2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Power Systems - Question 2 - 2016 - Paper 1
Question 2
2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V.
2.2 Draw a neat, labelled voltage phasor diagram that represe... show full transcript
Worked Solution & Example Answer:2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Power Systems - Question 2 - 2016 - Paper 1
Step 1
Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
In a delta-connected system, the phase voltage (
V_{ph}) is equal to the line voltage (
V_L).
Therefore, the phase voltage is:
Vph=VL=380extV
Step 2
Draw a neat, labelled voltage phasor diagram that represents a three-phase delta-connected system.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The voltage phasor diagram for a delta-connected system consists of three vectors (
V_L1, V_L2, V_L3) separated by 120°.
It can be represented as follows:
Vector V_L1 is at 0°.
Vector V_L2 is at 120°.
Vector V_L3 is at 240°.
Each vector has a magnitude of 380 V.
Step 3
Current delivered by the alternator at full load
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the current delivered by the alternator, use the formula:
I_L = rac{S}{ ext{sqrt}(3) imes V_L}
Substituting the values:
S=20imes103extVA,VL=380extV
Thus,
I_L = 30.39 ext{ A} $$
Step 4
Power rating of the alternator
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The power delivered can be calculated using:
P=extsqrt(3)imesVLimesILimesextCOS(θ)
Given the power factor (p.f.) is 0.87, we can substitute:
P = 17.4 ext{ kW} $$
Step 5
State the function of a kilowatt-hour meter.
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The function of a kilowatt-hour meter is to measure the amount of energy consumed by a consumer over a period of time.
Alternatively, it measures the energy usage of electrical systems.
Step 6
State TWO methods used to improve the power factor of a resistive inductive load.
97%
121 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Adding power factor correcting capacitors in parallel with the load.
Using synchronous motors to improve the overall power factor.
Step 7
Calculate the power consumed by the load.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the power consumed by the load, use the formula:
Pn=P1+P2
Substituting the given values:
Pn=120+50=170extW
Step 8
TWO advantages of using the two-wattmeter method when measuring power of a balanced load.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Only two watt meters are required instead of three for balanced loads.
The power of a balanced and unbalanced load can be measured accurately.