I want formula to calculate cable size as per load given in
kw & amp.I searched many sites but didn't right answer.Plz
reply me asap.
Answers were Sorted based on User's Feedback
Answer / engr. abed alraheem
dear you can calculate the size of the cable as follow:
first you need to measure the current capacity and it's low as follow:
Kw/(P.F * Voltage that we wanna use)= current
so when you get the current capacity for the load you can get the size from catalogs there is tables gives you the rating of current and how much the size for every rate for this current.
and this current will be ok if we dont have a voltage drop in line of distance between the source and the load.
actually all the distance between the load and the source have a voltage drop so we have to use this law in order to measure the percentage Voltage drop and it has be 2.5% from the load to panels:
V.D = {((x sin@ + Y cos @) x I x L)/ V } x 100%
I is the current that you measured.
L is the distance between the load and the panel.
V is the voltage that we use.
X is the AC resistance.
y is the AC reactance but sometime we ignore it.
X,Y it is depends on the size of cable so when you chose size this factors will control the voltage drop and you can change it sometimes to get the voltage drop rate as 2.5%
Is This Answer Correct ? | 17 Yes | 8 No |
Answer / inudeen
FIRST YOU CALCULATE VOLTAGRDROP
VD=1.732*V*L*I*Z/1000
%VD=VD/RATED VOLTAGE*100
LV CABLE MINIMUM=5%
MV CABLE MINIMUM=3%
V=VOLTAGE
L=LENGTH FOR FEED
I=AMPS
Z=IMPEDANCE
Is This Answer Correct ? | 10 Yes | 1 No |
Answer / antony ngigi
What is the correct size of cable[4core armoured cable]for
aload of 15kva.
Is This Answer Correct ? | 11 Yes | 4 No |
Answer / anubaby
Voltage drop Calculation for Exaple 2Run cable
=mV/A/M(Full load current/2)length
for single Run
length x Full load current x mV/A/M/1000
mV/A/M find from cable Catalogue
this is only for Voltage drop
Is This Answer Correct ? | 7 Yes | 1 No |
Answer / saran
hey hi he needs formula for cable calulation without refer
any catloge
Is This Answer Correct ? | 9 Yes | 4 No |
Answer / berialay
i have a load 230 ampere and the length to the MDP is 285
meter the is going to mdp in a pvc duct under ground i need
the formula
Is This Answer Correct ? | 6 Yes | 1 No |
Answer / nour travis
Voltage drop works on Ohms Law.Multiply the route length in
metres by the the number of ampers of current that the
appliance draws,and then multiply by the number of milivolts
per ampere metre (mV/A.m)
Formula: Vd = L x I x Vc /1000
Where: Vd is the voltage drop measured in volts
L is the length of the cable in metre
I is the circuit current in Ampere
Vc is the millivolts per amper per metre droped
along the cable
1000 is a correction factor applied because Vc is in
millivolts and Vd is in volts.
Is This Answer Correct ? | 37 Yes | 33 No |
Answer / engr wasim khan
Oh Dear..Its so easy...Follow these simple steps (below
link)...
http://www.electricaltechnology.org/2013/10/How-to-determine-the-suitable-size-of-cable-for-Electrical-Wiring-Installation-with-Solved-Examples-in-both-British-and-SI-System.html
Titled as "How to determine the suitable size of cable for
Electrical Wiring Installation with Solved Examples (in both
British and Si System) "
Is This Answer Correct ? | 5 Yes | 1 No |
Answer / engr.muhammad younas khan
CABLE SIZING FORMULAS:-
1ST STEP CALCULATE THE VOLTAGE DROP (Vd).
(1 ph.) Vd= 2 x I x L x P.F/(K x Area mm2)
(3 ph.) Vd=1.732 x I x L/(KxArea mm2)
where:-
I:- Load current
L:- Length of the conductor.(in meters)
K:- 58 for copper conductor and 36 for aluminum conductor.
2 for single phase and sqrt 3 for three phase.
____________________________________________________________
2nd step.
AREA mm2 = 2 X I X L /(K x Vd) (FOR SINGLE PHASE)
AREA mm2 = 1.732 x I x L/(K x Vd)(for three phase)
____________________________________________________________
FORMULAS DESIGNED BY
ENGR.
MUHAMMAD YOUNAS KHAN
M.Sc.Engg.,MIEE(UK)
NEAR KEEKER WALI MASJID.
MOHELLAH LAKER MANDI
WAZIRABAD CITY,PUNJAB
PAKISTAN
Is This Answer Correct ? | 5 Yes | 1 No |
Answer / santhosh
as one of our friend said above
for Al its 1.2 amps and for Cu its 5.0 amps per 1 Sq.mm
but actually its
1 .0 amp for Al and 8-12 amps for Cu [per 1 Sq.mm].
the basic thing is that suppose a cable for load of 100 amps is to be laid then
for Al 100 amps/3 phases=33.33 amps/phases,
as we don't have 33 amps cable take 36 Sq.mm cable (near and above the value) i.e,3 C*36 Sq.mm.
for Cu consider 8 amps for Cu
then 8 Amps*5 Sq.mm =40 Amps per core or phase
i.e,for 100 amps load u need to lay
a cable Al= 3 Core *36 Sq.mm
and for Cu =3 core *5 Sq.mm
Is This Answer Correct ? | 4 Yes | 1 No |
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