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PROBLEMS

8.4-1.

Heat-Transfer Coefficient in Single-Effect Evaporator. A feed of 4535 kg/h of a 2.0 wt % salt solution at 311 K enters continuously a single-effect evaporator and is being concentrated to 3.0%. The evaporation is at atmospheric pressure and the area of the evaporator is 69.7 m2. Saturated steam at 383.2 K is supplied for heating. Since the solution is dilute, it can be assumed to have the same boiling point as water. The heat capacity of the feed can be taken as cp = 4.10 kJ/kg · K. Calculate the amounts of vapor and liquid product and the overall heat-transfer coefficient U.

A1: Ans. U = 1823 W/m2 · K
8.4-2.

Effects of Increased Feed Rate in Evaporator. Using the same area, value of U, steam pressure, evaporator pressure, and feed temperature as in Problem 8.4-1, calculate the amounts of liquid and vapor leaving and the liquid outlet concentration if the feed rate is increased to 6804 kg/h.

A2: Ans. V = 1256 kg/h, L = 5548 kg/h, xL = 2.45%
8.4-3.

Effect of Evaporator Pressure on Capacity and Product Composition. Recalculate Example 8.4-1 but use an evaporator pressure of 41.4 kPa instead of 101.32 kPa abs. Use the same steam pressure, area A, and heat-transfer coefficient U in the calculations.

  1. Do this to obtain the new capacity or feed rate under these new conditions. The composition of the liquid product will be the same as before.

  2. Do this to obtain the new product composition if the feed rate is increased to 18 144 kg/h.

8.4-4.

Production of Distilled Water. An evaporator having an area of 83.6 m2 and U = 2270 W/m2 · K is used to produce distilled water for a boiler feed. Tap water having 400 ppm dissolved solids at 15.6°C is fed to the evaporator operating at 1 atm pressure abs. Saturated steam at 115.6°C is available for use. Calculate the amount of distilled water produced per hour if the outlet liquid contains 800 ppm solids.

8.4-5.

Boiling-Point Rise of NaOH Solutions. Determine the boiling temperature of the solution and the boiling-point rise for the following cases:

  1. A 30% NaOH solution boiling in an evaporator at a pressure of 172.4 kPa (25 psia).

  2. A 60% NaOH solution boiling in an evaporator at a pressure of 3.45 kPa (0.50 psia).

A5: Ans. (a) Boiling point = 130.6°C, boiling-point rise = 15°C
8.4-6.

Boiling-Point Rise of Biological Solutes in Solution. Determine the boiling-point rise for the following solutions of biological solutes in water. Use the figure in (P1), pp. 11-112.

  1. A 30 wt % solution of citric acid in water boiling at 220°F (104.4°C).

  2. A 40 wt % solution of sucrose in water boiling at 220°F (104.4°C).

A6: Ans. (a) Boiling-point rise = 2.2°F (1.22°C)
8.4-7.

Effect of Feed Temperature on Evaporating an NaOH Solution. A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water to a product of 50% solids. The pressure of the saturated steam used is 42 kPa (gage) and the pressure in the vapor space of the evaporator is 20 kPa (abs). The overall heat-transfer coefficient is 1988 W/m2 · K. Calculate the steam used, the steam economy in kg vaporized/kg steam, and the area for the following feed conditions:

  1. Feed temperature of 288.8 K (15.6°C).

  2. Feed temperature of 322.1 K (48.9°C).

A7: Ans. (a) S = 8959 kg/h of steam, A = 296.9 m2
8.4-8.

Heat-Transfer Coefficient to Evaporate NaOH. In order to concentrate 4536 kg/h of an NaOH solution containing 10 wt % NaOH to a 20 wt % solution, a single-effect evaporator is being used, with an area of 37.6 m2. The feed enters at 21.1°C (294.3 K). Saturated steam at 110°C (383.2 K) is used for heating and the pressure in the vapor space of the evaporator is 51.7 kPa. Calculate the kg/h of steam used and the overall heat-transfer coefficient.

8.4-9.

Throughput of a Single-Effect Evaporator. An evaporator is concentrating F kg/h at 311 K of a 20 wt % solution of NaOH to 50%. The saturated steam used for heating is at 399.3 K. The pressure in the vapor space of the evaporator is 13.3 kPa abs. The overall coefficient is 1420 W/m2 · K and the area is 86.4 m2. Calculate the feed rate F of the evaporator.

A9: Ans. F = 9072 kg/h
8.4-10.

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 boiling-point 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?

8.4-11.

Evaporation of Tomato Juice Under Vacuum. Tomato juice having a concentration of 12 wt % solids is being concentrated to 25% solids in a film-type evaporator. The maximum allowable temperature for the tomato juice is 135°F, which will be the temperature of the product. The feed enters at 100°F. Saturated steam at 25 psia is used for heating. The overall heat-transfer coefficient U is 600 btu/h · ft2 · °F and the area A is 50 ft2. The heat capacity of the feed cp is estimated as 0.95 btu/lbm · °F Neglect any boiling-point rise if present. Calculate the feed rate of tomato juice to the evaporator.

