For a water dispenser, you only need to know 1 kilowatt hour of electricity. It has nothing to do with voltage, frequency and current. It also has efficiency and how many degrees the temperature rises in plateau areas. boiling water can be prepared at 50 degrees.
Suppose the water heats from 20 degrees to 90 degrees. Regardless of the efficiency loss, the efficiency is 1. Then the work done by 1 kilowatt hour of electricity is 1 kW*1h=1000 w*3600 s=3600 kj.
Q=cm△t, the specific heat of water is C=1, m is the mass of water and △t is the temperature difference.
Then m=3600/70=51.4kg.
In other words, 1 kilowatt hour of electricity can heat 51.4 kilograms of water from 20 degrees to 90 degrees.
How much electricity does it take to boil a square of water?
First, it is important to determine the temperature of the tap water.
Assume that the temperaturee of tap water is 25℃ and it is heated to 90℃, then the temperature difference is 90-25 = 65℃
The specific heat capacity of water is 4.2x1000[J /Kg·℃)]
65 times 4.2x1000[J/Kg·℃)]=The amount of heat needed to heat a kilogram of water from 25℃ to 90℃ (ignoring energy loss, i.e. under ideal conditions) = 273KJ/Kg
One kilowatt hour of electricity = 1KW·h = 3600KJ
3600 /273=13.20Kg, that's the answer
As for the quantity needed, the time depends on the amount of water you boil. Suppose you want to boil n kilograms of water, then
9KW=9KJ/s
273×n÷9=30.33×n seconds
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The calculations above are all theoretical. In real conditions, the energy loss must be taken into account, that is, the boiling speed of water is slower than calculated. or water boiled by one degree of electricity is less than calculated.
Purely hand-made, hope to adopt it~~~~~
How much water does the chiller consume per kW of cooling capacity?
The specific heat of water is c=4.2kj/kg·℃ If domestic water is considered at normal temperature, that is, it is calculated at 20 ℃, and after boiling, it is calculated as 100. ℃ (regardless of different regions and different local atmospheric pressure) Effect on temperature), △t=80℃; the weight of 1 cubic meter of water: m=1t, Q=c×△t×m=336000kj,
Finally, convert kj to KW·h (1kw·h is equal to 1 kilowatt hour), 1kw·h=3600kj, so 336000kj=93.333kw·h (degree)
Conclusion: how much electricity is used to boil a square of water requires 93.333 kilowatt hours of electricity
< p>Friend:\x0d\ You are asking what is the average 1kw cooling capacity of the chiller, how much cooling water is needed, right? Or how much water islied? But you need to provide the specific COP of the cooler. Generally, the COP of the chiller is 6.0, and the temperature difference between the inlet and outlet water is 5 degrees. Calculate that the cooling capacity per 1 kW is approximately: \x0d\ Cooling. water: 200 kg/h\x0d\ Chilled water: 171 kg/h