FUNDAMENTALS OF HVAC Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 1
INTRODUCTION Terminology: HVAC: Heating, ventilating, air-conditioning and refrigerating. Air conditioning is the process of treating air so as to control simultaneously its temperature, humidity, cleanliness and distribution to meet the requirements of the conditioned space. (ASHRAE) 2
COURSE OBJECTIVES BURSA TECHNICAL UNIVERSITY (BTU) 1. To teach students the basic concepts of HVAC engineering. 2. To teach students to choose the most appropriate indoor and outdoor design conditions. 3. To teach students to calculate the heating and cooling load. 4. To teach students to estimate the energy consumption. 3
COURSE LEARNING OUTCOMES BURSA TECHNICAL UNIVERSITY (BTU) 1. Identify the important parameters for a given HVAC system. 2. Identify the parameters that effects the cooling and heating load. 3. Identify the parameters that effect the human comfort in a room. 4. Find the heating and cooling load for a given space. 5. Estimate the energy consumption for a given system. 4
Basics Of Heat Insulation Project Preparation -Increasing energy efficiency and reducing the initial investment cost can be achieved by providing heat insulation considering environment conditions and necessities. -For this purpose, TS 825 regulation was developed by Ministry of Environment and Urbanization. -Nowadays, application of TS 825 regulations for new buildings is mandatory. Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 5
TS 825 TURKISH STANDARD BURSA TECHNICAL UNIVERSITY (BTU) The aim of this standard is to limit the energy amounts used in the heating of the buildings in our country, therefore to increase the savings and to determine the standard calculation method and values to be used during the calculation of the energy need. This standard can also be used for the following objectives: By applying the calculation method and values explained in this standard to the various design options belonging to the building to be constructed newly, determining the design option that will provide the ideal energy performance. Determining the net heating energy consumptions of the current buildings. 6
TS 825 TURKISH STANDARD BURSA TECHNICAL UNIVERSITY (BTU) Before applying a renovation project to a current building, determining the saving amounts that the energy saving precautions that can be applied will provide. By calculating the energy need of the various buildings that will represent the building sector, estimating the energy need in the future in the building sector at the national level. Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 7
IMPLEMENTATION FIELDS Residences Management buildings Business and service buildings Hotel, motel and restaurants Education buildings Theatres and concert halls Prisons and detention houses Museum and galleries Airports Hospitals Swimming pools BURSA TECHNICAL UNIVERSITY (BTU) 8
Factors affecting the heating energy need of the building Building characteristics: Heat losses realized via the transmission, convection and ventilation (if any heat recovery) and thermal capacity. Characteristics of heating system: Duration of responding to the changes in the heating energy need by especially the control systems and heating system. Internal climate conditions: Temperature value that the ones using the building want, changes in these temperature changes at the different sections of the building and the different times of the day. External climate conditions: External air temperature, direction and magnitude of the prevailing wind. 9
