-13b. 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 = 0.20t= 10.86 hrsb.)X1 = Xt1 X* = 0.29-0.04 = 0.25X = Xt X* = 0.09-0.04 = 0.05x1 = 12.7/ 2 (1000) = 6.35 x 10-3 m

Equation 9.9-6

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:

Required: total time

Solution:

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 kgH2O/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=12.7Z=63.5V=0.914T=821.2CH=0.01kgH2O/kgd.a.P=1522kg/m3E=0.35X1=0.90X2-0.45

Required:tCRP

Solution:

tCRP =Ps (X1-X2) ah (T-Tw)

setting X2=Xc ; z=X1m@T =82.2 C & H= 0.01 from P.C. Tw=32.2 C

@82.2C u=0.019x10-3

Basis:1 kg d.a.

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, =0.99 kg/m3

NRE=DPVP/u

=604.83 > 350

Gt=(0.914m/s)0.99 =0.905kg/m2s

=0.853W/m2K

a=6(1-E)/DP =307.09

@Tw=32.2 C ; w =2425.27kJ/kg(by interpolation)

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 15.6oC and leaves at 60oC. The gas enters at 87.8oC 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=15.6C X1=0.04 Ts2=60CX2=0.002 TG2=87.8CH2=0.01 TG1=32.2C CpA=4.187 kJ/kgK CpB=1.465 kJ/kgK

Required:a.)Gb.)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=87.8C & To=0CHG2= CS(TG2-To) + H2 =[1.005 + 1.88(0.01)](87.8-0) + 0.01(2501) =114.9 kJ/kgd.a.

For the exit gasHG1=CS(HG1-To) + H1 =[1.005 + 1.88H1](32.2-0) +H1(2501) =32.36 +2561.54H1

For the entering solid,HS1=Cps(Ts1-To) + X1Cpa(Ts1-To) =1.456(15.6-) + 0.04(4.187)(15.6-0) =25.33kJ/kgd.a.HS2=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,GHG2+LsHS1=GHG1+LsHS2+QG(114.9)+453.6(25.33)=G(32.36+2561.54H1)+39853.3+10551.6G(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=389.15-38915.21H183076.16H1=1812.14So,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 X1=0.4133lb H2O/lb dry solid is to be dried to X=0.0374 lbH2O/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 119oF and remains essentially constant in temperature in the dryer. The saturation humidity at 119oF from the humidity chart is Hw=0.0786 lb H2O/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 Xc=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=119FX1=0.4133 Hw=0.0786lbH2O/lbdaX2=0.0374 A/Ls=0.30ft2/lbdsG=13280lbda/h Xc=0.0959TG2=203F KyMB=30.15lbmair/hft2H2=0.0562lbH2O/lbda

Requireda.)tCRPb.)tFRP

Solution:

Material Balance:Ls(Xc-X2)=G(Hc-H2)700(0.0959-0.0374)=13280(Hc-0.0562)Hc=0.0593lbH2O/lbda

GH2 + LsX1 = GH1 + LsX2

13280(0.0562)+700(0.4133)=13280H1+700(0.0374)H1=0.076 lbH2O/lbda So,

tCRP=4.204h

For Falling rate zone,

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 25.6C and has a humidity of H = 0.007 kg H2O/kg d.a. The moist leaves the dryer at 37.8C and H = 0.020 and part of it is recirculated and mixed air enters the dryer at 65.6C 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 airb. % air leavingc. heat added (heater)d. heat loss (dryer)

Solution:

H2O Bal. On the heaterG1H1 + G6H2 = ( G1 +G6 ) H4G1 (0.007) + G6 (0.020) = G1 (0.010) +0.01G60.01 G6 = 0.003 G1 .(1)

H2O Bal. On the dryer

(G1 +G6)H4 + LsX1 = (G1 + G6)H2 +LsX2

for LsLs = L ( 1 X ) = 907.2 (1 0.27) = 662.256 kg d.sdry basis:since G3 = G4 = G1+G6 = 41728.75 d.a/hat junction AG1H1 + G6H6 = G4H4

G1 = 32,099.15 kg fresh d.a /h

From eqn 1:

% recycled = 23.06% recycled air

Q = 440.605 KW

Amount of Heat loss from the dryer

Heat bal.

Q = 32.06 KW

MC CABE

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 1000oF with a humidity of 0.03 and wet bulb temperature of 150oF. The solids have a specific heat of 0.48 Btu/lboF; they enter the dryer at 70oF 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:

Solution:

L = 35.842 ft

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?

