Quote:
Originally Posted by t vago
Using your example of 1034 kPa and 700 K (or 427 C, if you will), and assuming that the liquid water was near 100 C, we find that:
The mass of your mystery gas in that closed volume is unknown. However, for a simplified exhaust gas containing 80% diatomic nitrogen, 11 percent carbon dioxide, and 9 percent steam, it's 0.2 grams.
It holds about 150 J of thermal energy. Of that, 91 J is available that can be used to vaporize water.
That gram of water needs 2260 J to completely turn into steam.
Roughly 4% of that gram flashed into steam, bringing the temperature of your closed system to that of the water, i.e. 100 C.
The pressure dropped, too. Ooooh, lookie - 101 kPa.
And you still have 0.96 grams of liquid water to deal with.
So, now you have 39 ccs of mystery gas, 1 cc of water, and it's all at 100 C and 101 kPa. Oh, excuse me - 212 F and 14.7 psia.
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Not about the point you were making itself ... just wondering what I am missing about this result you work out here ... please let me know what I'm missing or what error I've made in the bellow thoughts.
#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?
#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?
#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 ... 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.. 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.
So I was guessing the first two were skipped for simplicity ... but I don't see what I am missing for the 3rd?
Please clarify... thanks