Ubungo Msewe is place which situated in Dar-Es-Salaam region at Ubungo district. Ubungo Msewe is the place which provides residential services for both students and other people. This place is near University of Dar-Es-Salaam which make it to be good place for renting business to students of University of Dar-Es-Salaam .
PROBLEM STATEMENT
People and students at Ubungo Msewe face the situation of sharing energy meter which results to conflict problem between the tenants. This is due to the tenants with low energy consumption are required to pay the same amount of consumption costs compared to those with high energy consumptions
So due to this problem , The designing of a Single energy meter with multiple output the situation of sharing energy consumptions between tenants will be removed
OBJECTIVES
MAIN OBJECTIVES
To remove conflicts between tenants by introducing energy meter with multiple outputs which will enable each tenant to have his /her own energy consumptions costs.
SPECIFIC OBJECTIVES
Design a separated output circuit energy meter circuit
Collect appropriate tools and equipment.
Build a prototype
To test prototype
SIGNIFICANCE OF PROJECT
If the project will succeed it will enable each tenant at Ubungo Msewe to have his/her own energy consumption costs according to his/her own consumptions
To minimize the conflict between tenants
METHODOLOGY
Literature review
Data collection
Data analysis
Designing of circuit
Build the prototype
Testing the prototype
Report writing
LITERATURE REVIEW
Literature review include the existing system and proposed system
EXISTING SYSTEM
The existing system have single output from the microcontroller of the meter to all users(tenants)
DISADVANTAGE OF EXISTING SYSTEM
All tenants are required to pay the same amount despite of their consumption and results to conflict between tenants
ADVANTAGE OF PROPOSED SYSTEM
Each tenant will be able to have his or her own electrical energy consumptions costs depending on his or her own usage.
Also it will remove the problems which appear when tenants with low consumptions are forced to pay the same amount with tenants with high consumptions.
DATA COLLECTION
Analysis coverage.
This analysis will cover the following:
Dummy load at each tenant side.
Relay switching circuit.
ACS712 current sensor IC.
Mains voltage measurement unit.
User-system interface (the keypad)..
PROPOSED CIRCUIT DIADRAM
Part 1) Dummy load at each tenant’s side.
The dummy load is used to represent each tenant’s power usage.
To show the difference in power usage, one tenant will be represented by 3 light bulbs while the other will be represented by 1 light bulb.
The light bulb.
A light bulb is a device that converts electrical energy to light energy. It operates by heating up the filament inside it hence heat is produced along the way.
Qualitative analysis.
Light bulb used is a 220V 200W power.
Power in kWh = energy in kW x time in hours
Thus: in 10 minutes:
for a tenant with 3 light bulbs:
Power in kWh = 3 x 0.2kW x (10/60)hours
= 0.1kWh.
For a tenant with 1 light bulb:
Power in kWh = 0.2kW x (10/60)hours
= 0.033kWh.
The electrical current:
From power = current x voltage
Thus current = power/voltage
For 1 light bulb:
current = 200W / 220V
current = 0.91A.
For 3 light bulbs:
current = (3 x 200W) / 220V
current = 2.73A.
Part 2) Relay switching circuit.
This part enables the microcontroller to control the flow of electricity to a tenant’s home.
It consists of:
Relay.
Transistor switching circuit.
The relay: Qualitative analysis
From the results of the load side, the relay must be able to handle the electrical quantities calculated.
Thus, the relay must be able to bear the following quantities.
Current greater then 2.7A so a 5A or greater relay is selected.
Load side voltage greater than 220V so a 250V or greater relay is selected.
As the relay must handle a currents, a mechanical relay is selected.
For reliable operation, a relay with coil voltage of 12V is selected as it is better when switched by a transistor than a 5V one. Since the switching is for one customer, a single pole relay is selected.
The transistor: qualitative analysis.
For the purpose of relay switching, a Bipolar junction transistor is to be used.
The supply voltage is 12V and the coil resistance is 240Ω.
Thus, the collector current Ic is:
From Vcc – IcRc – Vce = 0
But when transistor is fully on, Vce ≈ 0.3V
Ic = (Vcc – Vce) / Rc
Ic = (12V – 0.3V) / 240Ω
Ic = 48.75mA.
Now the base current required idL:
From the relation between the collector current Ic and the base current Ib of a BJT:
Ic = Hfe x Ib
Where Hfe is the current gain which is typically 200 for collector current of 48.75mA.
Thus the base current is:
Ib = Ic / Hfe
Ib = 48.75mA / 200
Ib = 0.243mA.
the switching voltage from the microcontroller is 5V.
Since the bias configuration used is base resistor bias, the value of the base resistor Rb taking a base emitter Vbe voltage of 0.7:
Letting a voltage from microcontroller be Vin.
Vin – Vbe – IbRb = 0
Rb = (Viin – Vbe) / Ib
Rb = (5V – 0.7V) / 0.243mA
Rb = 17.695kΩ
Therefore a market available value of 18kΩ is selected,
Basing on the following calculated parameters
Typical current gain of 200.
