CHEMICAL E 321 _ MASS TRANSFER TUTORIAL Gas Absorption, Humidification, Drying. Q&A Plus Worked Solu

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MASS TRANSFER TUTORIAL Gas Absorption, Humidification, Drying

HUMIDIFICATION

A stream of air at 100˚C and 5260 mmHg contains 10% water by volume. Calculate the dew point. a. 90˚C b. 72˚C c. 83˚C d. 65˚C

A stream of air at 100˚C and 5260 mmHg contains 10% water by volume. Calculate the percentage of the vapor that condenses if the air is cooled to 80˚C at constant pressure. a. 20% b. 30% c. 25% d. 35%

A stream of air at 100˚C and 5260 mmHg contains 10% water by volume. Calculate the percentage of the vapor that condenses if the air is compressed isothermally to 8500mmHg. a. 12% b. 18% c. 15% d. 13%

relative humidity of the water vapor in the air is 25.

If partial pressure of water vapor when air is saturated with vapor at 300K is 3.6kPa, calculate the partial pressure of the water vapor in the a. 0.2 vessels. b. 0.5 c. 0.7 d. 0.9

In a vessel at 101.325kPa and 300K, the percentage relative humidity of the water vapor in the air is 25.

If partial pressure of water vapor when air is saturated with vapor at 300K is 3.6kPa, calculate the a.absolute humidity of the air.

0.006 b. 0.003 c. 0.001 d. 0.008

In a vessel at 101.325kPa and 300K, the percentage relative humidity of the water vapor in the air is 25.

If partial pressure of water vapor when air is saturated with vapor at 300K is 3.6kPa, calculate the a.humid 0.352 volume. b. 0.109 c. 0.856 d. 0.264

In a vessel at 101.325kPa and 300K, the percentage relative humidity of the water vapor in the air is 25.

If partial pressure of water vapor when air is saturated with vapor at 300K is 3.6kPa, calculate the a.percentage humidity.

31% b. 46% c. 10% d. 24%

The absolute humidity of a carbon dioxide–water vapor mixture at 310K and 100kPa is measured to be 0.022. Calculate the molal humidity. a. 0.0267 b. 0.0336 c. 0.0537 d. 0.0011

The absolute humidity of a carbon dioxide–water vapor mixture at 310K and 100kPa is measured to be 0.022. Calculate the percent relative saturation. a. 82.2% b. 80.4% c. 81.5% d. 84.9%

The absolute humidity of a carbon dioxide–water vapor mixture at 310K and 100kPa is measured to be 0.022. Calculate the percent saturation. a. 82.4% b. 80.4% c. 81.2% d. 84.9%

The absolute humidity of a carbon dioxide–water vapor mixture at 310K and 100kPa is measured to be 0.022. Calculate the temperature to which the gas is to be heated at constant pressure to reduce its saturation percent to 30%. a. 337.5K b. 327.5K c. 345.5K d. 461.5K

A gas mixture contains 0.0083 mol of water vapor per mol of dry methane at a temperature of 27˚C and a total pressure of 200kPa. Calculate the % relative saturation of the mixture. a. 46% b. 33% c. 58% d. 60%

A gas mixture contains 0.0083 mol of water vapor per mol of dry methane at a temperature of 27˚C and a total pressure of 200kPa. Calculate the % saturation of the mixture. a. 65% b. 77% c. 46% d. 50%

A gas mixture contains 0.0083 mol of water vapor per mol of dry methane at a temperature of 27˚C and a total pressure of 200kPa. Calculate the temperature (˚C) to which the mixture must be heated at 200kPa in order for the relative saturation to be 0.2. a. 53 b. 64 c. 37 d. 42

In a process in which it is used as a solvent, benzene is evaporated into dry nitrogen. At 297K and 101.3kPa, the resulting mixture has a percentage relative humidity of 60. It is required to recover 80% of the benzene present by cooling to 283K and compressing to a suitable pressure. What should this pressure be? The vapor pressure of benzene is 12.2kPa at 297K and 6kPa at 283K. a. 391kPa b. 353kPa c. 345kPa d. 386kPa

30,000m of cool gas (measured at 289K and 101.3 kPasaturated with water vapor) is compressed to 340kPa pressure, cooled to 289K and the condensed water is drained off. Subsequently the pressure is reduced to 170kPa and the gas is distributed at this pressure and 289K. What is the percentage humidity after this treatment? a. 26.5%The vapor pressure of water at 289K is 1.8kPa. b. 49.73% c. 31.25% d. 54.91%

