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Exercise E1 Multiphase systems
(written test, approx. 4 % of a 120-minute written test, SS2021)

A symmetrical three-phase generator in a delta connection shall be considered in the following.
A voltage with the RMS value $U_{\rm RMS} = 110~\rm V$ is applied between the terminals of each winding.
Through each of the windings, there is a current with an RMS value $I_{\rm RMS} = 5 ~\rm A$ and a phase shift of $\varphi = +25°$ compared to the voltage.

a) Draw the circuit diagram.

Result

ee2:ezrkjzifcegttcpc_solution1.svg

b) Specify the RMS value of the phase voltage $U_\rm L$ and the string voltage $U_\rm S$.

Path

Since the given voltage of $U_{\rm RMS} = 110~\rm V$ is applied between the terminals of each winding, this is also the string voltage $U_\rm S$.
For delta configuration, the phase voltage $U_\rm L$ is equal to the string voltage $U_\rm S$.

ee2:ezrkjzifcegttcpc_solution2.svg

Result

  • $U_{\rm S} = 110~\rm V$
  • $U_{\rm L} = 110~\rm V$

c) Specify the RMS value of the phase current $I_\rm L$ and the string current $I_\rm S$.

Path

Since the given current of $I_{\rm RMS} = 5~\rm A$ is running through each winding, this is also the string current $I_\rm S$.
For the phase current $I_\rm L$, one has to consider that at each node the sum of all the notes must be zero: $\sum_i I_i =0$.
By this (and showing in the example in the image below), One can see, that $I_\rm L= \sqrt{3} \cdot I_{\rm RMS} = \sqrt{3} \cdot 5~\rm A$

ee2:ezrkjzifcegttcpc_solution3.svg

Result

  • $I_{\rm S} = 5~\rm A$
  • $I_{\rm L} = 8.66~\rm A$

d) Determine the active power.

Path

The active power $P$ is given by: $P = 3 \cdot U \cdot I \cdot \sin(\varphi)$ \begin{align*} P &= 3 \cdot 110{~\rm V} \cdot 5{~\rm A} \cdot \sin(25°) \\ &= 1'610.888... ~\rm W \end{align*}

Result

$W = 1.61 ~\rm kW$