How many tons of saturated steam at 0.5 MPa are required to evaporate one ton of water? 20 points as reward

This post was last edited by artartar on 2012-1-15 at 22:54. It should be 657.328 kilograms, I think. My calculation is as follows: the specific heat of water is 1, so it takes 1X1000 kilogramsX1000 gramsX(100-0) = 100000000 calories for 1 ton of water to go from 0 degrees to 100 degrees. According to tables, 0.5 MPa corresponds to around 152.131 calories. Subtracting 100000000 from this value gives 657328 grams, which is equal to 657.328 kilograms. I’m not sure if this is correct

This post was last edited by come_now on 2012-1-16 at 11:30. The original poster missed one condition: what is the evaporation pressure, that is, what is the pressure of the secondary steam? And also, how many stages of evaporation are there? The more effective numbers, the less steam is used. If the poster is using steam at 0.5 MPa (absolute pressure) for single-effect evaporation at atmospheric pressure, the approximate calculation process is as follows: It takes 4.17 kJ/(kg·°C) of heat to raise 1 ton of water from 0°C to its boiling point of 100°C at atmospheric pressure. K) Heat required: 1000×4.17×100 = 417,000 kJ. For 1 ton of water to boil and vaporize under normal pressure, the latent heat of vaporization is 2259 kJ/kg; thus, the heat required is 1000×2259 = 2,259,000 kJ. Total energy consumption: 417,000 + 2,259,000 = 2,676,000 kJ. The latent heat of vaporization for saturated steam at 0.5 MPa is 2113 kJ/kg; therefore, the amount of steam required is 2,676,000/2113 = 1266 kg, which is equivalent to 1.26 tons. This is the theoretical calculation for single-effect evaporation. After being heated by steam, the water becomes hot water at a temperature of over 150 degrees Celsius, and this heat can also be utilized.

According to the tables, the enthalpy value of water vapor at 100 degrees Celsius is q=639.4 kcal/kg. The heat required to raise 1 ton of water from 0 to 100 degrees Celsius is Q=1000×639.4=639400 kcal； According to the tables, the phase change heat for saturated steam at 0.5 MPa (which should be 0.6 MPa in terms of absolute pressure) is 499.9 kcal/kg (for isobaric condensation) ； Steam required: 639400/499.9 = 1279 kg = 1.297 tons. For reference only

If the poster is using steam at 0.5 MPa (absolute pressure) for single-effect evaporation at atmospheric pressure, the approximate calculation process is as follows: It takes 4.17 kJ/(kg·°C) of heat to raise 1 ton of water from 0°C to its boiling point of 100°C at atmospheric pressure. K) Heat required: 1000×4.17=4170 kJ. This calculation is incorrect; it should be 1000*4.17*100=417000 kJ

You didn’t include the step of heating the water to the required temperature, right?

Going back upstairs: The enthalpy of water vapor at 100 degrees Celsius is q=639.4 kcal/kg. This value already includes the heat required for the water to go from 0 to 100 degrees Celsius; in other words, the enthalpy of water is 100 kcal/kg, while the phase change enthalpy at 100 degrees Celsius is 539.4 kcal/kg. Therefore, 100 + 539.4 = 639.4 kcal/kg

Reply to 5# Pfz2867: Haha, I’m sorry; I really did overlook it. Correction made

If it is three-effect evaporation, how should it be calculated? Thank you

Reply to 9# Pfz2867: Simply put, it involves three single-effect units connected in series; the secondary steam from the second effect is used by the third effect, and the secondary steam from the second effect of the third unit is utilized as well. Generally, a vacuum pump is needed at the end of the third effect to create a vacuum, otherwise the temperature difference between the various effects will be too low, resulting in poor efficiency. To roughly estimate the evaporation rate, after determining the operating pressure of the third effect, it is possible to consider the three effects based on equal pressure drops or equal temperature differences, in order to preliminarily determine the operating pressure (secondary steam pressure) for each effect. Using this value as a basis, the phase change heat of the secondary steam in each effect can be obtained by consulting tables. Then, perform a heat balance in a single-effect manner, taking into account the heat from flashing as well. For the detailed process, please refer to Section 9-6 on evaporation in the Chemical Engineering Handbook.

Reply to 2# artartar: Are you from Taiwan? or **?