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The density of steam at 5 kilograms is 2.6 KG/M3. Therefore, V=1000/2.6=385 M3. Using the equation P1V1=P2V2, with P1=5 atmospheres, P2=1 atmosphere, V2=22.4 KMOL/M3, and V1=4.48 KMOL/M3, we get m=V/V1=86 KMOL, which equals 1289 KG. I’m not sure if this is the correct way to calculate it!
It’s the law of conservation of mass, unless the condensation is incomplete.
Are the temperatures of 5 kilograms of water vapor and condensed water the same? Why was the isothermal formula used?
The density of steam at 5 kilograms is 2.6 KG/M3. Therefore, V=1000/2.6=385 M3. Using the formula P1V1/423=P2V2/273, we find that P1=5 atmospheres, P2=1 atmosphere. The volume per mole is V2=22.4 KMOL/M3 and V1=6.94 KMOL/M3. Thus, m=V/V1=55.5 KMOL, which corresponds to 1000 kilograms. I was really careless; thank you for your guidance!
1 ton of steam (H2O) will naturally yield 1 ton of water upon condensation (ignoring details).
It changes from a gas (423 K) to a liquid (273 K); why then use the equation of state for an ideal gas? Just use conservation of mass! Unless you have non-condensable gases in your outlet.
Whose trap is so good that it has no air leakage? Well, high-quality traps do indeed have very low levels of air leakage; if we ignore minor details, they are capable of producing 1 ton of condensed water. It’s just that the air has leaked out; it’s still water.
The amount of condensate water produced by 1 ton and 5 kilograms of steam is approximately 1 ton, roughly speaking.
According to PV=nRT, for 1 ton of steam at a pressure of 5 kg, n=53.3 kmol, and the weight of water is 957 kg; 1 ton at a steam pressure of 3 kg, specific gravity of 1.65, temperature of 133.5, n=53.8 koml; weight of water: 968 kg ; 1 ton at a steam pressure of 1 kg, specific gravity of 0.59, temperature of 99.6, n=54.7 kmol, weight of water: 985 kg ; No matter how it’s calculated, it seems like just checking the values of formulas. It’s still the law of conservation of mass – 1 ton of steam represents the weight of 1 ton of water that has been completely vaporized. Hehe. This post was last edited by huqingyu1981 on 2009-3-27 at 11:14
Theoretically, it should be, as long as the sealing is good and there are no losses.
Surely, if there’s no leakage, the quality will remain unchanged. What is the relationship between quality and phase transition? .
I only learned about the cooling of circulating water during my university studies, using cooling towers for that purpose, but I don’t know how to do the calculations; I’ve always found this topic confusing
Surely, if there’s no leakage, the quality will remain unchanged:victory: :victory:
According to the law of conservation of mass, in a closed system, the amount of condensate water produced from one ton of steam is definitely one ton
According to the law of conservation of mass, 1 ton of steam must condense into 1 ton of water; this is a fundamental principle that warrants no doubt or discussion. And this is true regardless of the steam pressure. The problem is that determining 1 ton of steam is not an easy task; calibration methods must be used for steam flow meters, and often the most accurate way is to measure the condensate water. Strictly speaking, 1 ton of steam still contains a small amount of inert gases, and it is not possible for 100% of them to be condensed during the condensation process.
1 ton of steam – no matter what the pressure is – its mass remains unchanged; whether it condenses into water or turns into ice, its mass is still 1 ton. If the mass changes, the change in energy will be astonishing! Unless nuclear fusion occurs!
Of course, 1 ton of condensed water is produced, unless the law of conservation of mass is not upheld.
In terms of quality, 1 ton of steam yields 1 kilogram of condensed water; the efficiency of the steam trap needs to be taken into account
Can you use the formula P1V1/T1 = P2V2/T2? By doing so, you’re assuming conservation of mass in these two states; the result will definitely be 1T. That’s simply due to the law of conservation of mass~~~