🚀 go-pugleaf

RetroBBS NetNews Server

Inspired by RockSolid Light RIP Retro Guy

Article View: sci.electronics.basics
Article #50

Re: phasors and power calcs

#50
From: John Popelish
Date: Thu, 26 May 2005 03:37
35 lines
3295 bytes
jan@janhurst.com wrote:
> Hello,
> 
> I am studying an introductory electronics course and have been given the 
> following problem:
> 
> A small factory has the following loads operating from the 240Vrms 50Hz 
> supply
> a) Forty 50 watts incandescent lamps with unity power factor
> b) Thirty-five 40 watts fluorescent lamps with a lagging power factor of 
> 0.9
> c) One 2400 watts air conditioning system with a lagging power factor of 
> 0.65.
> 
> I am asked to draw the complete phasor diagram and determine the total 
> peak current drawn and overall power factor of the factory.
> 
> I'm a little stuck :( I have some phasor diagrams put together which I 
> am a little unsure of, but I don't really know how to determine peak 
> current from this, can anyone shed some light or point me in the 
> direction of some material to explain this sort of problem? I've been 
> digging through a number of text books and am not really progressing.

Break each of the individual load current phasors into a real (in 
phase with voltage) and imaginary (90 degrees shifted with respect to 
voltage) components.  Add all the real components together as the real 
component of the total load phasor, and add all the imaginary 
components together as the total imaginary component of the total load 
phasor.  Then put these two components start to end and draw the 
diagonal sum that represents their resultant.  Of course, if you have 
a compass and protractor, you could just add the individual load 
phasors one to the next and get the same resultant.  Breaking them 
into components just makes it easer to do the sums, graphically 
without those tools  The divisions on the graph paper is the only tool 
needed.
Technical Details
Message-ID: <RLudnTCXu7nH3AjfRVn-iw@adelphia.com>
Path: arc1-ca.pugleaf.net!g2news1.google.com!news3.google.com!news.glorb.com!border1.nntp.dca.giganews.com!nntp.giganews.com!local01.nntp.dca.giganews.com!nntp.adelphia.com!news.adelphia.com.POSTED!not-for-mail
References: <42948abb$0$28838$5a62ac22@per-qv1-newsreader-01.iinet.net.au>

Show Full Headers
Complete Article Headers
X-Google-Thread: 111d0a,f6267b6883c0c47a
X-Google-Thread: 103407,f6267b6883c0c47a
X-Google-Attributes: gid111d0a,gid103407,public
X-Google-Language: ENGLISH,ASCII-7-bit
Path: arc1-ca.pugleaf.net!g2news1.google.com!news3.google.com!news.glorb.com!border1.nntp.dca.giganews.com!nntp.giganews.com!local01.nntp.dca.giganews.com!nntp.adelphia.com!news.adelphia.com.POSTED!not-for-mail
NNTP-Posting-Date: Wed, 25 May 2005 22:24:10 -0500
Date: Wed, 25 May 2005 23:37:52 -0400
From: John Popelish <jpopelish@rica.net>
Organization: This space not available for advertising.
User-Agent: Mozilla Thunderbird 1.0.2 (Windows/20050317)
X-Accept-Language: en-us, en
MIME-Version: 1.0
Newsgroups: sci.electronics.basics,aus.electronics
Subject: Re: phasors and power calcs
References: <42948abb$0$28838$5a62ac22@per-qv1-newsreader-01.iinet.net.au>
In-Reply-To: <42948abb$0$28838$5a62ac22@per-qv1-newsreader-01.iinet.net.au>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: 7bit
Message-ID: <RLudnTCXu7nH3AjfRVn-iw@adelphia.com>
Lines: 35
NNTP-Posting-Host: 69.168.0.152
X-Trace: sv3-88vTNQRW6waZqvyVXPcp7QJ3woggqOtuAxZRS8Pcxw81LygAX2GQunFZI/fg5GHtVkz3k7Rb6wwVRWv!A6o6bDAdylf72Gj9Ny1aVwsaT7nI59lLBpJQtGL7FjzrNasQ1kc+2rMf6RyD4xBVEZJBskePX/oF!OX7vZ0NFIFem5dP7
X-Complaints-To: abuse@adelphia.net
X-DMCA-Complaints-To: copyright@adelphia.net
X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers
X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly
X-Postfilter: 1.3.32