Case 14, afternoon 2013
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14. Wet air at normal pressure, with a temperature of 25°C, a humidity of 0.025 kg/(kg of dry air), and a flow rate of 6.5 kg/h, is heated to 95°C using a preheater; the preheater is supplied with heat by saturated steam at 120°C, and the condensate is discharged at the saturation temperature. When asking for the amount of steam required for heating (kg/h), which value should be used? (Hint: The latent heat of vaporization of water at 25°C is 2490 kJ/kg, and the heat of condensation of steam at 120°C is 2205 kJ/kg.) (A) 52.2 (B) 58.9 (C) 60.4 (D) 62.9 I calculated that I0 = 88.675 KJ/kg and I2 = 162.665 KJ/kg. The dry gas amount is L = 6.34 kg, and the amount of water vapor is 6.34 * (162.665 – 88.675) / 2205 = 0.2127 kg/h. There is no answer! No, the numbers in this question were copied wrong again!
You’re right! The number was copied wrong! Believe in yourself!
This post was last edited by zwp997 on 2015-6-10 18:03. Is it the question that’s wrong or the answer?
Lift it up, let’s seek together. . . . . . . . .
According to unofficial information, the flow rate should be 0.5 kg/s; let’s do the calculations together
Water vapor = 6.34 * (162.665 – 88.675) / 2205 = 0.2127 kg/h; the unit here should be kg/s, and 3600 must also be multiplied by this value
Water vapor = 6.34 * (162.665 – 88.675) / 2205 = 0.2127 kg/h. Here, 6.34 represents the amount of dry air; but shouldn’t the enthalpy calculated earlier refer to the enthalpy of wet air? Additionally, the wet air flow rate in this problem is actually 6.5 kg/s. I think the solution to this problem is as follows: I0 = (1.01 + 1.88H)t0 + 2490H —— (1) I1 = (1.01 + 1.88H)t1 + 2490H —— (2) (I1 – I0)*m = M*R; that is, (1.01 + 1.88H)*(t1 – t0) = M*R —— (3) By substituting the values, we get: M = (1.01 + 1.88*0.025)*(95 – 25)*6.5*3600/2205 = 785.2 kg/h. But there’s no answer given. . . It might be a mistake made while editing the title~~ Alternatively, the specific heat capacity of moist air can be calculated using the formula CH=1.01+1.88H, and then substituting this value into Q=CH*m*(t1-t0)=M*R will yield the same result
The last edit to this post was made by 2005180005 on 2017-7-30 at 12:13. The enthalpy of moist air refers to the enthalpy value per kg of dry air along with the water vapor it contains; therefore, it is sufficient to multiply by the flow rate of dry air. Additionally, the value 2490 should not be used in the enthalpy calculation – it should be 2500 (the latent heat of vaporization of water at 0°C). At 0°C, water vaporizes, and as the temperature of the water vapor rises from 0°C to t…
I have a question: why doesn’t the humidity change? The question doesn’t mention this condition