The empirical value is 13 kilowatt hours of electricity. The algorithm is: start heating from tap water temperature and continue heating to the set temperature of 60°C.
Heat required = c*m* (T1-T2), where c is the specific heat of water 4.2, m is the mass of water 1000 kg, T1 is 60° C and T2 is 15°C. The calculation result is 189,000 KJ and one kilowatt hour is tens of millions of hours, or 3,600 seconds * 1,000 watts = 3,600 KJ.
Dividing the two numbers equals 52.5 kWh. This is the amount of electricity needed for pure electric heating.
The energy efficiency ratio of air power is approximately 4, that is, one kilowatt hour of electricity can produce hot water equivalent to four kilowatt hours of electric water. Therefore, if air energy is used, the air energy is approximately equal to 4. the conselectricity consumption is 52.5/4 = 13.125 kilowatt hours.
That is to say, heating 1 ton of water requires approximately 13 kilowatt hours of electricity.
However, the energy efficiency ratio of air energy will decrease in winter when the outdoor temperature is relatively low (because it relies on absorbing heat from the air). Taking this factor into account, we calculate the decrease in the energy efficiency ratio in). winter at 50%.
Electricity consumption can then double to around 25 kWh.
In summary, several key points to consider when purchasing overhead power are:
1. Is there enough installation space?
2. If the climatic conditions (mainly temperature) are good.
3. If you produce air power for 1 ton of water, it is recommended to directly customize a stainless steel insulated water tank, the price is less than 1,500. The reswater supply provided by the air power manufacturer will cost a lot.
The electric water heater has a power of 1500W, the hot water temperature is about 80℃, two buckets of water per day and it is plugged in 24 hours a day. How much electricity. can we consume in 24 hours?
Using an electric kettle to boil water uses about 1,500 W. If it can fill two 5-pound hot water bottles, it's usually about 1,200 W ; if it is an electric kettle that can fill two to three and a half bottles of water, it is usually around 1,500 W. The calculation of energy consumption is that every 1 kW consumes 1 kilowatt hour in one hour (i.e. per kilowatt/hour). You should be able to see that the electrical rating of the electric kettle is listed on the brand. Based on this, you can know its power. measured energy consumption.
If the powerance of the electric kettle is 3000W, then the power consumption of continuous use for one hour is 3 degrees and the boiling time of the 3000W electric kettle is 5 minutes, then it is 3/60. =0.05*5 =0.25 If each kilowatt hour of electricity is equal to 1 yuan, then the price of using a 3000W electric kettle to boil a pot of water is 0.25 yuan.
Detailed information
Calculation analysis:
1. Check the amount of water in the pot and the initial temperature. Calculated by boiling 1.5 liters of water at 25 degrees in the pan, the total energy required is 4200*1.5*(100-25)=472500J. The amount of electricity required for such a large amount of energy is 472500/3600000=0.13.125 kilowatt hours of electricity. Due to some energy loss, the actual electricity consumption will be slightly higher.
2. Check the powernce on the water heater nameplate. Generally, the power is 1500 W to 2000 W. Take a pot of water and calculate the time needed to boil the water. It takes to boil a pot of water. Calculated in 5 minutes, the power consumed is 1.5*5/60=0.125 kWh. The 1.5KW in this formula is the power of the kettle, 1500W, 5 is the time to boil the water. 5 minutes, and 60 equals 60 minutes per hour. If you boil the kettle, the power is different from the time it takes to boil a pot of water. Simply change these two numbers to real values to calculate energy consumption.
The amount of electricity consumed in 24 hours is also directly related to the volume of the water heater, the temperature of the tap water and the ambient temperature.
For example, let's take an electric storage water heater with a capacity of 80 liters, a temperature oftap water of 20 degrees and an ambient temperature of 20 degrees to roughly estimate:
The power consumption of two barrels of hot water: 0.08 *(80-20)*1.163 = 5.6 degrees
Power consumption for heat preservation: 80 degrees of hot water, the power consumption for 20 hours of heat preservation at an ambient temperature of 20 degrees, it is about 2 degrees .
According to this calculation, the electricity consumption for 24 hours a day is approximately 8 degrees.