Air Conditioning System Design Calculation

Cooling Load Calculation CLTD/CLF method is used to calculate the cooling load of the building. Since window type air conditioning system is used in the residential building and the size are usually determined by the occupants, the CLTD/CLF method will only apply to LG/F to G/F which are central air-conditioned. air-conditioned. Assumption made 1. Indoor design condition : 25 °Cdb, 50 %Rh Outdoor design Condition: 33 °Cdb, 28°Cwb 2. Cooling Cooling load load at August 15:00, 16:00 and 17:00 3. External/thin wall :U value = 2.79 W/m 2K , Group D

Thick wall : U value = 2.10 W/m 2K , Group B 5. Lighting CLF CLF based on : a = 0.55, Group C Floor 6. No infiltration 7. Fresh Fresh air requirement requirement : 10 l/s l/s person person 8. Window type : Sterling Sliver Van-Tran on Bronze

BUIL BUILDI DING NG TYPE TYPE WALLS

WINDOW

LIGH LIGHTI TING NG AND AND ELECTRICAL EQUIPMENT

30 - STOR STOREY EY OFFIC OFFICE E BUI BUILD LDIN ING G External Walls: 25 mm mm stucco, 20 20 mm mm plaster, 10 100 mm concrete, 20 mm plaster, 13 mm finish. Group D , U value = 2.79 W/m2K Thick Walls : 25 mm stucco, 20 mm mm plaster, plast er, 300 mm concrete, 20 mm plaster, 13 mm finish. Group B , U value = 2.10 W/m 2K Partition Parti tion : 15 mm gypsum board, 50 mm air layer, 15 mm gypsum board. U value = 1.71 W/m 2K Sterling Sliver Van-Tran on Bronze, U value = 5.7 W/m 2K Shading coefficient = 0.33 Window-to-wall ratio = 0.52 No external shading device Ligh Lighti ting ng powe powerr inte intens nsit ity y = 8 W/m W/m2 Equipment power intensity = 15 W/m 2K

I-1

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After the detailed calculation, the result is shown as below:

Location

Area

No. of people

Sensible Load (W)

Latent Load (W)

Total Cooling Load(W)

SHR

Elderly Centre Child Care Centre Restaurant Clinic Study Room Game Room I Game Room II Multi-purpose Room 1 Mulit-purpose Room II BMS Room Gym. Security Room Shop 1 Shop 2 Shop3 Shop 4 Shop 5 Shop 6 Shop 7 Shop 8 Shop 9 Shop 10 Shop 11 Shop 12 Shop 13 Shop 14 Shop 15

21 117

6 30

4064 15611

2447 10586

6511 26197

0.62 0.6

100 32 25 25 30 78

25 8 7 7 8 10

17283 4382 3296 2915 3899 9111

13030 2960 2398 1369 2643 6342

30313 7342 5694 4283 6542 15453

0.6 0.6 0.58 0.59 0.6 0.6

45

10

5416

4012

9428

0.59

25 135 10 28 28 8 8 8 8 8 8 8 8 8 8 24 21 18

4 15 3 7 7 2 2 2 2 2 2 2 2 2 2 7 6 5

2735 17022 1754 4008 4234 2540 2322 2333 2127 2169 1951 2572 2300 2958 2147 3510 3111 2998

1630 14113 1269 2787 2787 88 2 88 2 88 2 88 2 88 2 88 2 88 2 88 2 1323 88 2 2446 2246 2105

4365 31135 3022 6795 7020 3422 3204 3215 3009 3051 2833 3454 3182 4281 3029 5956 5357 5103

0.63 0.58 0.58 0.6 0.6 0. 0.74 0.72 0.73 0.71 0.71 0.69 0.74 0.72 0.69 0.71 0.6 0.59 0.6

Total

842

84431

213196

185 128768







Air Side Selection Diffuser Selection Design Procedure: (a) Volume flow rate for individual zone = Peak sensible load for individual zone/ (1.2 x 1.02 x ∆T) (b) Select and determine the number and location of diffuser used in each zone (c) Determine the room’s characteristic length, L (d) Select the recommended recommended TV/L ratio from the table below (e) Calculate TV from values of T V/L and L Selection Base on the effective volume of the false ceiling and the appearance, Circular ceiling diffusers are selected. The table below show the characteristic of the circular ceiling diffuser. Room Load (W/m2) 250 190 125 65

