Chapter 8
8.1:
340
= 240.4 V
2
a.
V ph =
b.
Vll = 240.4 3 = 416.4 V
c.
I ph =
d.
I l = I ph = 70.7 A
e.
ω = 377 = 2 π f
f.
pf = cos (0.5236 − 0.87266) = 0.94 Leading
g.
P = 3 V ph I ph cosθ = 3 * 240.4 * 70.7 * 0.94 = 47.93 kW
h.
Q = 3 V ph I ph sin θ = 3 * 240.4 * 70.7 * 0.341 = 17.387 kVAR
i.
Z=
100
= 70.7 A
2
:
f =
377
= 60 Hz
2π
Vll
3 416.4
=
= 10.2 Ω
70.7
Il / 3
Z = 10.2 (cosθ + j sin θ ) = 10.2 (0.94 + j 0.341) = 9.59 + j 3.478 Ω
8.2:
a.
Vab = Vca ∠ − 120o = 480 ∠ − 180o V
Van =
Vab
∠ − 30 o = 277 ∠150 o V
3
b.
I a = I b ∠120o = 20 ∠240o = 20 ∠ − 120o A
c.
θ = ∠Van − ∠I a = 150o − 240o = − 90o
d.
P = 0 , since the power factor angle is 90o
8.3:
Fundamentals of Energy © M. A. El-Sharkawi, 2003
31
480∠0
= 277.1∠0
3
I a = 10∠ − 90 (Same angle as that of Vbc )
Van =
a
Z=
Van
= 27.71∠90
Ia
Z Van
=
= 27.71∠90
3 Ia
Z = 83.13∠90
b
8.4:
V ph = Vll = Vll ∠0 o
I ph =
Il
I ∠ − 50 o
I
∠30 o = l
∠30 o = l ∠ − 20 o
3
3
3
The power factor angle is the angle θ between the load voltage and the load current. Hence,
θ = − 20 0
8.5:
a.
b.
Il =
(173 / 3 )
= 19.97∠ − 36.86
4 + j3
pf = cos(36.86) = 0.8 lagging
173
× 19.97 = 5983.9VA
S = 3×
3
P = S × cos(36.86) = 4787.125W
Q = S × sin(36.86) = 3590.3VARS
c.
Vca
Ic
36.87o
Ib
Ia
Vab
Vbc
8.6:
a
b
I l = 20∠ arccos(0.9) + 30∠ − arccos(0.8) = 43.01∠ − 12.4 A
pf = cos(12.4) = 0.9766 lagging
Fundamentals of Energy © M. A. El-Sharkawi, 2003
32
P = 3 × 400 × 43.01× 0.9766 = 29100.924W
c
8.7:
340 ∠20 o
∠ − 30 o =196.3 ∠ − 10 o V
3
3
o
Van =Vbn ∠120 =196.3 ∠ − 10 o ∠120 o =196.3 ∠110 o V
Vbc
Vbn =
a
∠ − 30 o =
vbc
van
vab
n
30 0
20o
Reference
vbn
vcn
vca
Vbn 196.3 ∠ − 10 o
=
=19.63 ∠ − 70 o A
Z
10 ∠60 o
b
Ib =
c
The neutral current is zero for balanced three-phase system
5
* 0.746 = 1.243 kW
3
8.8: The power per phase is
8.9:
I a = I c ∠ − 120 o = 10 ∠140 o ∠ − 120 o = 10 ∠20 o
a
Va =
Vab
3
∠ − 30 o =
A
200 ∠50 o
∠ − 30 o = 115.47 ∠20 o V
3
θ = ∠I a − ∠Va = 0 o
P = 3 Vll I l cosθ = 3 200 *10 cos 0 o = 3.464 kW
b
8.10:
Convert Z Y to delta, then
Add both loads in parallel
Zt =
=
Z ∆ 2 = 3 Z Y = 15 ∠40 0 Ω
Z∆ Z∆2
Z∆ + Z∆2
(10 ∠ − 25 ) (15 ∠40 ) = 7.0570 + j 0.0294 = 7.0571∠0.24
(10 ∠ − 25 ) + (15 ∠40 )
o
o
o
o
Compute the load current
o
Ω
I ab
Fundamentals of Energy © M. A. El-Sharkawi, 2003
33
I ab =
Vab
208 ∠0
=
= 29.47 ∠ − 0.24o A
o
Z t 7.0571∠0.24
Calculate the line current
I a = 3 I ab ∠ − 30o = 51.04 ∠ − 30.24o A
8.11:
Convert delta into Y
ZY =
9∠30o
= 3 ∠30o Ω
3
Ib =
120∠0
= 40 ∠ − 30o A
3 ∠30o
I a = I b ∠120o = 40 ∠90o A
P = 3 Vl I l cos θ = 3
(
)
3 120 40 cos 30o = 12.47 kW
Fundamentals of Energy © M. A. El-Sharkawi, 2003
34