The air supply for a dryer has a dry-bulb temperature of 70˚F and a wet bulb temperature of 60˚F. It is heated to 200˚F by coils and blown into the dryer. In the dryer it cools along an adiabatic cooling line and leaves the dryer fully saturated. What is the dew point of the initial air?

a. 62 b. 35 c. 40 d. 54

The air supply for a dryer has a dry-bulb temperature of 70˚F and a wet bulb temperature of 60˚F. It is heated to 200˚F by coils and blown into the dryer. In the dryer it cools along an adiabatic cooling line and leaves the dryer fully saturated. What is its humidity?

a. 0.0066 b. 0.009 c. 0.008 d. 0.0052

The air supply for a dryer has a dry-bulb temperature of 70˚F and a wet bulb temperature of 60˚F. It is heated to 200˚F by coils and blown into the dryer. In the dryer it cools along an adiabatic cooling line and leaves the dryer fully saturated. What is its % relative humidity?

a. 33% b. 44% c. 55% d. 66%

The air supply for a dryer has a dry-bulb temperature of 70˚F and a wet bulb temperature of 60˚F. It is heated to 200˚F by coils and blown into the dryer. In the dryer it cools along an adiabatic cooling line and leaves the dryer fully saturated. How much heat is needed to heat 100ft3 to 200˚F? a. 234BTU b. 107BTU c. 168BTU d. 240BTU

The air supply for a dryer has a dry-bulb temperature of 70˚F and a wet bulb temperature of 60˚F. It is heated to 200˚F by coils and blown into the dryer. In the dryer it cools along an adiabatic cooling line and leaves the dryer fully saturated. How much water will be evaporated per 100ft3 of entering air? a. 0.262lb b. 0.333lb c. 0.185lb d. 0.497lb

The air supply for a dryer has a dry-bulb temperature of 70˚F and a wet bulb temperature of 60˚F. It is heated to 200˚F by coils and blown into the dryer. In the dryer it cools along an adiabatic cooling line and leaves the dryer fully saturated. At what temperature (˚F) does the air leave the dryer? a. 86 b. 93 c. 72 d. 61

Air having a dry bulb temperature of 37.8˚C and a wet bulb of 26.7˚C is to be dried by first cooling to 15.6˚C to condense water vapor and then heating to 23.9˚C. What is the initial humidity and percentage humidity? a. 0.0157, 40.9% b. 0.0157, 90.4% c. 0.0175, 40.9% d. 0.0175, 90.4%

Air having a dry bulb temperature of 37.8˚C and a wet bulb of 26.7˚C is to be dried by first cooling to 15.6˚C to condense water vapor and then heating to 23.9˚C. What is the final humidity and percentage humidity? a. 0.011, 59% b. 0.011, 95% c. 0.013, 59% d. 0.013, 95%

Which of the following processes may always be regarded as a constant-humidity process? a. Cooling of air b. Air conditioning c. Water-cooling in a cooling tower d. Heating of air

Wet bulb and dry bulb temperatures become identical at a. 100% saturation curve b. 78% saturation curve c. 50% saturation curve d. None of these

Steady state temperature reached by a small amount of liquid evaporating into a large amount of unsaturated vapor-gas mixture is a. Dry bulb temperature b. Wet bulb temperature c. Dew point d. Adiabatic saturation temperature

In the semiconductor plant at the Food Terminal Export Zone, an adiabatic dryer is used where air enters at 160°F. If the air has a dew point of 68°F and it picked up 0.08 lb water per 100 ft3, how saturated a. 50% is the air coming out of the dryer? b. 70% c. 60% d. 80%

It is desired to condition saturated atmospheric air at 70°F with entrained 0.0008 lb water per cu ft air to hot air at 200°F dry bulb and 115°F wet bulb temperatures. The air is passed thru a heater, then thru an adiabatic humidifier, then thru a reheater. The air, as it leaves the adiabatic humidifier, has a humidity of 90%. Calculate the temperature of the air (˚F) as it leaves the adiabatic humidifier. a. 107 b. 115 c. 137 d. 124

