Question 6 in the morning session of the 2013 case study
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6. A certain factory uses N2 to dry the adsorbents; N2 is first introduced into a preheater before being sent to the dryer. The preheater is heated using saturated water vapor, and the condensed water is discharged as saturated water. The adsorbent to be dried contains 15% (wt) water, and it is required that the water content of the dried adsorbent be no more than 3% (wt). After drying, the other components contain 0.25 kg of water per kilogram of N2. If heat loss is not considered, what is the minimum amount of heating steam (in kg) required to dry 3 tons of adsorbent? (Known: The enthalpy values of N2 at the inlet and outlet of the preheater are 50 kJ/kg and 120 kJ/kg respectively) ； The latent heat of heating steam is 1915 kJ/kg. (A) 21.54 (B) 45.34 (C) 54.26 (D) 74.14. This question provides quite a few known values, doesn’t it? I can’t figure it out! I feel that the humidity H1 at the dryer inlet is missing.
This question is a basic example of material balance and heat balance calculations! All necessary conditions are provided. The correct solution process is as follows: X1=0.15/(1-0.15)=0.17647, and X2=0.03/(1-0.03)=0.0309. The amount of dry material present is G=3000*(1-0.15)=2550 kg. Therefore, the amount of water that needs to be removed after drying is W=G*(X1-X2)=2550*(0.17647–0.0309)=371.20 kg. The amount of nitrogen required is L=W/0.25=1484.8 kg. As for the amount of steam needed, it is V=L*(H2-H1)/Hv=1484.8*70/1915=54.27 kg. The correct answer is (C). Keep working hard! The right mindset is what truly matters!
Since what is being sought is the minimum amount of heating steam required, it is assumed that H1=0, so that the amount of steam used is minimized
In the question, it is stated that the gas emitted after drying contains 0.25 kg of water per kilogram of N2. Does this refer to a dry basis or a wet basis?
Is it necessary to ignore the enthalpy changes of the adsorbent and water for this problem?