Coeffient Of Performance (C.O.P.)
The C.O.P. is a factor used to give the performance of a heat pump when working in a heat cycle. Its value is:
Heat output obtained from the condenser (Kcal./h.) C.O.P. = Electrical power absorbed by the equipment x 860 Kcal./Kw.
As an example we can see that the C.O.P. (or calorific performance in relation to the absorbed energy), of a pure electrical resistance is equal to 1, because for each Kw. absorbed from the electrical network we obtain 860 Kcal. Applying the above formula we have:
860 Kcal./h.C.O.P. electrical resistance = =1 1 Kw./h. x 860 Kcal./Kw.
In an air-to-air heat pump equipment that works in a heating cycle, C.O.P. of up to 3 can be obtained, which means, in this case, that for each Kw. consumed by the equipment we can obtain a heat power three (3) times higher than what we would obtain with a pure electrical resistance that absorbed the same electrical power from the network. In other words, in this equipment that has a C.O.P. of 3 for each Kw. absorbed, 860 x 3 = 2,580 Kcal./h are obtained.
The C.O.P. values are provided by the equipment manufacturer and vary directly with the outside temperature. The lower the outside temperature, the lower the C.O.P. that a heat pump provides.
Two types of C.O.P. should be distinguished:
1. The instant C.O.P. It is the one that provides the equipment for fixed operating conditions. This data is the one that is usually provided in the technical cataaogos of the manufacturers.
However, over the course of a heating season, the outside temperature conditions areconstantly changing. Subsequently, the C.O.P. provided by the equipment is also fluctuating, following variations in the outside temperature.
In order to be able to calculate the C.O.P. of any equipment operating in a given installation, the seasonal C.O.P. must be calculated over an entire heating season.
2. The seasonal C.O.P. It is the actual one of an installation during a certain period of time.
To proceed with its calculation, the following system can be used.
Calculation of the seasonal C.O.P. of an installation.
To proceed with this, the following must be known:
1.º The number of annual hours that a fixed temperature gives in the place where the equipment is installed.
2.° The thermal power necessary to maintain the comfort conditions within the premises that concerns us for each of the exterior temperatures that will occur in the place of the installation. (The line of loads of the room must be drawn as a function of the outside temperature).
3.º The calorific performance of the heat pump for each outside temperature. (Data provided by the manufacturer).
4.º The power absorbed by the equipment, for each of the conditions of section 3 above that are considered.
With the above data, a table must be made (see figure 9) that once completed will give us the value of the seasonal C.O.P. of a specific equipment applied to a specific installation.
Example of the calculation of the Seasonal C.O.P.
The table in Fig. 9 consists of 10 columns, each of which indicates:
Col. 1. Outside temperature °C. This column indicates the outside temperature values at which the equipment will operate during the entire interval considered (a winter season, for example). Logically, the more outside temperature values we consider, the more accurate the final result will be. However, taking intervals of 5ºC. gives sufficient accuracy.
Col. 2. Kcal./h. needed. This column indicates the heat output required in the conditioned room to maintain the design conditions. It is a function of the outside temperature and is a data that is taken from the line of charges of the local considered.
Col. 3. Kcal./h.B.C. This column notes the calorific yields of the heat pump for each of the outside temperature values considered. This information is provided by the computer manufacturer.
Col. 4. Power absorbed by the B.C. (Kw.). This column notes the values of the power absorbed by the heat pump. (Data provided by the manufacturer.) It is important to remember that the absorbed power that must be considered is not only that of the compressor of the equipment but also must be added to the previous one all the power consumed by the fans, etc., that mounts the conditioner, as well as that consumed during the de-icing of the external battery.
Col. 5. Operating hours. This column notes the hours that during the period of time that is considered, the outside temperature considered will occur at thesite of the installation. This data should be obtained from the tables provided by the appropriate bodies.
Col. 6. Operating factor. This factor is the result of dividing the data in column 2 (Kcal./h. needed) by the data in column 3 (Kcal./h. provided by the heat pump). It tells us the % of the heat output of the heat pump that is required to maintain the design conditions inside the room when the outside temperature is considered.
Col. 7. Kw./h. consumed. It is the result of multiplying the value of column 4 (Absorbed power) by column 5 (Operating hours) and by the value of column 6 (Operating factor). The resulting value is the power consumed by the heat pump conditioner during the periods in which the considered outside temperature is given. The total sum of all the partial values (obtained for each outside temperature), of this column gives the power consumed by the equipment during the entire period of time considered.
Col. 8. Kcal./h. complementary. As stated above, the heat output provided by a heat pump decreases with the decrease in outside temperature. Consequently, there may be areas where for low outdoor temperatures, the heat provided by the equipment is not sufficient to maintain the design conditions of the premises under consideration. In these cases, the heat pump equipment is provided with complementary electrical resistances that are only put into operation when the outside temperature is such that the equipment considered alone, can not maintain the design conditions. These resistors can only be connected when a thermostat located outside allows it and are commanded by the thermostat located inside the conditioned space. This column notes the Kcal./h. needed to complement the heat output provided by the equipment for a given outside temperature. It is the result of subtracting the value of column 2 (Kcal./h. needed in the room) from that of column 3 (Kcal./h. provided by the pump).
Col. 9. Kw factor. It is the result of dividing the value found in column 8 by 860 Kcal./h.
Col. 10. Kw./h. of complementary resistance. It is the result of multiplying the value of column 5 (Operating hours) by that of column.9 (Kw factor). It tells us the value of the Kw./h. used by the complementary resistors to help the heat pump maintain the design temperature for a fixed outside temperature. The sum of all the partial values in this column gives us the total value of the Kw./h. consumed by the complementary resistors during the operating period considered.
Suppose now that you want to know the seasonal -C.O.P. of a piece of equipment installed in a climate where the following temperatures occur annually:
900h.year-15°C. 700h.year-10°C. 400h.year-5°C. 50h.year-0°C.
For the temperatures considered, the thermal needs of the premises are as follows (to maintain 21° C. inside):
Para15°C.-3.158Kcal./h.Para10°C.-5.790Kcal./h.Para5°C.-8.420Kcal./h. Para0°C.-11.053Kcal./h.
The heat output provided by the selected heat pump equipment and its absorbed power are:
For 15° C. exterior provides 10,200 Kcal./h. - 3,6 Kw. For 10° C. exterior provides 9,200 Kcal./h. - 3,4 Kw. For 5° C. exterior provides 7,800 Kcal./h. - 3,2 Kw.Para 0° C. exterior provides 6,500 Kcal./h. - 3,0 Kw.
With these data we will proceed to prepare the aforementioned table, remaining as indicated in fig. 10.
From this table it can be deduced that:
a) The annual heat specified by the premises (at the temperatures considered), is the sum of the values of column 2 multiplied by their corresponding hours of operation (column 5) and has a value of:
(3,158 x 900) + (5,790 x 700) + (8,420 x 400) + (11,053 x 50) = 10,815,000 Kcal./h.
b) The total consumption of the heat pump for this period that we have considered (one year) is the sum of the values of columns 7 and 10.
3,933 + 552=4,485 KWA.
c) The seasonal C.O.P. (annual in this example) by definition is the result of dividing the heat output provided by the heat pump by the power absorbed by it to provide this heat (expressed this divider into units of heat).
10,815,000 Kcal./hC.O.P. seasonal = =2.84,485 Kcal./h. x 860 Kcal./Kw.
This result indicates that with this equipment a thermal performance 2.8 times higher has been achieved to condition the premises considered than if pure electrical resistances had been used. In other words, this heat pump has a performance of 280% over an electrical resistance.
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