T0.25/L for Max.ADPI 0.8 0.8 0.8 0.8

Maximum ADPI 76 83 88 93

For ADPI Greater than 70 80 80 90

Range of T0.25/L 0.7-1.3 0.7-1.2 0.7-1.5 0.7-1.3

Performance Data for Circular Ceiling Diffuser Flow rate (L/s) Hori. Throw T0.25(m)

105

125

155

180

205

230

255

310

2 .1

2.4

3.1

3.4

4.0

4.6

4.9

6 .1

Worked Example - elderly center From previous calculation, the required volume flow rate = 0.37m 3/s. Base on the room geometry, 3 diffuser diffuser is used ⇒ Max. ADPI = 83 For ADPI greater than 80, Range of T 0.25/L = 0.7-1.3 The characteristic length (L) is 3m ∴ Range of T 0.25 = 2.1m - 3.6m and so the selection is satisfy Also, supply air flow rate for one diffuser = total air flow rate / number of diffusers

∴

Supply air flow rate for one diffuser diffuser = 0.37 / 3 = 0.12m3/s. = 210cfm







From the above procedure, the number of diffuser and their corresponding flow rate can be found. Location

Room Area Room Load Maximum Number of Load (kW) (W/m^2) ADPI Diffuser Elderly Centre 3.9 21 186 80 3 Child Care Centre 16.9 117 144 80 14 Restaurant 22.5 100 225 70 5 Clinic 4.3 32 134 80 4 Study Room 3.1 25 124 80 2 Game Room I 2.9 25 116 50 2 Game Room II 3.9 30 130 80 2 Multi-purpose Room 1 7.4 78 95 80 6 Mulit-purpose Room 6.1 45 136 80 4 II BMS Room 2.7 25 108 80 2 Gym. 21.9 135 162 80 11 Security Room 1.8 10 180 80 1 Shop 1 4.2 28 150 80 4 Shop 2 4.4 28 157 80 4 Shop 3 2.2 8 250 70 1 Shop 4 2.0 8 250 70 1 Shop 5 2.00 8 250 70 1 Shop 6 2.00 8 250 70 1 Shop 7 2.00 8 250 70 1 Shop 8 2.00 8 250 70 1 Shop 9 2.00 8 250 70 1 Shop 10 2.00 8 250 70 1 Shop 11 2.50 10 250 70 1 Shop 12 2.20 10 220 70 1 Shop 13 3.30 24 138 80 2 Shop 14 3.10 21 148 80 2 Shop 15 2.90 18 161 80 2 80 Total 846

Flow rate per Diffuser m^3/s (cfm) 0.1(210) 0.1(210) 0.35(735) 0.1(210) 0.17(357) 0.148(315) 0.29(609) 0.22(256) 0.2(420) 0.14(294) 0.16(336) 0.18(376) 0.1(210) 0.1(210) 0.31(651) 0.31(651) 0.31(651) 0.31(651) 0.31(651) 0.31(651) 0.31(651) 0.31(651) 0.31(651) 0.31(651) 0.18(378) 0.16(336) 0.15(315)

Noise Level Level Acceptable HVAC Noise Levels in Unoccupied Rooms Rooms Occupancy Study Room Gymnasium Elderly Center Restaurant Muilt-purpose room

Noise Criteria (NC) NC 30 NC 40 NC 40 NC 40 NC 40

From the tables above, 8 in. diffuser with neck velocity at 600ft/min. is chosen because it can meet the demand of required required flow rate, throw at T0.25 and NC.

