Question 26 from the Professional Morning Exam in 2010
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26. In a single-effect evaporator, the liquid flow rate is 5000 kg/h. An NaOH aqueous solution at 140°C and a concentration of 20% (wt) was evaporated and concentrated to 50% (wt). It is given that the vapor pressure at heating temperature is that of saturated steam at 0.4 MPa; the condensed vapor is removed at the saturation temperature, and the enthalpy of this saturated steam is 2138.5 kJ/kg. The average pressure in the evaporation chamber is 0.055 MPa, the boiling point of the solution is 130 °C, and the enthalpy of this saturated steam is 2292.2 kJ/kg. The specific heat capacity of the feed solution at the initial and final concentrations is 3.47 kJ/(kg °C). The heat loss is 6% of the amount of steam used for heating. Note: Which of the following values is the required heating steam consumption (kg/h)? A. 2709 B. 2726 C. 4500 D. 2661 Solution process
Is the initial temperature of the feed solution 140?
The original poster is not familiar with the book. The temperature of the raw material solution in the title should be incorrect. There are two very important energy balance equations in evaporation. One ignores the concentration heat, the other does not. It isn’t mentioned in this question, but based on the given conditions, only the former method can be used for calculation. Dr*(1-0.06)＝Wr＇+Fcp(t1-t0). Just plug in the data.
The enthalpy of saturated steam is 2138.5 kJ/kg; the question is incorrect – it should be the heat of vaporization
There should be no error in the question; this problem can be solved by directly applying the formula. When the heat of dilution is not ignored, the equation is: DR = Wr + (F – W) * c1 * t1 – F * c0 * t0 + Q_loss. The amount of water that evaporates is W = 5000 * (1 – 0.2/0.5) = 3000 kg/hr. Using the heat balance equation: D * 2138.5 * (1 – 0.06) = 3000 * 2292.2 + (5000 – 3000) * 3.47 * 130 – 5000 * 3.47 * 140. This gives D = 2661 kg/hr