Factors affecting the heating energy need of the building Internal heat gain resources: Various devices and people expanding heat to the environment and used with the aims of internal heat resources having contribution to the heating, food cooking, obtaining hot water, illumination other than the heating system. Solar energy: Solar energy amount directly reaching to the heated place from the transparent building materials such as window. In the calculation method mentioned in this standard, the heat losses realized via the transmission, convection and ventilation and the internal heat gains and the solar energy gains have been considered. 10
Heat Insulation Project BURSA TECHNICAL UNIVERSITY (BTU) -Heat insulation projects are prepared according to the calculation methods given in TS 825. -Calculated -results must not be bigger than the limit values. -A heat insulation project must contain following informations; 1. Heat losses, heat gains, ratio of gain/loss, gain factor, monthly and annual heating energy demand values should be given according to TS 825 and they should be presented tabular. In addition, it should be shown that calculated annual heating energy demand is bigger than the limit value. 11
Heat Insulation Project BURSA TECHNICAL UNIVERSITY (BTU) 2. Glass type which used in windows and case type, areas of windows for each direction and heat transmission coefficients should be indicated. 3. Ventilation type should be specified. 4. Amount of condensing which occurs at the surfaces, should be compared with the standard values given by TS 825 regulation. 5. Structural elements located at outside of building such as colon, beam must be isolated. 6. While heating energy calculations are performed for attached buildings, the areas which contains the attached walls, should be evaluated as a inside wall. 12
An Example of Heat Insulation Project Calculate the annual heating energy need for the following building and determine the insulation thickness is suitable or not? 12 m 20 m 8 m Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 13
In order to calculate the annual heating energy demand, following steps should be followed. 1. Annual heating energy demand per volume of building should be calculated Q. 2. Restricted heat energy demand per volume of building should be calculated Q. 3. Finally, Q and Q are compared. If Q < Q then, insulation thickness is suitable otherwise insulation thickness is not feasible. 14
1. Calculation of overall heat transfer coefficient a. Thermal conductivity resistance of the basement floor (U t ) Material Slag λ W/mK 0.23 Tile 1.30 Blockade 0.70 Leveling concrete 1.40 Concrete 1.74 1 U t = 1 α i + n i=1 d i λ i + 1 α o 1 α i = 0.13 m 2 K/W α i =0 (land- contact) From Table 1 = 0.13 + 0.03 U t 1.3 + 0.05 1.4 + 0.15 1.74 + 0.25 0.23 + 0.1 0.7 = 1.484m2 K/W U t 0.674 W/m 2 K 15
1. Calculation of overall heat transfer coefficient b. Thermal conductivity resistance of the wall (U D ) Material λ W/mK Internal Plastering 0.870 Hollow brick 0.450 Polyurethane Foam 0.035 External Plastering 1.40 3 cm 4 cm 19 cm 2 cm 1 α i = 0. 13 m 2 K/W 1 α o = 0. 04 m 2 K/W From Table 1 = 0.13 + 0.02 U D 0.87 + 0.19 0.45 + 0.04 0.035 + 0.03 1.4 + 0.04 = 1.779m2 K/W U D 0.561 W/m 2 K 16
1. Calculation of overall heat transfer coefficient c. Thermal conductivity resistance of the ceiling (U disc ) Material λ W/mK d(m) Fiberglass 0.05 0.1 Reinforced concrete 2.10 0.12 Internal Plastering 0.87 0.02 1 α i = 0. 13 m 2 K/W 1 α o = 0. 04 m 2 K/W From Table 1 = 0.13 + 0.1 U disc 0.05 + 0.12 2.1 + 0.02 0.87 + 0.04 = 2.169m2 K/W U disc 0.461 W/m 2 K U p : thermal conductivity resistance of the window U p 2.8 W/m 2 K From Table 17