8.4-12.

Concentration of Cane Sugar Solution. A single-effect evaporator is being used to concentrate a feed of 10 000 lbm/h of a cane sugar solution at 80°F and containing a sugar content of 15° Brix (degrees Brix is wt % sugar) to 30° Brix for use in a food process. Saturated steam at 240°F is available for heating. The vapor space in the evaporator will be at 1 atm abs pressure. The overall U = 350 btu/h · ft2 · °F and the heat capacity of the feed is cp = 0.91 btu/lbm · °F The boiling-point rise can be estimated from Example 8.5-1. The heat of solution can be considered negligible and neglected. Calculate the area required for the evaporator and the amount of steam used per hour.

A12: Ans. Boiling-point rise = 2.0°F (1.1°C), A = 667 ft2 (62.0 m2)
8.5-1.

Boiling Points in a Triple-Effect Evaporator. A solution with a negligible boiling-point rise is being evaporated in a triple-effect evaporator using saturated steam at 121.1°C (394.3 K). The pressure in the vapor of the last effect is 25.6 kPa abs. The heat-transfer coefficients are U1 = 2840, U2 = 1988, and U3 = 1420 W/m2 · K, and the areas are equal. Estimate the boiling point in each of the evaporators.

A13: Ans. T1 = 108.6°C (381.8 K)
8.5-2.

Evaporation of 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 – 2.35 x, 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.

A14: Ans. Area A = 99.1 m2, steam rate S = 8972 kg/h
8.5-3.

Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a double-effect evaporator with reverse feed. The feed enters at 100°F and is concentrated to 25% solids. The boiling-point rise can be considered negligible, as can the heat of solution. Each evaporator has a 1000-ft2 surface area, and the heat-transfer coefficients are U1 = 500 and U2 = 700 btu/h · ft2 · °F. The feed enters evaporator number 2 and steam at 100 psia is fed to number 1. The pressure in the vapor space of evaporator number 2 is 0.98 psia. Assume that the heat capacity of all liquid solutions is that of liquid water. Calculate the feed rate F and the product rate L1 of a solution containing 25% solids. (Hint: Assume a feed rate of, say, F = 1000 lbm/h and a value of T1. Calculate the area and T1. Then calculate the actual feed rate by multiplying 1000 by 1000/calculated area.)

A15: Ans. F = 133 800 lbm/h (60 691 kg/h), L1 = 10 700 lbm/h (4853 kg/h)
8.5-4.

Concentration of NaOH Solution in Triple-Effect Evaporator. A forced-circulation triple-effect evaporator using forward feed is to be used to concentrate a 10 wt % NaOH solution entering at 37.8°C to 50%. The steam used enters at 58.6 kPa gage. The absolute pressure in the vapor space of the third effect is 6.76 kPa. The feed rate is 13 608 kg/h. The heat-transfer coefficients are U1 = 6246, U2 = 3407, and U3 = 2271 W/m2 · K. All effects have the same area. Calculate the surface area and steam consumption.

A16: Ans. A = 97.3 m2, S = 5284 kg steam/h
8.5-5.

Triple-Effect Evaporator with Reverse Feed. A feed rate of 20 410 kg/h of 10 wt % NaOH solution at 48.9°C is being concentrated in a triple-effect reverse-feed evaporator to produce a 50% solution. Saturated steam at 178.3°C is fed to the first evaporator and the pressure in the third effect is 10.34 kPa abs. The heat-transfer coefficient for each effect is assumed to be 2840 W/m2 · K. Calculate the heat-transfer area and the steam consumption rate.

8.5-6.

Evaporation of Sugar Solution in Double-Effect Evaporator. A double-effect evaporator with reverse feed is used to concentrate 4536 kg/h of a 10 wt % sugar solution to 50%. The feed enters the second effect at 37.8°C. Saturated steam at 115.6°C enters the first effect and the vapor from this effect is used to heat the second effect. The absolute pressure in the second effect is 13.65 kPa abs. The overall coefficients are U1 = 2270 and U2 = 1705 W/m2 · K. The heating areas for both effects are equal. Use boiling-point-rise and heat-capacity data from Example 8.5-1. Calculate the area and steam consumption.

8.6-1.

Water Consumption and Pressure in Barometric Condenser. The concentration of NaOH solution leaving the third effect of a triple-effect evaporator is 50 wt %. The vapor flow rate leaving is 5670 kg/h and this vapor goes to a barometric condenser. The discharge water from the condenser leaves at 40.5°C. Assuming that the condenser can maintain a vacuum in the third effect corresponding to a saturated vapor pressure of 2.78°C above 40.5°C, calculate the pressure in the third effect and the cooling-water flow to the condenser. The cooling water enters at 29.5°C. (Note: The vapor leaving the evaporator will be superheated because of the boiling-point rise.)

A19: Ans. Pressure = 8.80 kPa abs, W = 306 200 kg water/h


  

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