Question about the vaporization rate of sulfur dioxide!
Thread Content
Our company uses a process in which liquid sulfur dioxide is vaporized to produce sulfurous acid. I would like to calculate the release rate of sulfur dioxide from steel cylinders; I’m not sure where to start. Could anyone give me some guidance? At 0°C, the density of liquid sulfur dioxide is 1434 kg/m3, and the mass of sulfur dioxide released from each steel cylinder is 800 kg. The density of gaseous sulfur dioxide is 2.927 kg/m3, with a heat of vaporization of 394 kJ/kg; therefore, 800 kg of sulfur dioxide requires 3.152×105 kJ of energy to be vaporized.
I forgot to add that the pressure inside the sulfur dioxide cylinder is 15 MPa. This parameter is not required for the calculations; only a critical pressure of 7.87 MPa is needed. The value of 15 MPa is used merely to ensure that all sulfur dioxide is released, so that the pressure inside the cylinder can reach the critical pressure. The velocity has already been determined using Bernoulli’s equation, without having to account for pipe losses; there’s no need to reply to this post, thank you. This post was last edited by tongle on 2009-4-3 10:44.]
I’ve used a certain process in which sulfur dioxide is introduced into a solvent to dissolve it; it involves using a single gas cylinder along with hot water for spraying. The vaporization process seems slow, and there is still liquid remaining in the buffer tank during winter. You might try introducing sulfur dioxide directly from a higher elevation into the liquid – that’s what I do now, and it speeds up the process significantly