DESIGN CIRCUIT FOR MONITORING AND CONTROLLING ELECTRICAL FAULT PEDIATRIC COMPLEX (NPC) BUILDING AT MUHIMBILI HOSPITAL

 Problem of the statement

To prevent people and property from damage or injury, electrical faults in a power system must be cleared fast. In the early days of electrical power systems the fault clearing was administered by the maintenance staff, who visually detected the fault and manually operated a switch to clear the fault. As fault currents became larger and the operating requirements of the electric power system became more stringent, the need for automatic fault. 

The fault clearing system uses various protection devices such as fuses, protective relays and circuit breakers to detect and clear the fault.

At NPC building the occurrence of fault is high and there is intensive care unit for children  and other medical services like clinic and wards so the much delay in tracing fault lead also in delay of service.

The main problem is; when electrical faults occur, it takes long time for the technicians to troubleshoot the fault which causes delay in giving treatments to patients and sometimes can result to loss of life

Main objective

• To design a circuit which will monitor and inform the technician the circuit which leads to fault .
• Specific objectives
•  Program the microcontroller based circuit to monitor and control fault
• To design alarm and indicator system to inform the presence of the fault
• To design and implement the proposed circuit

methodology

1. Literature review and Consultation
2. Data collection
3. Data analysis
4. Design of the circuit
5. Simulation of the circuit
6. Building the prototype
7. Testing the circuit 
8. Report writing 

Literature  review

Different types of fault and their protection.

How the circuit breaker works once a fault occurs.

How microcontroller can be programmed to monitor electrical faults.

How electrical fault can be indicated etc.

                                                                circuit diagram

Power supply

Transformer Specifications Since the transformer is of 240Vac-12Vac then turn’s ratio is obtained as follows:

Np/Ns=Vp/Vs=240/12=20=20:1

The turn ratio of the transformer is 20:1

Vrms=12Vac

Vdc(peak)=√2×Vrms

Vdc(peak)=√2×12=16.97V

Vdc(peak)=16.97V

But,

                             V_(D.C)=V_(D.C(peak))−1.4V(drops in diodes) 

V_(D.C)=16.97−1.4V=15.57V

V_(D.C)=15.57V 

V ripple=    (I load)/2fc        and      Y=1/(4√3FRC)

Where       Y=ripple factor

F=frequency of AC mainly 50 Hz

R = resistance

But R=V_(D.C(peak))/𝐼𝐶

IC= Transformer current 500𝑚𝐴

R=16.97/0.05

R=33.9ohm

Let ripple factor be 0.04

C=1/(4√3𝐹𝑅𝑌)

C=1/(4√3𝑋50𝑋33.9𝑋0.04)

C=2128.9 μF

But available standard capacitor in market is 2200𝜇𝐹

The total resistance in series circuit equal to summation of the resistors in the circuit:

 R total=R1 + R2       

The total current according to ohm's laws is:

I =(𝑉 𝑡𝑜𝑡𝑎𝑙)/(𝑅   𝑡𝑜𝑡𝑎𝑙     )

 

 The voltage across 7.5KΩ resistance:

V2= I * R2 

 From substituting Eq. (2) in Eq. (3) get the voltage on 7.5KΩ:

V2=R2*(𝑉 𝑡𝑜𝑡𝑎𝑙)/(𝑅 𝑡𝑜𝑡𝑎𝑙) 

 

The supply voltage to the relay is 12VDC = VCC

Coil resistance = 200Ω, which is the load resistance Rc in this case.

At saturation, 

Vcc of the transistor = 0

Vbe = 0.7(silicon).

The transistor in this case is operating as a switch so voltage drop across it is considered to be negligible.

From;

Vcc = Vce = IcRc 

But Vce = 0