Siemens SINAMICS S150 - 6.3 Basic information about the drive system

Fundamental Principles and System Description
Engineering Information
SINAMICS Engineering Manual – November 2015
Ó Siemens AG
173/528
Calculation example 2:
A SINAMICS S150 converter in degree of protection IP20 is to be operated as a centrifuge drive on a 400 V supply
system at a maximum ambient temperature of 40 °C at a maximum installation altitude of 1000 m. The motor current
I(t) and motor frequency f(t) over time are shown in the diagram below. In continuous batch process without
downtimes, therefore, the converter operates according to a regularly recurrent, periodic load duty cycle. The
converter is to be operated at the factory-set pulse frequency in vector control mode (drive object of Vector type).
T
1
: Fill centrifuge
T
2
: Accelerate to spin speed
T
3
: Spin at spin speed
T
4
: Decelerate to emptying speed
T
5
: Empty centrifuge
T : Load cycle duration (T = T
1
+ T
2
+ T
3
+ T
4
+ T
5
)
Absolute value of motor current |I(t)| and motor frequency f(t)
1. Select the converter:
According to the diagram shown above, the converter must be capable of supplying a short-time current I
ShortTime
of
520 A. It therefore requires a base load current I
H
of at least
I
ShortTime
/ 1.5 = 520 A / 1.5 = 347 A.
For this reason, a converter rated for V = 400 V and I
rated
= 490 A is selected which, according to catalog D 21.3, has
a base load current I
H
of 438 A and thus a short-time current I
ShortTime
of 1.5
*
I
H
= 657 A. The factory-set pulse
frequency for vector control mode is 2 kHz. We shall now use a calculation to determine whether the selected
converter is suitable for the required periodic load duty cycle operating on the factory-set pulse frequency under the
conditions specified above.
2. Determine the current derating factor k
D
:
· k
Temp
: 1.0 (ambient temperature ≤ 40 °C, installation altitude < 1000 m, degree of protection IP20),
· k
Pulse
: 1.0 (factory-set pulse frequency)
· k
Parallel
: 1.0 (not a parallel connection of S120 Motor Modules)
· k
IGBT
: This factor must be taken into account because of the periodic load duty cycle. According to the
diagram above, the following applies:
- Current ratio ΔI = I
ShortTime
/ I
BaseLoad
= 520 A / 115 A = 4.52 ≈ 4.5.
- Load duty cycle duration T = 300 s.
With ΔI ≈ 4.5 and T = 300 s, the derating factor is according to derating characteristic 3 which is
applicable to the SINAMICS S150 converter with V = 400 V and I
rated
= 490 A: k
IGBT
= 0.8.
The current derating factor k
D
is therefore calculated as follows:
8080010101 ..... =×××=×××=
IGBTParallelPulseTempD
kkkkk .
3. Determine the permissible short-time current:
With a base load current I
H
= 438 A and a derating factor k
D
= 0.8, the permissible short-time current is
AAIkI
HDTimeShort
5254388.05.15.1 =××=××=
.
This value is slightly higher than the required maximum motor current of 520 A and is thus permissible.