Consultation on partial replacement of wiring in an old apartment

Michael asks:
Good time of day! Petersburg, the house of pre-revolutionary construction (1914), gas, wiring in the Al risers with cross-section (I won’t say exactly), but I think 4mm2. In the shield on the stairwell - AB at 20A. In theory, therefore, 4400-4600 watts are issued per apartment - max for an hour can be consumed until 6380 - 6670 watts are turned off.

Question 1: how to properly convert Amperes to W (by multiplying) by 220 or 230
It is planned to make a partial replacement of aluminum wiring in the apartment, namely:
forward 2-3 lines of 2.5 mm2 copper to energy-consuming consumers, and leave the light and other outlets on old aluminum wiring. Replacing the input cable to the apartment is not planned yet (we are waiting for the meter to be replaced by the energy supplying organization, and not by our own forces, it seems like since July 2020 this is their task and responsibility)

Question 2: is it sufficient to use a 4mm2 copper wire in the manufacture of the shield, if we take into account that in the future there will be a replacement for the input lead to 4mm2

Question 3: Are the current characteristics selected correctly:
• AB 20A (on the stairs) of IEK (standing at the moment)
• lead-in cable old Al 3-4mm (without replacement)
• counter (counter without replacement old 1960gv)
• switch 2P to 25A (to turn off the power in the apartment)
• AB 2P - 20A (duplicate)
• group AB 16A for connecting new lines
• connection of old Al wiring (via terminals with connection to AB 16A)

I know about the installation of the Ural Federal District, the Dif, but in this case, so far without them.

Regards, Mikhail

The answer to the question:
Hello!
And how do you consume 6.3-6.6 kW with an allocated power of 4.4-4.6 kW?
Questions:
1. You need to multiply by 230 - such a standard, but in general, depending on what technique - if some kind of heater (TEN), then its power depends on the voltage and you need to multiply the current by the voltage that you have in the network.
2. In the sense of "in the manufacture of the shield"? Do you mean the wiring in the shield to do with such a wire? Yes, it is quite possible, you can also 2.5 mm² in the shield, if the input is 20A.
3. The rated currents are chosen in principle correctly. They do not put more than 16A on sockets, and often 6-16A on light.
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One comment

  • Dmitry

    Al wires, dismantle and discard.
    There are formulas, they are from the course of physics, this is about the translation.
    And about 230, what is it all about? this is Un. What standard are you talking about?

    For questions to Michael:
    Proceed on the basis of which meter, and by which automatic machine it is limited, to take maximum power (if I understand correctly, you have 20A it is 4.4 kW. And how you can take more than 6 kW. I don’t understand, explain ....)
    Wiring should be done only with copper (forget about Al group networks, but there is no question about the mains supply network (cable)).
    1.5 mm. holds (Khoring) 19 A.
    2.5 mm. holds 25 A.
    4.0 mm. holds 35 A.

    In the shield, be sure to separate the PEN, and from the shield lead 3x2.5 mm. or 3x1.5 mm. (depending on the load).
    Put machines in the shield .... well, if there is room, then you can divide everything, if not, then group.
    I didn’t set differential protection myself, at the factory, yes, but at home, especially in the old house, a leakage current will be present, tormented with false positives.

    And so, I would start from the head, would demand from your Energy Supply Organization the outgoing machine to the maximum and then I would dance from here, what are the consumers in the apartment, what are the prospects, etc.

    To answer

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