Answer:
A. ) 591.7 v
B.) 991.7v
C.) 59.7%
D.) 47.9 Kw
E.) 247925 W
F.) 59.7 %
Explanation:
Given that a simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The transmission line is 100 km copper wire of 3 cm diameter. If the voltage at the load side is 400 V,
Let first calculate the resistance in the wire.
The resistivity (rho) of a copper wire is 1.673×10^-8 ohm metres
Resistance R =( L× rho)/A
Where Area = πr^2 = π × 0.015^2
Area = 0.00071 m^2
R = (100000 × 1.673×10^-8) / 0.00071
Resistance in wire = 2.367 ohms
Then let calculate the resistance in the load.
Also, since Power P = V^2 /R
Make R the subject of formula
R = V^2/ P
R = 400^2/100000
Resistance in load = 1.6 Ohms
Current l = V / R
I = 400/1.6 = 250 Ampere
a.) Voltage drop across the line V line will be achieved by using Ohms law.
V = I R
V = 250 × 2.367
V = 591.7 v
B.) Voltage at the source side Vsource will be
V = V line + V load
V = 400 + 591.7
V = 991.7 v
C.) Percentage of the voltage drop Vline /Vsource
591.7/991.7 × 100 = 59.7%
D.) Line losses
P = I V
P = 250 × 591.7
P = 147925 W
Power loss = 147925 - 100000
Power loss = 47,925 W
Power loss = 47.9 Kw
E.) Power delivered by the source
P = IV
P = 250 × 991.7 = 247925 W
F.) System efficiency
Efficiency = power line / power source × 100
Efficiency = 147925 / 247925 × 100
Efficiency = 59.7 %
