Is this correct method to calculate the Ah of the battery ?

[danishdeshmuk]

Is this correct method to calculate the Ah of the battery ?

First u need to calculate the total power rating of the equipments u need to operate. for eg: number of fans(1 fan=60W approximate) , lights, computers etc.. As an example lets take the total power rating =750W.

Then divide this by the power factor say 0.8 =937 VA. So ur UPS or inverter rating should be more than 937VA so lets take it to be 1KVA or 1000VA.

The next step is to know how much time it should support ur loads or equipments. Lets take 3 hrs. Then multiply the VA rating of ur inverter or UPS with the no: of hours.

1000*3=3000. So ur battery should be able to support this much Ah. To know the Ah rating of ur battery , divide this by the nominal rating of the battery which is usually 12V or 24V or 48V.

If i take the battery nominal voltage to be 24V , then Ah rating of the battery would be 3000/24 = 125Ah . If 125Ah is not available go for the next available standard ie 130Ah or 150Ah.

Have your say on it ?

thanks

 

[goldsmith]

Then divide this by the power factor say 0.8 =937 VA. So ur UPS or inverter rating should be more than 937VA so lets take it to be 1KVA or 1000VA.

Dear danishdeshmuk
Hi
What do you mean by that ? when you say , va , it means that you are referring to the supply power .and no needed to use power factor .
Ps=((pr^2)+(Pa^2)) under radical . and Pa = ps*cos teta and pr= ps* sin teta .
Do you know what is the meaning behind Ampere per hour ? let me describe it with an example :
if we have a simple 100A/h battery , it means if we get 100 ampere from that battery , it can supply that current for an hour . but if you get 50 ampere , it will give you that current for 2 hour ! and if you get 200 A from that it will give that current for half hour !
Best Wishes
Goldsmith

 

[kak111]

First u need to calculate the total power rating of the equipments u need to operate. for eg: number of fans
(1 fan=60W approximate) , lights, computers etc.. As an example lets take the total power rating =750W.

( Device power = 240V * 3.125A = 750W )
OK.

Then divide this by the EFFICIENCY say 0.8 =937 W.
This is inverter efficiency for conversion battery voltage to 240Vac
So our UPS or inverter rating should be more than 937VA
so lets take it to be 1KVA or 1000VA.

( Inverter power from battery = 24V * 41.67A = 1000VA ) (in case 24V battery)
OK

The next step is to know how much time it should support ur loads or equipments. Lets take 3 hrs.
Then multiply the VA rating of our inverter or UPS with the no: of hours.

OK
1000VA*3h=3000VAh = 3kVAh
OK

So our battery should be able to support this much VAh.
To know the Ah rating of ur battery , divide this by the nominal rating of the battery which is usually 12V or 24V or 48V.

OK

If i take the battery nominal voltage to be 24V , then Ah rating of the battery would be 3000/24 = 125Ah .
If 125Ah is not available go for the next available standard ie 130Ah or 150Ah.

OK

Check calculations 41.67A * 3h = 125Ah and 125Ah * 24V = 3000VAh

Your calculations are right.

 

[danishdeshmuk]

means this is correct method to calculate the required AH of the battery for a particular UPS or inverter ??? Right

 

[enjunear]

means this is correct method to calculate the required AH of the battery for a particular UPS or inverter ??? Right

Agree w/ kak111; your math looks good. If you want to throw additional margin in, you could also assume some non-ideal power conversion efficiency on the DC-to-AC stage... say 90%. That would bump your required VAh need from 3000 to 3000/0.9=3333 VAh. 3333 VAh / 24V = 138.9 Ah

If you started with 937 VA * 3 hrs = 2811 VAh, and 90% efficiency = 3123 VAh. You're still very close to the 3000 VAh that you we're initially planning on. You're definitely on the right path.

 

[goldsmith]

Hi again

Then divide this by the power factor say 0.8 =937 VA. So ur UPS or inverter rating should be more than 937VA so lets take it to be 1KVA or 1000VA.

But i'm disagree with this ( power factor !! for supply power ??!!)
Best Lucks
Goldsmith

 

[kak111]

In post #3

Then divide this by the EFFICIENCY say 0.8 =937 W.
This is inverter efficiency for conversion battery voltage to 240Vac
So our UPS or inverter rating should be more than 937VA
so lets take it to be 1KVA or 1000VA.

 

[pico]

https://en.wikipedia.org/wiki/Ampere_hour

 

[mister_rf]

Is this correct method to calculate the Ah of the battery ?

No, that’s only in theory, in practice need to consider also the battery efficiency. Let me explain.
..........
Manufacturers rate the capacity of a battery with reference to a discharge time when discharged at a rate that will fully discharge the battery in 20 hours.
For this example, if a battery might be rated at 125 Ah, in 20h the nominal discharge current would be 6.25 amperes.
But there is a catch called Peukert Law which means the capacity changes depending on the discharge rate. If a battery is discharged in a shorter time, with a higher current, the delivered capacity is less.
Peukert's law is as follows:
Cp= I ^k * t
Cp = is the capacity according to Peukert, at a one-ampere discharge rate, expressed in Ah.
I = the discharge current, expressed in Amperes
k = the Peukert constant,
t = the time of discharge, expressed in hours
The Peukert constant increases with age for any of the battery types above, but generally ranges from 1.1-1.25 for gel batteries, and 1.2-1.6 for flooded batteries.
=> The new corrected amperage usage
I^k
And for the above example, 3h for a 1000W load, the 24V batteries discharged at ~ 41Amps.
In this case the corrected amperage usage obtained
(need to consider some batteries k=1.2)
I = (41)^1.2= 86 Amps
the new requirements for the 24V batteries to obtain the 3000Ah
C > 3h * 86 = 256Ah

 

[danishdeshmuk]

what if for a 12V , 600W UPS, 3 amps total load ...... for 15 minutes back up time ........ battery capacity required ?

