Uploaded by Philip Raicer Ramos

FPSAD52 ELEP03 PROBLEM SET 2 RAMOS

advertisement
RAMOS, PHILIP RAICER R.
FPSAD52/ ELEP03
PROBLEM SET 2
10/03/2021
1.Asystem operates at 220 kVA and 11 kv. Using these quantities as base values, find the
base current and base impedance for the system.
Given
Shase 210 EVA
Vpase
kV
Rea'd
Inse: ?
Zbase
Solution:
Ibagebase 120
kA
kV
20A
Vbase
Voase
hae IIE - 550a
Zbase
2.
Using
220 kVA and 11 kV
as
values
base values, express 138 kV, 2 MVA, 60 A, and 660 N
Ip.u. acdual
Gven
per-unit
COA
Le
Sbace 22O kVA
base
VaaseIlkV
Sactual 2 MVA
Vactual
136 kV
Taciaal
GOA
Zacku
Con
Req d
as
Zp.1. Zaotu.
base
p. u values
Solution
Spu. * Sackal
Sbase
2MVA
210 kVA
.0 pu
Vactual
Nase
TTRY25B p-u.
Vp
220 EVA
Lbase Sease
Vbose
lV
-
Sse
Snace
=
20 A
550
GGO -\.1 pu
2okVA
RAMOS, PHILIP RAICER R.
FPSAD52/ELEP03
PROBLEM SET 2
10/03/2021
3. If 25 0 and 125 A are the base
the base kVA and base
voltage.
Given
and base current,
respectively,
for
a
system, find
SpaseIbo ) (Vpnse)
(125A) (3.115kV)
Zbase 25
oase
impedance
125A
Sbage
90. 615 k VA
Req'd
Sbase ?
Vbase?
Solution
Vase Iose) (2 base)
(115A) (25a)
Vbae
3.125 kV
4, The
percent values of the voltage, cCurrent, impedance, and
volt-amperes for a given power
system are 90, 30, 80, and 150 percent, respectively. The base
current and base impedance
are 60 A and 40 2,
Calculate
respectively.
the actual values of the voltage, current,
impedance, and volt-amperes.
Given
Vp.u
9O % = o.9 pu.
Ipa. bo, : 0.p:4.
Zp.u.
86, = 0.5p.u.
Sp.u. 150%,
\L5p.u.
Tbase"GOA
Zachal 31
Spse (ase) (Zb)
(6OA (4oa)
144 kVA
Sacual (Sp u.)(Spase)
Req'd: acual values
nex 2bqase
6 0 x40)»2.4 kV
Vochual (Vpu.)(Vonze)
(0.pu.)(a-4kv)
Nachel21Go V
Tacua (Ipa) Tsco)
(0.3p.tu.)( Go A)
|Iachual I5A,
C0.B pu)(40a)
Sbase
Zase 40
Soluhon
ZauatZp.u)(zbage)
(5p-u) ( 144EVA)
Sactual: 21G kVA
RAMOS, PHILIP RAICER R.
FPSAD52/ ELEPO3
PROBLEM SET 2
10/03/2021
5, A single-phase transmission line supplies a reactive load at a lagging power factor. The load
draws 1.2 p.u. current at 0.6 p.u. voltage while drawing 0.5 p.u. (true) power. If the base
voltage is 20 kV and the base current is 160 A, calculate the power factor and the ohmic
value of the resistance of the load.
Given
Vp-. O.Gp-
Zbase
Ip.u. 11p-u
VpaseLOE
Toase
Vp.u . p.u
T.u.Tpu.
Pp-
5pu.
P-
Vbase
2okV
Zactual (7pu.) (2base
Ibase
P4
(OSp-u) (125.a)
16OA
Zactual
Rea '
25A
G2.5
P.Fp.4.?
Zactual
2
Solution
P TVcos
P.F.
cose
P-F-p-upu(p-u
(o 5 p
(1.2p)(0-Gp:u.
P.Fp.u. 0.G14 p-u
impedance and base voltage for a given power system
respectively. Calculate the base kVA and the base current.
6. The base
Given
Zbase
l6n
Vpase
4ov
Spase IneVase)
(40A) 400v)
bce1GkVA|
Req'
Spnse
Ibase?
Solu Hon
Tbose 2base
Aoov
base 40A
are
10 Q and 400 V,
RAMOS, PHILIP RAICER R.
FPSAD52/ ELEPO3
PROBLEM SET 2
10/03/2021
7. The base current and base
respectively.
voltage of a 345-kV system are chosen to be 3000 A and 300 kV,
Determine the per-unit voltage and the base
impedance for the system.
Given
base
e
Lbag
base0o0A
SOkV
Vpase= 3Oo kV
3oo0 A
Vachuala 345 kV
ZbaseTooa]
Vp-y.
Zbase ?
Solution'
Vp.. Vactual
Vpase
345kV
300 kV
p-u.1.15.u.
8.If the rating of the system of Problem 2.2 is 1380 MVA, calculate the
to the base of Problem 2.2.
Given
Sactual 1330 MVA
Vochal
345 ky
Viose
3o0kV
base 3000 A
Req: Ip .u
?
Solu tien'
Lactel Sactua
Vacua
Ip.u. actua
Lase
400 OA
000 A
Tp-u. 1.93 p
13go MVA
345kV*4000 A
per-unit current referred
RAMOS, PHILIP RAICER R.
FPSAD52/ELEP03
PROBLEM SET 2
10/03/2021
9.
Express 100-0 impedance, a 60-A current, and a 220-V voltage
to the base values
of Problem 2.1.
achual
O0A
base
GOA
Vactual 210y
10
borse
400V
pase
A4OA
per-unit quontities referred
Ipu5Lactial
achual = GOA
Zbase
as
40A
-u.15pu
Vachual
Wp-u. V,ase
belGEVA
2oV
Req'd per unit vaues of 2, T,L V
Soluton
Zp.u. = Zacal
400V
p-u.O. S5pu
Zhase
10
pu.Op-u
10. A single-phase,
10-kVA, 200-V generator has internal impedance Z, of 2Q. Using the ratings
of the generator as base values, determine the generated
per-unit voltage that is required
to produce full-load current under short-circuit conditions.
Given
Sbase O k VA
Load
Vpase 0 O V
IFL
Let
Vp.a.ll load current
Soluion
full locd
Sbase
okVA
200 V
FL
50 AA
2 base
(50A)(2L)
Zbase 2
ReqJ
curreut under shof
l00 V
(0o V as Vackyal
V -u.
Vactal
Vbase
0OV
100 V
Vp-u. O p.u
cireuit
Download