Sizing of Fan-coil Unit







Worked Example - Shop 1

Worked Example - Elderly Centre 4064 = ms (1.012)(26-17) ms = 0.45kg/s vs = 0.45(273+17) /349 = 0.37 m s/s

∴ VF = 8 X 0.01 = 0.08 M 3/s VR = 0.37 - 0.08 = 0.29 L/S By mmhm = m F h F + m Rh R 0.37hm = 0.08(45) + 0.29(56) hm = 53.5J/kg Tm = 24oC db, 19.5 oC wb, 63%rh ∴

∴ The FCU cooling load = 0.45(1.012)(53.5-45) = 3.87kW

∴ Carrier Model 006 is selected.

2540 = ms (1.012)(25-17) ms = 0.314kg/s vs = 0.314 (273+17) /349 = 0.26 m s/s Also, total required fresh air = no. of people x 10 l/s/person

∴ VF = 4 X 0.01 = 0.04 M 3/s VR = 0.26 - 0.04 = 0.22 L/S By mmhm = m F h F + m Rh R 0.26hm = 0.04(45) + 0.22(53) hm = 52kJ/kg Tm = 23.5 oC db, 18.5 oC wb, 60%rh ∴ 0.314(1 .012)(52.45) 2.45) ∴ The FCU cooling load = 0.314(1.012)(5 = 2.24kW Refer to the catalogue, Carrier Model 004 is selected.

By the same method, the capacity of each FCU can be calculated.







Loca Locati tion on

Requ Requir ired ed Capa Capaci city ty Numb Number er of FCU FCU of FCU (kW) Elderly Centre 3.9 1 Child Care Centre 16.9 2 Restaurant 22.5 2 Clinic 4.3 1 Study Room 3.1 1 Game Room I 2.9 1 Game Room II 3.9 1 Multi-purpose Room 1 7.4 2 Mulit-purpose Room II 6.1 1 BMS Room 2.7 1 Gym. 21.9 3 Security Room 1.8 1 Shop 1 4.2 1 Shop 2 4.4 1 Shop3 2.3 1 Shop 4 2.3 1 Shop 5 2.50 1 Shop 6 2.30 1 Shop 7 2.10 1 Shop 8 2.00 1 Shop 9 2.00 1 Shop 10 2.00 1 Shop 11 2.90 1 Shop 12 2.10 1 Shop 13 3.30 1 Shop 14 3.10 1 Shop 15 2.90 1 Total 137.8 32

Sizing of PAU

Sele Select cted ed Mode Modell (C a r r i e r ) OO6 O12 O14 OO6 OO4 OO4 OO6 OO6 OO8 OO3 O1O OO3 OO6 OO6 OO4 OO4 OO5 OO5 OO5 OO5 OO5 OO5 OO5 OO5 OO5 OO5 OO5

Tota Totall Capa Capaci city ty kW 4.5 17.8 24.5 4.5 3 .2 3.2 4.5 9 6 .2 8 2 .9 7 .2 2 .9 4.5 4.5 3.2 3.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2









Total number of people in the air-conditioned air-conditione d space = 185 ∴ The total required fresh air flow rate = 185 x 0.01 = 1.85 m3/s So, the required capacity of the PAU (Q T) the required sensible load (QS) SHR

= 1.85 x 1.2 x (89-45) = 100 kW = 1.85 x 1.2 x (33-17) = 33.5 kW = 0.34

With regard from that, a PAU can be selected based on the following procedure: Unitsize selection

Form the Quick Selection Chart, Chart, for f or SHR < 0.6, The maximum face velocity = 2.5m/s The The air air supp supplly flow low rate rate = 1.85 1.85m m3/s = 6666m3/s Assumption: Air Entering Temperature : 27oCdb, 19 oCwb Water entering /leaving temp. : 7oC, 12oC Coil distance : 2.1 mm From the catalogue, Interklima, unitsize 20 is used Number Number of coil rows :6 Tota Totall cool coolin ing g capa capaci citty : 106. 106.5 5 Sensible Sensible cooling cooling capacity : 81.7 Water flow : 18151 l/hr Water pressure drop : 13.4kPa Air pressure drop on coil coil : 103 Pa Pressure drop on PAU

Mixing Box : 31 3 1 Pa Plenum : 12.5 Pa Bag Fil Filter terss (85%): (85%): 125 125 Pa Pre Filters : 50 Pa Coil Coi l : 103 Pa Total

