Approximately 1725 MBH, assuming the steam is all condensed, since the heat of vaporization is approximately 1000 BTU/lb. Heat of vaporization varies with steam pressure, so the answer will vary somewhat, depending on the steam pressure delivered, but that will usually get you within 5% or so.
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.
The number of pounds in a gallon of water changes with temperature. At 100 degrees Celsius, there is about 8 lbs. per gallon of water. So with one thousand lbs. of steam, we would have around 125 gallons of water.
To change 10 lbs of ice at 20°F to steam at 220°F, you would need to consider the heat required for each phase change: Heating the ice from 20°F to 32°F (melting point) - specific heat of ice Melting the ice into water at 32°F - heat of fusion of ice Heating the water from 32°F to 212°F (boiling point) - specific heat of water Vaporizing the water into steam at 212°F - heat of vaporization of water Heating the steam from 212°F to 220°F - specific heat of steam
a difference in pressure (differential pressure). For example putting air in a tire (I know its pneumatics and not hydraulics but it is the same concepts.) the air tank is holding 110 lbs of pressure and when you connect the hose to a tire with 25 lbs the air is going to flow from the tank to the tire. or from high pressure to low pressure.
80 inch lbs is equal to 6.67 foot lbs. This conversion can be done by dividing the number of inch lbs by 12 to convert to foot lbs.
100 lbs of steam. The volume that the steam will occupy will depend on the pressure.
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.
Well, it depends on what type of locomotive, diesel steam or electric, a standard diesel locomotive weighs about 425000 lbs, and a standard steam locomotive is about 778000 lbs
2000 lbs
Coal releases about 14-15,000 btu's of heat per pound. This will on average produce about 12 lbs of high quality superheated steam. With a **steam rate of an engine at about 4 lbs of steam per kilowatt produced, this coal plant can produce about 3 kilowatts per pound of coal burned. So to produce 10 MW of electricity this hypothetical plant will burn about 3,334 lbs. of coal per hour. In a day the coal consumed will be about 80, 000 lbs or 40 tons. *Actual fuel conditions, combustion efficiency, boiler efficiency and engine efficiency would need to be determined for any particular power plant to determine actual coal usage in that plant. **steam rates of different power plants can range from about 3.5 lbs steam/ KW to more than 10 lbs of steam/ KW.
The number of pounds in a gallon of water changes with temperature. At 100 degrees Celsius, there is about 8 lbs. per gallon of water. So with one thousand lbs. of steam, we would have around 125 gallons of water.
You need to look at a steam table first then Multiply lbs/hr steam x latent heat of evaporation in BTU/lb @ the operating pressure.
4000 lbs In 2012 Ron Wallace had his 2009 pound pumpkin weighed at the Topsfield Fair. Ron grew his record ONE TON pumpkin on the 1725 Harp seed and crossed it with the 1409 Miller.
The efficiency of a steam turbine is just the ratio of power out to power in, but if you want to be able to calculate it from the basic mechanical design, this is a specialised topic. In the link below is a general description of steam turbines, in the references and additional reading list there are some references that may help you.
To calculate the BTUs removed per hour, you can use the formula: BTUs = flow rate (lbs/min) × temperature change (°F) × 1. If the flow rate is 10 lbs/min and the temperature change is 15°F, the calculation is: 10 lbs/min × 15°F × 1 BTU/lb°F = 150 BTUs/min. To find the hourly rate, multiply by 60 minutes, resulting in 9,000 BTUs per hour.
To change 10 lbs of ice at 20°F to steam at 220°F, you would need to consider the heat required for each phase change: Heating the ice from 20°F to 32°F (melting point) - specific heat of ice Melting the ice into water at 32°F - heat of fusion of ice Heating the water from 32°F to 212°F (boiling point) - specific heat of water Vaporizing the water into steam at 212°F - heat of vaporization of water Heating the steam from 212°F to 220°F - specific heat of steam
A lot.