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Old 17-07-2004, 02:22 AM   #1
BarraCuda™
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Default Is yr 250watts MH really drawing 250watts?

Many take it for granted that the wattage rating on their equipment is the same wattage that they use. You are very very wrong ... most equipment like pumps/ballasts state their "true power" (e.g 250watts) That means a min of 250watts is needed to work but is the total wattage drawn from the powersupply the same 250watts???

Most likely not, the power (apparent power) drawn from the powersupply is often higher than what is needed. E.g The 250watts may actually draw 330watts from the powersupply.

In the next few days .. I will share with you guys. What is powerfactor and how correcting powerfactor can give you significant savings.

Btw .. I just measured my resun MD55 is operating at 207watts but the actual power used is only 130watts! That is abt a waste of 70watts!
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Old 17-07-2004, 04:11 AM   #2
ReDDeviLs
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WHAT!!!!!!!!!!!!!!! I just got a md55!!!!! HELP!!!!
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Old 17-07-2004, 12:49 PM   #3
spsman
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Default hi barra

hi barra,
great. pls inform us on the power factor and how to save electricity bills.

in layman terms,
mh requires high temp to burn. due to gas in the tube. and high temp is derived by wattage consumption. the ballast basically steps up. the ignitor just starts the bulb.
and the capacitor basically acts as a circuit protector.

The old and new bulbs are burning at different temps thus a discrepancy has incurred.

so the way to temper the entire lighting system could be by
1) tempering the ballast

there is this commercial device on sale on using the power factor to minimise cost. it is said to minimise all electrical appliance consumption. someone offered to sell it at $135 at pasir ris resort. well, i have seen it but not experimented on it yet. becos i am really unfamiliar with electrical works.
but seriously, after seeing explosive electric kettle and thermos flask, not forgetting the burnt overhead filter incident in S,pore that burnt the house last year. it really makes me wonder how things really work.
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Old 17-07-2004, 01:00 PM   #4
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i came across this too,infact my mum spotted that.. it claim to cut down like 30% of total electricity if my memory doesnt fail me.
but i dont really trust that. anyone came across that?
if really can work,i know where to get.
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Old 17-07-2004, 01:14 PM   #5
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The ballast is just a current limiter, as MH bulbs got negative resistance so if you operate the bulb without a ballast, the MH will blow up due to the excessive current.

The ignitor is to provide the striking voltage from 2000V to 4500V needed to ignite the bulb. It also mantain the voltage needed to operate the bulb.

The capacitor is the key thing here, its a shunt capacitor. It works by countering the inductive load with reactive load. The overall result is lower apparent power and higher powerfactor. This also means that less power consumed and less heat.
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Old 17-07-2004, 02:23 PM   #6
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Default i see

so are u manipulating the capacitor..? sure intd to learn from u.

i will follow this thread. thks pal
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Old 17-07-2004, 04:05 PM   #7
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Default Result is out!

I measured the ampere of a single 250watts using a 20uF shunt capacitor.

1.67A x 230V = 384.1watts !!

Instead of drawing only ~1.1A (250watts) .. its drawing an actual 1.67A .. that is a wastage of 134.4watts!!


To calculate the power factor .. we take true power / apparent power = 250/384.1 = 0.65.
This meant that 35% of the power drawn are not used for the lights but to heat the wiring/ballast!


Here are the new and old capacitors
On the left is the 25uF and right is the old 20uF capacitor. I'm going to use new 25uF capacitor for a comparison.









After changing the capacitor, the ampere drawn is 1.35A

1.35A x 230V = 310.5watts

384.1watts - 310.5watts = 73.6watts
Just by changing the capacitor .. I saved 73.6watts!


I took the total power consumption by the 2x250watts MH


2.58A x 230V = 593.4watts

To get the power factor - 500watts/593.4watts = 0.84

So I have corrected the powerfactor from 0.65 to 0.84 .. what does it mean??


Using 20uF capacitor
384.1watts x 2 bulbs = 768.2watts


Using 25uF capacitor
593.4watts

reduction in power consumption
768.2watts - 593.4watts = 174.8watts



Now covert all these figures into $ savings
Assuming 10hrs of light per day and 30days per month

174.8watts x 10hrs = 1748Wh = 1.748kWh
1.748kWh x 30 days = 52.44 kWh

Tariffs now is at $0.1685/kWh
52.44kWh x $0.1635 = $8.57/month



This might not be much savings .. but what if its yr high powered chiller or pump??
Next .. I will try to correct the powerfactor of the resun MD55
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Old 17-07-2004, 04:17 PM   #8
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Default Atco Capacitors

Do not type in caps!! - Edited by BarraCuda™
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Old 17-07-2004, 06:02 PM   #9
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wow! so by changing to capacitor we can save money?? will the light intensity be compromised?? Lowering the current might be lowerring heat output and light output simultaneously, thus decreasing the PAR.. how do we know that it is only lowering the heat output?? for all we know the light may be less bright.. but maybe it's not visibly apparent but PAR may have dropped???
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Old 17-07-2004, 07:56 PM   #10
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Quote:
Originally Posted by SGRay
wow! so by changing to capacitor we can save money?? will the light intensity be compromised?? Lowering the current might be lowerring heat output and light output simultaneously, thus decreasing the PAR.. how do we know that it is only lowering the heat output?? for all we know the light may be less bright.. but maybe it's not visibly apparent but PAR may have dropped???
Nope, I did not reduce the current intake. All I did is to correct the powerfactor by changing the impedence. The reactive power will counter with the apparent power to shift the powerfactor closer to 1.

The actual intake is 380watts while the bulb only consume 250watts .. so where the excess power go? It goes into heating the ballast and wiring.

The reason why I did not aim for a PF 1 its because the ballast, ignitor and capacitor also consume power, if you over do it .. the PF will be reduced.
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