I think this is:the kWh of the losses is simply calculated as following:
Power losses cost for one day=0.07* (5+3.5)*12= 7.14 $/day
Annual losses cost = 7.14 * 365day=2606.1 $/year
Your cost is given in kWh. It's way easier to multiply what you're already given "kW * hours" for your kWh5-KW copper losses is 5000 joules/ second. how come we use this number to times number of hours directly?
no we don't take the no load losses during no operation periodI think this is:
Power losses cost for one day = 0.07* (5+3.5)*12 + 0.07*(3.5)*12 = $10.08/day
Annual losses cost = 10.08 * 365day=3679.2 $/year
The core losses are occurring with load and no load.
What is the actual answer? I'm curious
no we don't take the no load losses during no operation period
please see the following in which only the time of operation is considered.
View attachment 23150 View attachment 23151
View attachment 23150 View attachment 23151But I am also confused about that , to be honest with you I found some sites on the internet speak about adding these no load losses for all the time.
This is the correct solution.the kWh of the losses is simply calculated as following:
Power losses cost for one day=0.07* (5+3.5)*12= 7.14 $/day
Annual losses cost = 7.14 * 365day=2606.1 $/year
When the transformer is at full load both copper loss and core loss occur. But core loss is negligible compared to the copper loss as there will be a huge surge of current passing through it as it is in short circuit.But I am also confused about that , to be honest with you I found some sites on the internet speak about adding these no load losses for all the time.
Thanks SamThis is the correct solution.
When the transformer is at full load both copper loss and core loss occur. But core loss is negligible compared to the copper loss as there will be a huge surge of current passing through it as it is in short circuit.
So its 12 hours at full load, then 12 hours at no load = 24 hours