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Power consumption help

TUSSFC Avatar
6y, 5m agoPosted 6 years, 5 months ago
We are looking to get an air conditioning unit which uses the following power (in the spec):

RATED POWER.............1600W
MAX INPUT POWER..............1800W

Anyone know how i can calculate how much electricity I could expect it to use? For example, amount per hour of usage.

No idea where to start.
TUSSFC Avatar
6y, 5m agoPosted 6 years, 5 months ago
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1 Like #1
You could try this (you can put 1600 or 1800W as your power) and then how much per kW which is on your bill.

http://www.ukpower.co.uk/tools/running_costs_electricity/
banned 1 Like #2
about 20p per hour
#3
Thanks. No as bad as I thought it would be then :-)
#4
Are you looking to buy a mobile or a split system. A split heat pump will give you 300% efficiency on the heat cycle, great for the winter.
ie 1000w of power in gets 3000 watts of heat out,

As with most things you get what you pay for, a cheap mobile is usually noisy and not very reliable, the stuff sold by the likes of B&Q et al are impossible to get parts for.

Try your local Climate Center.
#5
How can you get over 100% efficiency? You can't create energy, mearly change it from one form to another?
#6
Its only for the REALLY hot days as my other half feels the heat badly. So it will be used for a few hours on occasional days. Therefore we're looking for something cost effective (just watching eBay at the moment).
#7
Benjimoron;8861797
How can you get over 100% efficiency? You can't create energy, mearly change it from one form to another?

+1

lol at 1000w of power creating 3000w of heat! If someone could do that they would be a billionaire overnight.
#8
The person who said you cannot create energy is correct, nevertheless they are 300% efficient as a heater as they use electrical power to take heat from the air outside and raise the temperature inside. Check it out on the internet (that's what it is for).
Having been in the air conditioning business for many years I do know what I am talking about having supplied similar systems for ultra efficient buildings and low carbon footprint systems.

Quote below from Wikipedia
When comparing the performance of heat pumps, it is best to avoid the word "efficiency" which has a very specific thermodynamic definition. The term coefficient of performance (COP) is used to describe the ratio of useful heat movement to work input. Most vapor-compression heat pumps utilize electrically powered motors for their work input. However, in most vehicle applications, shaft work, via their internal combustion engines, provide the needed work.

[COLOR="DarkGreen"]When used for heating a building on a mild day, a typical air-source heat pump has a COP of 3 to 4, whereas a typical electric resistance heater has a COP of 1.0. That is, one joule of electrical energy will cause a resistance heater to produce one joule of useful heat, while under ideal conditions, one joule of electrical energy can cause a heat pump to move much more than one joule of heat from a cooler place to a warmer place.[/COLOR]

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