Approximately 1 lb of steam at 212°F contains 1 lb of water.
It takes 970.4 BTUs to convert 1 lb of water at 212°F to 1 lb of steam at 212°F. This process is known as the latent heat of vaporization.
1 pound of water at sea level can produce approximately 26.8 cubic feet of steam.
The number of mmBTU in 1000 lbs of steam can vary depending on the specific heat content of the steam. On average, 1 lb of steam contains about 1,200 BTU, which would equate to 1.2 mmBTU for 1000 lbs of steam.
A 210 lb person would displace approximately 26.25 gallons (around 100 liters) of water based on the assumption that 1 pound of body weight equals roughly 0.125 gallons of water.
The amount of heat required to convert 1 kg of steam to water at its boiling point is known as the latent heat of vaporization. For water, this amount is approximately 2260 kJ/kg.
It takes 970.4 BTUs to convert 1 lb of water at 212°F to 1 lb of steam at 212°F. This process is known as the latent heat of vaporization.
It would take 1 ton of water to create 1 ton of steam.
If the water content of the steam is 5% by mass, then the steam is said to be 95% dry and has a dryness fraction of 0.95.Dryness fraction can be expressed as:ζ = ws / (ww + ws) (1)whereζ = dryness fractionww = mass of water (kg, lb)ws = mass of steam (kg, lb)GAJANAN Nalegaonkar
1 pound of water at sea level can produce approximately 26.8 cubic feet of steam.
First convert 1 lb of water to lb-moles which is 0.055 lb-moles (you'll need this later). This problem can be broken into 3 steps:(1) You need to detemine how much heat is needed to raise room temperature water (68oF) to 212oF. This can be used using the heat capacity of water which at room temperature is 1 Btu/lboF. So the amount of heat needed for this is:Q1 = m*Cp*ΔT= (1 lb)*(1 Btu/lboF)*(212 - 68oF)= 144 Btu(2) Next you need to account for the phase change. The water changes to steam at 212oF. You use the heat of vaporization which you can look up in any Chemistry or Chemical Engineering Handbook. The Hvap that I found is 17493.5 Btu/lb-mole.Q2 = n(lb-moles)*Hvap= (0.055 lb-moles)*(17493.5 Btu/lb-mole)= 972.64 Btu(3) Next you need to find out how much heat is needed to raise the temperature of the steam from 212 to 213oF. You can look up the heat capacity of steam at 212oF to be 0.485 Btu/lboF.Q3 = m*Cp*ΔT= (1 lb)*(0.485 Btu/lboF)*(213-212oF)= 0.485 BtuTo find the total heat needed add Q1+Q2+Q3 (144+972.64+0.485) =1117.12 Btu
16 ounces in 1 lb. of water.
1 M3 of water will make 1 tomme of steam
It takes 1 BTU to raise 1 lb of water per degree Fahrenheit.
1 ton of steam. how much steam is vague question. steam can be at different pressures and saturations. At 212°F, 14.7 psia, liquid water has a specific volume of 0.016716 ft3/lbm and steam has a specfic volume of 26.80 3/lbm, which is a volume ratio of ~1603 : 1 of steam:water.
Find the total weight of the water and divide by the total about of air. ((100lb air X .10lb water) + (50lb air x .02lb water)) Divided by (100lb air + 50lb air)=.073lb water per lb of air
The answer will depend on the starting temperature of the water. It will also depend on the pressure.
When 1 cc of water is converted into steam, it can expand to approximately 1,700 cc (or 1.7 liters) at 100°C and standard atmospheric pressure. This significant increase in volume occurs because steam occupies much more space than liquid water due to the increased energy and separation of water molecules in the gaseous state.