Heat transfer problems
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I have 2000 kg of gas, with an average specific heat capacity of 6.4 KJ/Kg·°C. It takes 0.6 MPa of steam to heat this gas from 20 degrees to 140 degrees. How many kilograms of steam are required? I’m not sure if it’s because I don’t know how to calculate this, or because I haven’t taken all the properties of steam into account; my calculated result differs significantly from that of a colleague. I would appreciate some guidance from someone who knows more about this................ Thank you in advance. When calculating the heat required for steam, should one use its enthalpy value or its latent heat of vaporization? ? ? Furthermore, is there only one purpose to using steam with high pressure as a heat source? It’s all to increase the temperature of the medium being heated. There’s another question – I would appreciate it if an expert could explain this: The final temperature of the medium being heated in my case reaches 140 degrees; shouldn’t the water coming out of the steam condensation tube be saturated water at that point? It should be steam at 140 degrees, right? So can the latent heat of steam still be used as calculation data? This post was last edited by lcpinging on 2009-2-18 00:03]
For reference only: The heat released is the latent heat of condensation as steam turns into water at 100°C. The latent heat of condensation for 0.6 MPa steam is approximately r=300 kcal/kg; assume D kilograms of steam are required per hour. Q of condensation heat release = Dr = 300D, with no heat loss considered. The heat absorbed is: the heat absorbed by 2000 kg of gas as it rises from 20°C to 140°C. Q absorbed = Cm(t2-t1) = (6.4/4.18) × 2000 × (140-20) = 367,464 kilocalories. 310 D = 367,464; therefore D = 1185 kg
Thank you, Floor 2. My calculation method is the same as yours, and the results are similar. The latent heat of saturated steam at 0.6 MPa is actually 2071 Kj/Kg; therefore, the amount calculated by me is less than that calculated by Floor 2. However, one of our colleagues calculated that only 400 Kg is needed, which I don’t understand. I’m not sure if there’s an error in the calculation somewhere. This post was last edited by lcpinging on 2009-2-17 at 23:59