It is desired to condition saturated atmospheric air at 70°F with entrained 0.0008 lb water per cu ft air to hot air at 200°F dry bulb and 115°F wet bulb temperatures. The air is passed thru a heater, then thru an adiabatic humidifier, then thru a reheater. The air, as it leaves the adiabatic humidifier, has a humidity of 90%. Calculate the temperature of the air (˚F) as it leaves the heater. a. 228 b. 193 c. 239 d. 215

1000 cfm of air (A) at 95˚F dry bulb, 74˚F wet bulb is mixed with 2000 cfm of air (B) at 65˚F dry bulb, 54˚F wet bulb. Determine the dry bulb temperature (˚F) for the mixed stream. a. 83 b. 90 c. 75 d. 85

1000 cfm of air (A) at 95˚F dry bulb, 74˚F wet bulb is mixed with 2000 cfm of air (B) at 65˚F dry bulb, 54˚F wet bulb. Determine the wet bulb temperature (˚F) for the mixed stream. a. 61 b. 55 c. 50 d. 68

If 100 lb of dry air at 50˚F db and 50˚F wb is mixed with 300 lb of dry air at 110˚F db and 80˚F wb, the resulting mixture will have a dry bulb (˚F) of a. temperature 80 b. 90 c. 85 d. 95

If 100 lb of dry air at 50˚F db and 50˚F wb is mixed with 300 lb of dry air at 110˚F db and 80˚F wb, the resulting mixture will have a wet bulb (˚F) of a. temperature 85 b. 73 c. 65 d. 70

The degrees of freedom associated with a humidification process is a. 1 b. 2 c. 3 d. 4

Which of the following remains during sensible cooling process? a. Humidity b. Partial pressure of vapor c. Both a and b d. Neither a nor b

constant

During sensible cooling process, a. Relative humidity decreases b. Wet bulb temperature increases c. Relative humidity increases d. Both b and c

Air in an amount 1000 cfm at 150˚F, 20% RH is passed over a refrigerated coil and thereby brought to 60˚F, 90% RH with the condensed moisture withdrawn at 55˚F. The air is then reheated by means of an electric heating coil to 150˚F. Compute the moisture removed in lb/min. a. 1.45 b. 2.08 c. 3.67 d. 4.19

Air in an amount 1000 cfm at 150˚F, 20% RH is passed over a refrigerated coil and thereby brought to 60˚F, 90% RH with the condensed moisture withdrawn at 55˚F. The air is then reheated by means of an electric heating coil to 150˚F. Compute heat removed by the refrigerated coil, expressed as tons of refrigeration. a. 11 b. 15 c. 17 d. 13

Air in an amount 1000 cfm at 150˚F, 20% RH is passed over a refrigerated coil and thereby brought to 60˚F, 90% RH with the condensed moisture withdrawn at 55˚F. The air is then reheated by means of an electric heating coil to 150˚F. Calculate the wattage of the heating coil required. a. 38kW b. 16kW c. 24kW d. 44kW

DRYING

A batch of wet solid was dried on a tray dryer using constant drying conditions and a thickness of material on the tray of 25.4mm. Only the top surface was exposed. The drying rate during the constant rate period was R=2.05kg/h-m 2. The ratio of mass of solid per area used was 24.4kg/m2 exposed surface. 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=0.30 using the same drying conditions but a thickness of 50.8mm with drying from the top and bottom surface.

a. 0.298hr b. 0.829hr c. 0.982hr d. 2.089hr

The portion of water in the wet solid that cannot be removed by the air in its contact is called a. Free moisture content b. Final moisture content c. Equilibrium moisture content d. Critical moisture content

Moisture contained by a substance in excess of the equilibrium moisture is called a. Unbound moisture b. Critical moisture c. Free moisture d. Bound moisture

The critical moisture content in case of drying indicates a. Beginning of falling rate period b. Beginning of constant rate period c. End of falling rate period d. None of these

In drying at the falling rate period, the rate at which moisture is removed from the solid material a. Increases with time b. Remains constant c. Decreases with time d. No more drying takes place

Heat sensitive materials like certain pharmaceuticals and food stuff can be dried a. In indirect tray dryer b. By freeze drying c. In spray dryer d. Festoon dryer

conditions from 25% db to 10% db in 4.17 hours. The critical moisture content is 15% db and its equilibrium moisture content is 5% db. How long will it take to dry the solid from 30% db to 8% db under the same drying a. 6.7hr conditions? b. 2.7hr c. 4.4hr d. 1.4hr