GiventT = 7 hrs1st condition2nd conditionx1= 35x1 = 35x2 = 10x2 = 5xc = 20x* = 4

Required: tT @ second condition

Solution:Basis: 100 total moistureTotal moisture equilibrium moisture = Free moisture

1st conditionX1: 35 4 = 31X2: 10 4 =6Xc: 20 4 = 16

2nd conditionx2 = 5 4 =1at:

Rc = 4.3848 kg H2O/ h m2

At 2nd condition:X1 =31 ; X2 = 1

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 = 575Cp for dry sand = 0.21 cal/gCCp for dry air = 0.238 cal/gCCp 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 + LsX2G(0.007) + 10 x3(0.5) = GH1+10 x 3(0.003)0.007G + 4.7 x103 = GH1 (1)

HG2 = Cs(TG2- To) + H2 = [1.005 + 1.88(0.007)](120-0) + 0.007(2501) HG2 = 139.6862 kJ/kg d.a.

For the exit gas,

HG1 = Cs(TG1- To) + H1 = [(1.005 + 1.88H1)(45-0) + H1 (2501)HG1 = 45.225 + 2585.6 H1

For the entering solid,

Hs1 = Cps(Ts1-To) + X1 Cpa(Ts1-To) = 0.221(25-0) + 0.5(0.48)(25-0) Hs1 = 11.25 cal/g Hs2 = Cps(Ts2-To) + X2Cpa(Ts2-To) = 0.21(400-0) + 0.03(0.48)(40-0) Hs2 = 8.976 cal/g

Substituting into Eq,

G HG2 +Ls Hs1 = GHG1+Ls Hs2 + QG(139.6862) +10(11.25) = G(45.225 + 2585.6 H1) + 10(8.976) +594.4612G 17.74 = 2585.6GH1 (2)

Using (1) &(2),G=159140.7017g/h~159.1407 kg/hH1=0.03653 gH2O/g air

EVAPORATION

1. It is a unit operation which means to concentrate a solution considering a non-volatile solute and volatile solvent.a. condensationb. evaporationc. precipitationd. extraction2. It is conducted by vaporization of a portion of the solvent to produce a concentrated solutiona. condensationb. evaporationc. precipitationd. extraction3. Which of the following is not a processing factor in evaporation?a. solubilityb. temperaturec. frothingd. specific heat4. 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 interfaceb. The standard temperature difference is based on the temp of the liquid at the liquid interfacec. The log mean temperature difference is based on the temp of the boiling liquid at the liquid vapor interfaced. The log mean temperature difference is based on the temp of the liquid at the liquid interface5. The heat required to vaporize 1lb of the solvent is taken as the ____ at the exposed surface temperature of the solutiona. latent heat of vaporizationb. specific heatc. heat transferd. specific heat of condensation

