I can’t do it anymore. Heroes, please help! Thank you!
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Water with a flow rate of 10 T/H needs to be heated from 40 degrees to 60 degrees. Steam with a mass of 0.6 kilograms is used, at a temperature of 160 degrees; it is used for direct mixed heating. How many tons of steam are required? It would be best to include the formulas. Thank you, waiting online
First, you need to determine what the operating pressure for mixed heating is Additionally, you need to find several parameters: first, the average heat capacity of water between 40 and 60 degrees; second, the latent heat of steam at the operating pressure, that is, the phase change heat and the phase change temperature; and third, the average heat capacity of water between 60 degrees and the phase change temperature. Once these parameters are determined, the amount of steam required can be calculated using the law of conservation of energy
It’s too complicated. I hope an expert can provide the formula and the step-by-step process. Thank you!
This post was last edited by ylb913 on 2015-6-16 at 12:50. P (absolute pressure), Temperature, Latent heat /MPa /℃ /(kal/kg): 0.7, 165, 494. The enthalpy change when steam turns into water at 60 degrees is equal to 494 + 165 – 60 = 599 kcal/kg; the enthalpy value of water at 165 degrees can be found in tables as 167, so it would be 601 kcal/kg – pretty much the same. 600W = 10000 * (60 – 40) = 200000; thus, W is approximately equal to 330 kg.
Haha, thank you so much! Mmmmm~ I got the same result: ÷(685-60) = 334.5 = 0.3345 T/H
Yours should be 661-60=601kcal/kg.