480V about 20A
240V about 40A
look at the nameplate
duty cycle and efficiencies vary
when starting a 3 phase induction motor,first start with star connection,because reduced current is being applied and after start change it to delta connection then. full line current is applied to the motor.
The transition from star connection to delta connection requires that the current flow through the motor is interrupted. This is termed "Open Transition Switching" and with an induction motor operating at partial speed (or full load speed), there is a large current and torque transient produced at the point of reconnection. This transient is far worse than any produced by the DOL starter and can cause severe damage to equipment and the supply.
The starting current of a motor is usually much higher than the normal current of that motor's full load condition. So for the protection purpose, in case of selecting fuse, the fuse has to be rated 2-3 times the full load of the motor so it does not blow on start. An induction motor will have a starting current of around 6 times FLC.
Full load starting current is typically in the region of 5or 6 times the full load motor current;.
whenever the load increases,the current drawn by the motor to do or to fulfill the required energy to the load. so the current will increase generally.Increase in load will cause the full utilization of motor,so speed of rotor will decrease.
when starting a 3 phase induction motor,first start with star connection,because reduced current is being applied and after start change it to delta connection then. full line current is applied to the motor.
What is the continuous current of 10HP star delta motor
The ratio is between 30% to 40 %
To calculate the current in a Star-Delta configuration for a 30 kW, 415 V, 3-phase induction motor, you first determine the full load current (FLC) in Star configuration using the formula: ( I = \frac{P}{\sqrt{3} \times V \times \text{PF}} ), where ( P ) is power (30,000 W), ( V ) is line voltage (415 V), and PF is the power factor (assumed to be around 0.8 if not specified). For the Delta configuration, the current would be calculated as ( I_{Delta} = I_{Star} \times \sqrt{3} ). In the Star configuration, the current will be approximately 41.6 A, and in Delta, it will be around 72 A.
A star-delta ladder refers to a type of electrical circuit configuration used to start and operate three-phase induction motors. In this configuration, the motor initially operates in a star (Y) connection to reduce the voltage and current during startup, minimizing inrush current and mechanical stress. After a brief period, the motor switches to a delta (Δ) connection for normal operation, allowing it to run at full power. This method improves efficiency and prolongs the motor's lifespan.
A wye-delta starter is an electrical control method used to reduce the starting current of a three-phase induction motor. Initially, the motor is connected in a wye (Y) configuration, which reduces the voltage across each winding to about 58% of the line voltage, thereby limiting the inrush current. After a preset time, the starter switches the motor to a delta (Δ) configuration, allowing it to run at full voltage and speed. This transition helps protect the motor and electrical system from damage due to high starting currents.
Slip Ring Induction Motor and Squirrel Cage Indution Motor
This type of motor starting is used to reduce the inrush current to a lower level. Once the motor is operating at the lower level of run current it changes to a delta connection which allows the motor to operate at its full load amperage rating.
The transition from star connection to delta connection requires that the current flow through the motor is interrupted. This is termed "Open Transition Switching" and with an induction motor operating at partial speed (or full load speed), there is a large current and torque transient produced at the point of reconnection. This transient is far worse than any produced by the DOL starter and can cause severe damage to equipment and the supply.
Full load current ofthe motor x 0.58
Since (in North America, at least), the horsepower of a motor describes its rated output power, you need to know its efficiency in order to determine its input power, then its rated voltage, to determine its full-load current.
The inrush current to the starting motor is lower due to the the motor coils being in series across the line creating a higher impeadence. When the starter connects to delta connection the motor is already turning and it does not have to go through the locked rotor phase. This system is being phased out now with the use of electronic soft starts and by-pass contactors.To start a motor direct on line (DOL) in delta requires approximately 6 to 8 times the full load current of the motor and delivers full torque. Delta places 415v across each of the windings on the motor. A Star-Delta starter starts the motor in Star which places 240v across each winding on the motor. This results in a lower starting current and also a lower starting torque.