On Sun, Jun 26, 2011 at 5:19 PM, Mark Iverson <zeropo...@charter.net> wrote:
> Joshua wtote on Saturday, June 25, 2011 11:49 PM: > > Okay, due to my randomly selecting an unrealisticly low flow-rate of > 10g/sec, I can see where it > could be confusing. Let me try to clear things up... > 10 g/s is neither unrealistic, nor low. It is perfectly easy to generate that flow rate. But in the Krivit run, and most of the steam producing runs, it is more like one or two grams per second. In the secret run, it is much higher than 10 g/s. I don't see how choosing the wrong hypothetical flow rate should make an explanation confusing. Why shouldn't it work for any flow rate? We BOTH agree that the saturation mass of water vapor is 600g/m^3 for 100.1C > steam at ambient > pressure. Well, I'm not sure what you mean by "saturation mass of water vapor". The density of steam at 100C and one atmosphere pressure is .6 kg/m^3. That's the amount of water vapor per unit volume. My original example used 10g/sec which is no where near saturation, that is obvious, so I > can see where that may have been confusing. These are two different things. One is a flow rate, the other is a density. How can you compare them? 10 g/s input can produce dry steam or wet steam depending on the power in the reactor. > I'll revise the example to be close to the saturated > condition in the e-Cat tests... > > Here's a scenario which is just BELOW saturation... > - 594g/sec liquid water in (this is 594/600 = 99% of saturation flowrate) > Sorry, I don't follow. One second is an arbitrary unit of time. There is no reason the mass of water entering in one second should match the amount of steam in one cubic meter in order to have "saturation". More importantly, this flow rate is about 300 times higher than used in the Krivit demo. If you think the actual value of the flow rate has some relevance to the Krivit run, which is under discussion, then this example doesn't have any. > - assume that the power level of e-Cat is such that 100% of the water is > vaporized, but little or > no excess power, so now the entire mass of water (594g/sec) is vaporized > every second, but not > superheated. > OK, but have you thought about the power needed for this? It works out to 1.5 MW. This would already be better than Rossi's promised 1 MW plant. But ok, whatever, let's see where it leads. > - according to the methodology, the pressure inside the chimney is the > same as ambient, thus, the > flowrate of the dry steam out the chimney is equivalent to 594g/m^3. No. The mass flow rate, in this case, is 594 g/s, or the volume flow rate is 1 m^3/s, which by the way is 1000L per second; that would be something to see. Flow rate and density (or humidity) have different units. You can't just interchange them like that. We do NOT need to measure the > flow rate. If we know all the water is being converted to steam, then no, we don't need to measure the flow rate. But we don't know the steam is dry. That's what we're trying to determine. Don't you see, you're arguing in a circle. You know the flow rate is 594 g/s because the steam is dry, and you know the steam is dry because the flow rate is 594 g/s. What you need to tell me is how you know the steam is dry from the measurement of the humidity of the steam, which is always 600 g/m^3, even if the *steam* flow rate is 10 g/s or 100 g/s or 500 g/s. But those 3 cases represent very different levels of steam dryness.