6. What is the usual solvent in an evaporation process?a. waterb. alcoholc. benzened. ether7. For feed of organic salts in water, the heat capacity may be assumed ___ to that of water alonea. greaterb. lesserc. equald. none of the above8. The effect of boiling point rise due to _____ is commonly neglected in evaporation calculationsa. heat transferb. temperature changec. hydrostatic headd. change in velocity9. It is the difference in temperature of the liquid at the heat source and the liquid at the top evaporating liquida. temperature changeb. heat transferc. boiling point rised. log mean temperature10. When two or more evaporators are connected, it is calleda. single effect evaporatorb. double effect evaporatorc. multiple effect evaporatord. multiple effect forward evaporator11. ___ 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 bodya. economyb. capacityc. percent recoveryd. efficiency12. ___ of the evaporator is the total amount of evaporation it is capable of producing per unit timea. economyb. capacityc. percent recoveryd. efficiency13. This can be determined by the measurement of the total heat transferred to the evaporating mixturea. economyb. capacityc. percent recoveryd. efficiency14. It is the most important single factor in evaporator design, since the heating surface represents the largest part of the evaporator cost.a. temperatureb. temperature differencec. heat transferd. pressure15. The greatest increase in steam economy is achieved by reusing the vaporized solvent, which is done in the___a. single effect evaporatorb. double effect evaporatorc. multiple effect evaporatord. multiple effect backward evaporator16. 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 evaporatorb. single effect evaporatorc. forced circulation evaporatord. crystallizing evaporator17. ___ which is the growth on body and heating surface walls of a material having a solubility that increases with increase in temperaturea. saltingb. scalingc. crystalsd. fouling18. It is the deposition and growth on body walls, and especially on heating surfaces, of a material undergoing an irreversible chemical reaction in the evaporatora. saltingb. scalingc. crystalsd. fouling19. It is the formation of deposits other than salt or scale and maybe due to corrosiona. saltingb. scalingc. crystalsd. foulingThis evaporator is suitable for the widest variety of evaporator applicationsa. multiple effect evaporatorb. Short-tube vertical evaporatorsc. forced circulation evaporatord. crystallizing evaporator20. This is one of the earliest types still in widespread commercial uses, and its principal use is in the cane industrya. multiple effect evaporatorb. Short-tube vertical evaporatorsc. forced circulation evaporatord. crystallizing evaporator21. When the ratio of feed to evaporation is low, it is desirable to provide fora. recirculation of productb. steam recyclingc. used of multiple effect evaporatord. high powered evaporator22. This is normally the cheapest per unit capacity type of evaporatora. long-tube evaporatorb. single effect evaporatorc. short-tube evaporatord. falling film long tube evaporator23. The principal problem in the falling film long tube evaporator is thea. feed distributionb. temperature controlc. high pressured. heat loss24. The falling film long tube evaporator is widely used for concentrating ___a. heat sensitive materialsb. highly soluble materialsc. high boiling point rised. corrosive materials25. 26. this usually results from the presence in the evaporating liquid of colloids or of surface-tension depressantsa. splashing lossb. entrainment lossesc. heat lossesd. foaming losses27. ___ are usually insignificant if a reasonable height has been provided between the liquid level and the top of the vapor heada. splashing lossesb. entrainment lossesc. heat lossesd. foaming losses28. These are frequently encountered in an evaporator if feed is above the boiling pointa. splashing lossesb. entrainment lossesc. heat lossesd. foaming losses29. These are frequently used to reduce product lossesa. knitted wire meshb. mechanical thermo compressorsc. steam jacketd. entrainment separators30. This employ reciprocating, rotary positive-displacement , centrifugal or axial flow directiona. knitted wire meshb. mechanical thermo compressorsc. steam jacketd. entrainment separators31. the requirement of low temperature for fruit-juice concentration has led to the development of an evaporator employing a ___a. steam economyb. multiple effectc. temperature distributiond. secondary fluid32. 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 feedb. multiple effectc. temperature distributiond. secondary fluid33. The ___ of a multiple effect evaporator will increase in proportion to the number of effectsa. steam economyb. multiple effectc. temperature distributiond. secondary fluid34. 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 indicateda. heat sensitivityb. corrosionc. fouling factord. scaling35. ____ usually range from a minimum of about 1.2 m/sa. horizontal tube velocityb. tube velocityc. tube rated. none of the above36. s solutions are heated and concentration of the solute increases, ___ limit of the material may be exceeded and crystals may form.a. b. crystallizationc. seedingd. solubilitye. frothing37. 38. The simplest form of evaporator consists of an open pan in which the liquid is boileda. open kettleb. vertical type natural circulationc. falling film type evaporatord. agitated film evaporator39. A variation of the long-tube evaporatora. forced circulation type evaporatorb. vertical type natural circulationc. falling film type evaporatord. agitated film evaporator40. This is done in a modified falling film evaporatora. forced circulation type evaporationb. vertical type natural circulationc. falling film type evaporationd. agitated film evaporation41. In this operation, the fresh feed is added to the first effect and flows to the next in the same direction as the vapor flowa. Feed-forwardb. Feed-backwardc. Parallel-feedd. None of the above42. Involves the adding of fresh feed and the withdrawal of concentrated product from each effecta. Feed-forwardb. Feed-backwardc. Parallel-feedd. None of the above 43. The fresh feed enters at the coldest effecta. Feed-forwardb. Feed-backwardc. Parallel-feedd. None of the above44. 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 useda. Enthalpy concentration chartb. Duhring chartc. Equilibrium chartd. Mollier chart45. When a multiple effect evaporator is at steady state operation, the flow rates and rate of evaporation in each effect are ___a. greaterb. lesserc. constantd. equal46. In commercial practice, the areas in all effects for a multiple effect evaporator area. greaterb. lesserc. constantd. equal 47. 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. increasedb. decreasedc. held constantd. neglected48. ___ of the vapor is recovered and reuse in a multiple effect evaporatora. heat transferb. latent heatc. specific heatd. sensible heat48. The increase in steam economy obtained by using multiple-effect evaporator is at the expense of ___.a. reduced capacityb. product qualityc. evaporators efficiency d. percent yield49. Evaporation is also sometimes called as___a. Thermal recompressionb. Liquid compressionc. Water distillationd. Vapor compression50. In a ____ system, the vapor is compressed by acting on it with high-pressure steam in a jet ejector.a. b. Thermal recompressionc. Liquid compressiond. Water distillatione. Vapor compression

