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, compute the following: a. Voltage drop across the line Vline b. Voltage at the source side Vsource c. Percentage of the voltage drop Vline /Vsource d. Line losses e. Power delivered by the source f. System efficiency

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 %

lhmarianateixeira lhmarianateixeira · Answer 2 · 2022-03-15T13:20:04+01:00

In this exercise we have to use the circuit knowledge of an electrical system and calculate the characteristics so we have to:

A. ) 591.7 v

B.) 991.7v

C.) 59.7%

D.) 47.9 Kw

E.) 247925 W

F.) 59.7 %

Organizing the information given in the statement we have that:

power is 100 kW.
line is 100 km
3 cm diameter
voltage at the load side is 400 V
The resistivity is 1.673×10^-8 ohm metres

Calculating the resistivity we find that:

[tex]R =( L* \rho)/A\\A= \pi r^2 = 0.00071 m^2\\R = (100000 * 1.673*10^{-8}) / 0.00071\\R = 2.367 ohms[/tex]

Then it becomes simpler to calculate the power and current, we have:

[tex]P = V^2 /R= 1.6 Ohms\\l = V / R = 250 Ampere[/tex]

A) With the above information, we can calculate the voltage as:

[tex]V = I R\\V = 250 * 2.367\\V = 591.7 v[/tex]

B) Now calculating the source voltage, we find that:

[tex]V = V line + V load\\V = 400 + 591.7\\V = 991.7 v[/tex]

C.) The percentage will be calculated as the division of the two previous values, like:

[tex]591.7/991.7 * 100 = 59.7\%[/tex]

D.) Like any imperfect circuit, losses occur, so the loss will be calculated as:

[tex]Power loss = 147925 - 100000\\Power loss = 47,925 W\\Power loss = 47.9 Kw[/tex]

E.) Power delivered by the source, can be:

[tex]P = IV\\P = 250 * 991.7 = 247925 W[/tex]

F.) System efficiency, will be:

[tex]Efficiency = power line / power source *100\\Efficiency = 147925 / 247925 * 100\\Efficiency = 59.7 \%[/tex]

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