Question 26 in the morning session of professional knowledge in 2010: A question about evaporation
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This post was last edited by Pingdan Shi on 2015-1-25 at 14:19. 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) is 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 condensate of this steam 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 raw material 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 Answer: C. Please provide the solution process, thank you!
This post was last edited by simpse on 2012-6-6 at 19:36. 2138.5 kJ/kg should be the enthalpy of vaporization of saturated steam at 0.4 MPa; this can be confirmed by referring to a steam table. w=5000(1-0.2/0.5)=3000; D=/2138.5/(1-6%)=2661.3. Choose D. I made a mistake with one of the values earlier; it has been corrected.
2138.5 kJ/kg should be the enthalpy of vaporization of saturated steam at 0.4 MPa; this can be verified by referring to a steam table. t5 u K w=5000(1-0.2/0.5)=30003 ?+ v" N5 S9 ?9 \! V( Z5 \ D=/2138.5/(1-6%)=2609.6) H2 ^ Choose D. What does 3000*3.47*140 in 3000(2292.2-3.47*140) mean?
It seems that using 2138.5 kJ/kg is incorrect; instead, (2138.5–589.08) should be used. This difference represents the latent heat of vaporization. 2138.5 kJ/kg is the enthalpy. I think it should be calculated as follows: From tables, the enthalpy of water at 140 degrees is 589.08 kJ/kg. Therefore, the value is 3000(2292.2–3.47*140)+5000*3.47*(130–140)]/(2138.5–589.08)/(1–6%) = 3601. No answer available
Everyone is thinking too much; don’t worry about the validity of the data. By simply plugging in the data given in the question into the formula, the result can be calculated; the answer is D. Dr=WH`+(F-W)h1-Fh0+QL
Shouldn’t the formula be “F-W”? The heat loss is 6% of the amount of steam used for heating; does that mean 6%D? Is there any expert who can provide an answer? Q
Can 2138.5 be used directly as the latent heat of vaporization? It should be necessary to subtract the enthalpy of saturated water, right? Please clarify. Thank you!
I calculated it the same way, but there’s no answer; it’s really frustrating