Answer keyEvaporation (objective questions)1. b2. b3. d4. a5. a6. a7. c8. c9. c10. c11. a12. b13. b14. c15. c16. d17. a18. b19. d20. c21. b22. a23. a24. a25. a26. d27. a28. b29. d30. b31. d32. c33. a34. b35. b36. c37. a38. c39. d40. a41. c42. b43. b44. c45. d46. a47. b48. a49. c50. a

Unit Operations Economics

1. ____ is the money returned to the owners of capital for use of their capital.a. Interestb. Cashc. Loand. credit2. Any profit obtained through the uses of capital a. Interestb. Cashc. Loand. credit3. 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 timea. simple interestb. compound interest c. nominal interest d. interest on loan4. When an interest period of less than 1 year is involveda. simple interestb. ordinary interest c. nominal interest d .compound interest5. This interest stipulates that interest is due regularly at the end of each interest period. a. simple interestb. ordinary interest c. nominal interest d.compoundinterest6. In ______interest rate, other time units are employed butthe interest rate is still expressed on an annual basis. a. simple interestb. ordinary interest c. nominal interestd.compoundinterest7. 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.a. continuous Interestb. Cash flowc. Loand. credit8. 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.a. principalb. Cash flowc. capitald. credit9. An ____is a series of equal payments occurring at equal time intervals. a. principalb. interestc. capitald. annuity10. 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. principalb. interestc. capitald. annuity11. Common type of ____which involves payments, which occur at the end of each interest period a. Perpetuity b. Present worth of annuity c. capitald. annuity12. Defined as the principal which would have to be invested at the present time at compound interest rate to yield a total a. Perpetuity b. Present worth of annuity c. capitald. annuity13. An annuity in which the first payment is due after a definite number of years. a. Perpetuity b. Present worth of annuity c. deferred interestd. deferred annuity14. It is an annuity in which the periodic payments continue indifferently. a. Perpetuity b. Present worth of annuityc. deferred interestd.deferredannuity15. Defined as the original cost of the equipment plus the present value of the renewable perpetuity.a. Perpetuity b. Present worth of annuityc. deferred interest d.capitalizedcost16. An ____design could be based on conditions giving the least cost per unit of time or the maximum profit per unit of production.a. optimum economicb. plant c. simple economicd.plant economic17. 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. a. optimum b. plant design c. simple d.plant economic18. The ___has one distinct advantage over the analytical method.a. optimum b. graphicalc. simple d.economical19. 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. a. optimum b. break-even pointc. annuityd. balanced20. The total product cost per unit of time may be divided into the two classifications of operating costs and ____costs. a. optimum operating costsb. Organization costsc. annuity costsd. superproduction costs21. 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. a. operating costsb. Organization costsc. annuity costsd. superproduction costs22. ______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.a. operating costsb. Organization costsc. annuity costsd. superproduction costs23. Extra expenses due to increasing the rate of production are known as ____.a. operating costsb. Organization costsc. annuity costsd. superproduction costs24. In a true ___operation, no product is obtained until the unit is shut down for discharging.a. batchb. continuousc. semi-batchd. semi-continuous25. In _____cyclic operations, product is delivered continuously while the unit is in operation, but the rate of delivery decreases with time.a. batchb. continuousc. semi-batchd. semi-continuous26. _____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 operationa. batchb. continuousc. semi-batchd. semi-continuous27. It is a mathematical technique, for determining optimum conditions for allocation of resources and operating capabilities to attain a definite objective.a. economicsb. programmingc. linear programmingd. dynamic programmingUnit Operations Economics

