Some calculations for modified sine wave inverter

Hi ,

I need some help with calculations for modified sine wave inverter (square wave with deadband).
I have to power a variable frequency drive system / induction motor which is rated for 7.5HP ( 5500W ) @ 470V ; measured with multimeter.This is to supply the drive when the 415V phase to phase mains voltage is lost.
It will run for a maximum of 2 minutes.
The possibilities are for using
A) 12V (7.6AHr battery) x 4 = 48V
B) 12V (7.6AHr battery) x 6 = 72V
Is there any way to get a good estimate of the current requirement from the battery for both the possibilities ?

Using raw calculations...

To get 5500 W from battery pack A:
5500 / 48 = 115 amps. This may not be possible from small batteries as you describe. Even if they can provide 115 A it will deplete them in less than 4 minutes.

From battery pack B:
5500 / 72 = 76 amps. This will deplete the batteries in six minutes.
To get 76A requires that the primary power loop have no greater than 1 ohm impedance. Each battery can be allowed to have only the tiniest internal resistance.

In any case, to start the motor going may require surge current (although I don't know if this is the case with a variable frequency drive system / induction motor). The batteries and inverter may be called on to supply 2 or 3 times normal current for a few seconds.

Furthermore when driving the motor with square waves, you lose some efficiency as compared to a true sine wave. Expect to use 10 percent more power than you measure when using mains power.

I agree with BradtheRad. It is even worse, you need some reserve for accelerating the motor + load, and you have loss. so I would calculate with about 7kW (inclusive loss).

You need large batteries to supply about 100A for two minutes. When using starter batteries, life expectancy will reduce significantly. For such porpuse you need many Ah of starter batteries, or "deep cycle" batteries (expensive).

The effective Ah rating of batteries drops when they have to supply large current. You may check the websites from battery manufacturers for application notes.

The B pack has preference w.r.t to current, but note that most safety standards consider this a dangerous voltage.

Gentlemen,thanks for replying.

Actually I have seen a couple these units working.

1)One of the units drive the motor directly.This is the one with 48V battery pack.The motor is driven with a frequency of 10Hz,slower speed,not the usual 50Hz.Since the speed is lesser, for the same torque, the power is lesser.Probably that is why it works.

2)The other unit with 72V drives the motor through a VFD also at a slower speed than the usual running speed.In the case of a VFD there is a ramping time which can be programmed,which probably reduces the starting torque needed.
In both cases it is a full square wave with a center tapped primary transformer.
These batteries are the sealed maintenance free type.I could not get the battery datasheet,so I do not know how much would be their internal resistance.

Hello,

We based our calculations on your info. If you provide us wrong, insufficient or misleading info, you don't get a useful answer and we are wasting our time.

I fail to see which part of my info was wrong or misleading, maybe insufficient in hindsight.

Since, BradtheRAd's reply said that it is unlikely that a battery of 48V can sustain a current of 115A ,the possible reason for it to work occurred to me.I just gave some supplementary info as it struck me that it could be relevant.My second post was to ask if my reasoning was correct.

It is definitely not my intention to waste anyone's time.

Going down from 50 Hz to 10 Hz reduces the input power significantly (with almost factor 5), hence the current consumption and Ah product reduce with same factor. In addition a battery has better discharge efficiency under less current. This makes it feasible with relatively small batteries.

The actual Ah rating for the batteries depends on the battery type and the number of outages/year.

phaedrus said:

Gentlemen,thanks for replying.

Actually I have seen a couple these units working.

1)One of the units drive the motor directly.This is the one with 48V battery pack.The motor is driven with a frequency of 10Hz,slower speed,not the usual 50Hz.Since the speed is lesser, for the same torque, the power is lesser.Probably that is why it works.

2)The other unit with 72V drives the motor through a VFD also at a slower speed than the usual running speed.In the case of a VFD there is a ramping time which can be programmed,which probably reduces the starting torque needed.
In both cases it is a full square wave with a center tapped primary transformer.
These batteries are the sealed maintenance free type.I could not get the battery datasheet,so I do not know how much would be their internal resistance.

Click to expand...

Running the motor at lower rpm may help to some degree, although the W-to-HP formula cannot be evaded.

Your quoted 5500W rating on the label must apply to a 7.5HP load. (The formula is 746 W per 1 HP.) Perhaps however you will attach a load which draws less than 5500W (less than 7.5HP) from the motor. Then battery draw goes down correspondingly. In that case your battery packs can be adequate, the 48V string and 76V string.

Thanks for replying.Now it is clearer to me.

H.P. = 2*3.14*N*T/4500 which is the mechanical formula.Since the torque needed remains the same,if the same load has to be hoisted,at a lower RPM, lesser power would be needed.Which should correspond to lesser current,I suppose.

Perhaps however you will attach a load which draws less than 5500W (less than 7.5HP) from the motor.
The load remains the same,since this is a back up kind of operation.