Number of Thermal Time Constants nTTC Abstract Thermal Time Constant (TTC) indicates the time required for change of 63.2% of temperature difference. This work develops useful application of TTC for determination of temperature change in buildings. Higher TTC results in slower change of room temperature, contributing to thermal comfort, to quality of the building and to energy conservation. The procedure developed in the work is for both purposes: to find the temperature change in the period of time and to find the period of time of specific temperature change. Thermal Time Constant – TTC The term "Thermal Time Constant – TTC" in relation to building element is clearly described in publication [1]. Thermal Time Constant (TTC) is a measurement of the time required for a building element to change 63.2% of the difference between its initial and final body temperature when subjected to a step function change in temperature, under zero power conditions. The unit of TTC is hour (TTC is expressed in hours). TTC may be determined using publication [2]. The publication [1] includes a formula for "Time Constant Value", the above mentioned 63.2%. Number of Thermal Time Constants – nTTC by Joseph Nowarski Formula 1 – Time Constant Value 1 – 1/e = 0.632 e 1 Euler’s Number, approximately 2.7182818284590452353602874713527 Euler's Number can be generated in MS1Excel applying the following formula: Formula 2 – Calculation of Euler Number in MS1Excel =EXP(1) Formula 3 – Determination of Time Constant Value in MS1Excel =1 1/EXP(1) Calculation of Time Constant Value using MS1Excel results in 0.632120558828558. The publication [1] includes also an example demonstrating the use of Thermal Time Constant. In the example the building's TTC is 40 hours. The room temperature of the building was kept as 18°C for long time. At the starting point, all internal and external energy sources were switched off, while the ambient temperature is 8°C. For the purpose of this example, the ambient temperature remains constant day and night 8°C and there is no solar radiation. The room temperature in the building will gradually decrease. This is not linear process. Theoretically, the room temperature will reach 8°C at endless time (never), but will get close to it slower and slower. The initial difference between 18°C room temperature of the building and 8°C ambient temperature is 10°C. Number of Thermal Time Constants – nTTC by Joseph Nowarski Formula 4 – Initial temperature difference ∆T0 = Tin0 Tout AT0 = 18°C 1 8°C =10°C AT0 Initial temperature difference, °C Tin0 = 18°C Initial room temperature, °C Tout = 8°C Ambient temperature, °C The TTC of the example is 40 hours. TTC of 40 hours means that the initial temperature difference will be decreased by the Time Constant Value (63.2%) in 40 hours. Formula 5 – Temperature difference after one TTC T1TTC = ∆T0 * (1 – (1 – 1/e)) T1TTC 1 – 1/e Temperature difference after one TTC, °C Time Constant Value = 63.2% T1TTC = ∆T0 * 1/e In the example case: T1TTC = 10°C * 1/2.718 = 3.68 °C The temperature difference was reduced by 6.32°C, which means that the new temperature difference (after 40 hours) is 3.68°C and the new room temperature (after 40 hours) is 8°C + 3.68°C = 11.68°C Number of Thermal Time Constants – nTTC by Joseph Nowarski Formula 6 1 New room temperature after one TTC Tin1TTC = Tout + ∆T0 * (1/e) Tin1TTC = 8 + 10 / 2.718 = 11.68°C Tin1TTC New room temperature after one TTC, °C e Euler Number Two Thermal Time Constants The above analysis was related to one TTC for example of TTC=40 hr. This section includes considerations related to 2 TTCs, 2 x 40 hr = 80hr. In the first TTC the temperature difference was reduced by Time Constant Value (63.2%). During the second TTC, the new temperature difference will be reduced again by the Time Constant Value (63.2%). Formula 7 – Temperature difference after two TTCs T2TTC = T1TTC * (1 T2TTC = (1 – 1/e)) T1TTC * (1/e) T1TTC = ∆T0 * (1/e) T2TTC = ∆T0 * (1/e) * (1/e) T2TTC = ∆T0 * (1/e)^2 T2TTC = ∆T0 /(e^2) T2TTC 1 – 1/e Temperature difference after two TTCs, °C Time Constant Value = 63.2% Number of Thermal Time Constants – nTTC by Joseph Nowarski In the example case: T2TTC = 10°C * (1/ 2.718)^2 = 1.35°C The temperature difference was reduced by 63.2% during the first TTC reaching the 3.68°C value. This new temperature difference was further reduced during the second TTC period by additional 63.2% reaching 1.35°C value. The new temperature difference after two TTCs (40 hours and additional 40 hours) is 1.35°C and the new room temperature (after 80 hours) is 8°C + 1.35°C = 9.35°C Formula 8 1 New room temperature after two TTCs Tin2TTC = Tout + T1TTC * (1 Tin2TTC = Tout + (1 – 1/e) T1TTC * 1/e Tin2TTC = Tout + (∆T0 * 1/e) * 1/e Tin2TTC = Tout + ∆T0 * (1/e)^2 Tin2TTC = Tout + ∆T0 / (e^2) Tin2TTC New room temperature after two TTCs, °C In the example case: Tin2TTC = 8 + 10 / (2.718^2) = 9.35°C Number of Thermal Time Constants – nTTC by Joseph Nowarski Number of Thermal Time Constants The above procedure can be repeated for any number, based on general formula: Formula 9 – Temperature difference after n TTCs TnTTC = ∆T0 /(e^n) n number of TTCs TnTTC temperature difference after n TTCs Considering 7 TTCs, which is 7 x 40 hours = 280hr in the example case: T7TTC = 10 /(e^7) = 0.009°C Formula 10 1 New room temperature after n TTCs TinnTTC = Tout + ∆T0 / (e^n) n number of TTCs TinnTTC New room temperature after n TTCs, °C Considering 7 TTCs, which is 7 x 40 hours = 280hr in the example case: Tin7TTC = 8 + 10 / (e^7) = 8.009°C Number of Thermal Time Constants – nTTC by Joseph Nowarski Fraction of Thermal Time Constant The above formulas may be used also for fraction of Thermal Time Constants. Considering 15% of TTC, which is 40 hours * 15% = 6hr, in the example case: TnTTC = ∆T0 /(e^n) n = 15% T15%TTC = 10 /(e^0.15) = 8.61°C TinnTTC = Tout + ∆T0 / (e^n) n = 15% Tin15%TTC = 8 + 10 / (e^0.15) = 16.61°C The meaning of the above example for building having 40hr TTC is that after 15% of TTC, which is 6 hours, the room temperature will drop from 18°C to 16.61°C. Room Temperature Calculations Based on TTC The above formulas allow calculations of period of time required for specific room temperature change. The formula for temperature difference depends on number of TTCs: TnTTC = ∆T0 /(e^n) The question is which value of n should be applied to receive AT of 1°C. Number of Thermal Time Constants – nTTC by Joseph Nowarski The above formula should be displayed in the following form to find the number of TTCs: y = b^n This allows determination of "n" using logarithm: n = log(b,y) The above formula may be displayed in the required form as follows: TnTTC = ∆T0 /(e^n) (e^n) = ∆T0 / y = ∆T0 / TnTTC TnTTC y = e^n n = log(e,y) Formula 11 – Number of TTCs to reach specified room temperature n = lan(y) y = ∆T0 / ∆TnTTC ∆TnTTC = Tin Tout n number of TTCs AT0 initial difference between room temperature and ambient temperature, °C, in the example case = 10°C Tin specified room temperature for which the number of TTCs is calculated, °C Tout ambient temperature, °C, in the example case = 8°C Number of Thermal Time Constants – nTTC by Joseph Nowarski Table 1 – Time required for room temperature change for TTC=40hr Room Temp Tin 18 17 16 15 14 13 12 11 10 9 8.5 Temp Drop Tin0 Tin 0 1 2 3 4 5 6 7 8 9 9.5 y ∆TnTTC 10 9 8 7 6 5 4 3 2 1 0.5 ∆TnTTC / ∆T0 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 ∆T0 / ∆TnTTC 1.00 1.11 1.25 1.43 1.67 2.00 2.50 3.33 5.00 10.00 20.00 No of TTCs n 0.00 0.11 0.22 0.36 0.51 0.69 0.92 1.20 1.61 2.30 3.00 Time to reach hr 0.00 4.21 8.93 14.27 20.43 27.73 36.65 48.16 64.38 92.10 119.83 The temperature drop of the first 1°C, from 18°C to 17°C, is relatively fast and takes 0.11 TTC. In case of the example, the building TTC is 40 hours, it means that the temperature drop from 18°C to 17°C will take 4.21 hours. The temperature drop of the next degree will take already more time. Temperature drop from 18°C to 9°C, which is 90% of the initial temperature difference between the room temperature and the ambient temperature, will take already 2.3 TTC, more then 90 hours. TTC as part of Energy Rating of Buildings system As demonstrated on the above example, TTC allows easy determination of room temperature change of the building, indicating temperature comfort. In the above example, the temperature drop from comfortable 18°C to not comfortable 17°C, will take about 4 hours for TTC = 40hr, about 2 hours for TTC = 20hr and 11 hours for TTC = 100hr. Similarly, the temperature increase in hot summer from comfortable 26°C to not comfortable 27°C will take the same time periods as calculated above, considering 36°C hot ambient temperature. Number of Thermal Time Constants – nTTC by Joseph Nowarski TTC is very useful and very important information for the home tenant. Higher TTC results in slower change of room temperature, contributing to thermal comfort, to quality of the building and to energy conservation. That's why the TTC parameter should be included in energy rating of buildings system, building energy labeling system and green buildings standards. References 1. Thermal Time Constant – TTC, Joseph Nowarski, Academia.edu https://www.academia.edu/28374502 2. Thermal Time Constant – TTC, Joseph Nowarski 2016, ASIN: B01F18XGQK * * *