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Bridging Loans

The dc supply to a vacuum cleaner motor is realised via full-bridge rectification and smoothing of a single ph?

The dc supply to a vacuum cleaner motor is realised via full-bridge rectification and smoothing of a single phase, 230Vrms, 50Hz, ac supply. The motor is a 2-pole separately excited brushed dc machine which has its field supplied directly from the dc supply. The machine parameters are as tabulated below. The armature coil is connected to the dc supply via a pwm power semiconductor switch that facilitates armature voltage control during starting/run-up of the motor. The maximum torque presented by the vacuum system is 1.754Nm. Draw the circuit schematic of the drive-system and state the voltage matrix and torque equation for the machine.
(b)If the motor no-load speed is 13,500rpm, calculate the armature current, motor speed and efficiency at full-load torque.

Vacuum cleaner motor model parameters

Field resistance, Rf=650.0Ω
Field inductance, Lf=1.0H
Armature resistance, Ra=7.0Ω
Armature inductance, La=18.7mH

Assuming your are referring to a phase width modulator control circuit:

Have to use PLC type ladder diagram on this:

-------(SW)------------------------------------------|PWM|-----

--------[ PWM ]--------------------------(motor armature)-------
.
--------[ field rectifier]----------------------------(motor field)-----------

PWM: module

--[ phase firing circuit ]-----------{SCR bridge rectifier}---------

OR:
-----[phase firing cicuit]----------[SCR birdge]-------[filter]-----

1.754 Nm = .394 lbf. foot pounds

1 horsepower is = 33,000 lbf.

general formula for hp from torque:

( 2 x L x pi x F x RPM) / 33,000
since Length not given I used unit circle of 1
(3.1416 x 0.394 x 13,500) / 33,000 = .506= .5hp.

746 watts per horsepower:

746 x .5 = 377.478 watts

now use watts law to get the current:

W= E x I
.
W = (I x R ) x I = I^2 x R

given: E = 230V:

derived W = 377.478W

W = E x I
divide both sides by E

W / E = (E x I ) / E

W / E = I
in standard position'

I = W /E
plug in numbers:

I = 377.478 / 230
I = 1.642 = 1.6 amps running:

field impedence: XL = 2 * (pi) * F * L

XL= 2 * 3.1416*50* 1 = 314.16 ohms if half wave rectifier used

impedence = square root ( XL^2 + R^2)

Z = Square root ( 314.16^2 + 650^2)

Z = 721.9 ohms:

if full wave bridge rectifier used:

XL= 2(pi)FL

XL = 2 x 3.1416 x 100 x 1

XL = 628.31

Z= square root ( 628.31^2 + 650^2)

Z = 904.03 0hms = 904 ohms

field current::

I = 230 / 904 = 0.25 amps: or 1/4 amps:

if well filtered with not ripple acting on field disreguard inductance:

I = 230 / 650 = .35 amps or .4 amps or 2/5 amps:

Armature current: inductance has to be taken into account because the SCR bridge will not be filtered:

XL = 2(pi)FL

XL = 2 x 3.1416 x 100 {full wave} x 18.7 x10^-3

XL= 11.749 = 11.75 ohms:

Z = square root ( 7^2 + 11.75^2)

Z = 13.677 ohms:

armature current at start up or locked rotor:

I= 230 / 13.677

I = 16.82 amps locked rotor:

1.6 amps running:

armature CEMF:

CEMF = 230 - (1.677 x 13.677)

CEMF = 208.1168 = 208.12 V.

hope that helps

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