Konutlarda yıllık enerji ihtiyacının modellenmesi
Başlık çevirisi mevcut değil.
- Tez No: 66818
- Danışmanlar: PROF. DR. AHMET ARISOY
- Tez Türü: Yüksek Lisans
- Konular: Enerji, Energy
- Anahtar Kelimeler: Enerji modelleri, Konut, Energy models, Dwelling house
- Yıl: 1997
- Dil: Türkçe
- Üniversite: İstanbul Teknik Üniversitesi
- Enstitü: Fen Bilimleri Enstitüsü
- Ana Bilim Dalı: Enerji Ana Bilim Dalı
- Bilim Dalı: Belirtilmemiş.
- Sayfa Sayısı: Belirtilmemiş.
Özet
ÖZET Binalarda yıllık enerji ihtiyacının belirlenmesi, daha iyi tasarımlar yapabilmek, bina için uygun işletme stratejisini belirlemek ve bir takım standartları oluşturmak gibi konularda faydalı olacaktır. Bina kütlesinin ısı depolama kapasitesini göz önüne alan dinamik modeller gerçeğe çok daha yakın sonuçlar vermektedir. Bu çalışmada ülkemizdeki konutlar için kullanılabilecek, dinamik modellemeye dayalı bir enerji simülasyon programı geliştirilmesi hedeflenmiştir. Yapılan program yalnız tek zon kabulü yapılabilen konutlarda ve ısıtma için kullanılabilmektedir. Program örnek bir takım bina ve ısıtma sistemlerine uygulanmış sonuçlar tablo ve grafik halinde düzenlenmiştir. Bu verilerden elde edilen ilk sonuç hem konfor şartlarının sağlanması hem de lüzumsuz yakıt sarfiyatının önlenmesi için bir otomatik kontrol sisteminin gerekliliğidir. Bunun yanı sıra ısıl direnç ve ataletleri farklı yapı malzemelerinin yakıt sarfiyatı ve iç ortam sıcaklığı üzerindeki etkileri gözlenmiştir. Kesintili veya tam gün çalışma gibi değişik işletme stratejilerinin etkisi gözlenebilmektedir. Böylece yakıt sarfiyatı ve konfor şartlan açısından en uygun durum tesbit edilebilmektedir. Kazan ve brülör kapasitelerinin lüzumsuz yere büyük seçilmesinin yakıt sarfiyatına etkisi gözlenmiş ve ortam sıcaklığım arttırmanın yakıt sarfiyatı olarak maliyeti incelenmiştir. V1U
Özet (Çeviri)
SUMMARY Building energy requirement estimatings are important for tree reasons. Design, economic optimisation and for defining some building standarts. There are two kind of energy estimating methods. Steady state and dynamic methods. Steady state models neglect the mass, in other word the heat capasity of the building. They suppose that the heat gain and the heat loss of the building is equal. In dynamic models the heat storage ability of the building mass is taken in care. Stady state methods are easy to use. Calculations are more simple and take less time but the results are less acurate.lt is possible to get more accurate results by using dynamic methods but they require more solution time. For that reason dynamic methods are generally used with computers. After the improvement of the computers dynamic methods began to be used widely instead of static methods. The effects of the radiative heat sources, like sun, lights or cooking items can be observed by only dynamic models. Because the absorption coefficient of air is be near zero so the radiative heat sources do not cause an increase in the temprature of the inner air directly. Radiation, comes from these kind of sources causes an increase in the temprature of the building mass. So if the suface temprature of the mass, like walls or other items inside, is higher than air temprature there is a heat transfer from them to air. If the temprature is smaller, the temprature increase of the surfaces prevents the high amount of heat loss from the inner air. So we can say that the radiation effects the inner air with a time lag. It is impossible to see this effect by the stady state models. There are two kinds of dynamic models. The first is more fundemental one, it is based on solving the main energy conservation equations. The equations are solved by analytic or numeric methods or by transfer functions.Transfer function method in ASHRAE can give an example to this method. The second method called weightig coefficients method is more easier but less acurate one. In this method heat gain or loss is calculated and the effect of the mass taken in care by some time lag coefficients. CLTD or TEAD/TA methods in ASHRAE are the examples of this method.In this study the first way is prefered. Differantial equations are established for the building and solved by finite difference method. Today it is possible to find lots of commercial energy estimating programs. But most of them do not give much importance to heating system, they generally specialized on cooling. Even the ones which include heating, make solutions for air or fan coil systems. It is very difficult to find programs for heating systems with radiators. But this case is the most existing and important one for Türkiye. In adition some of the assumptions and climatic