Collector current of 48.75mA.
Base current of 0.243mA.
An 2N3904 NPN BJT transistor is selected.
Part 3) current measurement section.
This part deals with measuring electrical current consumption of a particular tenant. In measuring current we will use a hall effect sensors.
Hall effect sensor is the transducer that varies its output voltage in response to magnetic field. The sensor operate as analog transducer
Transducer is the device that converts energy from one form to another
In figure 1 below, a characteristic curve shows relation between the flowing current and the output voltage.
Gradient, g = ∆y / ∆x
= (3.5 – 2.5) / (5 – 0)
= 0.2
Henceforth, the output voltage is always 0.2 times the flowing current at that instant.
Part 4) Mains voltage measurement unit.
The main principle is to use a value at the output of a full bridge rectifier to reflect the ac mains voltage input.
The output of a full bridge rectifier is to be arbitrarily 10V.
This voltage is spliced into two parts one of which goes to the microcontroller for measurements.
This is arbitrarily chosen to be 2.5V which is midway between minimum(0V) and maximum(5V) measurable dc voltages so as to give room for measuring a mains even if it fluctuates.
Qualitative analysis.
From the equation of a potential divider,.
Vout = (R2 / (R1 + R2)) x Vin
Taking R1 fixed at 10kΩ,
2.5 = (R2 / (10,000 + R2)) x 10
R2 = 3.3kΩ
From the equation of a full bridge rectifier with filter capacitor (here it is fixed at 1000µF) ,
Vdc ≈ (1 – (1 /(2fRLC))) x Vrectified
Therefore Vrectiiied = Vdc /(1 – (1 / (2fRLC)))
Vrectified = 10 / ( 1- (1 / (2 x 50 x (10,000 + 3,300) x 1000 x 10-6)))
Vrectified = 10V.
Now, to get the peak rectified voltage
From Vrectified = (2 x Vpeak(rectified)) / П
Vpeak(rectified) = (10 x П) / 2
Vpeak(rectified) = 15.7V
This peak rectified dc voltage relates to the peak ac secondary voltage by he following equation/
Vpeak(secndary) = Vpeak(rectified) – 1.4V
Henceforth, Vpeak(secondary) = 15.7V + 1.4V
Vpeak(secondary) = 18.1V (a.c peak)
Now the peak input ac at the primary is
Vpeak(primary) = Vrms(mains) x √2
Vpeak(primary) = 220 x √2
Vpeak(primary) = 311.1V
Hence, the transformer ration required is;
r = Vpeak(primary) / Vpeak(secondary)
r – 311.1V / 18.1V
Transormer ratio r = 17.2:1
Therefore, a 2.5V at the input terminal of the microcontroller will represent 220Vrms ac mains supply.
Transfer ratio between these two quantities is
transfer ratio = mains voltage / voltage read by microcontroller
= 220 / 2.5
= 88.
This means that if the microcontroller reads 2V then the mains supply is 176V rms.
Transformer power calculations.
Total power of the circuit is equal to the sum of the power consumed by:
Two transistor switching circuits.
The two relays
Arduino Mega board.
LCD display.
The ACS712 current sensor IC.
The significant power to be taken into account is that dissipated by the transistor switching circuits and that dissipated by the relays.
Power dissipated by the remaining components is so low thus neglected in the calculations.
In their datasheets they are just stated that they are low power devices and their power dissipation is not given as well.
Power consumed by relays is,
power = I2R ,
since relays dissipate electrical energy as heat.
Total power by two relays = 2I2R.
= 2 x (48.75 x 10-3)2 x 240Ω
= 1.14W.
Power dissipated by one transistor is:
power = (Vce x Ic) + (Vbe x Ib)
power = (0.3 x 48.65 x 10-3)+(0.7 x 0.243 x 10—3)
= 0.0148W
The total power consumed by both transistors is then,
total power = 2 x 0.0148W
= 0.0296W
Hence, the total power to be supplied by the transformer is
= 1.14W + 0.0296W
= 1.2W.
To account for random unaccounted power losses in the transformer windings and for all the remaining parts a 5W transformer with a step down factor of 17 is employed for this purpose.
CONTROL UNIT
Arduino AT Mega 2560 will be used as a control unit due to the following properties
Properties of Arduino AT Mega 2560
Operating voltage 5V.
Number of Digital 1/0 pins are 54 in which 16 pins are used as Analog input pins
It has 256KB of flash memory for storing code ( of which 8KB are used as boot loader)
Clock speed is 16MHz
Proposed circuit after data analysis
CONCLUSION
The design system met my expectation. So it can be concluded that the project objective and its designing consideration have been met.
RECOMMENDATION
I recommend that the idea of this project should be modified and implemented to the society so as to enable people(tenant) to buy/pay for electricity according to their usage