2.4-9. Throughput of a Single-Effect Evaporator. An evaporator is concentrating \( F \mathrm{~kg} / \mathrm{h} \) at \( 311 \mathrm{~K} \) of a \( 20 \mathrm{wt} \% \) solution of \( \mathrm{NaOH} \) to \( 50 \% \). The saturated steam used for heating is at \( 399.3 \mathrm{~K} \). The pressure in the vapor space of the evaporator is \( 13.3 \mathrm{kPa} \) abs. The overall coefficient is \( 1420 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \) and the area is \( 86.4 \mathrm{~m}^{2} \). Calculate the feed rate \( F \) of the evaporator. Ans. \( F=9072 \mathrm{~kg} / \mathrm{h} \) :2.4-10. Surface Area and Steam Consumption of an Evaporator. A single-effect evaporator is concentrating a feed solution of organic colloids from 5 to \( 50 \mathrm{wt} \% \). The solution has a negligible boilingpoint elevation. The heat capacity of the feed is \( c_{p}=4.06 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \) \( \left(0.97 \mathrm{btu} / \mathrm{lb}_{\mathrm{m}} \cdot{ }^{\circ} \mathrm{F}\right) \) and the feed enters at \( 15.6^{\circ} \mathrm{C}\left(60^{\circ} \mathrm{F}\right) \). Saturated steam at \( 101.32 \mathrm{kPa} \) is available for heating, and the pressure in the vapor space of the evaporator is \( 15.3 \mathrm{kPa} \). A total of \( 4536 \mathrm{~kg} / \mathrm{h}(10 \) \( 000 \mathrm{lb} \mathrm{m} / \mathrm{h} \) ) of water is to be evaporated. The overall heat-transfer coefficient is \( 1988 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\left(350 \mathrm{btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\right) \). What is the required surface area in \( \mathrm{m}^{2} \) and the steam consumption?
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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 square meters. 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.
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For concentrating the dilute solution from 9.5% solid to 40.5% solid, a single-effect evaporator is used. The concentrated product is produced with 425 kg solid per hour. The feed enters the evaporator at 60 °C and steam is supplied at 2 atm gauge pressure (saturation temperature 135 °C, latent heat of vaporization 517 kcal/kg). The operating pressure of the evaporator is 100 mm Hg absolute (boiling point is 51 °C, latent heat of vaporization 567 kcal/kg). Data: overall heat transfer coefficient = 420 kcal/h m^2 °C, specific heat of the dilute solution = 0.8 kcal/kg °C. Use only the data given in the problem. Neglect the boiling point elevation. Calculate the following. a) Heat load of the evaporator (in kcal/hr) Answer 1 b) Steam consumption rate (kg/hr) Answer 2 c) Surface area of the evaporator (in m^2) Answer 3 d) New value of capacity (in kg/hr) if the pressure of the evaporator raised to 300 mmHg (boiling point is 77 °C, latent heat of vaporization 555 kcal/kg)
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