: 321.5 Pa







Water Side Pressure Drop

From the catalogue, for water flow rate=18151 l/h, the water pressure drop a t coil is 14kPa. Fan Performance Curve

The total pressure drop = 446.5 Pa, and also the volume flow flow rate is 1.85m 3/s

∴ Based on the catalogue, cat alogue, two fan are selected and their performance index index are shown as below: Model Efficiency (%) Fan Power (kW) Fan speed (rpm) Noise Level Level ( dB)

H-560 0.79 0.9 800 48

H630 0.67 0.95 680 56

From the above table, the performance including efficiency, power and noise level of H-560 are better than that of H-630. Therefore, H-560 are selected as the PAU pump. Dimension

Size of Fan : Modal H-630 = 790mm(L) x 790mm(W) Size of PAU : Unitsize 20 Height = 1600mm Weight = 1600mm Leng Lenght ht = Len Lengt gth h of of fan fan sect sectio ion n + Leng Lenght ht of coil coil seti setion on + len lengt gth h of of filter + length of mixing box = 1900 + 700 + 300 + 1000 = 3600 mm







Supply Air Duct Sizing Supply Air Side

Ducting sizing sheet + fan sizing No de

Type of fitting

Air flow(l/s)

Duct Size (mm)di a

Duct Velocit Velocit Duct Size y y length (mm*mm) Pressure

Fitting loss coefficie nt

(m) 1 duct 2 branch-T 2 duct(gym.) 3 branch-T 3 duct(gym.) 4 branch-T 4 duct(gym.) 5 branch-T 5 duct 6 branch-T 6 duct 7 elbow(game) 7 duct 8 branch-T 8 duct 9 branch-T 9 duct 10 branch-T 10 duct(gym.) 11 elbow(game) 11 duct 12 branch-T 12 duct 13 branch-T 13 duct 14 branch-T 14 duct(game) 15 branch-T 15 duct 16 branch-T 16 duct(study) 17 branch-T 17 duc 18 branch-T 18 duct 19 b hT

2000 80 80 80 80 80 80 750 750 350 350 80 80 270 270 80 80 150 150 70 70 350 350 300 300 70 70 270 270 80 80 200 200 70 70 140

575 800*400 180 200*150 180 200*150 180 200*150 180 200*150 180 200*150 180 200*150 450 500*300 450 500*300 450 500*300 450 500*300 190 200*150 190 200*150 250 350*150 250 350*150 19 190 200*150 19 190 200*150 210 250*150 210 250*150 190 200*150 190 200*150 450 500*300 450 500*300 300 350*240 300 350*240 190 200*150 190 200*150 255 300*240 255 300*240 190 200*150 190 200*150 250 360*200 250 360*200 190 200*150 190 200*150 250 300*240

7.75 3.6 3. 3.6 3.6 3. 3.6 3.6 3. 3.6 4.75 4.75 4.75 4.75 3.6 3.6 4.25 4.25 3.6 3.6 3.6 3. 3 .6 3.6 3.6 4.75 4.75 4.75 4.75 3.6 3. 3.6 4. 4 .5 4.5 3.6 3. 3.6 4.25 4.25 3.6 3.6 4 25

36.2 8.2 8.2 8.2 8.2 8.2 8.2 13.6 13.6 13.6 13.6 7.8 7. 7.8 10.9 10.9 7. 7.8 7.8 8.2 8.2 7.8 7. 7.8 13.6 13.6 14.4 14.4 7. 7.8 7.8 13.2 13.2 7. 7.8 7.8 10.9 10.9 7. 7.8 7. 7.8 10 9

Duct Press drop (Pa/m) 1

3 0.85 5

1 0.85

5

1 0.85

5

1 1

3

1 1

3

1 1

3

1 0.45

1.8

1 1

3

1 0.85

5

1 1

3

1 1

3

1 0.195

4.5

1 1

3

1 0.195

2

1 1

3

1 0.2

5

1 1

3

1 02

Pressur Section e al Drop Pressur e (Pa) (P (Pa) 3 3 7.0 5 6.7 7.0 5 6.7 7.0 5 6.7 24.0 3 25 25.7 24.0 3 25 25.7 7.8 3 10.8 4.9 1.8 6.7 7.8 3 18 7.0 5 6.7 7.8 3 10.8 24.0 3 25 25.7 2.8 4.5 18.3 7.8 3 18 2.6 2 6.7 7.8 3 18 2.2 5 20 7.8 3 10.8 22







