Understanding three phase power consumption

[mfacen]

I work on a shrimp farm as an electronics engineer, I dont have a lot of experience in power electronics, just wiring three phase motors required by the farm as aerators.
We have a big diesel generator 100 KW of power, today the grid went out and the gen kicked in, after all motors were back to running I read the following at the generator control board:
Voltage phase L1 to L2 = 220 V
Voltage phase L1 to Ground = 127 V
Current L1=157 A L2 = 161 A L3 = 160 A
Power 45 KW

I wonder how can the power be 45 KW total with that consumption.

 

[zsolt1]

hi,
your system is slightly unbalanced so you can consider the general formulas for balanced systems in star connection like below
**broken link removed**
with U=220V I ~=159A , so you have :
S= sqrt(3)UI= 60.5 kVA
P= sqr(3)UI cosfi = 45 kW => cosfi = 0.74 , not unusual power factor for only motors in the circuit... Still an improoved power factor of cosfi=0.82 would be better
For very precise measurement you have to get phase voltages and calculate S for each phase (since your system is slightly unbalanced )

 

[FvM]

your system is slightly unbalanced

I would rate 2 percent imbalance as "well balanced".

The current is well within generator rating, but I wonder if your plant doesn't have an automatic VAR compensator to improve the power factor?

 

[zsolt1]

maybe "slightly unsymmetrical " Come on FVM jou live in Germany , you like perfection
now seriously , if there are mono-phase induction motors you should check their capacitors . Usually single phase im has 1 cap for startup & 1 for running (maybe centrifugal switch to decouple the startup cap) , depending from it's power the motor could also have only one capacitor . It happened to me that i changed a capacitor for 0.75kw motor and it's cosfi improved from 0.62 to 0.8 (probably was dry because it was an old motor) . Also if you have single phase motors you could move some from L2 and L3 to L1 (that only for the seek of precision ;-) )

 

[KlausST]

Hi,

S= sqrt(3)UI= 60.5 kVA

this is true.

For me it is easier to understand (more visible) that there are three phases each supplying its own apparent power of U(RMS) x I(RMS)
Where U(RMS) is the voltage against the star point (in your case also: round)

Then the apparent power in one phase is 127V x 160A = 20.3kVA.
Here all three phase current are almost the same, so you may multiply this by 3: 20.3kVA x 3 = 60.9kVA.

You can calculate also with individual currents and add all three.

I rather use the formula: P(app) = 3 x U(RMS-Star) x I(RMS)

Klaus

 

[FlapJack]

zsolt1, your attachment does not load.

 

[zsolt1]

sorry , the attachment was only a drawing in paint representing the voltage & current vectors .. ... anyway things are clear your system is ok , if you try to achieve the cosfi neutral of 0.82 it would be even better , Since this is only a backup supply (UPS) that works rarely don't bother , i bet that the normal power supply comes from a transformer post that has automatic cosfi improvement equipment

 

[FvM]

Since this is only a backup supply (UPS) that works rarely don't bother.

It bothers well if the generator kVA rating would be exceeded due to the bad power factor and the circuit protectors trip.

 

[zsolt1]

yes , but he didn't say that :grin:

- - - Updated - - -

S = sqrt(3)UI is considered for symetrical star connection , where U is line voltage and I is line current which is equal to phase currents due to star connection . Because your system is symmetrical (or just ''slightly asymmetric") i assumed for all 3 phases line voltage U~220V and line current I~ 159A .
Like suggested , for ( ) precision you can determinate apparent power separately for all 3 phases using phase values instead of line values and then add them .
Of course the bad power factor indicates some trouble , just check the motors .. maybe you have a bad one

 

[mfacen]

Thank you for all the replies, I'm sorry we didnt have internet at the farm for several days due to a storm that damaged the antenna. I forgot to mention this farm is located in the souther mexican pacific.
So the discrepancy I see is due to the Power Factor from what I understood from the replies, and it is not very efficient as it is. The weird thing is that there is a capacitor bank installed that was supposed to rectify the power factor and make the setup more efficient. 90% of the power is used to run three phase motors that are anywhere between 2 and 5 horse power. It is important for us to make things more efficient as we are working at the limit of our energy supply from the utility company and the yield of the shrimp depends on how much oxygen we can inject into the water ( we use blade airators and blowers ). Do you recommend talking to the engineers about the power factor correction ? It would be a big deal as they are in Mexico City and its quite away where the farm is located.
Thanks again for your input on this matter.

 

[FvM]

Power factor isn't necessarily a problem for you, as long as the generator has sufficient kVA reserve to handle the reactive power (which is apparently the case), and you're not charged for kVArh by the utility company (don't know about your price plan).

The 15 kVAr provided by your diesel generator don't cost you much, a few 100 W additional losses, less than 1 percent of the total fuel consumption.

 

[schmitt trigger]

What I would be concerned is the electric motor's startup current, which can be several times its running current.

If you have to start your fans and aereators from the diesel generator, make sure that you don't start them all simultaneously. You could trip the generator's breaker.

Rather, start the first motor, wait for the current to stabilize, then start the second.....repeat until all motors have started.
Then you can start the remaining loads (computers, lights, etc)

 

[KlausST]

Hi,

Just to clarify... A power factor of 0.7 does not mean that 30% of the energy is wasted..

The current in the wires is 30% higher than in optimal state and there will be a bit more power loss in the cables, but not 30%.

An example.
Let's say the cable resistance is 1 ohms, the load takes 2W of real power.
To ease the calculation let's s some a single phase system with 130V at the load (in either case ).

W ith a power factor of 1.. The current is 2000W/ 130V = 15.38A.
The power loss in the cable is I x I x R = 15.38A x 15.38A x 1 Ohms = 237W.
This means the company has to supply 2237W.
(The input voltage at the company is 145.4V)

Now with a power factor of 0.7 the current is 15.38A / 0.7 = 22A.
The new cable loss at 22A is 484W.
So the company has to supply 2484W. This is the true working power.
It is 237W more than with a power factor of 1.
(The apparent power a at the customer is 130V x 22A = 2860VA)

Klaus

 

[zsolt1]

yep, talking with engineers would be good idea ... You can check yourself some things also. If you have a capacitor bank built in , probably the engineers who designed the system , had sized correctly the capacitor bank . Condensators dry out in time. So if it is available, you could measure separately the condensators from the capacitor bank. If there is a big difference between what you measure and what is on the nameplate , the condensator needs to be changed.
Indeed in ac you have : active power [kW], reactive power [kVAr] and aparent power [kVA];
You pay for the active power [W] (actually the active energy consumed [kWh].....) , The bad thing is that most consumers consume the active power [kW] and also generate reactive power [kVAr], This is the 'bad' power since is mostly not consumed (i call it ping-pong power), it is transported in power lines from one place to an other and it just over charges the wires with amps . Actually this what damages your power factor in the end .
And actually you are paying for how bad you damage the power factor also. How?
Since energy is integral of P=UIx cosfi , you may have low cos fi , so you could say good for me, my counter will count less P in the given period :thumbsup: .. The trick is that due to reactive power you have increased current in the power line (see example from Klaus) including the counters current coil , so it keeps rotating . You have current flow in the circuit that is useless .

PS: motors (generally 'coils') and capacitors generate reactive power Q . By the difference of their nature, inductiv and capacitiv ,they can compensate each other . So if you have Q generated by coils you put capacitors and if you have capacitor like consumers you put coils . Indeed capacitive like consumers are rare , for example at large factory or power plant , the smoke needs to be filtered in ionisator chambers also . Those are like huge capacitors and smoke passes through them.