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I tried to study your networks, but it is difficult without the parameters (voltages, impedances).

It is certainly an interesting exercise to write a program for load flow and short circuit calculations. But is is worth of the effort? It may be, if the program shall have some special properties that are not available in existing programs. But it might take a long time to write such a program...

You can model a single wire in an three-phase network with the help of a double open-line fault (phases b and c open, or broken, for example), see the attached picture. In practice, this is very much the same as a single-line-to-earth...

A similar problem is discussed in thread238-376120

If you are interested in calculating voltages and currents in three-phase networks, you might have look at a program called elplek, see http://pp.kpnet.fi/ijl . Elplek is a simple, freeware load flow and short circuit program, that is easy to use (in my opinion:). The program can also do simple...

The parameters of the line and load have not given by the OP. I have tried to guess the parameters, as Load: 111 MVA, pf 0.92 Line: 100 km, modeled in 10 sections. X = 0.44 Ohm/km, R = 0.09 Ohm/km, B = 7.7e-5/Ohm km (realistic, or typical ? ) With these parameters the pf = 0.84 at the source...

See thread797-339131

As waross wrote, either use phase-phase or phase-neutral variables, but not a mixture. The complex power in an impedance is equal to: voltage over the impedance multiplied by the conjugate of the current through the impedance. Your equation for the power using phase-phase voltage is not...

As davidbeach observed, there is not such a thing as a negative or zero sequence subtransient reactance (let's call them X2" and X0"). The d-axis subtransient reactance X"d is used to calculate the initial short circuit current for a 3-phase fault. Similarly, the initial sequence currents for...

I think the explanation is in the circulating current of the delta-windings, see the attached single phase diagram. The fault current returns from earth to the neutral point of the Y-windings of the transformer, and splits there in three equal components in the a-,b-,and c-windings. The a,b,c-...

See elplek at http://pp.kpnet.fi/ijl/ It is free.

The first picture, with Z = 0.1 Ohm. The second one is included in the main message.http://files.engineering.com/getfile.aspx?folder=385d1333-f9be-4eea-9f43-4948b028bc79&file=LLE-0.1.jpg

A short circuit calculation typically gives only the phase currents in the case of a LLE-fault. The current to earth must be calculated separately from the current balance. Consider an example: V = 1000V, Z1 = Z2 = Z0 = 0.1 Ohm (first picture). The A- and B-phase currents and the current to...

Directional overcurrent relay, maybe http://www.ece.uidaho.edu/ee/power/ECE525/hws/lab2/Directional_OC_Chapter-5_AMPS.pdf

Try primary litium cells with a nominal voltage of 1.5V. The voltage is in practice about 1.7V.

The coefficient h contains all factors, such as Stefan-Boltzmann constant, area, emissivity. (I just picked "h" to mean all this.) The equations can certainly be refined. I only wanted to explain the basic idea.

The light output from an incandescent lamp depends on the temperature of the filament, as stated above by others. The calculation of the voltage needed for a given filament temperature is straightforward, in principle (but not in practice :). In steady state the temperature assumes a value where...

Jeah, I have written the program. The downloadable file motdatpr.zip contains the program and a brief description of the calculation method. It is a simple and straightforward solution of the parameters, nothing to write a paper about. For name and e-mail see above.

Motdatpr, maybe (not motdat)? See http://pp.kpnet.fi/ijl/prog.htm But it is not a simulator. It can be used to prepare induction motor data for the ATP-transient simulator. For contacts, use the feedback form on the "main" page http://pp.kpnet.fi/ijl/ The program "elplek" on the main page is...

The problem is an unconventional load flow-problem: Power and power factor of the generator are given, but not the voltage. Hints for solving: Unknowns: The current from the generator, the current to the load, and the voltage at the 11kV bus, all as complex variables. Three equations needed...