Wet solid is dried in a batch dryer under constant drying conditions from 35% to 10% in 7 hours. The critical moisture content is 20% and the equilibrium moisture is 4%. All moisture contents are on dry basis. How long should it take to dry the same solid from 10% to 5% moisture under the same drying conditions. a. 6.5hr b. 3.3hr c. 5.1hr d. 4.8hr

constant drying conditions. The critical moisture is 20% and the equilibrium moisture content is 7%. All moisture contents are on a dry basis. The time needed to dry from 15% to 5% free moisture under the same drying a. 4.28hris conditions b. 2.67hr c. 1.85hr d. 2.03hr

A porous material dries in the open air at a rate approximately proportional to its moisture content.

If a sheet hangs in the wind losses half its free moisture in the first hour, when will it have lost 99%, conditions remaining the same? a. weather 6.6hr b. 7.5hr c. 4.7hr d. 8.1hr

The initial moisture content of a food product is 77% (wet basis), and the critical moisture content is 30% (wet basis). If the constant drying rate is 0.1/m2-s, compute the time required for the product to begin the falling rate drying period. Neglect equilibrium moisture content. The product has a cube shape with 5cm sides, 3 and the initial product density is 950kg/m . a. 23.4s b. 48.5s c. 34.6s d. 53.2s

Ipil-ipil leaves will be dried in a moving train of trays dryers. The wet leaves containing 75% water (wet basis) is to be dried to 15% (wet basis) in tray containing 1mx1.5m. The average specific gravity of the leaves is 0.75. The wet leaves are spread out in the tray to a uniform thickness of 8cm. Calculate the number of trays needed to produce 1MT of the dried leaves. a. 20 b. 38 c. 45 d. 17

Ipil-ipil leaves will be dried in a moving train of trays dryers. The wet leaves containing 75% water (wet basis) is to be dried to 15% (wet basis) in tray containing 1mx1.5m. The average specific gravity of the leaves is 0.75. The wet leaves are spread out in the tray to a uniform thickness of 8cm. Calculate the amount of water removed per MT of product. a. 2400kg b. 5800kg c. 3300kg d. 4700kg

An adiabatic tunnel dryer handles 100kg of banana chips per batch reducing the moisture content from 50% and 12% wet basis.

Drying takes 10hrs to complete by flowing air at 82˚C and 5% RH. The air leaves the dryer saturated. Assuming that the rate of drying is constant, calculate the volume of air blown in m3/hr. a. 223 b. 267 c. 250 d. 208

Detergent solution is dried to a powder in a a. Spray dryer b. Tunnel dryer c. Spouted bed dryer d. Pan open to atmosphere

Calcium ammonium nitrate (a fertilizer) is dried in a a. Rotary dryer b. Tunnel dryer c. Vacuum dryer d. None of these

Rotary dryer cannot handle a. Free-flowing materials b. Sticky materials c. Dry materials d. Granular materials

A wet solid is to be dried in a tray dryer under steady-state conditions from a free moisture content of X1=0.4 to X2=0.02. The dry solid weight is 99.8kg dry solid and the top surface area for drying is 4.645m2. The drying rate curve can be represented by the figure below. Calculate the time for drying using graphical a. 6.67hr integration in the falling-rate period. b. 7.78hr c. 8.89hr d. 9.88hr

A wet solid is to be dried in a tray dryer under steady-state conditions from a free moisture content of X1=0.40 to X2=0.02.

The dry solid weight is 99.8kg dry solid and the top surface area for drying is 4.645m2. The drying rate curve can be represented by the figure below. Calculate the time for drying using a straight line a. 2.953hr through the origin for the drying rate in the falling rate period. b. 3.952hr c. 5.239hr d. 9.235hr

GAS ABSORPTION

In a gas absorption tower, mass transfer occurs primarily a. From the liquid phase to the gas phase b. From the liquid phase to the packing c. From the gas phase to the liquid phase d. From the gas phase to the packing

Separation of a mixture of two gases by absorption in the liquid solvent depends upon the difference in their a. Viscosity b. Solubility c. Density d. Relative volatility

In gas absorption, one solute is being absorbed while the rest of the components are assumed to be ______ and the liquid is ________. a. Inert, volatile b. Non-soluble, non-volatile c. Non-soluble, volatile d. Soluble, volatile

2

1,000kmol/h of flue gases containing 50% by volume CO2 and the balance being inert gases, is fed into the column. The lean gas leaves the absorber with a 2% by volume CO2. Calculate the quantity of CO2, in kg/h, that is recovered from the regeneration of a. liquor.

