Electrical Engineering

The permanent magnet moving coil instrument is the most accurate type for D.C. Measurements. The working principle of these instruments is the same as that of the d'Arsonval type of galvanometers, the difference being that a direct reading instrument is provided with a pointer and a scale

(Fig) Permanent magnet moving coil instrument

Construction of PMMC Instruments

• The constructional features of this instrument are shown in Fig.
• The moving coil is wound with many turns of enameled or silk covered copper wire.
• The coil is mounted on rectangular aluminum former, which is pivoted on jeweled bearings.
• The coils move freely in the field of a permanent magnet.
• Most voltmeter coils are wound on metal frames to provide the required electro-magnetic damping.
• Most ammeter coils, however, are wound on non-magnetic formers, because coil turns are effectively shorted by the ammeter shunt.
• The coil itself, therefore, provides electro magnetic damping.

Magnet Systems

• Old style magnet system consisted of relatively long U shaped permanent magnets having soft iron pole pieces.
• Owing to development of materials like Alcomax and Alnico, which have a high co-ercive force, it is possible to use smaller magnet lengths and high field intensities.
• The flux densities used in PMIMC instruments vary from 0.1 Wb/m to 1 Wb/m.

Control

• When the coil is supported between two jewel bearings two phosphor bronze hairsprings provide the control torque.
• These springs also serve to lead current in and out of the coil. The control torque is provided by the ribbon suspension as shown.
• This method is comparatively new and is claimed to be advantageous as it eliminates bearing friction.

Damping

• Damping torque is produced by movement of the aluminium former moving in the magnetic field of the permanent magnet.

Pointer and Scale

• The pointer is carried by the spindle and moves over a graduated scale.
• The pointer is of lightweight construction and, apart from those used in some inexpensive instruments has the section over the scale twisted to form a fine blade.

• This helps to reduce parallax errors in the reading of the scale. When the coil is supported between two jewel bearings two phosphor bronze hairsprings provide the control torque.
• These springs also serve to lead current in and out of the coil.

Torque Equation.

The torque equation of a moving coil instrument is given by

As the deflection is directly proportional to the current passing through the meter (K and G being constants) we get a uniform (linear) scale for the instrument.

Errors in PMMC Instruments

The main sources of errors in moving coil instruments are due to

• Weakening of permanent magnets due to ageing at temperature effects.
• Weakening of springs due to ageing and temperature effects.
• Change of resistance of the moving coil with temperature.

Advantages and Disadvantages of PMMC Instruments

The main advantages of PMMC instruments are

• The scale is uniformly divided.
• The power consumption is very low
• The torque-weight ratio is high which gives a high accuracy. The accuracy is of the order of generally 2 percent of full-scale deflection.
• Using different values for shunts and multipliers may use a single instrument for many different current and voltage ranges.
• Since the operating forces are large on account of large flux densities, which may be as high as 0.5 Wb/m, the errors due to stray magnetic fields are small.
• Self-shielding magnets make the core magnet mechanism particularly useful in aircraft and aerospace applications.

The chief disadvantages are

• These instruments are useful only for D.C. The torque reverses if the current reverses. If the instrument is connected to a.c., the pointer cannot follow the rapid reversals and the deflection corresponds to mean torque, which is zero. Hence these instruments cannot be used for a.c.
• The cost of these instruments is higher than that of moving iron instruments.

Just remember that

(Watts)2 + (VAr)2 = (VA)2

This formula links the actual power in watts to the volt-amps which is the volts times the amps, and the bit left over is the VAr or volt-amps reactive.

For example you have a 250 v load of 400 watts that draws 2 amps. The

power factor is 0.8 because

VA = 250 x 2 = 500, and 0.8 x 500 gives the power of 400 watts.

The VAr formula is:

(VAr)2 = (VA)2 - (Watts)2

so calculate 5002 - 4002 and the answer is 90000 and take the square root. That means that there are 300 VArs that need to be got rid of.

So if we put a capacitor of 300 VAr across the load, that will take out the VArs and the overall current taken drops to 1.6 amps, which will make you one of the supply company's star customers. Actually they don't mind with a 400 watt load but if it was 400 kW they would be pleased.

300 VArs means that the capacitor must draw 300/250 amps or 1.2 amps and the formula for current in a capacitor is 2.pi x freq x C x V so that the capacitance must be 1.2 / (2.pi x 50 x 250) for a 50 Hz system so the capacance is 0.0000153 Farads or 15.3 microfarads. On a 60 Hz system it would be less, 12.7 microfarads.

The above assumes that the load has a 'lagging' power factor, because the phase of the current lags behind the voltage, which is usual for induction motors and many other types of load. If the load already had a leading power factor you would have to put an inductor on to correct it, but that would be exceptional and the supply company might even be happy to let you continue without PF correction because you would be putting 'leading' VArs into the system.

Answer

Power-factor capacitor banks are expressed in reactive volt amperes (var), not in terms of their capacitance (farads). The capacitance of power-factor capacitor banks are of academic interest only, because they are always labelled in reactive volt amperes. To determine the necessary reactive volt amperes (of the capacitor bank) you do the following.

Step 1: Using the existing power factor, determine the existing reactive power of the load.

Step 2: Using the desired power factor, determine the required reactive power of the load.

Step 3: Subtract the required reactive power (step 2) from the existing reactive power (step 1). and the answer will give you the reactive power of the necessary capacitor bank.

Horsepower is the Imperial unit of measurement for power. In North America, the horsepower is still used to define the rated output power of a motor, whereas the watt is used to define its corresponding input power. Because no device is 100% efficient, a motor's rated output power is always lower than its corresponding input power.

Elsewhere in the world, horsepower is considered to be an obsolete unit of measurement and, so, the watt is used to define both the rated output power and the corresponding input power.

p= V*I*COS FIEE SAY COS PYE IS UNITY

10*1000=230*I*1

I= 10000/230

I=43.4 A

10 kW is the output power of the machine. You will need to know the machine's efficiency in order to determine its input power before applying the above equation.