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ENERGY EFFICIENCY AND AIR CONDITIONING SYSTEM DESIGN &

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Presentation on theme: "ENERGY EFFICIENCY AND AIR CONDITIONING SYSTEM DESIGN &"— Presentation transcript:

1 ENERGY EFFICIENCY AND AIR CONDITIONING SYSTEM DESIGN &
ENERGY EFFICIENT OPERATION OF AIR CONDITIONING SYSTEMS

2 4.4 METHODS OF ESTIMATING BUILDING HEAT LOADS
Building heat load is the sum of: Building envelop heat gains including heat gains through building structures and outdoor air ventilation and air infiltration. Building internal loads. This includes heat loads generated by the building equipment and human occupying the buildings.

3 The methods used for estimating building heat loads can be grouped into the following fwo categories: Manual methods. Computer based simulations. The accuracy of both methods depends on the methodologies used in estimating these loads.

4 4.4.1. MANUAL METHOD OF ESTIMATING BUILDING HEAT LOAD
Calculating the building heat gain in this method is based on estimating the following building heat loads: Estimating building heat gains through building structure. The Overall Thermal Transfer Value (OTTV) and Roof Thermal Transfer Value (RTTV) methods are used to estimate building heat gains through building walls and roof. Estimating building heat load generated through outdoor air ventilation and air infiltration. Estimating this heat load requires some understanding of the local conditions and quality of controlling outdoor ventilation and infiltration.

5 Estimating heat load generated by building occupants.
Estimating heat load generated by equipment used in the air conditioned areas. This includes lighting, office equipment and other equipment run inside the air conditioned areas, such as pumps, heaters, driers and others. The following paragraphs address the methods of estimating the building structure heat loads. The estimation of other building heat loads has been addressed in sufficient details before. The reader needs to refer to the relevant sections for estimating other building heat loads.

6 4. 4. 1. 1. CALCULATING HEAT LOAD TRANSFERRED
CALCULATING HEAT LOAD TRANSFERRED THROUGH BUILDINGS WALL STRUCTURE The method of calculating the heat load transferred through building walls is called Overall Thermal Transfer Value (OTTV). The OTTV requirement, which applies only to air conditioned building is aimed at achieving the design of building envelope so as to cut down extremal heat gain and reduce the cooling load of the air conditioning system. The general formulae used is as below: Where Ao1, Ao1, Aon are gross area of wall for each orientation

7 According to MS 1525 (Code of Practice on Energy Efficiency and Use of Renewable Energy for Non-Residential Buildings), it is recommended that for the overall thermal transfer value as given equation above shall not exceed 50 W/m2 for an airconditioned building having a total cooling load of 100 kW and above. The overall thermal resistance (overall R-value) of a building element consisting of n layers can be calculated as follows:

8 k-value is the thermal conductivity of the material given in engineering references. After identfiing the values of R for various building material layers, then the total thermal transmittance (i.e. U-value) can be calculated as follows:

9 4.4.1.2. CALCULATING HEAT LOAD TRANSFERRED THROUGH BUILDING ROOFS
The method used for calculating the heat load transferred through building roofs is called Roof Thermal Transfer Value (RTTV). What is the general formulae used in calculating RTTV for a roof with skylight?

10 Note: CF is the correction factor with reference to:
the orientation of the roof and the pitch angle of its skylight. BUILDING HEAT LOAD CALCULATION CASE STUDY One storey building with dimension shown in Figure below is used an office area. The office is non-smoking office building. Table 2 shows design parameters used in estimating air condition heat load.

11 Figure above Office Floor PIan and Cross Section

12 Table above was Office Building Design Parameters

13 4.4.1.3.1. BUILDING STRUCTURE HEAT LOAD
WALL HEAT LOAD CALCULATIONS Calculatinq U-Value of The Walls: For RC Beam Using k-values and air film resistance shown in Table A1-l & Al-z and the dimension of the RC beam shown in Figure bellow, the calculation of the RC beam U-value is depicted in Table bellow.