2. Calculation of heat transfer area a. Area of windows and doors (A p ) For four floors A p = 1 1.30 5 + 2 1.3 + 0.6 1.3 0.6 0.6 + 0.8 2.2 3 4 A p = 15.52 4 = 62.08 m 2 b. Area of outer wall (A D ) A D = A brüt A p A D = 12 8 2 + 20 12 2 A p A D = 609.92 m 2 c. Ceiling area (A T ) A T = 20 8 = 160m 2 d. Floor/Flooring area sitting on the ground(a t ) A t = 20 8 = 160m 2 18
3. Specific Heat Loss of Building (H) In case there is 1 K temperature difference between the internal and external environments, it is the heat energy amount lost in the unit time via ventilation and transmission from the external shell of the building. Its unit is «Watt/Kelvin». H = H i + H h H i : Heat loss realized via the transmission and convection (W/K) H h : Heat loss realized via the ventilation (W/K) 19
3. Specific Heat Loss of Building (H) a. Heat loss by conduction (H i ) H i = A U + I U i A U = U wall A wall + U window A window + 0.8 U c A c + 0.5 U bf A bf +U ow A ow A U = 0.561 609.92 + 2.8 62.08 + 0.8 0.461 160 + 0.5 0.674 160 A U = 624.21 W/K If there is no heat bridge in our building, so that term of I U i can be neglected. H i = 624.21 W/K 20
3. Specific Heat Loss b. Heat loss by ventilation (H h ) BURSA TECHNICAL UNIVERSITY (BTU) H h = ρ c V = ρ c n h V h = 0.33n h V h TS 825/2.2.1.2 V h = 0.8 V brüt = 0.8 1920 = 1536 m 3 V h Ventilating volume (m 3 ) n h = 1 (Chosen) H h = 0.33 1 1536 = 506.88 W/K H = H i + H h Finally, specific heat loss can be calculated as; H = 624.41 + 506.88 H = 1131.29 W/K 21
4. Monthly Average Heat Energy Gain Φ i,ay Φ i,ay = A n 5 TS 825/2.2.2 A n = V brüt 0.32 TS 825 from Chart 3 Φ i,ay = V brüt 0.32 5 = 3072 W 5. Monthly Average Solar Energy Gain Φ g,ay Φ g,ay = r i,ay g i,ay I i,ay A i r i = 0.6 (Chosen) TS 825/2.2.3 g = 0.75 TS 825/2.2.3 For each floor A south 4.36 m 2 I south,jan 72 W/ m 2 A north 4.20 m 2 I north,jan 26 W/ m 2 A east 4.36 m 2 I east,jan 43 W/ m 2 A west 2.60 m 2 I west,jan 43 W/ m 2 g,jan = 0.8 0.75 = 0.6 r i = Monthly average ghosting factor of the transparent 22 surfaces in the direction of i
Φ g,ay = r i,ay g i,ay I i,ay A i Φ g,ay = 0.6 0.6(72 4.36 + 26 4.2 + 43 2.6 + 43 4.36) 4 Φ g,ay = 1040.256 W 6. Monthly Gain Usage Factor KKO ay The rate of contribution of the total of the internal heat gains and solar energy gain to the heating of the environment. It is unitless. Gain Loss KKO ay = Φ g,ay + Φ i,ay / H T i,ay T d,ay TS 825/2.2.4 T i,ay = 19 Monthly average value of the internal temperature TS 825/0.5 T d,ay = 8 Monthly average value of the external temperature TS 825/Appendix-2 23
KKO ay = 1050.256 + 3072 / 1131.29 19 9 KKO ay =0.33 η ay = 1 e 1/KKO m TS 825/2.2.4 η ay = 1 e 0.33 = 0.95 1 7.Total Monthly Heating Energy Need Q ay Q ay = H T i,ay T D,ay η ay (Φ i,ay + Φ g,ay ) t Q ay = 1131.29 19 8 39123 86400 30 Q ay = 22114555 kj for January Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 24
If the same calculations are realized for other months, total annual heating demand is calculated as follows; Q yıl = 81875071 kj = 22761 kwh 8.Evaluation of the Results Q= Q yıl /A n A n : Building usage area m 2 A n = 0.32 V brüt = 0.32 1920 A n = 614.4 m 2 Q = Q yıl /A n Q = 22761/614.4 =37.046 kwh/m 2 Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 25
Permitted annual heating energy need Q From Appendix 1-B for 1.region A V = 992 1920 Q = 46. 62 A V + 17. 38 Q = 41. 36 kwh/m 2 Q= 22761/614.4 =37.046 kwh/m 2 Q = 41.36 kwh/m 2 Q < Q 37. 046 < 41. 36 Insulation is suitable. 26
