Questions regarding Association **Question Set 5-41
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The problem is as follows: In a continuous counter-current dryer, a certain solid material is dried using hot air. Known air conditions: Air humidity upon entering the dryer: 0.01 (kg/kg dry, omitted later); enthalpy: 120 kJ/kg. The temperature of the wet air exiting the dryer is 38 degrees Celsius. The material condition is as follows: the moisture content upon entering and exiting the dryer is 0.04 (kg/kg of dry material; the unit is omitted here) and 0.002 respectively. The inlet and outlet temperatures are 27 degrees Celsius and 63 degrees Celsius respectively. Dry material flow rate: 450 kg/hr. Heat capacity of completely dry material: 1.465 kj/(kg·K). Assuming a heat loss of 5 kW from the dryer, determine the air flow rate. My question is this: *The answer provided at the end of the exercise set calculates H2, which is the humidity at the wet air outlet. The method I used was to ignore this humidity level and instead apply the mass balance and energy balance directly. Mass balance: W (amount of water vaporized) = GC(X1-X2) = V(H2-H1); this calculation yields the amount of water that is dried, which is 17.1 kg/hr. Energy balance: The heat released by the wet air in the dryer is used for: 1. vaporizing water, 2. heating the completely dry material, and 3. covering heat losses. Thus, we have: V(1.01+1.88H1)(t1-t2)=17.1*(r0+Cv*t2-Cl*27)+450*(1.465+X2*Cl)*(63-27)+5*3600. Here, V represents the volume of dry air; r0 is the heat of vaporization of water, Cv is the specific heat of water in gas form, Cl is the specific heat of water in liquid form, and X2 represents the moisture content at the outlet of the material. The calculated value of V is 1036, which is still half the value given as the answer, namely 1486. This problem has been bothering me for two days. I know my question is long, so I thank every friend who took the time to read my post. If you have also done this problem and can offer some guidance, I would be extremely grateful! I’ll give extra points.
This post was last edited by zwp997 on 2015-9-9 10:40. For thermal equilibrium, V(1.01+1.88H1)(t1-t2)=17.1*(r0+Cv*t2-Cl*27)+450*(1.465+X2*Cl)*(63-27)+5*3600; the value on the left side is incorrect. Were you performing moist heating at that time? The first two items on the right side have ready-made formulas for wet materials. That’s right. So H2 was still obtained. r0 = the heat of vaporization of water; Cv = the specific heat of water in gas form. The values for these two parameters should be between 27 and 63. .
First of all, thank you very much for your reply; I fully understand what you mean. I know where I went wrong. When I did the calculations, I directly used the model of continuous drying from the chemical engineering textbook, assuming that there is a preheater before entering the dryer. As a result, the condition at the dryer outlet is incorrect. It’s just wrong, in short!
It’s a post from almost two years ago; I’ve forgotten it too. But I wish you success in the Juhua exam. I’ve already passed it, hehe