data are not convinient to Türkiye' s conditions IXand it is not possible to change them by entering the program. Also they are so expensive to use for most of the people. Because of all these reasons it is decided to make an energy estimating computer program based on dynamic modelling for Türkiye. Program has tree parts, fist part is space load modeling. Second is system simulation part and the last one is heat supply model. This tree parts generaly exist in most of the energy programs. Some of them also have a forth part for economic analysis, but it is not existing in this program. The first part, space load modelling is the most important part. In this part heat losses are calculated by solving the main energy conservation equations. It is assumed that the building has only one zone and the inner temprature is the same in every where. Heat losses are the transmission loss from walls, roof and floor and infiltration loss from windows. Heat gains are radiation of the Sun, inner heat sources like people or cooking devices and the heat given to building by radiators. The shape below is shown a wall, roof or a floor. For calculating the heat loss or gain from the inner air, the temprature of the surface 1 must be known. For this the transmission equation for solids can be used. To S Qrl Td It2 dT = k d t p.Cp <£t OX2 For solving this equation we must write energy equivalent equation for the surfaces 1 and 2. At surface 1 there is heat radiation comes from inner heat sources and Sun (Sun light comes from windows). And there is heat convection between air and this surface. At the surface 2 there is also a sun radiation and heat convection between outer air and this surface. The total heat comes to two surface is stored in the wall. We can write the equation as an example for surface 1. h. A. (T0- Td ) + Qrl. A = - k. dT. A + p.cp. A. dx. ÖTAt the equation above the first term shows heat convection, second is the heat radiation comes from Sun and inner sources. Third term is the heat flow inside the wall and forth term shows the heat which is stored in the part has a thick of dx. For getting the temprature of the surface 1, we must solve that equation. It is very difficult to solve it analyticaly. It is better to solve it by numerical methods. Finite difference method is convinient for solution. If we devide the wall to n particles we can easily get the solution. n Ax Ax At the shape above the temprature of the middle point represents the temprature of each part. If we make Ax smaller we can get more accurate results. By using finite difference method we can write the equation again. h. (Tpo-Tp0 + Qr = -k. (T2-T1) + p.Cp. Ax. (T^1 - Tp) Ax At Ti1*1 represents the temprature of the part 1 at time t +1. This temprature depends on the temprature of this part at time t and the temprature of the part 2 at time t. If we know the tempratures at time t we can find the tempratures of all the parts at any time later. We can put Ti^1 alone at one side of the equation. TV*1 = At. (( p.Cp Ax/2 - k - h ). Tip + k. T?p + h. Tnp + Qr ) ) p.cp(Ax/2) At Ax Ax We can not select At and Ax independently because of the second law of thermodynamics. For to obey second law the constraints below can be written. p.cp(Ax/2) - _k_ - h >= 0 At Ax XIo.cv(Ax/2) >= _k_ + h, J_ >= _k_ + h At Ax At Ax p.cp(Ax/2) At <= p.cp(Ax/2) _k +h Ax For the tempratures of the inner parts, transmission equation for solids is used. If we write the equation by using finite difference method we can get the equation below. TJ*1 - Tmp = oc. ( Tm+1p + Tm.!p - 2. Tmp ) At Ax2 As can be seen from the equation the temprature of the part m depends on the tempratures of two part near it and the temprature of this part at time t. We can put Tm1*"1 alone at one side of the equation. TnT1 =Aj^.(Tm+1p + Tm.,p) + At(l-2jx)Tmp Ax2 At Ax2 Also there is some constraints for this equation. For definig the temprature of inner air we must write an energy equivalent equation again. Ph. cPh. Vh. 5T = input energy - output energy öt Here Vh represents the volume of the inner air. If we use finite difference method again we can get the equation below. Ph. cPh. Vh. ny*1 - Tift = input energy - output energy At XllAt the equation above input energy is the heat comes from inner heat sources, the heat from radiators and the heat passes from the walls which has a higher temprature from air. Output energy is the heat loss by infiltration, and heat goes to walls which has lower temprature from air. Second part of the program is the modelling of the heating system. As it is mentioned before it is assumed that radiators are being used. We can write the equation