# PE power 148

### Help Support Engineer Boards:

#### yaoyaodes

##### Active member
Can anyone explain this to me? The question says that the transformer has 5-KW copper losses and 3.5-KW core losses. my understanding is that one watt is one joule (unit of energy) per second. does the 5-KW and 3.5-KW need to time (60*60) to get KWh?

#### Attachments

• IMG_3981.jpg
29.9 KB · Views: 17

#### usamoh2002

##### New member
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

#### yaoyaodes

##### Active member
5-KW copper losses is 5000 joules/ second. how come we use this number to times number of hours directly?

#### Sam_

##### Member
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
I 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

What is the actual answer? I'm curious

#### Sam_

##### Member
5-KW copper losses is 5000 joules/ second. how come we use this number to times number of hours directly?
Your cost is given in kWh. It's way easier to multiply what you're already given "kW * hours" for your kWh

#### usamoh2002

##### New member
I 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

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.

#### usamoh2002

##### New member
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

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.

#### Sam_

##### Member
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.
View attachment 23150 View attachment 23151

If I have 10 transformers in my house, plugged in and running at no load all YEAR...….Will I not get a bill? I have no load at all on them. Yes, because those core losses are still drawing power

#### Sam_

##### Member
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
This is the correct solution.

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.
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

#### usamoh2002

##### New member
This 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
Thanks Sam
so the no load losses will be taken in account only during the operation period
good