Coal and steam cannot be easily converted.
However, there is indeed a link between these two substances, which is the energy balance, which means that it must be converted:
The calorific value of coal ( kJ/ kg) * coal Weight (kg)*efficiency = volume of steam (kg)*enthalpy of steam (kJ/kg)
It should be explained that different types of coal have different calorific values, different steam parameters, and large differences in enthalpy values, so the conclusion you get is reliable if it satisfies the formula above, otherwise it is unreliable.
As it is standard coal, the calorific value is 7000kcal/kg=29308kJ/kg
Based on the three steam parameters you provided, I I checked and found that the enthalpy values are: 3219kJ/kg;3474kJ/kg;3396kJ/kg;
Considering that the boiler efficiency is around 90%, we havens basically calculated that it can produce up to about 8t of steam.
Also, your question is mainly about the relationship between steam consumption and coal consumption. Personally, I think it has a greater relationship with steam settings. For 1 kg of coal to generate 3 kilowatt hours of electricity, for units over 300 MW, I agree; but for a steam consumption of 4.5 kg to produce 1 kilowatt hour of electricity, I think it is possible, but the different units should be different, like the steam of a 1000 MW unit. The throughput is about 3,000 tons and 600 MW is about 2,000 tons.
I would like to know the impact of changes in water temperature on the coal consumption of boilers. For example, for every 1 degree the feed water temperature drops, how much will my coal consumption increase?
Question 1,
1, find the heat that must be absorbed by the vaporization of a ton of water: Q absorption = λm water = 2260000J/kg =Q suction/q=2260000000J/( 29270000J/ kg)=77.2kg
Note: 1, λ=2260000J/kg is the heat of vaporization of water;
2, q=29.27 Megajoule/kg is the calorific value of standard coal;
3. This is simply a calculation of the conversion of water at 100 degrees Celsius to steam at 100 degrees Celsius. It doesn't calculate that water needs to absorb heat to warm up and that steam does. to absorb heat for heating. Heat loss is not taken into account.
Question 2,
1. Electric energy to be obtained: W=10000kWh=3.6x10^10J;
2. : m coal=Q put/q=W/q=(3.6x10^10J)/(2.927x10^7J/kg)=1230kg
Note: heat loss is not taken into account counts here.
How many kilowatt hours equal 1 ton of steam
Whenever the water temperaturefeed drops by one degree, the coal consumption will increase by about 0.14 grams, the water volume *water temperature difference*0.14/29270 is the coal consumption.
The reasons why the boiler feed water temperature cannot be increased excessively:
1. When the feed water temperature is high, if the feed water temperature is increased again, the efficiency of the generator set will not increase much, and the investment and maintenance of equipment will increase considerably.
2. Excessive increase in feed water temperature significantly increases operator safety risks; >3. Excessive increase in feed water temperature can easily cause excess steam in the boiler, resulting in overflow and burns for personal safety.
Your questionn is inaccurate, because the power plant units have different steam parameters (pressure, temperature), and the steam quality is also different, so this way there is no way to compare and measure, and there is no answer to this question.
Power plants use coal consumption to measure and compare. In fact, the question should be “How much coal is consumed per kilowatt hour of electricity?” » ". This value is now generally between (313 and 385 kg).
That is to say, 313 to 385 kilograms of coal (steam formed after combustion) can produce one kilowatt hour of electricity.