below between radiators and air. ms(Tg-Tç) = KF(a&±Tfi)-Th) 2 As an automatic control system five different systems are being used. First is thermostat control of the boiler. In fact this is not a real control system and this represents the case which an automatic control system does not exist. Second one is the system with outer air compansation device. At this system temprature of the water in the boiler is changing according to the temprature of outer air. We can represent this case with the equation Tg = f (T0). Third system is four way valve system. In this system the boiler temprature is 90°C but the temprature of the water going to system is changing according to outer air temprature. At the fourth system there are thermostatic valves on the radiators with a thermostat control boiler. Fifth control system also has thermostatic valves on radiators but this time with a outer air temprature compansation device. By using finite difference method equation can be written as below: Tçp = (ms-KF/2).Tg + KF. Thp (ms+KF/2) At the last two system thermostatic valves can be set to any temprature. Water flow decreases or increases for getting the desired air temprature. Of course there are many valves on every radiators and every of them has a different flows. But in the model it is assumed that there is only a big valve on the boiler and it controls the flow of the boiler. After determining of the energy requirements it is necessary to simulate heat source for determining fuel consumption. Thirt part of the program includes heat source modelling. In this study it is assumed a cylindircal steel boiler with a fuel oil burner is used. We can write the energy equvialent equation at the burner and get working time of the burner for one period. The equation is: tç= Qd / ( qt>r qs- qÇk ) Here Qd is the heat, stored in the system, qbr is heat flow at the burner, qs is system energy requirement and qÇk is heat loss of boiler while it is working. XlllEquation for stand by time of boiler is: td=Qd/(qs+qdk) qak is the heat loss of boiler while it is not working. After getting of the working and standing times the equations below for fuel consumption and boiler efficiency can be written as: B= tç.qbr/ (tç+td).Hu Eff=(tç+td).qs/tç.qbr The assumptions below are made for this study: 1- Building is assumed as one zoned. 2- Every mounth represents by a typical day which has the average tempratures of one mounth 3- Outer temprature, Sun radiation etc. are assumed to be constant for one hour. 4- Particles of the building like walls, floor, roof are assumed to be made from one homegeneous material. 5- Building is assumed to have a rectangular shape. 6- It is assumed that every side of the building is open and there is not any shade effect. Program applied to a sample building and took some results. Sample building jr floors and eight flats. Total area of one floor is 442 with a fuel oil burner. Boiler has a capasity of 80.000 kcal/h. 9 has four floors and eight flats. Total area of one floor is 442 m and there is a boiler The most important thing which is seen from the results is the essential of using a control system. The most efficient system is thermostatic valve control system. By using this control type it is possible to save 40 % of fuel. Outer air compansation device makes 30 % fuel economy. For to see the effect of building mass two building was compared. Both of buildings were made from the materials which have same thermal resistance but different densities. Results showed that effect of the mass of the building depends on the type of control system. For example there is not much fuel consumption difference between light and heavy buildings for the first control system. But there is about 10 % difference in fuel consumption when thermostatic valves are used, in adition to this saving in heavy building temprature fluctation during night is less than the light one. One of the other important thing is the management of the heating system. It is possible to get 10 % fuel saving by stopping the boiler during night. Also temprature drop is not much. Temprature drops to 17 °C maximum during night. xivEvery one degree increase in inner air temprature causes 10 % more fuel consumption. This is quite much amount. So it is very important even one degree. This sensitivity in setting the temprature makes thermostatic valves more valuable. As a conclusion for getting economy from energy it is very important to know the effects of all factors. This is possible by using dynamic modelling only.So it is very essential to develop this kind of software. XV
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