Example-2 By taking a residence built as two floors load-bearing construction having width of 9 m, length of 10 m and height to 5.5 m from exterior to exterior and being located in the 3rd degree day region, calculate the heating energy need for January? The thickness of insulation material which is λ = 0,035 W/mK from the external side on the brick and reinforced concrete on the walls is 5 cm. The windows are multi glazing glass. The thickness of insulation material which is λ = 0,040 W/mK on the ceiling is 12 cm. On the flooring, the thickness of insulation material which is λ = 0,030 W/mK is 6 cm. 27
Example-2 Window area, A p = 20 m 2 Reinforced concrete area, A bet = 10 2 + 9 2 0.12 3 A bet = 13.7 m 2 External wall area, A D = 9 5.5 2 + 10 5.5 2 A D = 175.3m 2 Ceiling area, A T = 9 10 = 90 m 2 External door area, A k = 2 m 2 V brüt = 9 10 5.5 = 495 m 3 Total area, A top = 389 m 2 A n = 0.32 V brüt = 0.32 495 = 158.4 m 2 Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 28
Heat Transfer Coefficients BURSA TECHNICAL UNIVERSITY (BTU) U D = 0.471 W/ m 2 K U Dbet = 0.575 W/ m 2 K U p = 2.4 W/ m 2 K U T = 0.305 W/ m 2 K U t = 0.432 W/ m 2 K U k = 4 W/ m 2 K Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 29
Heat Loss by transmission and convection (H i ) H i = 173.3 0.471 + 13.7 0.575 + 20 2.4 + 0.8 90 0.305 + 0.5 90 0.432 + 2 4 H i = 186.9 W/ K Heat Loss by ventilation (H h ) V h = 0.8 V brüt = 0.8 495 = 396 m 3 n h = 0.8 H h = 0.33 0.8 396 = 104.54 W/ K H = H i + H h H = 186. 9 + 104. 53 = 291. 43 W/ K 30
Heat Gains Monthly Average Heat Energy Gain Φ i,ay As the building will be used as residence, the internal heat gains can be taken as 5 W/m 2. Φ i,ay = A n 5 = 158.4 5 = 792 W Monthly Average Solar Energy Gain Φ g,ay For the ghosting factor to be used during the calculation of the solar energy gains, by considering that the building has less than 3 storeys and its surroundings are open, the value of r i,ay = 0,8 is selected (Chart 5). Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 31
As in the window system, the multi glazing glass has been used, the value of g has taken as 0,75 (Chart 6). In this example; g,ay = 0. 80 0. 75 = 0. 60 A i values, namely the total window areas for all directions are calculated. In the building selected as an example, the below given window area have been calculated. A güney = 8 m 2 A kuzey = 4 m 2 A doğu = 4 m 2 A batı = 4 m 2 I i,ay values are taken from the Annex C for each month. To be an example for the month of January, the monthly solar radiation intensity values obtained from the Annex C are as follows. I güney,ocak = 72 W/ m 2 I kuzey,ocak = 26 W/ m 2 I batı,doğu,ocak = 43 W/ m 2 32
Φ g,ay = r i,ay g i,ay I i,ay A i Φ g,ay = 0.8 0.6 72 8 + 0.8 0.6 26 4 + 0.8 0.6 43 4 + 0.8 0.6 43 4 Φ g,ay = 492 W To calculate the gain usage factor, firstly the KKO ocak is calculated according to the following equation KKO ay = Φ g,ay + Φ i,ay / H T i,ay T d,ay KKO ay = 792 + 492 / 291.44 19 ( 0.3) =0.23 As the building will be used as residence T i,ocak, is taken as 19 from the Annex B.1. T d,ocak is taken from the Annex B.2. This value for the third degree day regions is -0,3. 33
The gain usage factor η ocak is calculated as follows according to the equation η ay = 1 e 1/KKO m η ocak = 1 e 1 0.23 = 0.99 In this case, the heat gains for the month of January are found as follows Q ay = H T i,ay T D,ay η ay (Φ i,ay + Φ g,ay ) t Q ocak = 291.44 19 ( 0.3) 1271.2 86400 30 10 3 Q ocak = 11284.510 kj Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 34
Prof.Dr. Yusuf Ali KARA Res.Asst. Semih AKIN 35