28. The concept of this is based on converting an overall decision situation involving many variables into a series of simpler individual problems with each of these involving a small number of total variables.a. economicsb. programmingc. linear programmingd. dynamic programming29. When quality constraints or restrictions on certain variables exist in an optimization situation, a powerful analytical technique is the use of :a. Lagrange Multipliersb. Geometric Programmingc. steepest Ascent or Descentd. dynamic programming30. For the optimization situation in which two or more independent variables are involved, response surface can often be prepared to show the relationship among the variables one of the early methods proposed for establishing optimum conditions from response surfaces is known as the method:a. Lagrange Multipliersb. Geometric Programmingc. linear programmingd. steepest Ascent or Descent31. 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 costa. Lagrange Multipliersb. Geometric Programmingc. linear programmingd. steepest Ascent or Descent32. ____(CPM) and program evaluation and review technique) have received particular attention and have shown the desirability of applying mathematical and graphical analysis a. critical path methodb. critical planning methodc. careful planning and managementd. none of the above33. The word ____is used as a general term for the measure of the amount of profit that can be obtained from a given situation. a. profitabilityb. incomec. net incomed. sales34. It is the common denominator for all business activities before capital is invested to a project or enterprisea. profitabilityb. incomec. net incomed. sales35. ____on investments is ordinarily expressed on an annual percentage basis. The yearly profit divided the by the total initial investment necessary represents the fractional return:a. rate of returnb. interest incomec. profitabilityd. annuity36. ___is defined as the difference between income and expense. a. profitb. gross incomec. operating expensesd. sales37. This is a function of the quantity of goods and services produced and the selling price.a. salesb. gross incomec. operating expensesd. profit38. The method of approach for a profitability evaluation is by ____, which takes into account the time value of the money and is based on the amount of the investment that is unreturned at the end of each year during the estimated life of the project. a. marketing analysisb. gross incomec. discounted cash flowd. law of demand39. In the net present worth study, the procedure has involved the determination of an index or interest rate which discounts the annual ____to a zero present value when properly compared to the initial investment.a. marketing analysisb. gross incomec. cash flowd. selling price40. The ____concept is useful for comparing alternatives which exist as possible investment choices within a single overall project.a. marketing analysisb. sales inventory c. discounted cash flowd. capitalized cost profitability41. ____ is defined as the minimum length of time theoretically necessary to recover the original capital investment in the form of cash flow to the project based on total income minus all cost except depreciation.a. deadlineb. due datec. payout periodd. maturity date 42. The term ___ refers to a special type of alternative in which facilities are currently in existence and it may be desirable to replace these facilities with different ones.a. reservesb. replacementc. substituted. none of the above43. In the formula, P = F(1+I )-N , where I = interest rate per period and N the number of interest periods is known asa. sinking fund factorb. single payment present worth factorc. uniform seriesd. capital recovery factor44. Which of the following relationships between compound interest factors is not correct?a. Single payment compound amount factor and single payment present worth factor are reciprocalsb. Capital recovery factor and uniform series present worth are reciprocalsc. Capital recovery factor equals the sinking fund factor plus the interest rated. capital recovery factor and sinking fund factor are reciprocals45. An annuity is defined asa. Earned interest due at the end of each interest periodb. Cost of producing a product or rendering a servicec. Amount of interest earned by a unit principal in a unit of timed. A series of equal payments made at equal interval of time46. Which one of the following method is not a method of depreciating plant equipment for accounting and engineering economic analysis purposes?a. double entry methodb. fixed percentage methodc. sum-of-years digit methodd. sinking fund method47. ___ is a science which deals with the attainment of the maximum fulfillment of societys unlimited demands for goods and services.a. Political scienceb. Economicsc. Engineering Economyd. Social Science48. It is the branch of economics which deals with the application of economics laws and theories involving engineering and technical projects or equipments.a. Political scienceb. Economicsc. Engineering Economyd. Social Science49. This refers to the products or services that are directly used by people to satisfy their wants.a. Producer goods and servicesb. Consumer goods and servicesc. Luxury Productsd. Supply50. Those that are used to produce the goods and services for the consumer are calleda. Producer goods and servicesb. Consumer goods and servicesc. Machineries and Technologyd. Supply51. This refers to the satisfaction or pleasure derived from the consumer goods and services.a. Product Satisfaction b. Consumers contentmentc. Luxury Productsd. Utility52. These are products that have an income-elasticity of demand greater than one.a. Basic Commoditiesb. Consumer goods c. Luxury Productsd.Supply53. If the income elasticity of demand is greater than one, this means thata. as income decreases, more income will be spent on the basic commoditiesb. as income increases, more expenses will be made on the basic commodity productsc. as income increases, more income will be spent on luxury itemsd. as income decreases, less will be spent on the basic commodity products54. It is the amount of goods and products that are available for sale by the suppliersa. Producer goods and servicesb. Consumer goods and servicesc. Machineries and Technologyd. Supply55. The want or desire or need for a product using money to purchase it.a. Supplyb. Consumer goods and servicesc. Luxury Productsd. Demand56. The law of supply and demand, states thata. When free competition exists, the price of the product will be that value where supply is equal to the demandb. When perfect competition exists, the price of the product will be that value where supply is equal to the demandc. When free competition exists, the price of the product will be that value where supply is greater to the demandd. When perfect competition exists, the price of the product will be that value where supply is greater to the demand57. This is a form of market structure where the number of suppliers is used to determine the type of market.a. Market Structuresb. Competitionc. Free competition marketd. Perfect Competition58. This is a market situation wherein a given product is supplied by a very large number of vendors and there is no restriction of any additional vendor from entering the market.a. Market Structuresb. Competitionc. Free competition marketd. Perfect Competition59. A place where the vendors or the sellers and vendees or the buyers come together is calleda. Market Structuresb. Marketc. Free competition marketd. Perfect Competition market60. This is defined as the interest on a loan or principal that is based on the original amount of the loan or principal.a. Ordinary Simple Interestb. Simple Interestc. Exact Simple Interestd. Compound Interest61. This is based on one bankers year.a. Ordinary Simple Interestb. Simple Interestc. Exact Simple Interestd. Compound Interest62. __ is based on the exact number of days in a given year.a. Ordinary Simple Interestb. Simple Interestc. Exact Simple Interestd. Compound Interest63. One bankers year is equivalent toa. 365 daysb. 350 daysc. 12 months and 30 daysd. 12 months and 15 days64. This is defined as the interest of loan or principal which is based on the amount of the original loan and the previous accumulated interesta. Ordinary Simple Interestb. Simple Interestc. Exact Simple Interestd. Compound Interest