15900 the fat b. 27700 c. 18500 d. 21600

In general, operating pressure should be ___ and temperature ___for an absorber, to minimize stage requirements and/or absorbent flow rate to lower the equipment volume required to accommodate the a.gas flow High, low b. Maximum, low c. Low, high d. Minimum, high

A gas mixture containing 1% acetone and 99% dry air by volume is contacted with pure water in a plate gas absorber in which 90% of the acetone is recovered. The inlet gas flow rate is 30kmol/hr and the inlet water flow rate is 90kmol/hr. The absorber operates at 27˚C and 1atm. The equilibrium curve is given by y*=2.53x. What is thea.concentration of the strong liquor?

0.003 b. 0.002 c. 0.001 d. 0.004

A gas mixture containing 1% acetone and 99% dry air by volume is contacted with pure water in a plate gas absorber in which 90% of the acetone is recovered. The inlet gas flow rate is 30kmol/hr and the inlet water flow rate is 90kmol/hr. The absorber operates at 27˚C and 1atm. The equilibrium curve is given by y*=2.53x. What is thea.number of theoretical plates required?

5.1 b. 6.2 c. 7 d. 7.9

plate column at 1 atm. The mixture enters the bottom of the absorber containing 2% acetone by volume. Pure water is used to remove 95% of the acetone. If the actual liquid flow rate is 1.2 times the minimum required and if the equilibrium relationship is represented by y*=2.53x, what is the mole fraction of acetone in a. 0.0015 the strong liquor? b. 0.0066 c. 0.0025 d. 0.0038

plate column at 1 atm. The mixture enters the bottom of the absorber containing 2% acetone by volume. Pure water is used to remove 95% of the acetone. If the actual liquid flow rate is 1.2 times the minimum required and if the equilibrium relationship is represented by y*=2.53x, what is the theoretical number of a. 9.16 stages? b. 12.75 c. 23.81 d. 28.43

Air from a solvent plant contains 2% n-pentane. The pentane concentration is to be reduced to 0.1% by contacting the gas, at 80˚F and 147psi with 6950lb/hr of an absorption oil that contains 0.05% pentane by weight; 100000ft3 of the gas, measures at 60˚F and 14.7psia, is handled per hour. The molecular weight of the oil is 220. The vapor pressure of pentane at 80˚F is 523mmHg. Raoult’s law and Dalton’s law are assumed to be valid at these conditions. a. 3.6Calculate the number of perfect plates required.

b. 7.8 c. 4.9 d. 5.1

Air from a solvent plant contains 2% n-pentane. The pentane concentration is to be reduced to 0.1% by contacting the gas, at 80˚F and 147psi with 6950lb/hr of an absorption oil that contains 0.05% pentane by weight; 100000ft3 of the gas, measures at 60˚F and 14.7psia, is handled per hour. The molecular weight of the oil is 220. The vapor pressure of pentane at 80˚F is 523mmHg. Raoult’s law and Dalton’s law are assumed to be valid at these conditions. a. 3.6Calculate the number of perfect plates required.

b. 7.8 c. 4.9 d. 5.1

Propane is to be stripped from non-volatile oil by steam in a countercurrent tower. 4mol of steam will be supplied at the bottom of the tower for every 100mol of oil-propane feed at the top. The oil originally contains 2.5mol% propane and this concentration must be reduced to 0.25mol%. The tower is maintained at 280˚F and 35psia. The equilibrium relationship is y=33.4x, where x and y are mole fraction units. How many equilibrium stages are required? a. 1

b. 3 c. 4 d. 2

Propane is to be stripped from non-volatile oil by steam in a countercurrent tower. 4mol of steam will be supplied at the bottom of the tower for every 100mol of oil-propane feed at the top. The oil originally contains 2.5mol% propane and this concentration must be reduced to 0.25mol%. The tower is maintained at 280˚F and 35psia. The equilibrium relationship is y=33.4x, where x and y are mole fraction units. If the pressure is increased to 70psi, how many a. 6 equilibrium stages are required?

