load calculation

describe yourself first !!! where do you live, what is your job ? etc...etc....

hang on guest, someone else may of more help...

Good answer Tomad,

but two items need mentioning, firstly the question of Peukerts exponent.

We cannot go into the maths of this as we do not know what type of battery he is using. However the poster will need to take into account that if he wishes to discharge his batteries over 2 hours he will need to look for a battery rating marked at C2. The following is the data for a heavy duty 2V cell designed for such applications. As can be seen the quicker the discharge the lower the available capacity.

If the poster uses 12V monoblocs he will find that they are usually rated at C20 and so will appear to have an even higher rating. To compound this possible mis-specification monoblocs are rarely rated for 70% discharge and to discharge this type of battery so deeply will lead to a very short service life.

The question of losses in the cables and contacts is unlikly to have any great effect on the capacity calcs but it is quite possible that the actual load is at 120Vac or 230V ac and thus fed from an inverter. In which case losses of 8% to 15% should be factored into the calculation.

If the question is a related to an anticipated project i would advise the poster to ask advice from an experienced marine electrician or a fork lift truck maintenance company, both of whom are dealing with this sort of capacity and loadings on a regular basis.

Regards

Chas

7 OPzS 490 = battery type

490Ah = Nominal Capacity

546 = C10 (10 Hour discharge capacity)

483 = C5

425 = C3

304 = C1