MCCB Categories and Shortcircuit Protection
1. Categories of A and B of MCCB ( Molded Case Circuit Breaker)
Without an international short time delay for choosiness for conditions of short circuit and hence without a short circuit withstanding rating of current. For this category, they are actually miniature circuit breaker which operates at the end of the socket outlets and the final distribution. If a short circuit occurs, they will trip immediately (Bird, 2012).
This category of MCCB has withstood breaking ability. Hence they do not essentially trip in case of any short circuit. This will make the downstream circuit breakers to switch off.
We always impose an upper limit on the earth fault loop impedance to ensure that all the faults are covered within that range (Hambley, 2017). With the upper limit most electrical appliances if not all are fully protected by the earth loop impedance protection. There is a different circuit with different loads and rating (socket outlets and lighting circuits’ protections are different). In cases where we cannot attain the upper limit, we employ the correct cable sizes calculations to ensure that a suitable and a perfect cable size is employed.
Discrimination is an act of choosing a protective device and adjusting their settings in order to limit interruption to electrical installations under fault conditions (Hesse, 2011). This can be checked through evaluation of current and time discrimination.
Time discrimination: This is checked as the circuit breaker are put to withstand electrodynamics effects and thermal effects of the current fault at the time of the delay (delay period).
Current discrimination: This has two kinds of discrimination (overload discrimination and short circuit discrimination). The overload is checked through a selection of an upstream device having a higher current rating and a pickup level for the next downstream device. While the short circuit is checked through setting a circuit breaker to a limit which doesn’t cause unnecessary trip (Ojwach, 2012).
Ambient temperature has a direct impact on the conductor sizing. The temperature to be considered is that of the air around the cables used during the installation. Depending on the ambient temperature set, the size of insulation will be chosen for a particular ambient temperature. For our case, the ambient temperature is 40^{0}and 30^{0} for the rest of the rest of the building. The PVC insulation which will be required is 0.77 mm allround the cable.
Grouping of circuits together will lead to a reduction current carry capacity and this is suitable for correction factors hence it will ensure that there is no overheating in the cables (PRASAD, 2014).
Sizing Cables for Overload and Shortcircuit Protection
Overheating occurs in a conductor when the current through the passing through the conductor is too high for the size of the cable, therefore, the correct size of the cable is obtained through calculation. Choosing the accurate cable during installation is vital to ensure an acceptable conductors’ life and insulation subjected to the thermal effects of carrying current for long periods of time in ordinary service (Smith, 2013).
And the calculation of the cable size is done using the below equation.
S = 
(Ia²t) 
………………………………………………………………………………1 
k 

where 
S is the minimum protective conductor crosssectional area (mm^{2}) 

Ia is the fault current (A) 

t is the opening time of the protective device (s) 

