Hey guys,

I'm stuck on a part of a tuts question.

Here is the question;
A small cabin needs 2.4 kWh per day, all of which is used in the evening from 7 pm to 11 pm. If you decide to supply only enough 12 V battery storage to cover each evening, what should the minimum rated Ah capacity of batteries be if they should not be discharged more than 80%? Assume the battery operating temperature to be 25°C.

What I have;
I get that 2.4kwh is needed over 4 hours at 80% discharge (rate of discharge is 80% over 4hours @ 25°C)

The solutions go on to say that, from the following graph, the rated battery capacity is C/5

I have no idea how to get this capacity of C/5 from the graph

The following example just confuses me further;

Can someone walk me though the process of using the graphs to determine the battery capacity rating?

thanks guys

 

You need to start somewhere. You know the energy draw over the 4 hour period is 2.4kWh. This should lead to an average load power consumption.....? You know the battery voltage is 12V and hence can determine the load current at 12V to meet the aforementioned power loading requirement. If you know the current and the duration of the loading you can find the essential lost Ampere-Hours to sustain the load. Then you can consider the other factors. So how many Ampere-Hours does the load consume over the 4 hours of discharge?

 

So, I = P/V = 2400/12 = 200Ah (over 4hrs) right?

 

And I'm guessing this leads to a 250Ah battery rating from 200/0.8???

 

And I'm guessing this leads to a 250Ah battery rating from 200/0.8???

Now you have to take into account the nominal discharge rate at 25C. What adjustment would you have to make in the nominal 250 Ah capacity to account for the effective capacity at the C/5 discharge rate at 25C? In other words, where does the Capacity/(Rated Capacity) % C/5 curve of choice intersect the 25C condition in the graph?

 

ok I can see that it would be approximately 75% right? But how did you select C/5?
I think I may be missing something obvious :\

 

ok I can see that it would be approximately 75% right? But how did you select C/5?
I think I may be missing something obvious :\

If you discharge the battery by 80% in 4 hours this is effectively equivalent to a full discharge in 5 hours - i.e. the C/5 rate. So it would seem to me a reasonable choice.

 

Ok I see now.

Now for the 2ed example, a C/20 rate is used.
I'm getting a much higher discharge time.

Here is what I have for the second example;
60% Max depth of discharge (From graph not included. Can include if asked)
This is over 2 days storage (48hrs)
So with a discharge rate of 60%/48hr = 1.25% per hr
Thus for 100% discharge => 100/1.25 = 80hrs (or C/80)
So pretty far of the solution of C/20