1. “I don’t know what voltage is needed to turn it”? It depends on the motor nameplate. What is its nominal voltage, this is the voltage that must be supplied to it. Of course, it can be 10 to 15% lower. It needs direct current to provide direct current, and it needs alternating current to provide alternating current.
2. “Start a single-phase motor with a power of 50 kW”. A 50 kW motor is a fairly small motor. Common motors are three-phase. It must be a special engine.
3. "I want to use batteries connected in series to start a single-phase motor with a power of 50 kW, but I don't know what voltage is needed to turn it." batteries, which are DC power supplies. Single-phase motors are powered by alternating current, which cannot be directly matched. You need an inverter to convert DC to AClternative, and it is high power, so the cost will not be low. The necessity and feasibility of this plan requires careful consideration.
4. The engine must be loaded to do meaningful work. This is of little importance if it can only be started without load. Starting an engine of such power requires special starting equipment, which is expensive and has to be taken into account.
5. The battery must not only have a certain voltage, but must also have sufficient capacity (amp-hour Ah), otherwise it cannot provide such a large current and the motor will not run. If the battery is 12 V, if it can run at full load (mechanical) for one hour, its electrical energy is 50 kWh/eta, and eta is the efficiency of the motor. If it is 0.9, 55 kWh are needed. Without taking into account losses from energy conversion such as inverters, 55 kWh = 55,000 VAh, the required capacityof the 12 V battery is 55,000 VAh/12 V = 4583 Ah. The capacity of ordinary car batteries does not exceed 200 Ah. If you think about it, you need 4583/200=23 batteries! Connect in parallel or series to match the inverter.
Okay, let's talk about it.
The generator output power of a small hydropower station is 50 kW, the output voltage is 350 V, and the total resistance of the transmission line is 4 ohms in order to lose power to the transmission line
1. If 50 kW generates electricity. If the unit has poor cooling, it is very likely that your unit's cooling system has not been cleaned in a long time. Under normal circumstances, the scale in the water tank should be cleaned. after the device has been used for six months or after a cumulative working time of 500 hours. If you exceed these values, not cleaning it over timeemps can cause your device to not cool properly.
2. If the performance of the diesel generator set fan is poor, such as the fan installation is not firm, the blades are loose, cracked, deformed, etc., it may cause the heat dissipation performance to be reduced. of the unit, thus causing poor cooling of the diesel generator set.
3. In addition, if
the water pump impeller of the 50kw generator set is damaged or the gap between the impeller and the hull is too large, it will cause insufficient circulation. water in the cooling system, causing the water temperature to be too high, resulting in poor cooling of the unit.
4. Dirty cooling water from the diesel generator set will also cause poor cooling of the unit. The cooling water for the unit must be fresh, clean water.re containing mineral salts such as calcium and magnesium. If conditions are limited. and only hard water is available, the water must be softened before use.
5. Additionally, a lack of cooling water in the radiator is also a cause of poor cooling of the unit. Therefore, during daily use of the unit, frequent inspections and timely replenishment should be carried out. the liquid level will be too low, which will affect the cooling effect of the unit, worsening the cavitation of the diesel engine. If added every day, check for water leaks in the cooling system.
Reference documents:
The calculation can only be carried out without taking into account losses in transformers and distribution lines.
We know that the output power of the power plantpower is 50 kW, line resistance is 4 ohms, line loss is 5%, or 2.5 kW, user power is 50-2.5 = 47.5 kW. , and the user voltage is 220 V.
Step-down transformer output current=47.5*1000/220=216(A).
The current on the transmission line can be calculated by the formula P=I^2*R, where R=4 ohms, P=2.5KW=2500W, and the calculated current is 25A.
The output voltage of the step-up transformer is calculated using P=UI and U=50/25=2(KV)
The voltage lost in the transmission line is 2KV *5%=100V
The input voltage of the step-down transformer is 2-100/1000=1.9 (KV)
(If you don't give me any wealth to such an annoying problem, I won't help you next time)
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