









Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Detailed lecture notes for a university lab session on single phase power, including calculations for finding voltage and current phasors, complex power, power angle, power factor, active and reactive power, and instantaneous power.
Typology: Lab Reports
1 / 16
This page cannot be seen from the preview
Don't miss anything!










ilolj {breul
peol
l u s l d
,r,r0lJu0llLuJJoJuI
u0!lelsqn
r r l s! 0p u e
lulsusJl
I
u0rlnq
u0rss
l0rlu03peol
l0rlu
u0!lelsqns
l0rlu
lueldiamod
8/9/2004 Lab II Lecture on Single Phase Power D. Niebur
.T l. ^ 0
,
r i l (^) : (^) - -. n ' r i a c u - j I
't cJ Uol a ;. ll., u,l L{,
(",1- zi
li
--tr"s xl
23 - s8 ,lt/
3 tu l
t,- (^) A L! (^) lLr vh ', Il IJ',,.l (^) b U ' :! q i t f p^
t : j t I
PowerPlant
PomrPlant
Slopup Translormer
c
CBOpen
SecondaryDistdbution Syslom
I
[t. !r," (^) 1 3 lI
. /! |' b o o! i t 6
6 ' I ot E 5
!
5 o .!
c I 5
D. Niebur Lab II Lecture on Single Phase Power 8/9/
The following single phase circuit is given:
Load
ZLine
v(t)
The time dependent ac voltage is
v t 339. 4 cos t 60º V
The line impedance is
Z Line ^5 j
and the load impedance is
Z Load 8. 66
Find
Is the power factor leading or lagging?
Solution:
V max 339. 4 V
| V | ^ V^ eff ^ V^ rms ^ V max /^2 ^ 339. 4/^2 ^ 240. 0
V 240 e j /3^ V
Z total
We thus have to find the total impedance of the circuit as the sum of 2 serial impendances Z line and Z Load and convert the complex impedance from cartesian to polar coordinates:
Z total Z line Z Load 8. 66 5 j
| Z | ^ 8. 66^2 ^25 ^ 10. 000
angle Z arctan 5
0. 523 6 rad 0. 5236 180/ 30. 0º /
Z total
I max (^) | I | 2
i t I max cos t angle Z
24 2 cos t 30º
33. 94 cos t 30º A
p t v t i t 240 2 cos t 60º 24 2 cos t 30º
5760 2 2
cos t^ ^ 60º^ −^ t^ −^ 30º ^ cos t^ ^ 60º^ ^ t^ ^ 60º
5760 cos30º cos 2 t 120º
5760 cos 1 6
cos 4 60 t 2 3
4988. 3 5760. 0 cos753. 98 t 2. 094 4
We have two ways to compute it:
0
T p t dt
with T (^22) 1201
0
120 cos 1 6
cos 4 60 t 2 3
dt