65. It is the amount of money or payment for the use of a borrowed money or capitala. Annuityb. Interestc. Simple Interestd. Simple Annuity66. This is defined as the basic annual rate of interesta. effective rate of annuityb. effective rate of interestc. nominal rate or annuityd. nominal rate of interest67. It is defined as the actual or exact rate of interest earned on a principal during1 year perioda. effective rate of annuityb. effective rate of interestc. nominal rate or annuityd. exact simple interest68. This refers to the difference between the future worth of a negotiable paper and its present worth.a. Depreciationb. Market valuec. Discountd. Fair value69. This refers to the sales of stock or share at reduced pricea. Depreciationb. Market valuec. Discountd. Fair value70. This is the deduction from the published price of services or goods.a. Depreciationb. Salvage valuec. Discountd. Fair value71. Annuity is simply defined asa. series of equal payments occurring at equal time interval of timeb. series of payments occurring at equal time intervalc. series of equal payments occurring at certain time intervald. series of divided payments at pay periods72. An annuity of a fixed time span is also called asa. Ordinary annuityb. Annuity Certainc. Annuity Dued. Deferred Annuity73. __ is the type of annuity where the payments are made at the beginning of each period starting from the first perioda. Ordinary annuityb. Annuity Certainc. Annuity Dued. Deferred Annuity74. An annuity where the payments are made at the end of each period beginning on the first perioda. Ordinary annuityb. Annuity Certainc. Annuity Dued. Deferred Annuity75. It is when the first payment does not begin until some later date in the cash flow

a. Ordinary annuityb. Perpetuityc. Annuity Dued. Deferred Annuity76. When an annuity does not have a fixed time span but continues indefinitely, then it is referred to asa. Ordinary annuityb. Perpetuityc. Annuity Dued. Deferred Annuity77. This refers to the rate of interest that is quoted in the bond.a. bondb. bond ratec. bond valued. interest rate of bond value78. It is the present worth of the future payments that will be receiveda. book valueb. bond valuec. fair valued. salvage value79. 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 sella. Book valueb. Market valuec. Use valued. Fair value80. The amount of the property which the owner believed to be its worth as an operating unit is calleda. valueb. Market valuec. Use valued. Fair value81. ___ is the amount obtained from the sale of the property.a. book valueb. bond valuec. market valued. salvage value82. It is the worth of the property determined by a disinterested person a. Book valueb. Market valuec. Use valued. Fair value83. 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 timea. Depreciationb. Depleting valuec. Discounted valued. Fair value84. __ is the money worth of an asset or producta. Priceb. Market valuec. valued. Fair value85. This refers to the present worth of all profits that are to be received through ownership of a particular propertya. Priceb. Market valuec. valued. capitalized cost86. Refers to the sum of its first cost and cost of perpetual maintenance of a propertya. Priceb. Market valuec. perpetual valued. capitalized cost87. __ is a long term note or a financial security issued by businesses or corporation and guaranteed on certain assets of the corporation or its subsidiaries.a. Contractb. Receiptc. bond valued. bond88. This value implies that the property will still be use for the purpose it is intendeda. salvage valueb. book valuec. use valued. depreciation89. A depreciation computation method which is also known as Diminishing Balance Methoda. Sinking Fund Methodb. Straight-line Methodc. Declining Balance methodd. Break-even analysis90. Also called as Constant- Percentage Method in depreciation methoda. Sinking Fund Methodb. Straight-line Methodc. Declining Balance methodd. Break-even analysis91. This is also known as Resale valuea. salvage valueb. book valuec. use valued. depreciation92. ___ is considered as the simplest type of business organization wherein the firm is controlled and owned by a single persona. sole-proprietorshipb. Partnershipc. Corporationd. Entrepreneurship93. __ is a firm owned and controlled by two or more persons who are bind to an agreementa. sole-proprietorshipb. Partnershipc. Corporationd. Entrepreneurship94. It is a firm owned by a group of ordinary shareholders and the capital of which is divided up to the number of shares. a. sole-proprietorshipb. Partnershipc. Corporationd. Entrepreneurship

