What occurs if the load is removed from an operating series DC motor?
It will accelerate until it falls apart.
When a substantial amount of current is applied to a DC motor at start - has no emf resistance (back emf), then the rotation will increase quickly, and without anything to slow it down, it will accelerate past the hardware tolerances, causing damage.
For more info, please see Sluiceartfair site for the article on What will happen to a DC series motor when its load is removed?
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If a shunt motor running with a load has its shunt field opened, how would this affect the speed of the motor?
It will speed up.
From: wp2ahg
A shunt wound motor has a high-resistance field winding (a 'shunt') in parallel with the motor.
So, it passes some of the circuit's energy around the motor and through the field winding (shunt) to limit the motor's speed.
Removing the shunt would remove that speed control causing an increase in the motor's rpms. It will speed up.
To fully understand the behavior of shunt DC motors the key concept is "armature" which is kind of like torque in a car.
It has key variables such as Pole Pitch (back and front pitch, resultant pitch and communicator pitch, Coil Span, Commutator Pitch. Please see Electrical 4 U site for article on Armature Windings: Pole Pitch, Coil Span And Commutator Pitch
And, for information on DC shunt motors, please see Electrical 4 U article on DC Shunt Motor: Speed Control, Characteristics & Theory
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The expression “voltage regulation” as it applies to a shunt-wound DC generator operating at a constant frequency refers to:
Voltage fluctuations from load to no-load.
Basically, the speed will increase, decrease, and increase. For information on DC shunt motors, please see Electrical 4 U article on DC Shunt Motor: Speed Control, Characteristics & Theory
From: wp2ahg
A shunt wound motor has a high-resistance field winding (a 'shunt') in parallel with the motor.
So, it passes some of the circuit's energy around the motor and through the field winding (shunt) to limit the motor's speed.
Removing the shunt would remove that speed control causing an increase in the motor's rpms. It will speed up.
To fully understand the behavior of shunt DC motors the key concept is "armature" which is kind of like torque in a car.
It has key variables such as Pole Pitch (back and front pitch, resultant pitch and communicator pitch, Coil Span, Commutator Pitch. Please see Electrical 4 U site for article on Armature Windings: Pole Pitch, Coil Span And Commutator Pitch
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What is the line current of a 7 horsepower motor operating on 120 volts at full load, a power factor of 0.8, and 95% efficient?
57.2 amps.
NOTE: One (1) horsepower in electricity is the same as 746 Watts. Specifically, 746 is in Newton meters per second.
For more info, please see Basic Electrical Engineering site for proof of 1 hp = 746 watts
From wp2ahg:
Amps = ((( (746 x hp) / V ) / E.F ) / P.F ) =
\[{\text{Amps}}=\frac{\frac{\frac{{746}\times{\text{ Hp}}}{\text{Volts}}}{\text{Efficiency}}}{Power}\]
((( (746 x 7hp) / 120 volts) / 95%ef / .8pf) = 57.2 amps
\[{\text{Amps}}=\frac{\frac{\frac{{746\text{ W/Hp}}\times{7\text{ Hp}}}{120\text{Volts}}}{0.95\text{ Efficiency}}}{.08\text{ Power}}\] \[\text{Amps}={57.2\text{ amps}}\]
or
\begin{align} \text{Ohm's law: }\\ I &= \frac{P}{E}\\ Amps &= \frac{Watts}{Volts}\\ \\ \end{align}
\[\frac{746\text{ Watts/Hp}\times{7\text{ Hp}}}{120\text{ Volts}\times{0.95\text{ Efficiency}\times{0.8\text{ Power Factor}}}}\]
\[\text{Amps}={57.2\text{ amps}}\]
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A 3 horsepower, 100 V DC motor is 85% efficient when developing its rated output. What is the current?
26.300 amps.
From wp2ahg:
\[\text{I}=\frac{746\text{ watts/Hp}\times \text{Hp}}{\text{ Volts}\times{\text{ Efficiency}\times{\text{Power Factor}}}}\]
[PF not given]
\[\text{I}=\frac{746\text{ watts/Hp}\times \text{3 Hp}}{100\text{ Volts}\times{0.85\text{ % Efficiency}}}\]
\[\text{I}=\frac{2,238\text{W/Hp}}{85\text{ Efficiency}}={26.3\text{ amps}}\]
NOTE: One (1) horsepower in electricity is the same as 746 Watts. Specifically, 746 is in Newton meters per second.
For more info, please see Basic Electrical Engineering site for proof of 1 hp = 746 watts
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The output of a separately-excited AC generator running at a constant speed can be controlled by:
The amount of field current.
The field current can compensate for mechanical lags, heat, hardware conditions, etc.
For detailed explanation, please see the Science Direct site for the article on Rotor Excitation, particularly on section 0.3.4 Effect of VAR Control on AVR/Rotor Field Current
For info on key concepts, please see the U.S. Nuclear Regulatory Commission's pdf document on 9.0 GENERATOR, EXCITER, AND VOLTAGE REGULATION
For an overall info on Electrical AC Engines, please see Brown University site for pdf document on BASIC AC ELECTRICAL GENERATORS
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