Bosch Power Tools 1375AK, 1375-01 specs

Fig. 8:

Cable cross-section as a function of voltage and lead length

Move horizontally from the left or the right, depending on the type of current, with the value of the power to be transmitted until the row intersects with the column for the voltage. Next, move vertically downwards until you intersect with the line for the lead length (simple length), then move hori- zontally again to the left or the right.

Fig. 9:

Cable cross-section as a function of voltage and performance factor

base line until you intersect with the frequency curve. Then move horizontally to the left or right.

The larger of the calculated values for the cable cross-section is decisive in determining the lead.

Inductive resistance is significant for large cross-sections. These, in turn, are necessary at low voltages or high frequencies. Calculation of the curve in

Fig. 11 was based on an assumed output factor cos f of 0.7 for the consumer.

For single phase alternating current installations with an output factor cos f = 1, inductive resistance can be ignored even for large cable cross-sections.

Example a:

Transmission of 4 kW, 72 V rotary current, cos f = 0.8, lead length (simple) 10 m. Cable cross-section calculated in accordance with Fig. 9: 2.75 mm2.

Cable cross-section calculated in accordance with Fig. 10: 4.8 mm2 (selected cross-section: 6 mm2)

The cable cross-section of 2.75 mm2 calculated on the basis of Figs. 9 and 10 is not sufficient; the cable would get too hot. Testing in accordance with Fig. 11 is not necessary since the cross-section is less than 10 mm2.

Example b:

Transmission of 3 kW, 220 V single-phase alternating current, cos f = 0.9, lead length (simple):

direct current single-phase alternating current

alter- nating current

rubber hose line

securely installed

power to be transmitted (AC)

Fig. 9

The cross-section calculated in Fig. 9 is now tested for tempera- ture rise. Move horizontally from the left with the value of the power to be transmitted until you intersect with the column for the voltage. Next, move vertically downwards until you intersect with the line for the output factor cos f; finally, move horizontally to the right to find the cross- section for the type of lead you are using.

Fig. 10:

Cable cross-section as a function of frequency and inductive resistance

If the cross-section for rotary current resulting from Figs. 9 and

10 exceeds 10 mm2, you must

then apply the precise calculated

value to Fig. 10 in order to take

the inductive voltage drop into

account. Next, move vertically

power to be transmitted (AC)

100m. Cable cross-section calcu- lated in accordance with Fig. 9:

4mm2. Cable cross-section calcu- lated in accordance with Fig. 10: 0.9 mm2. According to Fig. 9, a crosssection of 4 mm2 is required. This value is decisive since Fig. 10 yields a value of only 0.9 mm2 and there is no danger of overheating.

Example c:

As in “Example a”, but at 200 Hz rotary current with lead length of 100 m. The cable cross-section calculated in accordance with Fig. 9 is 27 mm2. This value must be tested in accordance with Fig. 11. In this example, the larger cross-section of 50 mm2 must be selected.

cross-section (with inductive resistance)

Fig. 10

upwards from the horizontal

Fig. 8

Bosch Customer Support Services is always available to answer questions on the use of high frequency tools and the area of high frequency technology in general.