95. It is defined as a distinct legal entity separate from the individuals who owns it and can engage in any business transaction which a real person could do.a. sole-proprietorshipb. Partnershipc. Corporationd. Entrepreneurship96. Also known as joint-stock company or a cooperativea. Incorporationb. Partnershipc. Corporationd. Entrepreneurship97. This refers to the situation where the sales generated are just enough to cover the fixed and variable cost.a. break-evenb. Break even chartc. break even pointd. none of the above98. The break even chart showsa. the relationship between income and fixed costb. the relationship between the volume and fixed cost, variable costs, and incomec. the relationship between the assets and fixed cost, variable costs, and incomed. the relationship between the liabilities and fixed cost, variable costs, and income99. The level of production where the total income is equal to the total expenses is known as a. break-evenb. Break even chartc. break even pointd. none of the above100. The sum of perpetuity is a/ana. book valueb. zero valuec. infinite valued. bond value

Board Problems:

1. Design based on conditions giving the least cost per unit time and maximum profit per unit productiona. battery limitb. break-even pointc. optimum economic designd. plant design2. A flow diagram, indicating the flow of materials, unit operations involved, equipment necessary and special information on operating temperature and pressure is aa. schematic diagramb. qualitative flow diagramc. process flow chartd. quantitative flow diagram3. In plant design implementation, soil testing is done to determinea. pHb. load bearing capacityc. porosityd, viscosityUnit Operations Economics

4. This includes all engineering aspects involved in the development of either a new, modified or expanded industrial plant is calleda. plant designb. optimum designc. process designd. engineering design5. A chemical engineering specializing in the economic aspects of design is calleda. plant engineerb. cost engineerc. design engineerd. process engineer6. This refers to the actual design of the equipment and facilities for carrying out the processa. process engineeringb. plant designc. process designd. optimum design7. 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 obtaineda. preliminary designb. quick estimate designc. firm process designd. detailed design estimate8. 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 calleda. pr5oject feasibility studyb. plant designc. project development and researchd. product development9. Discusses the nature of the product line, the technology necessary for production, its availability, the proper mix of production resources and the optimum production volumea. market feasibilityb. socio-economic feasibilityc. technical feasibilityd. management feasibility10. Discusses the nature of the unsatisfied demand which the project seeks to meet growth and the manner in which it is to be meta. financial feasibilityb. management feasibilityc. market feasibilityd. technical feasibility11. A multiple effect evaporator produces 10, 000 kg of salt from a 20% brine solution per day. One kg of steam evaporates 0.7 N kg of water in N effects at a cost of P25 per 1000 kg of steam. The cost of the first effect is P450, 000 and the additional effects at P300, 000 each. The life of the evaporator is 10 years with no salvage value. The annual average cost of repair and maintenance is 10% and taxes and insurance is 5%. The optimum number of effects for minimum annual cost is a. 3 effectb. 5 effectsc. 4 effectsd. 2 effects12. 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 equationF = 187.5 Lq 1.8 mc 0.20 in ft-lbs/lbm

d 0.20 Di 4.8

Cf = 1.44 Di 1.5 L , in p/yr

Where:L = length of straight pipe, ft.q = steam flow rate, cu. ft. per secmc = steam viscosity cp.d = steam density, lb per ft3Di = inside diameter of pipe, inchesThe optimum pipe diameter that should be used for transporting the above steam is a. 6 inb. 4 inc. 3 ind. 5 in 13. A smelting furnace operating at 2, 400oF is to be insulated on the outside to reduce heat losses and same on energy. The furnace wall consists of a 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 300oF. 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-ft2-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 ft2 of area, the optimum thickness of the outer insulation that should be installed isa. 8 in b. 12 inc. 10 ind. 6 in14. 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:

lbz = 1.5(1.1lbxlbz +1.3lbylbz lbxlby)0.5

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 isa. P0.3 per lbzb. P0.5 per lbzc. P0.12 per lbzd. P0.25 per lbz15. 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 R = (0.2/hr) Ca.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 isa. 45%b. 60%c. 50%d. 40%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:

