UEA Energy Conservation Home Page ENV-2D02

ENERGY CONSERVATION [ENV-2D02] - 2006

The following are questions asked by people regarding questions during revision. In some cases I have added extra questions of my own which might be supplementary questions.

Question about Degree-Days
Question about Degree-Day Formulae
I do not seem to have one of the Lecture Note Handouts
Worked example of CCGT from last week's revision session
Query over availability of Data Book
Question about air-exchange - interchange with adjoining rooms
Question about Primary Energy Ratio and Delivered Energy
Query over Ranges of Values - which to use
Question on Time Switching
Question on U-values of Roof
Further aspects on abosrption Heat Pumps following revision session on 13th May.
Question on Predicted Mean Vote

9th May 2004 - I am having a bit of trouble working through the worked example for Degree Days in the lecture notes, and it says to see previous exam questions, however, I cannot find one on Degree Days.
9th May 2004: I missed the revision session, and I understand that a sheet giving Degree Day formulae was given out - where can I get one
9th May 2004: I do not appear to have a handout for lectures X - Z can I have another copy?

I am out of handouts for most sections and will have to print one off - as I shall be away for 2 days this week, you will find it much quicker to access the relevant handout from the Course WEB Page.

11th May 2004: Do we have to bring our own Data Books
11th May 2004: Why does air exchange take place only with outside (with the air at the outside temperature (say 0 deg C)? Surely there will be exchange with adjoining rooms

12th May 2004: Can we have a copy of the example on CCGT you went through last week?

This is attached - you can access by clicking here

12th May 2004: In section 11.13. Example. I thought that the primary energy ration is the primary energy divided by the delivered energy. And that the delivered energy was that delivered to end use- the customer. But according to this example, the primary energy ratio incorporates all the losses up to the point where the fuel reaches the power station.. Is that what it is?

The Primary Energy is the content of the energy as it exists in the group before extraction. Delivered Energy is the content of the energy as it is delivered to the point of use and includes the losses in extraction and transport to poitn of use. It is related to the Primary Energy by the Primary Energy Ratio i.e. the proportion of energy which must be extracted to provide one unit of delivered energy. For gas it is around 1.06, oil, about 1.08, and coal 1.02. However, the figures do change slightly from time to time and country to country as electricity is almost always used in the extraction of primary fuels, and there is a feedback effect from this and the magnitude of the feedback is related to the fuel mix in paower station etc. However, none of the figures above have changed by more than 0.1 over the last 30 yeasr in the UK.
For a coal fired power station, the primary energy ratio will be 1.02 (from above) as would coal as delivered to a house. However, in the case of electricity, there are further conversions in the power station, so the overall primary energy ratio for electricity will be the ratio of the energy in the ground to that at the point of end use.
Since, the efficiency of electricity generation has improved significantly, particularly sicne the introduction of CCGT stations, the Primary Energy Ratio for electricity is constantly changing. In the 1930's it was around 6 - 6.5. In the example in section 11.13, the effeciency of the coal fired power station is 38.5%. So the overall amount of energy delivered to the home (taking account of 5% tranmission losses in this case) will be
1 x 0.385 x 0.95
--------------------- = 0.359 units of delivered energy
1.02
Thus overall the Primary Energy Ratio for electricity will be 1 / 0.359 = 2.79.

12th May 2004: General question: I am a bit worried about finding the U values of different materials in the data book. Some of them have ranges and I am not sure which value I should use. Will these be given to us in the exam question or will we need to find them in the data book ourselves?

It is normal practice not to give information which is in the data book in an exam, otherwise some questions could become overwhelmed with information already available. The use of the data book is to mimic real life where a person would look up relavant values for use. In the case of a range of values, this is a situation that often confronts a person, and part of the exercise is to exercise rational judgement in such instances. Remember, energy loss is an imprecise science anyway and at best one can get an estimate within 10%. Usually the range of values it will not matter overly which you use, but you should give some justification based on your experience.
For instance, it would be perfectly permissable to use the mean value, or if you suspect poor workmanship the worst case value. The important thing is that you explain your assumptions. Provided that your assumptions are rationally thought out then you are unlikely to be penalised much if at all. Where you might be penalised is in the case of single glazing where the U-values can range from around 4.5 to 5.7 W per sqm per deg C. The 4.5 value is more approapriate for south facing, and clearly if you had a north facing room and you used 4.5, then this would be inconsistent.