 

[mister_rf]

We assume that the output voltage 230V / 3Amps load
P= ~ 700W, the 12V batteries discharged at ~ 58 Amps.
The corrected amperage usage obtained for k=1.2
I = (58)^1.2= 130 Amps
the requirements for the 12V batterie
C > 1/4h * 130A = 32.5Ah

 

[danishdeshmuk]

KAK111 is this correct ?? what other experience man say about it

 

[enjunear]

KAK111 is this correct ?? what other experience man say about it

When you get above C/4 discharge rates on batteries (Pb-acid, for sure), you will certainly see this effect. You should take it into account, especially for fast-discharge scenarios. The closer you get to a C/10 discharge rate, the less pronounced the effect becomes (note exponential dependency on I).

 

[kak111]

Peukert's law, presented by the German scientist W. Peukert in 1897,
expresses the capacity of a lead–acid battery in terms of the rate at which it is discharged.
As the rate increases, the battery's available capacity decreases.

Like this....


For a lead–acid battery however, the value of k is typically between 1.1 and 1.3.
(1.2-1.6 for flooded batteries)

The Peukert constant varies according to the age of the battery, generally increasing with age.

Application at low discharge rates must take into account the battery self-discharge current.
At very high currents, practical batteries will give even less capacity than predicted from a fixed exponent.

Peukert's law becomes a key issue in a battery electric vehicle where batteries rated,
for example, at a 20 hour discharge time are used at a much shorter discharge time of about 1 hour.

pict. from batteryunversity.com


The equation does not allow for the effect of temperature on battery capacity.

Read more
http://en.wikipedia.org/wiki/Peukert's_law

This cannot be used with the usual battery data that is available.
This is because the only battery data available is usually something like the 20 hour discharge rate and Peukert's exponent.
For this reason we must modify the equation to T=C(C/R)n-1/In by introducing another term [(C/R)n-1]
that takes into account the battery hour rating and capacity.

Read more
http://www.smartgauge.co.uk/peukert_depth.html

For most renewable energy systems, the most important battery characteristics are the battery lifetime,
the depth of discharge and the maintenance requirements of the battery.
This set of parameters and their inter-relationship with charging regimes, temperature and age are described below.

**broken link removed**

 

[danishdeshmuk]

what a typical computer takes the amps ? with CRT monitor , ADSL Router .........

 

[danishdeshmuk]

what battery capacity & what UPS capacity is needed for 1 Amps load of computer & its accessories ? Including CRT monitor which takes the total amps from 1 amps to 2.7 amps for a second or two ...... should i consider it as 1 amps load ?

thanks

---------- Post added at 23:43 ---------- Previous post was at 23:32 ----------

i think if i consider 1 amps load then the calculations will be :

220 w = 220/0.8 = 275 VA , consider it as 300 VA
for half hour back-up , 300 VA * 0.5 hr = 150 VAh
for using 12 V battery , 150VAh / 12 V = 12.5 Ams.hr battery capacity , right ??

 

[kak111]

If we do not take into account the Peukert`s law
and the factor of 0.8 present 12Vdc to 220Vac conversion efficiency.
Right , Yes.

With Peukert´s

-current from 12V battery is (220/12) = 18.3A
if Peukert´s constant is 1.2 then with simple form...

For a one-ampere discharge rate, Peukert's law is often stated as:

C_p = I^k t

where:

C_p is the capacity at a one-ampere discharge rate, which must be expressed in A·h.
I is the actual discharge current relative to 1 ampere, which is then dimensionless.
t is the actual time to discharge the battery, which must be expressed in h.

Icorrected = 18.3^1.2 A = 32.7 A

and when time is 0.5h

then C_p is 32.7A * 0.5h = 16.4Ah
( 12Vdc to 220Vac conversion efficiency not included)

 

[danishdeshmuk]

Should haven't been consider the inrush current that is when i switch on the CRT monitor then total amps goes upto 2.7 amps ? right ???

why didn't you included the 12 Vdc to 220Vac conversion efficiency ? not needed or what's the reason behind it ?

b.t.w thanks very much , very informative indeed

 

[kak111]

Inrush current has a non-existent role in the total amount of the discharge and can be ignored.
The calculation has two non-exactly known variables ,
Peukert constant and efficiency values , that we do not exactly know,
so I presented in the above only Peukert laws effect to the dicharge value.

 

[danishdeshmuk]

so on a safe side 15 Amps .Hr or 20Amps.Hr battery would be enough for 1 Amps total load of my PC , ok, what do you say ?