Answer:
False. Apart from heat loss, there are some other energy loss factors in electric motors.
They are:
(1) Core losses
(2) Windage losses
(3) Eddy Current
(4) Stray Load
Explanation:
STATOR RESISTANCE/HEAT LOSS(Heat loss I^2*R), which is the product of the square of the current multiplied by the resistance of the stator winding. The rotor also experiences I^2*R losses in the squirrel-cage rotor bars, called rotor resistance loss (rotor I^2*R).
CORE LOSSES also occur, originating in the lamination steel. Core losses include hysteresis losses, which result from reorientation of the magnetic field within the motor’s lamination steel, and eddy current losses resulting from electrical currents produced between laminations due to the presence of a changing magnetic field.
These electrical currents occur in both stator and rotor cores, but primarily in the stator, as these losses are proportional to the frequency of the current. The frequency of current in the rotor bars is only a small fraction of the line frequency, as the rotor current frequency is proportional to slip (the difference between operating speed and synchronous speed). Both the stator and rotor laminations have an insulated coating to reduce shorting losses (eddy current) from adjacent laminations.
Friction losses are from the motor bearings and lubrication.
WINDAGE LOSSES combine losses from the rotor spinning in air that creates drag and those from cooling fans used on the motor, along with friction losses in the bearings.
STRAY LOAD losses also are present.
Answer:
patience
Explanation:
you can wait it out,
you can wait to solve,
you can wait for another opinion,
you can wait for an answer,
you can wait to get help,
or you can wait till it's not an issue.
Answer:
Explanation:
Given that:
From process 1 → 2
Process 2 → 3
The volume is constant i.e
Process 3 → 1
P = constant i.e the compression from state 1
Now, to start with 1 → 2
The work-done for the process 1 → 2 through adiabatic expansion is:
We know that 1 bar =
∴
For process 2 → 3
Since V is constant
Thus:
W = PΔV = 0
For process 3 → 1
W = PΔV
The net work-done now for the entire system is :
The sketch of the processes on p -V coordinates can be found in the image attached below.