12th May 2004: "The key thing to remember about saving with time switching is that the energy required is directly proportional to the mean internal temperature, and if this is kept higher - with a larger boiler or better insulation, then the saving will be proportionally less." I don't understand exactly why.
,In a steady state situation, the heat loss is proportional to the temperature difference between inside and outside, and thus if the mean temperature within a house is lower than another identical one, it will have a lower energy requirement.
On time switching, the internal temperature will start to drop as soon as the heating is switched off. In a typical well insulated house, the temperature might drop from say 20 deg C to around 16 - 17 deg C overnight. In a poorly insulated house, the temperature may drop o between 10 and 15 deg C. In a well insulated house, the heat loss rate might be 200 W per sqm per deg C and with timeswitching, the mean temperature might be around 18.5 deg C as opposed to the 20 deg C without timeswitching. If the external temperature is 5 deg C, the the heat requirement with no timeswitching is (20 - 5) * 200 = 3000 W. With time switching, the requirement would be (18.5-5) *200 = 2700 W a saving of 10%.
In the poorly insulated house with say 400 W per sqm per deg C, the loss at 5 deg external temperature would be (20 - 5) *400 = 6000W or twice that of the well insulated house. Now with time switching, the mean temperature will fall perhaps to 15, so the heat requirement with time switching will not be (15 - 5) * 400 = 4000W, still well above the well insulated house, but this now represents a saving of 33.3%.
Thus a poorly insulated house will always save PRORTIONALLY more energy from time switching than a well insulated house.
If there is a larger boiler, then the temperature will rise more quickly when the heating comes on, and over the day the mean temperature will be somewhate higher than with a smaller boiler. Comfort levels may be achieved quicker, but the saving will be less. In the case of the poorly insulated house, and say the boiler has a size of say 12 kW, it might take 3 - 4 hours to get up to thermostat temperature. If on the other hand the boiler is say 18kW, the temperature will reach the thermostat level much earlier and the mean temperature might now be 15.5 or even 16 deg C. If it is the latter then the new heat requirement would be (16-5) *400 = 4400 W, and so the proportional saving will now be reduced to 26.7%.

12th May 2004: Could you maybe explain why we have to multiply the resistance of the layers above the air space with cosA? I understand it has to do with the geometry of the roof but I cannot visualize exactly how such a roof would look.

The roof is shown in Fig. 6.8, and with regards to heat flow from the upstairs rooms, there is normal heat transfer through the ceiling. This includes the internal surface resistance typically 0.11 m^{2 o}C W ^-1, the resistance of the plasterboard (0.06), the reststance, if any of the insulation, the resistance of the cavity forming the loft space (0.18).
The resistance is thus the sum of these i.e. 0.11 + 0.06 + 0.18 = 0.35. To this value should be added the resistance of any insulation. In most cases there will be a pitched roof, and the total resistance of the pitched roof at an angle A will be the resistance of any felt (0.11), the resitance of the cavity between the felt and tiles - only if felt is present(0.12), the resistance of the tiles (0.04), and finally the external resistance (0.04)..Typical values of the resistances for these components are i Now since thes are at an angle A they will have a greater area than the reference area(i.e. the area of the ceiling below. This means that there will be proportionally more heat loss though this section and to bring it into equivalence we need to get the projected horizontal area of the pitched roof which will be the Area * cos A. These means that in this example the resitance of the pitched roof is [0.11 + 0.12 + 0.04 + 0.04] *cos A. Typically A will be 45 deg, so in this example the effective resistance of the pitch will be 0.219 m^{2 o}C W ^-1 (i.e. the evaluation of the previous expression)
Finally to get the overall resitance this will be (0.35 + 0.219 + any resistance from insulation). The U-value will of course be the reciprocal of this.

