Predicting Extraction for an Existing Tower with a Given Number of Steps. An existing tower contains 5.0 theoretical steps. It is desired to predict its performance under the following conditions using the system water–acetone–trichloroethane given in Example 27.4-4: An inlet water stream of 1000 kg/h containing 8.0 wt % acetone is extracted with 750 kg/h trichloroethane containing 1.0 wt % acetone in this countercurrent tower at 25°C. Determine the outlet concentration x1 in the water solution and in the trichloroethane solution using an analytical equation. Also, plot the equilibrium and operating lines, and step off 5.0 steps. Assume dilute solutions.

# Daily Archives: January 17, 2021

## Calculate the minimum solvent that can be used. [Hint: In this case, the tie line through the feed L0 represents the condition for minimum solvent flow rate.

Minimum Solvent and Countercurrent Extraction of Acetone. An aqueous feed solution of 1000 kg/h containing 23.5 wt % acetone and 76.5 wt % water is being extracted in a countercurrent multistage extraction system using pure methylisobutyl ketone solvent at 298–299 K. The outlet water raffinate will contain 2.5 wt % acetone. Use equilibrium data from Appendix A.3.

a. Calculate the minimum solvent that can be used. [Hint: In this case, the tie line through the feed L0 represents the condition for minimum solvent flow rate. This gives V1min. Then, draw lines LNV1min and L0VN+1 to give the mixture point Mmin and the coordinate xAMmin. Using Eq. (27.4-4), solve for VN+1min, the minimum value of the solvent flow rate VN+1.]

b. Using a solvent flow rate of 1.5 times the minimum, calculate the….

## Determine the break-point time, the fraction of total capacity used up to the break point, the length of the unused bed, and the saturation loading capacity of the solid.

Drying of Nitrogen and Scale-Up of a Column. Using molecular sieves, water vapor was removed from nitrogen gas in a packed bed (C1) at 28.3°C. The column height was 0.268 m, with the bulk density of the solid bed being equal to 712.8 kg/m3. The initial water concentration in the solid was 0.01 kg water/kg solid and the mass velocity of the nitrogen gas was 4052 kg/m2 · h. The initial water concentration in the gas was co = 926 × 10 –6 kg water/kg nitrogen. The breakthrough data are as follows:

A value of c/co = 0.02 is desired at the break point. Do as follows:

a. Determine the break-point time, the fraction of total capacity used up to the break point, the length of the unused bed, and the….

## What is the required surface area in m2 and the steam consumption?

Surface Area and Steam Consumption of an Evaporator. A single-effect evaporator is concentrating a feed solution of organic colloids from 5 to 50 wt %. The solution has a negligible boilingpoint elevation. The heat capacity of the feed is cp = 4.06 kJ/kg · K (0.97 btu/lbm · °F) and the feed enters at 15.6°C (60°F). Saturated steam at 101.32 kPa is available for heating, and the pressure in the vapor space of the evaporator is 15.3 kPa. A total of 4536 kg/h (10 000 lbm/h) of water is to be evaporated. The overall heat-transfer coefficient is 1988 W/m2 · K (350 btu/h · °F). What is the required surface area in m2 and the steam consumption?

## What is the required surface area in m2 and the steam consumption?

Surface Area and Steam Consumption of an Evaporator. A single-effect evaporator is concentrating a feed solution of organic colloids from 5 to 50 wt %. The solution has a negligible boilingpoint elevation. The heat capacity of the feed is cp = 4.06 kJ/kg · K (0.97 btu/lbm · °F) and the feed enters at 15.6°C (60°F). Saturated steam at 101.32 kPa is available for heating, and the pressure in the vapor space of the evaporator is 15.3 kPa. A total of 4536 kg/h (10 000 lbm/h) of water is to be evaporated. The overall heat-transfer coefficient is 1988 W/m2 · K (350 btu/h · °F). What is the required surface area in m2 and the steam consumption?

## what velocity of water flow is needed and what is the size range of the pure product?

Separation by Settling. A mixture of galena and silica particles has a size range of 0.075–0.65 mm and is to be separated by a rising stream of water at 293.2 K. Use specific gravities from Example 30.1-3.

a. To obtain an uncontaminated product of galena, what velocity of water flow is needed and what is the size range of the pure product?

b. If another liquid, such as benzene, having a specific gravity of 0.85 and a viscosity of 6.50 × 10 –4 Pa · s is used, what velocity is needed and what is the size range of the pure product?

## Calculate the amounts and compositions of the overflow V1 and the underflow L1 leaving the stage

1. Leaching of Oil from Soybeans in a Single Stage. Repeat Example 31.2-1 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.

2. Leaching a Soybean Slurry in a Single Stage. A slurry of flaked soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt % oil and 90 wt % solvent hexane. This slurry is contacted with 100 kg of pure hexane in a single stage so that the value of N for the outlet underflow is 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow V1 and the….

## Calculate the amounts and compositions of the exit streams and the total number of theoretical stages.

Countercurrent Multistage Leaching of Halibut Livers. Fresh halibut livers containing 25.7 wt % oil are to be extracted with pure ethyl ether to remove 95% of the oil in a countercurrent multistage leaching process. The feed rate is 1000 kg of fresh livers per hour. The final exit overflow solution is to contain 70 wt % oil. The retention of solution by the inert solids (oil-free liver) of the liver varies as follows (C1), where N is kg inert solid/kg solution retained and yA is kg oil/kg solution:

Calculate the amounts and compositions of the exit streams and the total number of theoretical stages.

## Calculate the surface area of each effect if each effect has the same area, and the steam rate.

Evaporation of a Sugar Solution in a Multiple-Effect Evaporator. A triple-effect evaporator with forward feed is evaporating a sugar solution with negligible boiling-point rise (less than 1.0 K, which will be neglected) and containing 5 wt % solids to 25% solids. Saturated steam at 205 kPa abs is being used. The pressure in the vapor space of the third effect is 13.65 kPa. The feed rate is 22 680 kg/h and the temperature 299.9 K. The liquid heat capacity is cp = 4.19 – 2.35x, where cp is in kJ/kg · K and x in wt fraction (K1). The heat-transfer coefficients are U1 = 3123, U2 = 1987, and U3 = 1136 W/m2 · K. Calculate the surface area of each effect if each effect has the same area, and the steam rate.

## Calculate the initial humidity and percentage humidity.

1. Humidity and Wet Bulb Temperature. The humidity of an air– water vapor mixture is H = 0.030 kg H2O/kg dry air. The dry bulb temperature of the mixture is 60°C. What is the wet bulb temperature?

2. Dehumidification of Air. Air having a dry bulb temperature of 37.8°C and a wet bulb temperature 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.

a. Calculate the initial humidity and percentage humidity.

b. Calculate the final humidity and percentage humidity. [Hint: Locate the initial point on the humidity chart. Then, go horizontally (cooling) to the 100% saturation line. Follow this line to 15.6°C. Then, go horizontally to the right to 23.9°C.]