ISBN 9786054735846

A. ARIFJANOV et al./ ISEM2016 Alanya – Turkey
In general, the equation describing the daily schedule of electric load has the form:
S (x) = Ui =1,24 Si (x)
M
i1
M
i
or S3(x) =
6hi
(xi 3 – 3xi 2 x + 3xi x 2 – x 3 ) +
6hi
(x 3 – 3x 2 xi-1 + 3x xi-1 2 – xi-1 3 ) +
2
2
2
2
f
f
i1
Mi
1
hi
fi
1
Mi
1
hi
fi
i
Mi
hi
Mi
hi
xi
xi
x
x
x
xi
1
x
xi
1
h 6h
h 6h
h h 6h
6h
i
( M
M
h
i
i
i
M
2
i
2
i
M
h
where:
M
a
a
0
1
1
2h
i1
3
x
)
6h
2
f ) x(
M
x
i
i
i1
i
6h
i
i
1
)
6h
M
i
i
i1
3
6h
a
i
i1
i0
i
(
M x
x
i
3
i
i
3
x a
M
M
x
i1
( M x M x ) 1
i1
i
i
i
i
x
i1
i1
2
i
x
3
i1
a
i2
) x
2
6f
i
(
M
i1
xa
i3
i
i
i
x
2
i1
x
6f
i
M
x
i1
i1
M
i
x
2
i
i1
2f
h
2
i
i1
x
i
i
M
a 1
2
a
2
2
2
2
2
( M
i
x i 1
M
i 1
x i M
i 1
h i M
i
h i 2 f
i 1
f
i
2 h
i
1 3
3
2
(
6 6
2
3
Mi
1
x
i Mi
x
i1
fi
1
xi
fi
xi
1
Mi
hi
xi
1
Mi
1
hi
xi
)
6hi
On the base of this algorithm, substituting the values Мi ,
fi
, hi, xi from the table, we obtain a
system from 24 equations in 3 rd order.
As a result, for the curve of the daily schedule load we obtain a system of equations, which has the
following general form:
S 3(x) = a i0 x 3 + a i1 x 2 + a i2 x+a i3,
where ai0, ai1, ai2, ai3 - coefficients of each equation.
Substituting the data from the daily schedule in these equations, we can obtain the curve segments
of the daily schedule, and combining all of the 24 segments, as a result, get the curve of daily
schedule, that is almost completely coinciding with the characteristic of daily schedule of consumer
loads.
For each electrical consumer or appliances construct its equation of the curve and own formula for
the maximum electrical load.
As a result, the calculation is obtained precise and more authentic and closer to the actual load,
taking into account the peculiarities of load conditions change, the time factor for each
appliance and consumer included in the city's power system.
Example 1. The characteristic daily schedule electrical load of the hospital, built using the Lagrange
polynomial for x0 = 1, x1 = 10, x2 = 16, x3 = 22; y0 = 10, y1 = 54, y2= 98, y3 = 10 is as follows
(fig.1) [6]:
i1
h
P(t) = 21.95057 – 15.91852 t + 2.72962 t 2 – 0.26084 t 3
)
2
i
175