Want this question answered?
Lightbulb(load),Battery(Power source),Switch,Copper wire
Since the breaker that is installed on the generator set will be sized to the output of the 30 kW generator, the load will stay connected until the thermal trip of the breaker trips the load off line. This will be in the matter of seconds before it happens. To handle a 75 kW load and depending if it is an inductive or a resistive load you will need at least a 80 to 100 kW generator.
The first part is that you need a power source. The second thing is that you need a load. The third thing is that you need the power cables.
There are two types of power - real power and reactive power. Real power is power that is used, such as the power used to light a light bulb. Reacitve power is power that is held and released by a reactive element (capacitor or inductor), thus is not actually used to do any work. The reactive elements cause a phase shift between voltage and current, which manifests itself as a change in power factor. Power companies must supply both reactive and active power. Total power is equivalent to sqrt( Reactive^2 + active^2). Not only do they need to supply the reactive power, their equipment must be sized to handle a larger total power. Reactive power is generated by installing extra equipment - capacitor banks or inductor banks - or by running generation in such a way that more reactive power is created (this will lower the power plants' real power output). Depending on who the customer is, they may not be billed for the reactive power, thus the oversizing of equipment, and the supplying of reactive power does not generate any revenue (but costs them). If power factor is bad enough, the power company will lose money, and thus require power factor correction, or will bill on both real and reactive power usage. It's simple economics (money)!
For a 3 phase motor, the voltage is 415. Being an inductive load the power factor is commonly .85 The constant for 3 phase circuit is 1.732 You need to run the motor on no load and you use clamp meter to take the no load current. In considering the fact that on no load the measured current should not be more than 0.5 of the rated. Assuming , under normal condition the no load current is 10 amps. Then the load current will be 20 amps. Then from above and in using the formula P=I*V*p.f*1.732 Therefore P= 20*415*0.85*1.732. P=12.3 kw Thank you. Bankole 2348035447224
Yes.
Typically you are referring to a pure resistive load and not an inductive load. To measure a resistive load you need an Ohm meter. You can buy cheap ones for $10 to $20 on-line or at a store like Radio Shack. Usually they are combined with a volt meter.
Need more info to give you the most usefull answer. However, generically speaking, sounds like what you need is a 120v relay. Put the relay output contacts in series with the fan (aka the load) then wire the relay coil in parallel to the light bulb power. This will make the load come on when the light is powered on. Tip, if the fan is driven by an induction motor (most are, if it has brushes it is NOT an induction motor) then your speed controller will have to be the kind for an inductive load ( NOT the kind for a resistive load, like a lightbulb ). Good luck.AnswerA solid state relay, or SSR should do it for you. SSRs take a broad range of input voltage as you require, and have snubbers to allow them to switch inductive loads such as the fan.
You should switch if there is a Failover Script for Dual. You can read more at blog.taragana.com/.../how-to-load-balancing-failover-with-dual-multi-wan-adsl-cable-connections-on-linux.
Lightbulb(load),Battery(Power source),Switch,Copper wire
According to soil condition, load, environmental conditions we need different types of foundations.......
no because the load is always going to weigh different and will need different force to use each time.
Each model is different. You need to state the model to get an answer.
Since the breaker that is installed on the generator set will be sized to the output of the 30 kW generator, the load will stay connected until the thermal trip of the breaker trips the load off line. This will be in the matter of seconds before it happens. To handle a 75 kW load and depending if it is an inductive or a resistive load you will need at least a 80 to 100 kW generator.
You would need a 20 amp switch when the load controlled by the switch can draw as much as 20 amps. If you have a 20 Amp breaker supplying the circuit then you need to size all switches and outlets on that circuit to 20 amps.
It depends on the voltage of the compressor. Two horse power (electric) is 1492 watts, but watts are volts times amps, so you need to know the voltage. Since the motor is an inductive load, you will also need to know the power factor, so as to compensate for true vs apparent power.
The way I hooked up the strobe lights in my truck, i did the following: [need a switch that has (+)(load)(-). Three prongs..] -run a wire from positive(+) of the battery to the positive(+) of the switch. -run a negative(-) from the battery to the negative(-) of the switch. -run the positive of the strobe to the (load) of the switch. -either run a negative(-) wire from the strobe to the switch, or mount it to the vehicle body somewhere.. Hope this helps.. Good luck