CHECAL PT.2 Flashcards
For the preparation of potassium nitrate, 10000 kg/h of a 20% KNO3 solution is mixed with a recycle stream and sent to an evaporator. The rate of evaporation is 1.5 times the rate of introduction of recycle stream. The concentrated solution leaving the evaporator contains 50% KNO3. This is admitted to the crystallizer which yields crystals containing 5% water. At the crystallization temperature the solubility is 50 kg/100 kg of water. The major part of the mother liquor leaving the crystallizer is returned to the crystallizer as recycle. Calculate the concentration of KNO3 in the stream entering the evaporator
25%
For the preparation of potassium nitrate, 10000 kg/h of a 20% KNO3 solution is mixed with a recycle stream and sent to an evaporator. The rate of evaporation is 1.5 times the rate of introduction of recycle stream. The concentrated solution leaving the evaporator contains 50% KNO3. This is admitted to the crystallizer which yields crystals containing 5% water. At the crystallization temperature the solubility is 50 kg/100 kg of water. The major part of the mother liquor leaving the crystallizer is returned to the crystallizer as recycle. Calculate the flow rate of recycle stream in kg/h
5143
For the preparation of potassium nitrate, 10000 kg/h of a 20% KNO3 solution is mixed with a recycle stream and sent to an evaporator. The rate of evaporation is 1.5 times the rate of introduction of recycle stream. The concentrated solution leaving the evaporator contains 50% KNO3. This is admitted to the crystallizer which yields crystals containing 5% water. At the crystallization temperature the solubility is 50 kg/100 kg of water. The major part of the mother liquor leaving the crystallizer is returned to the crystallizer as recycle. Calculate the rate of production of crystals
2007
Fresh air containing 4 mol% water vapor is to be cooled and dehumidified to a water content of 1.70 mol% H2O. A stream of fresh air is combined with a recycle stream of previously dehumidified air and passed through the cooler. The blended stream entering the unit contains 2.3 mol% H2O. In the air conditioner, some of the water in the feed stream is condensed and removed as liquid. A fraction of the dehumidified air leaving the cooler is recycled and the remainder is delivered to a room. Taking 100 mol of dehumidified air delivered to the room as a basis of calculation, calculate the moles of fresh feed
102 mol
Fresh air containing 4 mol% water vapor is to be cooled and dehumidified to a water content of 1.70 mol% H2O. A stream of fresh air is combined with a recycle stream of previously dehumidified air and passed through the cooler. The blended stream entering the unit contains 2.3 mol% H2O. In the air conditioner, some of the water in the feed stream is condensed and removed as liquid. A fraction of the dehumidified air leaving the cooler is recycled and the remainder is delivered to a room. Taking 100 mol of dehumidified air delivered to the room as a basis of calculation, calculate the moles of water condensed
2.4 mol
Fresh air containing 4 mol% water vapor is to be cooled and dehumidified to a water content of 1.70 mol% H2O. A stream of fresh air is combined with a recycle stream of previously dehumidified air and passed through the cooler. The blended stream entering the unit contains 2.3 mol% H2O. In the air conditioner, some of the water in the feed stream is condensed and removed as liquid. A fraction of the dehumidified air leaving the cooler is recycled and the remainder is delivered to a room. Taking 100 mol of dehumidified air delivered to the room as a basis of calculation, calculate the moles of dehumidified air recycled
290
An equimolar liquid mixture of benzene and toluene is separated
into two product streams by distillation. A process flowchart and
a somewhat oversimplified description of what happens in the
process follow:
Inside the column a liquid stream flows downward and a vapor
stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving
the top of the column, which contains 97 mole% benzene, is
completely condensed and split into two equal fractions: one is
taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains 89.2% of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which 45% of it is vaporized. The vapor generated in the reboiler (the boil-up) is recycled to become the rising vapor
stream in the column, and the residual reboiler liquid is taken off
as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation
yB (1-yB)
xB (1-xB) = 2.25
where and are the mole fractions of benzene in the vapor and liquid streams, respectively. Take a basis of 100 mol fed to the column, determine the molar amounts of overhead and bottom products
46 and 54 mol
An equimolar liquid mixture of benzene and toluene is separated
into two product streams by distillation. A process flowchart and
a somewhat oversimplified description of what happens in the
process follow:
Inside the column a liquid stream flows downward and a vapor
stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving
the top of the column, which contains 97 mole% benzene, is
completely condensed and split into two equal fractions: one is
taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains 89.2% of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which 45% of it is vaporized. The vapor generated in the reboiler (the boil-up) is recycled to become the rising vapor
stream in the column, and the residual reboiler liquid is taken off
as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation
yB (1-yB)
xB (1-xB) = 2.25
where and are the mole fractions of benzene in the mole fraction of benzene in the bottom
0.1
An equimolar liquid mixture of benzene and toluene is separated
into two product streams by distillation. A process flowchart and
a somewhat oversimplified description of what happens in the
process follow:
Inside the column a liquid stream flows downward and a vapor
stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving
the top of the column, which contains 97 mole% benzene, is
completely condensed and split into two equal fractions: one is
taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains 89.2% of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which 45% of it is vaporized. The vapor generated in the reboiler (the boil-up) is recycled to become the rising vapor
stream in the column, and the residual reboiler liquid is taken off
as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation
yB (1-yB)
xB (1-xB) = 2.25
where and are the mole fractions of benzene in the percentage recovery of toluene in the bottoms product
97%
Fresh orange juice contains 12.0 wt% solids and the balance water, and the concentrated orange juice contains 42.0 wt% solids. Initially a single evaporation process was used for the concentration, but volatile constituents of the juice escaped with the water, leaving the concentrate with a flat taste. The current process overcomes this problem by bypassing the evaporator with a fraction of the fresh juice. The juice is concentrated to 58 wt% solids, and the evaporator product stream is mixed with the bypassed fresh juice to achieve the desired final concentration. Calculate the amount of product (42% concentrate) produced per 100 kg fresh juice fed to the process.
28.6
Fresh orange juice contains 12.0 wt% solids and the balance water, and the concentrated orange juice contains 42.0 wt% solids. Initially a single evaporation process was used for the concentration, but volatile constituents of the juice escaped with the water, leaving the concentrate with a flat taste. The current process overcomes this problem by bypassing the evaporator with a fraction of the fresh juice. The juice is concentrated to 58 wt% solids, and the evaporator product stream is mixed with the bypassed fresh juice to achieve the desired final concentration. Calculate the fraction of the feed that bypasses the evaporator
0.099
A limestone analysis: CaCO3 – 92.89%, MgCO3 – 5.41% and unreactive – 1.70%. By heating the limestone, you recover oxides that together known as lime: How many pounds of calcium oxide can be made from 1 ton of limestone?
1041 lb
A limestone analysis: CaCO3 – 92.89%, MgCO3 – 5.41% and unreactive – 1.70%. By heating the limestone, you recover oxides that together known as lime: How many pounds of CO2 can be recovered per lb of limestone?
0.44
A limestone analysis: CaCO3 – 92.89%, MgCO3 – 5.41% and unreactive – 1.70%. By heating the limestone, you recover oxides that together known as lime: How many pounds of limestone are needed to make 1 ton of
lime?
3550
The synthesis of ammonia proceeds according to the following reaction:
N2 + 3 H2 à 2 NH3
In a given plant, 4202 lb of nitrogen and 1046 lb of hydrogen are fed to the synthesis reactor per hour. Production of pure ammonia from this reactor is 3060 lb per hour. What is the limiting reactant?
N2
