[t vago]

Originally Posted by IamIan

#1> You can't inject the liquid water into that system without the application of the liquid water being at a higher pressure than the system was before you added the water... unless that was ignored for simplicity?

Yes, that was ignored for simplicity. I did ask drmiller100 how he proposed to inject water into a 150 psia pressure vessel, while ensuring that that same pressure vessel at 2500 psia would not leak back, but he chose to ignore my question.

Originally Posted by IamIan

#2> Even before the temperature change takes effect ... displacing the volume of the added liquid water itself in a fixed volume will increase the pressure of the system ... plus the additional pressure ( energy ) added to the system that was needed to inject the water in the first place... unless that was ignored for simplicity?

That was also ignored for simplicity. However, I'll go over the math of how pressure and temperature would rise by injecting 1 gram of water into a 40 cc vessel.

It's true that the liquid water volume will displace the gas, effectively shrinking the gas to 39 cc. This is due to the fact that 1 gram of water has a volume of 1 mL, or 1 cc. Using the ideal gas law (P1 * V1 = P2 * V2), that should raise the gas pressure from 150 psia to 154 psia, which is negligible.

The adiabatic compression of the exhaust by the injected water will cause the exhaust gas temperature to increase as its pressure increases. Using T2 = T1 * (V2 / V1) ^ (k - 1), we find that the final temperature is 807 F instead of 800 F. Again, negligible. They do not change the final outcome, as discussed below.

Originally Posted by IamIan

#3> The end system result you gave looks like it is just the effect of the initial gas in the fixed volume ... which contracts as it cools from the transfer of energy to the liquid water ... where did you account for the expansion effect of the % of liquid water that is converted to steam in the system ... that liquid to steam expansion counters the initial gas contraction you showed

No, it doesn't. Look, the water that flashed into steam is saturated steam. It's right at the boiling point temperature. Assuming a temperature of 100 C for the steam, that means that its corresponding pressure is 101 kPa, or 14.7 psia. The 1700:1 so-called expansion factor that drmiller100 keeps bandying about is misleading in this, because it is for standard atmospheric conditions. In other words, it doesn't apply to exhaust gas at 800 F and 150 psia stuck in a little 40 cc volume.

Let's look at that 0.04 grams of water that flashed into steam. It formerly occupied 0.04 cc of volume, but now it occupies the entire 39.04 cc of volume occupied by the gas, because it is itself a gas now. For that tiny amount of liquid, we just saw a 975:1 expansion factor, which is not 1700:1. However, that water sucked out all of the available heat energy from the exhaust gas to do so, and it would have sucked out even more heat had the water been below 100 C when it was introduced into the pressure vessel. The end result is that the exhaust gas, steam from the liquid water that flashed, and the remaining liquid water all have the same temperature, which is the temperature of the introduced water.

Originally Posted by IamIan

... even just 0.04 grams of liquid water to steam is a significant increase in pressure for a fixed volume ... the if not contained in the fixed volume the initial 1cc of liquid water would be 0.96cc of liquid water + about ~64cc of steam..

Nope. Here, though, you'll find that I made a mistake in neglecting to show the exhaust gas final pressure. It turns out that, due to the ideal gas law, the final exhaust gas pressure is 82 psia, or 55% of the original pressure.

Using the law of partial pressures, we add the new steam at 14.7 psia to the existing exhaust gas at 82 psia. We come up with 96.7 psia total for the system final pressure, which is roughly 2/3 of the original exhaust gas pressure.

Like I said, this idea of spraying water into exhaust gas is merely a novel way to cool off said gas.

Originally Posted by IamIan

confined by the 40cc fixed volume and the other contents in that volume ... I don't see how you got the 0.04 grams of steam being that low of a final system pressure.

That's because I wasn't clear, and I apologize for the resulting confusion.

This just goes to show how complicated an ideal example is. Real-world modelling is even more complicated. There are steam tables that must be consulted. There are gas tables that must be consulted. The example here has been ridiculously simplified in order to make it as understandable as possible.