Ask the Expert
A panel of UKAEE volunteer energy management experts is available to provide quick advice to UKAEE members by email (see link below).
Accessing advice: please fill in the fom below and we will endeavour to respond to you as soon as possible.
Note: the exchanges below may have been edited for brevity
A.M. asked: is it legitimate for a supplier of LED lighting to claim that improved power factor will reduce the kWh energy consumption of our lighting installation which at present uses electromagnetic ballasts?
Response: Not if you are metered on the low-voltage side of your supply transformer. Improving the power factor merely reduces the current drawn for a given real power output. You will save some cash on demand charges if they are denominated in kVA, or if you are explicitly charged for poor power factor and the reduction tips you over a penalty threshold. If you are metered on the HV supply, however, you can expect some kWh savings through lower transformer losses.
D.S asked: We are utilising a version of ‘degree days’ mainly relating to heating, but I am looking for data on degree day for cooling AND a methodology for taking moisture content (i.e. external RH) into account.
Response: Using cooling degree days will deal with much of your requirement. They are available free from, for example, degreedays.net. To take ambient moisture into account you can integrate the ambient moisture excess and deficit through time, to create a separate humidification and dehumidification index in an analogous way to degree-days for dry-bulb temperature. Look for 'dehumidification' in the A to Z at www.vesma.com.
You do not need to compute outside versus inside enthalpy (although you could) because enthalpy only adds the sensible and latent heat together, the former accounted for by heating and cooling degree days and the latter by the humidification and dehumidification indices.
This process will only really work if you measure the weather parameters yourself locally.
P.S asked: we want to quantify an estimated cost saving from dropping pool water temp in an outdoor swimming pool per 1 degree. We just need a approx. figure on this so any thoughts on a general equation/rule of thumb that would work on this (eg 5% gas saving from 1 degree drop)
Response: I have attached the below links for you that I hope may be of some use in your investigation. Apologies that the main content is not in SI units. The SI unit example calculation at EngineeringToolbox for heat required is a basic equation but will be of interest as the idea is to reduce water temperature only. Substituting the inlet and desired water temperatures along with heat up times will give the required kW for different target temperatures.