How many tons of steam are needed to raise water at 32℃ to 56℃℃
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This post was last edited by liu530014416 on 2024-11-7 09:40. There is a injector on site where steam mixes with water; the water flow rate is 5 t/h, the water temperature is 32°C, and the steam has a pressure of 0.4 Mpa and a temperature of 152°C. After mixing with the water, the outlet water temperature is 56°C, and the total flow rate is equal to 5 plus the steam flow rate. Since there is no flow meter on the steam pipeline, could any expert calculate approximately how many t/h the steam flow rate is?
Solution: First, we need to calculate the specific heat capacity of water and the enthalpy of steam. Typically, the specific heat capacity of water, c_w, is about 4.2 kJ/kg·°C (kilojoules per kilogram per degree Celsius). For steam, we have saturated steam at 0.4 MPa, whose enthalpy value h_s is approximately 2745 kJ/kg (data obtained from steam tables). The total heat Q required to warm the water can be calculated using the following formula: Q = m_w * c_w * (T_final - T_initial). Here, m_w = 5000 kg (the given water flow rate is 5 tons per hour); T_initial = 32°C; T_final = 56°C. Substituting these values into the formula gives: Q = 5000 kg * 4.2 kJ/kg·°C * (56°C - 32°C) = 5000 * 4.2 * 24 = 504000 kJ/h. This is the amount of energy required to heat the water from 32°C to 56°C. Now we calculate the amount of steam. The total energy Q_s provided by steam can be calculated using the mass of steam m_s and the enthalpy value of steam h_s: Q_s = m_s * h_s, where h_s = 2745 kJ/kg. The mass of steam m_s can be determined by equating the energy provided by steam to the energy required by water: 504000 kJ/h = m_s * 2745 kJ/kg; thus, m_s = 504000 / 2745 ≈ 183.6 kg/h. Therefore, the steam flow rate is approximately 183.6 kg/h, which is equivalent to 0.184 tons per hour. So the required steam volume is approximately 0.184 tons per hour. .
This post was last edited by weilongwu on 2024-11-10 10:27
The water did not undergo a phase change, but the steam did. Should the temperature change from 100°C to 56°C be taken into account? I would like some advice
Since the final temperatures are the same, it’s necessary to calculate the temperature difference
Because the latent heat of phase change for steam is very large, it can be considered alone in approximate calculations.
The heat absorbed by the cold water as its temperature rises is equal to the heat released when the steam condenses. Therefore: Heat absorbed by water: Q = 5×103×(56–32) = 1.2×105 Kcal/h. Heat released by steam: Q = A×504 + A×(152–56) = 600A Kcal/h. 600A = 1.2×105; thus, A = 200 Kg/h
This post was last edited by uffss2000 on 2024-12-14 21:44. Strictly speaking, it needs to be calculated. When calculating the steam consumption, use the steam enthalpy minus the saturated water enthalpy; the difference in enthalpies can be used for the calculation