14 Using the calculate R-value shown in Table above, U-value can
RC Beam Cross Section Using the calculate R-value shown in Table above, U-value can be calculated as follows: Urcb = 1/R = 1/0.346 = 2.89 W/m2K

15 Brick Wall Cross Section
For Brick Wall Similar to the U-value calculation for RC beam, the U- value calculations for brick wall shown in Figure 8 are shown in Table above. Brick Wall Cross Section

16 Brick Wall U-Value Calculation
Using the calculate R-value shown in Table bellow, U-value can be calculated as follows: Ubw = 1/R = 1/1.839 = 0.54 W/m2K Brick Wall U-Value Calculation

17 For Glass Window Shading Coefficient (SC) of the glass window is 0.5 (given by the manufacturer). Let R = 0.15 for the glass window. Ugw = 6.66 W/m2K Area/Overall U-value Calculations For North-Facing Wall ⦁ RC beam Awl = 0.5 x 32 =16.0 m2 ⦁ Brickwall Aw2 = 1.7 x32 =54.4 m2 ⦁ Glass Af = 1.5 x 32 =48.0 m2 Gross wall area =118.4 m2 WWR =48/ =0.41

18 Overall Uw for Opaque wall can be calculated as follow:
Overall Uw: = [(16 x 2.89) + (54.4 x 0.54)]/( ): = W/m2 K Overall Uf = (48x6.66)/48 = 6.66 W/m2K i.) For South-Facing Wall Ignoring the shading effects of SE and SW facing corners on the south facing walls, the overall U-values of the south-facing walls can be considered equivalent to that for north-facing wall. ii.) For East-Facing Wall Ignoring shading effects of SE-facing comers on the east- facing wall, U-value of east-facing wall can be calculated as follows:

19 ⦁ RC beam Aw1 = 0.5 x 16 =8 m2 ⦁ Brickwall Aw2 =1.7 X 16 =27 .2 m2 ⦁ Glass Af =1.5 x 16 =24 m2 Gross wall area =59.2 m2 WWR =24/ =0.41 Overall Uw : = [(8 x 2.89) + (27.2 x 0.54)]/( ) =1.074 W/m2 K iii.) For West-Facing Wall (Areas same as East-facing wall)

20 OTTV Calculation i.) For North-Facing Wall Let the solar correction Factor (CF) for north-facing wall is CF = Substituting the above calculated U value and other defined parameters, OTTV of this wall is calculated as follows: OTTV = (12)(l )(1.074) + 3.4(0.41)6.66 + (211)(0.83) (0.41)(0.5) = W/m2

21 For South-Facing Walls
Given that the Solar Correction Factor (CF) for south-facing wall is CF = Accordingly, OTTV for these walls is: OTTV = (12)(l )(1.074) + 3.4(0.41)6.66 + (211)(0.85)(0.41)(0.5) = W/m2

22 iii) For East-Facing Walls
Let Correction Factor (CF) for east-facing wall is CF = Accordingly, OTTV for these walls is: OTTV = (12)(l )(1.074) + 3.4(0.41)6.66 + (211)(1.15)(0.41)(0.5) + = W/m2

23 iv) For West-Facing Walls
Given that the Solar Correction Factor (CF) for west- facing wall is CF = Accordingly, OTTV for these walls is: OTTV = (12)(l )(1.074) + 3.4(0.41)6.66 + (211) (1.14)(0.41)(0.5) = W/m2

24 v) For Whole Building: OTTV = [(118.4 x 52.79) + (118.4 x 53.65) + (59.2 x 66.63) + (59.2 x 66.19)] / ( ) = W/m2 (which is not within the specified value of 50 W/m2 ) The total heat gain through the 4 walls is: Total wall heat load = (118.4 x 52.79) + (118.4 x 53.65) + (59.2 x ?) + (59.2 x 66.19) = Watts = kW

25 (B) ROOF HEAT LOAD CALCULATIONS
Figure bellow show the roof sketch with the location of 3 skylights : Roof Sketch

26 U Value Calculation For Opaque Roof Section (see Figure bellow)

27 Opaque U-Value Calculation
Ur = 1/R = 1/ = (W/m2K)

28 Skylight Cross Section
ii) For Skylight U-value skylight can be calculated as shown in Table below: Skylight Cross Section

29 Skylight U-Value Calculation
Us = 1/R = 1/ = (W/m2K) SC = 0.5 (given)

30 Area of the whole roof = 32 x 16 – 2 x (7 x 7): 414m2
Area Calculation Area of the whole roof = 32 x 16 – 2 x (7 x 7): 414m2 Areas of three skylight: 3 x (3 x 3): 27 m2 Opaque area: : 387 m2 SKR = Skylight Ratio = (Areas of three skylights)/ (Area of the whole roof) = 0.065