b. 8 c. 4 d. 3

A certain manufacturing company produces 24 MT/day of exhaust gas containing 3.5% n-pentane. The pentane concentration is to be reduced to 5% of the original amount by contacting the gas with 24 MT/day of absorbing oil at 80˚F and 101.325kPa. Molecular weight of oil is 180. The vapor pressure of n-pentane at 80˚F is 53mmHg. Assume 24 hours continuous operation per day. What is the concentration of the exit liquid?

a. 2.23% b. 2.32% c. 3.22%

d. 22.3%

A certain manufacturing company produces 24 MT/day of exhaust gas containing 3.5% n-pentane. The pentane concentration is to be reduced to 5% of the original amount by contacting the gas with 24 MT/day of absorbing oil at 80˚F and 101.325kPa. Molecular weight of oil is 180. The vapor pressure of n-pentane at 80˚F is 53mmHg. Assume 24 hours continuous operation per day. What is the number of theoretical stages?

a. 6 b. 5 c. 4 d. 3

A certain manufacturing company produces 24 MT/day of exhaust gas containing 3.5% n-pentane. The pentane concentration is to be reduced to 5% of the original amount by contacting the gas with 24 MT/day of absorbing oil at 80˚F and 101.325kPa. Molecular weight of oil is 180. The vapor pressure of n-pentane at 80˚F is 53mmHg. Assume 24 hours continuous operation per day. What is the minimum solute free liquid flow rate (kmol/h)?

a. 0.18 b. 0.81 c. 1.08 d. 8.01

What could be the effect of lowering the temperature of stripping? a. Number of stages increases b. Number of stages decreases c. Either A or B d. Cannot be determined

The reverse is true for stripping. However, the operating pressure should not be too high and the operating temperature should not be too lowa.asVaporize to ___ the feed gas. b. Compress c. Condense d. None of these

A mixture of benzene vapor and flue gas contains 12.7 mole % benzene and is to be scrubbed continuously in a packed tower operated at atmospheric pressure at 430C. The tower is to be designed to treat 36000 ft3 per hour of entering gas and the exit gas is to contain 1.5 mole % benzene. Pure solvent will be supplied at the rate of 28 lbmol/hr. The solution of benzene in oil may be assumed to follow Raoult’s law and the vapor pressure of pure solvent at 430C is 0.263 atm. It is agreed that the maximum allowable superficial velocity of the gas is 1 ft/sec. Calculate the height of the tower required, assuming that the height a. of a6.5f transfer unit is 2 ft.

b. 7f c. 5.8f d. 9.3f

A packed tower with an inside volume of 300 ft is to be used for an absorption under such conditions that KGa=2.8lbmol/hr-atm-ft3 and Henry’s law is applicable to the ammonia solution. The pressure difference driving force at the top of the column (P–P*) is 0.009atm, and the pressure difference driving force at the bottom of the column is 0.09atm. If 550lb of ammonia, in addition to the inert gases, enters the tower per hour, what percent of the entering ammonia will be absorbed? a. 85%

b. 72% c. 54% d. 91%

reduced to 0.5mol% in a packed absorption tower at 293K and 1.013x105Pa. The inlet pure water flow is 68.0 kmol/h and the total inlet gas flow is 57.8kmol/h. The tower diameter is 0.747m. The film mass-transfer coefficients are k’ya=0.0739kmol/s-m3-mol fraction and k’xa=0.169 kmol/s-m3-mol fraction. Calculate the a. of 0.18 height the tower. b. 0.81 c. 1.08 d. 8.01

reduced to 0.5mol% in a packed absorption tower at 293K and 1.013x105Pa. The inlet pure water flow is 68.0 kmol/h and the total inlet gas flow is 57.8kmol/h. The tower diameter is 0.747m. The film mass-transfer coefficients are k’ya=0.0739kmol/s-m3-mol fraction and k’xa=0.169 kmol/s-m3-mol fraction. Calculate the a. of 0.18 height the tower. b. 0.81 c. 1.08 d. 8.01

Flooding results in a. High tray efficiency b. Low tray efficiency c. High gas velocity d. Good contact between the fluids

Operating velocity in a packed tower is usually a. Half the flooding velocity b. Twice the flooding velocity c. Equal to the flooding velocity d. More than the flooding velocity

Flooding in a column results due to a. High pressure drop b. Low pressure drop c. Low velocity of the liquid d. High temperature

Channeling in a packed tower results from a. High pressure drop b. Non-uniformity of packing c. Maldistribution of liquid d. Both b and c

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