# How to save energy: tips and solutions

## Recommendation points

This article will focus on simple and understandable ways to save energy in everyday life and during construction. The article will tell you about the most common problems in the operation of electrical appliances and the network, associated financial losses and solutions to these problems..

## The problem of invisible consumption

The longer the wire, the greater the resistance and volume of the conductive metal. Wires consume energy to create and maintain a magnetic field around the metal rod. This also applies to wires with excess cross-section..

Illustrative example. Observations have shown that the power loss in home networks is 0.5 W / r. m / mm2. That is, 1 running meter of a cable with a cross section of 1 mm2 (one core) when operating at design power (up to 300 W for 1 mm2) consumes 0.5 W / h on induction. Suppose that an LCD TV (250 W) and a table lamp (50 W) with a total power of 300 W are connected to the power point. With an extension cable length of 1 m, the losses will be:

• N1 = 0.5 / (300/100) = 0.16%, which is quite acceptable loss.

But with a cable length of 10 meters:

• Nten = N1 x 10 = 1.6% or 5 W / h

With continuous operation for 10 hours a day for 30 days, losses from only one cable 10 m long, with a cross section of 1 mm2 and a load of 300 W will be:

• P = 5 x 10 x 30 = 1500 W or 1.5 kW

Common household extension cords have a cross section of 1.5-2.5 mm2. For every additional 0.5 mm 40% idle flow is added. I.e. wire 1.5 mm2 will lose 0.65W per meter and the wire is 2mm2 – 0.9 W / m.

Decision:

1. When designing a repair or construction, calculate the trunk so that the length of the wires is minimal. This will reduce costs not only for material, but also for operation..
2. Choose the length and size of your portable extensions carefully.
3. Try not to use extension cords with a large margin of cable. This applies to the work of energy-intensive tools, machine tools, especially welding machines. A coiled bobbin is essentially an induction coil, which itself is a consumer (resistance).

## Standby problem

Appliances that are off but connected to the mains consume a small amount of energy. Approximate consumption of devices in “standby” mode:

1. Home computer with monitor – 5 W / hour.
2. Laptop (with a charged battery) – 2 W / h.
3. Air conditioner – 2.5 W / hour.
4. Music center, home theater – 2-6 W / hour (depending on power).
5. MFP (printer) – 1 W / hour.
6. Mobile phone / tablet charger – 3.5 W / h.
7. Router – 4 W / hour.
8. DVD player – 1.5 W / hour.

In total, all devices connected at the same time consume about 23.5 W / h. Per day comes out: 23.5 x 24 = 564 W / hour. Per month: 564 x 30 = 17 kW drain completely unnoticed from your pocket.

Decision:

1. Disconnect from the mains appliances that have not been working for a long time * (more than 3 hours), especially electronic ones – TV, computer, music center, speakers.
2. Use special controllers that automatically turn off the outlet at a set time (for example, during office hours or at night). The cost of such a device is about \$ 10. That is, while it pays off within 2 years.
3. In order not to wear out the outlet by constantly pulling out the plug, use surge protectors with an indicator shutdown button.

* This does not apply to refrigerators. Turning off for more than 10 hours is highly discouraged.

## Lighting

According to statistics, from 12 to 16% of electricity in Russia is spent on lighting. In order to verify this, you can make a simple calculation.

Illustrative example. A standard three-room apartment has 8 rooms that require lighting (3 rooms, 2 corridors, a bath, a toilet, a kitchen). For normal room lighting, 150 W * incandescent lamps are required, for other rooms – 100 W. To provide light in an apartment using incandescent lamps, you will need:

• P = 150 x 3 + 100 x 5 = 950 W / hour

Suppose that in maximum mode the light is on for 3 hours a day. Then the daily consumption will be:

• Rday = 950 x 3 = 2850 W or 2.85 kW

Per month it is: 2.85 x 30 = 85.5 kW.

* Installing notoriously weak lighting fixtures can cause discomfort, fatigue and impaired vision.

Wrong decision. Since childhood, we have been taught to leave the light behind us. This is useful for long-term (more than 3 hours) outages. Frequent switching on / off does not save electricity, but, on the contrary, consumes more of it, since a pulse of energy is required to heat up the gases or the lamp spiral. In addition, for this reason, lamps most often fail ahead of time. “Economical” lamps are especially sensitive to this process.

The correct solution is to gradually replace the lighting lamps with energy-saving ones. A modern 15 W fluorescent lamp is identical in luminous intensity to a 100 W incandescent lamp. This means that the effect of replacing all lamps in our example will be equal to:

• Rsber = 85.5 / 6 = 14.25 kW / month. or 85.5 – 14.25 = 71.25 kW net savings.

Although such lamps are more expensive than incandescent lamps, they last more than 3 times longer (1500 hours for a conventional lamp versus 5000 hours for an economical one).

Exact calculation of the effect of replacing lamps:

1. The cost of an incandescent lamp B10–100 W E27 (100 W) is 0.3 cu e.
2. The cost of a 15 W energy-saving lamp is 2.7 USD. e.
3. The cost of 1 kW of electricity is 0.1 USD. e.
4. The difference in the price of each lamp is 2.7 – 0.3 = 2.4 cu. e.
5. The difference in the price of all lamps: 2.4 x 8 = 19.2 cu. That is, we accept 20 cu. e.
6. Power consumption during the operation of energy-saving lamps in the mentioned mode: 15 W x 8 pcs. = 120 W / hour.
7. Energy consumption per month in the mentioned mode: 120 W x 3 hours per day x 30 days = 10800 W or 10.8 kW.
8. Difference in energy consumption: 85.5 – 10.8 = 74.7 kW.
9. Approximate difference in money: 74.7 kW / month x 0.1 USD e. / kW = 7.47 us. e. / month net savings.
10. Payback period for all lamps *: \$ 20 i.e. / 7.47 = 2.6 months.

That is, replacing all lamps will pay off about three months after installation..

* At a price of 1 kW = 0.1 c.u. e. and lamps of proper quality.

The most modern lamps are LED, or LED. When using them, lighting becomes practically free. The equivalent of a 100 W incandescent lamp is equal to a 4 W LED lamp. They cost about 10 cu. e. / piece and require the installation of additional power supplies.

## Saving in little things

Finally, it is worth mentioning some details that we usually do not pay attention to. Here are some quick tips to help improve the efficiency of your appliances and improve your home comfort:

1. Clean the filters of the vacuum cleaner as often as possible when cleaning. Clogged filter reduces suction power by up to 50% and consumes up to 30% more electricity.
2. Cleanliness is not only a guarantee of health, but also of economy. Lighting devices should be as transparent as possible. The same goes for windows. Their main function is to transmit natural light, which reduces the need for additional lighting. Contaminated incandescent appliances – kettle, toaster, stove – release energy less efficiently, which means that food takes longer to cook.
3. Use dimmers – dimmers – instead of rocker switches. They will allow you to adjust the light power in the desired mode. True, the dimmer does not work with “economical” lamps, but is ideal for LED.
4. Do not operate the washing machine with an empty or half-empty drum. The same kilowatts can be used to wash a full drum. Download it according to the instructions.
5. Choose a place for the refrigerator away from heat sources (stove, window with sunlight) – this reduces up to 30% of its power consumption.
6. Whenever possible, when purchasing equipment, give preference to devices with energy saving classes A, A +, A ++ (they consume energy up to 30-40% less).

Saving electricity is not only beneficial for your wallet, but also for the entire planet. After all, it not only gives us energy, but also utilizes it in the form of waste heat. The less energy we use, the cleaner the air in our cities.

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