31 Roof Thermal Transfer Value (RTTV)
CF = 1 RTTV =

32 Is it lower than the minimum recommended by MS 1525?
The total roof heat load is: = ?

33 The total building structure heat load is equal to:
Total Wall Heat loads + Total Roof Heat Loads =

34 4.4.1.3.2 TOTAL BUILDING HEAT LOAD
As discussed before, the total building heat load consists of the sum of: Building structure heat loads, ? kW (see the above) Heat gains through outdoor air ventilation and air infiltration. Building occupant heat load. Lighting load. Other indoor building equipment heat load. Building heat gains through building structures and outdoor air ventilation and air infiltration are estimated as follows:

35 Outside Air Ventilation Load
Total outside air ventilation = 500 l/s The estimated outside air infiltration is I .25 air change per hour (Ch/hr). This is equal to: = 414 m2 (floor area) x 3.70 m (height) x (Ch/hr) x (l/m3)/3600 (s/hr) = l/s Accordingly, the total outside air supplied through ventilation and infiltration is: = 500 l/s l/s = l/s

36 Using Psychrometric Chart design parameters, the following properties of the indoor and outdoor air:
Indoor Air Properties: Dry bulb temperature:24oC RH:60%o Enthalpy : kllkg Specific volume:0.857 m3/kg Outdoor Air Properties: Dry bulb temperature:33.5oC RH :65%o Enthalpy: kJ/kg Specific volume: m'lkg

37 The above indicates that the building gains 36. 27 kJ/g (=. 89
The above indicates that the building gains kJ/g (= kJ/kg kJ/kg) per I kg of outside air supplied through ventilation and infiltration. The total supplied outside air circulated is (l/s) or m3/s. Using the specific air volume m3/kg, then the amount of outside air circulated in the building is: = [ (m3/s)] /[0.899 (m3/kg)] = kg/s From the above, the building heat load added by the outside air is: = (kJ/kg) x 1.15 (kg/s) = kJ/s = kW

38 Building Occupant Heat Load
As the heat load per occupant is 150 Watts/person and the number of total occupants is 50, then the total occupant heat load is: = 150 Watts/person x 50 persons = 7500 Watts = 7.5 kW

39 Lighting Load As the lighting power density is 15 Watts/m2 and the total occupied area is 414 m2, then the total lighting heat load is: = 15 Watts/m2 x 414 m2 = 6210 Watts = 6.21kW

40 Other Indoor Building Equipment Load
According to Table 2, the total building equipment heat load is 32 kW. Table 3 shows the total building heat loads. Building Heat Loads

41 4.4.2. ESTIMATING BUILDING HEAT LOAD BY COMPUTER SOFTWARE
There are a number of softwares available in the market that can perform cooling load estimates. Softwares can make tedious manual calculations easier especially with analyses of multiple spaces and iterative calculations. One of the difficulties of estimating loads of a larger building with multiple spaces is the determination of the peak hour, i.e. the time when the peak loads occur.

42 Depending on how the rooms are zoned, the determination of the peak load and airflows can be tedious with iterative calculations to find the peak hour. This is where cooling load estimating software can be effective in the design process. Some software incorporates the performance of cooling coils in its code to provide a good estimate of the coil selection and performance parameters. This provides a better estimate of the airflow and cooling coil performance. Traditional methods of estimating this using "coil bypass factors" are subject to use input and experience.

43 Softwares can also facilitate with quick "what if' analyses of load design. Alternative materials, shading, internal loadings can be analysed quickly once the model is setup. The cooling load methodology used by softwares can vary and produce different results for the same input. The Transfer Function Method (TFM) uses advanced modelling of varying effects of thermal storage on the building structure to calculate an effective Cooling Load Temperature Difference (CLTD) for different types of building structures (walls, roofs) with time of day. This is adopted as the ASHRAE CLTD method of load calculation and is ideal for computer software calculations.

44 Even with computer software simulations, it is imperative that results be analysed for "reasonableness". As advanced as softwares are today, the results are only as good as the inputs. A check in the relative load components (envelope, airflow, internal and system loads) is necessary to ensure the calculations are "reasonable". Much of this relies on the experience of the designer or user of the software.

45 END….


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