14. a large car fire presents the possibility of
a. an explosion.
b. o toxic fumes.
c. both of the above

the answer would possibly be A. if not then C.

a. The magnitude of the line source voltage is

Vs = 4160 V

b. Total real and reactive power loss in the line is

Ploss = 12 kW

Qloss = j81 kvar

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

Ss = 540.046 + j476.95 kVA

Ps = 540.046 kW

Qs = j476.95 kvar

Explanation:

a. The magnitude of the line voltage at the source end of the line.

The voltage at the source end of the line is given by

Vs = Vload + (Total current×Zline)

Complex power of first load:

S₁ = 560.1 < cos⁻¹(0.707)

S₁ = 560.1 < 45° kVA

Complex power of second load:

S₂ = P₂×1 (unity power factor)

S₂ = 132×1

S₂ = 132 kVA

S₂ = 132 < cos⁻¹(1)

S₂ = 132 < 0° kVA

Total Complex power of load is

S = S₁ + S₂

S = 560.1 < 45° + 132 < 0°

S = 660 < 36.87° kVA

Total current is

I = S*/(3×Vload)   ( * represents conjugate)

The phase voltage of load is

Vload = 3810.5/√3

Vload = 2200 V

I = 660 < -36.87°/(3×2200)

I = 100 < -36.87° A

The phase source voltage is

Vs = Vload + (Total current×Zline)

Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)

Vs = 2401.7 < 4.58° V

The magnitude of the line source voltage is

Vs = 2401.7×√3

Vs = 4160 V

b. Total real and reactive power loss in the line.

The 3-phase real power loss is given by

Ploss = 3×R×I²

Where R is the resistance of the line.

Ploss = 3×0.4×100²

Ploss = 12000 W

Ploss = 12 kW

The 3-phase reactive power loss is given by

Qloss = 3×X×I²

Where X is the reactance of the line.

Qloss = 3×j2.7×100²

Qloss = j81000 var

Qloss = j81 kvar

Sloss = Ploss + Qloss

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

The complex power at sending end of the line is

Ss = 3×Vs×I*

Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)

Ss = 540.046 + j476.95 kVA

So the sending end real power is

Ps = 540.046 kW

So the sending end reactive power is

Qs = j476.95 kvar