Why is there a huge drop in the values of active and reactive power when the voltage of three phase voltage source is changed?

Hi Community,

I have modelled a grid using a programmable three phase voltage source and there are three loads attached to it. Model & Parameter file is attached. Each load is defined by the rated active and reactive powers (P& Q) as follows -

Load 1: Rated P = 25 kW; Rated Q = 5 kVA

Load 2: Rated P = 30 kW; rated Q = 10 kVA

Load 3: Rated P = 125 kW; rated Q = 10 kVA

The loads 2 and 3 are switched off at T = 2 secs and switched back on at T = 4 secs. I am also measuring the powers at each of the individual loads as well as the total power supplied by grid.

When I use the voltage source rating = 2300V, the P & Q values are nearly the same (but not exact match) as the above specified parameters.

When I use voltage source = 400V (with everything else remaining the same) there is a huge drop in the measured P & Q values and I am unable to follow why is this happening?

Q1) Why are the measured P&Q so low for V = 400V vs V = 2300V? Is there some limitation on the power this grid can deliver? If yes, then how is this being limited and how can I determine what is the maximum power rating of the load that can be connected across this grid source?

Q2) Additionally in the first scenario with 2300V why are the measured P&Q not the exact match of rated power of loads slightly lesser. I assume some power is lost in "losses" but I canot figure out where are these losses happening?

Answer 1

[Ankita]

For Q1) I have figured out that a low value of parameter "Short-circuit power" was causing the issue.

I was directly specifying the source internal impedance via R and L values, and so I did not pay attention to this parameter. However on some helpful pointers from MATLAB technical support team, I realised the below

I. Parametrize Source with RL Values

•Previously, I set source & line impedance manually (for both 2300V & 400V) as follows

• R = 2 ohms
• Line R = 0.2 ohms
• Line X/R ratio = 18.8
• Line L ~= 0.01 Henry

•For 2300V, this internally set theSSC power = 1.38*10^6 W. This is a high value so things were working fine.

•For 400V, This internally set the SSC Power = 4.17*10^4 W. This limited the power delivering ability of the load

II. Parametrize Source with SSC power & X/R ratio

•When we used parametrize source as a function of SSC Power & X/R ratio – things were working fine. Because SSC Power = 1*10^6 W

•I then looked into the calculation of SSC power. Back calculated R & L values for the above SSC (1e6) and now model is working fine at 400V as well. with R = 0.281 ohms & L = 0.017 Henry.

For Q2) @Vinay's comments helped.

Without any source & line impedances power values match the rated power (Ideal source, no internal impedance hence no losses).

With the source impedances in picture, reducing the rated power by the ratio of (V_terminal/V_rated)^2 gives me a very close match to the measured power values.

where,

• V_terminal = measured voltage at the terminal of the source, (i.e. after the drop across internal impedances)
• V_rated = rated voltage of loads.

Answer 2

[Vinay]

Hi @Ankita

The P and Q values deviates from the rated power of the load when rated voltage (2300V PP rms) is given from the source is due to voltage drop in the source impedance(Series RL type) .

Source voltage(Phase to Ground) =

There is a drop of around 25 V across the source impedance as the voltage after the source load is 1853V .This causes a drop in the power across the load.

The load impedance is paramatrized in terms of rated voltage and rated power which means load will consume rated power when rated voltage is provided across them.The power across the load is proportional to the square of voltage applied.

When source is changed from 2300 V to 400 V the power will be reduced by factor of approximate and can be observed in the measured power values.

I hope this helps!

[Ankita]

Hi @Vinay,

Thank you for your investigation and observations!

(1) Voltage Ratings: From your reply I notice you are talking about the rated voltage of the load in terms of Ph-Gnd. I was under the impression that rated voltage of each 3 phase component is specified in terms of Ph-Ph rms values. Is this not true?

I applied rated voltage of both the grid and the load as 2300V. With this

Ph-Ph, peak value of voltage = Ph-Ph, rms Voltage * sqrt(2) = 2300 * sqrt(2) = 3252 V.

And the voltage after source impedance block is ~3000V (with all 3 loads connected).

which would be 252 V dropped across the source impedance. From here, I know my loads demand P = 25+30+125 = 180 kW. How do I calculate actual delivered P to the loads to verify the drop I am seeing is indeed entirely across the source impedance?

(2) for your last point, when the source voltage changes from 2300 V to 400V, the rated voltage of the loads have also been changed to 400V. In this case apart from the drop across the source impedance contributing to some small power loss, should I still be seeing a massive reduction in Power?

[Vinay]

Hi @Ankita,

The Phase-2-Phase voltage are used for the calculation of the power.The drop in the voltage is occuring due to the following two parameters.

1. Three Phase Source Impedance
2. Source Impedance and Line X/R.

The issue can be resolved by removing the impedance line and making the 'Impedance' of the 'Programmable Three phase source' to zero.This will deliver 180 Kw for both 2300 V and 400V.

[Ankita]

Thank you @Vinay!

Your suggestions helped me a lot to clarify Q2 of my post.

Without any source & line impedances power values match the rated power (Ideal source, no internal impedance hence no losses).

With the source impedances in picture, reducing the rated power by the ratio of (V_terminal/V_rated)^2 gives me a very close match to the measured power values.

where,

• V_terminal = measured voltage at the terminal of the source, (i.e. after the drop across internal impedances)
• V_rated = rated voltage of loads.