First we have to assume...
We assume as follows:
- methane is non-liquefied
- methane is pure
- methane is at NTP (normal temperature and pressure)
Then the facts:
- I am calculating at SI-units or their derivatives
- 2800 cu.ft. is approx. 79,3 cubic meters
- energy content of methane is 9.9 kWh/m3
- that gives us energy of 785,07 kWh
We are comparing this energy to a flat of 50m2, heated by district heating. So, no methane is used for heating.
On average, flat of 50m2 consumes 3600 kWh per year, so 2800 cu.ft. is enough to cover energy need of this flat for 2.6 months. NB! No heating is calculated here, see above.
So, one cubic meter of methane (35,3 cu.ft) can be compared roughly to 1 liter of gasoil (diesel) for energy content.
There are 2.8 L in 2800 mL.
1 and three quatersish
There are 175 pints in 2800 ounces. One pint is equal to 16 ounces, so dividing 2800 by 16 gives you 175 pints.
Yes, 2.8 kVA is equal to 2800 watts. This is because 1 kVA is equivalent to 1000 watts, so 2.8 kVA would be 2800 watts.
To convert 2800 gallons of diesel to kilograms, you would first need to know the density of diesel, which is around 0.85 kg/L. Since 1 gallon is approximately equal to 3.785 liters, you would multiply 2800 gallons by the density to find the approximate weight in kilograms. In this case, 2800 gallons of diesel would be roughly 2800 gallons * 3.785 L/gallon * 0.85 kg/L ≈ 9545 kg.
The power of the engine and the maximum speed of the 1970 BMW 2800 are 197 hp and 230 km/h respectively.
If dry, it will weigh approximately 2000 lbs. If wet , approximately 2800 lbs. If dry, it will weigh approximately 2000 lbs. If wet , approximately 2800 lbs.
220 horse @2800 rpm
2800
70% of 2800= 70% * 2800= 0.7 * 2800= 1,960
5% of 2800= 5% * 2800= 0.05 * 2800= 140
36% of 2800 = 36% * 2800 = 0.36 * 2800 = 1008
56020% of 2800= 20% * 2800= 0.20 * 2800= 560
89% of 2800= 89% * 2800= 0.89 * 2800= 2492
% rate:= 7/2800 * 100%= 0.0025 * 100%= 0.25%
2800
2800/1