k is a factor depending on the conductor material and insulation, and the initial and maximum insulation temperatures (Tharaja, 2013). 
2. Option A
Draw a flow diagram to show how you would select a cable size that satisfies the BS 7671 protection requirements for Overload, Short Circuit, Indirect Contact and Voltage Drop
For the selection of the conductor size in order to avoid overheating the following are taken into account;
The type of the insulation which will define the maximum permissible temperature during the operation, and according to CIBSE the insulation of EPR can withstand 90^{0} C while the PVC can withstand 70^{0} C (Tonardo, 2013). Therefore from the formula of getting the overload thermal as
V=I^{2}R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
So when a given voltage is pushing a current then the resistance can be obtained, lets assume that the resistance is 0.2 ?
R= . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
And taking the conductor to be copper ? = 1.72 × 10 ^{8}
The length can be taken as 100 m
0.2=
A= 1176470588 m^{2}
A=
r= 1.9 mm
Cable size according to the voltage drop.
The cable size can be selected from the voltage drop, and this is done through some calculations (John, 2013).
Assuming that
Current rating= 250 A/ ampere
Calculating Voltage Drop and Cable Size Selection
R1= 0.5 m ?/ft
X1= 0. 4m ?/ft
The total length =72ft
Power factor = 0.9
The voltage drop
V= I (R1cos ? + X1sin ?)
Where V is the voltage drop in volts, I is the current in ampere/ phase, R1 is the conductive resistance in ohm/ft, X1 is conductor inductive in ohm/ft, ? is the angle obtained from power factor and L is one way length of the circuit source to load in kft (Couling, 2015).
Power factor = cos ?
?= cos^{1} 0.9
? = 25.84^{0}.
L=
L= 0.072 kft
V= I (R1cos ? + X1sin ?)
V= 1.73× 250 [ (0.5 × cos 25.84) + ( 0.4 × sin 25.84 )] × 0.072
V= 433× 0.6243× 0.072
V = 19.46 volts
Voltage drop 6 %
The size of the cable is obtained by the following equation
r=
r=
r=
r= 1.92 mm
D= 3.89 mm
3. The ground floor distribution is connected 2 meters while the other five floors have 4 meters each.
Current rating= 250 A/ ampere
R1= 0.38 m ?/m
X1= 0.13m ?/m
The total length is given by 2+ (4×5) = 22 m
The voltage drop along the length of Busbar 2 (East) can be given by the following equations
V= I (R1cos ? + X1sin ?) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Where V is the voltage drop in volts, I is the current in ampere/ phase, R1 is the conductive resistance in ohm/ft, X1 is conductor inductive in ohm/ft, ? is the angle obtained from power factor and L is one way length of the circuit source to load in kft (Katz, 2010).
Power factor = cos ?
?= cos^{1} 0.9
? = 25.84^{0}.
And from 1 ft= 0.3048m
R1= 0.38 ×0.3048
R1= 0.115824?/ft
X1= 0.13 ×0.3048
X1= 0.0396 ?/ft
L= 22 m
L= 22/0.3048
L=
L= 0.072 kft
V= I (R1cos ? + X1sin ?)
V= 1.73× 250 [ (o.1158 × cos 25.84) + ( 0.0396 × sin 25.84 )] × 0.072
V= 433× 0.12149× 0.072
V = 3.7875 volts
Voltage drop at the top 2 east is given as 3 %
The size of the cable is obtained by the following equation
Reducing Harmonics in Electrical Installations
r=
r=
r=
r= 0.9622 mm
D= 1. 92 mm
Assumption made
For this calculation, their current is assumed to be flowing uniformly in the crosssection of the conductor (no skin effect). The resistivity of the conductor is also taken to be equal through the conductor (Baessler, 2010).
4. The standby generator is installed beside the building and the power from the generator is then connected to the switch room where the main switch box. These generators are automatic and will sense when there is power blackout and pick up immediately to supply the power. The electrical characteristics of component parts of assemblies shall apply (Keisler, 2012). When the components are mounted in their enclosures, appropriate derating factors having been allowed for the effect of the enclosures, other components, and interconnections. The cables sizes would apply for supply as 6mm2. The standby generator would be able to supply the five floors (Philip, 2015).
Harmonics are currents or voltages with frequencies that are integer multiples of the fundamental power frequency. If the fundamental power frequency is 60 Hz, then the 2nd harmonic is 120 Hz, the 3rd is 180 Hz. This harmonic will hence result in overheating of equipment and conductors (Mitchell, 2013). Therefore it must be reduced in the design, the following designs would be employed to do away with the harmonics:
A suitable design to reduce harmonics would support a load of many PC like a call center, would specify the neutral wiring to be more than the phase wire capacity by a factor of 1.73. This would ensure that the harmonic is reduced hence there will be no overheating (Wilingstone, 2011).
 Use DC generators power supply which is not affected by harmonics
The internal power supplies are not energy efficient, and they generate substantial heat, which puts a costly burden on the room’s air conditioning system. Heat dissipation also limits the number of servers that can be housed in a data center. It could be worthwhile to eliminate this step by switching to DC power
 Use a harmonicmitigating transformer
A harmonicmitigating transformer (HMT) is designed to handle the nonlinear loads of today’s electrical infrastructures. This transformer uses electromagnetic mitigation to deal specifically with the triplen (3rd, 9th, 15th,…) harmonics.
 Use Krated transformers in power distribution components.
A standard transformer is not designed for high harmonic currents produced by nonlinear loads. It will overheat and fail prematurely when connected to these loads. When harmonics started being introduced into electrical systems at levels that showed detrimental effects (circa 1980), the industry responded by developing the Krated transformer.
 Use separate neutral conductors. .
On threephase branch circuits, instead of installing a multiwire branch circuit sharing a neutral conductor, run separate neutral conductors for each phase conductor. This increases the capacity and ability of the branch circuits to handle harmonic load.
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