Water Side Selection Chiller Selection From the previous calculation, the total cooling load is 213196 (61 tons). Because of the consideration of flexibility and the space available, 2 chillers is adopted ∴ Each chiller with about 31 tons is selected. 1. Chilled water flow rate By Q = m x CPW x T

where

Q = Coil load (kW) m = Chilled Chilled water flow rate (kg/s) C = Sensible heat capacity (kJ/kg) Chilled water temperature difference difference (K) ∆T = Chilled

The mass flow rate of the chiller chiller = 106.58 / {4200x(12.5-7)} {4200x(12.5-7) } = 4.6 l/s 2. Chilled selection Design Criteria: 1. The leaving chilled water temperature is 7 oC 2. The condenser entering air temperature is 32 oC 3. The chilled chilled water temperature rise is 5.5oC Type: Manufacturer: Model Cooling Capacity: Input power Chiller Pressure Drop Rated load current Fan motor size Fan speed Dimension Weight (kg)

Air-cooled liquid chiller TRANE CXAG 206 128kW 42kW 20 kPa 88 4 x 1 .4 13 2630(L) x 1840(W) x 2070 (H) 1590







Chilled Water Pipe Sizing Design Criteria 1. Pressure drop per meter pipe length length is within within the range 100 Pa/m Pa/m to 400 Pa/m. 2. Velocity is between the range 0.6m/s to 2.5m/s. 3. Refer to CIBSE Guide C4 the material used for piping are shown as follows: ~ Large size steel pipings are utilized for the diameter of pipes greater than 150mm, with water flow at 10oC in the pipes. ~ Heavy grades steel (black steel) pipings are utilized for the diameter of pipes less than 150mm, with water flow at 10 oC in the pipes. 4. The fitting loss coefficient are listed below: Code No. 1 2 3 4 5 6 7 8

9 10

Type of fitting

Tee Joint 80mm -80mm (dia) 80mm -100mm (dia) 80mm -125mm (dia) 80mm -150mm (dia) 80mm -175mm (dia) 80mm -200mm (dia) 80mm -225mm (dia) 80mm -250mm (dia) Bend Elbow

Fitting Loss coefficient 4 0.1 0.08 0.2 0.5 0.1 0.4 vertical /horizontal 0.3/1.3 0.4/1.4 0.55/1.45 0.7/1.48 0.85/1.55 0.9/1.6 0.95/1.62 1/1.65 0.3 0.1

5. Pipes Pressure Pressure Drop Calculation Calculation Pipe Pressure(Pa) = Pipe length (m) x Pipe Pressure loss per unit length (Pa/m)









Pump Selection To select a suitable pump pump for water side equipment, equipment, two data must be calculated: 1. The total pressure drop along the critical path. 2. The total volume flow rate of the system and the pumps 1. Criterical Path

Total pressure drop along the criteria path: The pipings and fittings loss = 65.2 kPa (on Critical Path) The PAU coi coil frictio ction n loss loss = 14 kPa kPa The chiller pressure drop = 20 kPa ∴ The to total pr pressure lo loss = 99 99.2 kPa For safety function, function, 10% safety margin are added. Hence, the total to tal pressure loss = 99.2 x 10% = 110kPa = 11 mH 2. Volume flow rate of each pumps

As calculated before, before, the total cooing c ooing load of the system is 61 tons and two 31 tons(106kW) are adopted. Therefore, By Q = m x C PW x ∆T, PW The volume flow rate of the chiller = 106.58 / {4200x(12.5- 7)} = 4.6 l/s = 16.6m 3 /s ∴ The volume flow rate of the each pump is 4.6 l/s.

Air Conditioning System Design Calculation

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