A + B ------- R +S where k = 500 cuft/(lbmole-hr).The amount of R required is 95 lbmoles/hr. Data:* in extracting R from the reacted mixture, A and B are destroyed* 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* capital and operating cost for backmix reactor is P0.10 per(cuft-hr).The optimum backmix reactor size is:a. 23, 900 ft3b. 24, 200 ft3c. 25, 900 ft3 d. 23, 500 ft3

17. April 1992. A unit of welding machine cost P45, 000 with an estimated life of 5 years. Its salvage value id P2, 500. Find its depreciation rate by straight-line method.a. 17.75%b. 19.88%c. 18.89%d.15.56%18. April 1997.A machine has an initial cost of P50, 000 and a salvage value of P10, 000 after 10 years. Find the book value after 5 years of using straight-line depreciation.a. P12, 500b.P30, 000c. P16, 400d.P22, 30019. October 1992.The initial cost of a paint sand mill, including its installation, is P800, 000. The BIR approved life of this machine is 10 years for depreciation. The estimated salvage value of the mill is P50, 000 and the cost of dismantling is estimated is estimated to be P15, 000. Using straight-line depreciation, what is the annual depreciation charge and what is the book value of the machine at the end of six years.a. P74, 500 ; P340, 250b. P76, 500 ; P341, 000c. P76, 500 ; P 342, 500d. P77, 500 ; P343, 25020. November 1997.The cost of the equipment is P500, 000.00 and the cost of installation is P30, 000. If the salvage value is 10% of the cost of equipment at the end of five years, determine the book value at the end of the fourth year. Use straight-line method.a. P155, 000b.P140, 000c. P146, 000d. P132,600 21. April 1998. An equipment costs P10, 000 with a salvage value of P500 at the end of 10 years. Calculate the annual depreciation cost by sinking fund method at 4% interest.a. P791.26b.P950.00c. P971.12d. P845.3222. November 1995.A machine costing P720, 000 is estimated to have a book value of P40, 545.73 when retired at the end of 10 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %? a.28b. 25c. 16d.3023. November 1998.AMD Corporation makes it a policy that for any new equipment purchased; the annual depreciation cost should not exceed 20 % of the first cost at any time with no salvage value. Determine the length of service life is necessary if the depreciation used is the SYD method.a.9 yearsb. 10 yearsc. 12 yearsd.19 years24. November 1996.At 6%, find the capitalized cost of a bridged whose cost is P250M and the life is 20 years, if the bridge must be partially rebuilt at a cost of P100M at the end of each 20 years.a. P275.3Mb.P265.5Mc. P295.3Md. P282.1M25. October 1995A 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 companys rate return on the money is 10% . Determine the capital recovery per years.a. P5, 115b.P4, 946c. P5, 625d. P4, 80526. October 1998.The annual maintenance cost of a machine shop is P69, 994. if the cost of making a forging is P56 per unit and its selling price is P135 per forged unit, find the number of units to be forged to break-even.a.886 unitsb. 885 unitsc. 688 unitsd.668 units27. October 1990.Compute for the number of locks that an ice plant must be able to sell per month to break even based on the following data:Cost of electricityP20.00Tax to be paid per block 2.00Real Estate Tax 3, 500.00per monthSalaries and wages 25, 000.00 per monthOthers 12, 000.00 per monthSelling price of Ice 55.00 per blocka. 1228b. 1285c. 1373d.131228. April 1998XYX Corporation manufactures bookcases that sell for P65.00 each. It costs XYZ Corporation P35, 000 per year to operate its plant. This sum includes rent, depreciation charges on equipment, and salary payments. If the cost to produce one bookcase is P50.00, how many cases must be sold each year for XYZ to avoid taking a loss?a. 2334b. 539c. 750d.2333

Answer KeyBoard Problems

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Answer KeyObjective Problems

1. a2. a3. a4. b5. d6. c7. a8. a9. d10. d11. d12. b13. d14. a15. d16. a17. a18. a19. b20. b21. a22. b23. d24. a25. d26. d27. c28. a29. d30. b31. a32. a33. a34. a35. a36. d37. c38. c39. d40. c41. a42. a43. b44. d45. d46. a47. b48. c49. b50. a51.