13th May 2004: Follow on aspects to question on absorption heat pumps in revision session

An absorption heat pump is shown schematically below

In an absorption Heat pump the absorber/desrober replaces the electricially driven compressor in a conventional heat pump. The key to operation is the abosrption of a gas into a fluid and its release on heating. Usually there is a small pump which increases the pressure in the same was as the compressor. However, unlike a normal heat pump, it is liquid that is being compressed, and since the energy required equals the pressure multiplied by the change in volume, and the volume change with pressure in a liquid is very small, very small amounts of electrical energy are actually needed. [in a conventional compressor where gas is the fluid, then increasing the pressure by a factor of 10 means that the volume reduces by approximately the same amount, and hence there is much more work done by the mechanical compressor].
The absorber is kept cool and absorbs the gas while after the pressure has been increase, heat imput causes the gas to boil off and also increase in pressure further [in some cases, this increase in pressure is sufficient to avoid the use of a pump altogether]. The solution with the gas removed is then recysled and cooled using a heat exchanger (which also helps to heat the high pressure fluid).
Coefficients of performance are usually in the range 1.2 - 1.6 for the heat pump mode and between 0.8 and 1.2 for the refrigeration mode. Though theses performances are low, they do open opportunities for effective energy conservation when used in conjunction with CHP or otehr waste heat sources (including solar energy). For instance, a limitation with CHP is the need to provide a heat load, and this can present a problem in summer. By using an absorption heat pump, waste heat arising during generation in summer can be used to chill water which can be used for air-conditioning or cooling purposes. Southampton Geothermal Scheme uses such an absorption chiller to provide chilled water in summer. There is also a proposal for a similar scheme at UEA - (see next section).
Interestingly, using CHP with an absorption chiller is a "Win-win" situation. At UEA, for example, there is an increasing demand for electricity for chilling scientific equipment. Equally, in sumemr we have a problem that we do not have sufficient use of heat in the summer. If we could utilise the waste heat to power an absorption heat pump not only would be able to generate electricity in a more environmentally friendly way, but we would also reduce our own demand for electricity - i.e we win on two accounts. Of course in winter, it does not make sense to use such chillers as we need the heat for space heating in buildings

13th May 2004: How do you estimate the Predicted Mean Vote?

To estimate the PMV you need measurements or estimates of
air-temperature
activity level - metabolic rate
clothing level
wind velocity
The above parameters are needed to start with - the following two are needed only if the mean radiant temperature differs from the air-temperature or the relative humidity is not 50%

mean-radiant temperature
relative humidity

from the first four parameters select the relevant table in the data book corresponding to the activitiy level
select the sub table corresponding to the clothing level
select the column corresponding to the wind-velocity - if indoors you can take this as zero
read off the PMV from the appropriate temperature line.
If the temperature is between values then interpolate in the noraml way. If the clothign level is between sub tables, the repeat for the sub table with clo level below and that above the actual clo value, and interpolate to get relvant value in between. If the activity level is between table values, then an interpolation between tables will be necessary, although it is very unlikely this would be set in an exam question.
The value obtained is the Predicted Mean Vote with MRT = air-temperature, and relative humidity equal to 50%. To correct for the MRT not equal to the air-temperature, there are graphs on which you can read off the correction to the PMV for each one degree temperature difference between MRT and air-temperature. These should be in the data book, however, due to a cock up though these appeared in edition 6.0, for some reason they did not in 6.1 nor in the latest version which will be available to you in the exam. Consequently, unless the appropriate graphs were provided separately for you, you would not be able to do this correction, and you can assume that the pMV is as determined above. The same issue also applies to relative humidity.

This page is maintained by Keith Tovey (e-mail: k.tovey@uea.ac.uk).