combinepdf (13)

Continue………… 0.029
d) Calculate latent heat gain from ventilation
W = 0.029 – 0.009
Q latent = 4840 x CFM x W = 0.02 lb/lb

W is a grain of moisture (lb/lb) 0.009
Must Plotting in psychometric chart
55 95
Use the design conditions
a) Outdoor design condition; 95F
b) Outdoor design humidity; 80%
c) Saturated Coil design ; 54F

Continue…………. SUMMARY Total CFM Air Required
e) Calculate latent heat gain from ventilation
Space cooling load (CFM) = 750 CFM
Q latent = 4840 x CFM x W
= 4840 x 100 CFM x 0.02 Ventilation Load (CFM) = 100 CFM
= 9680 Btu/hr
Total cooling load CFM = 850 CFM

SUMMARY SUMMARY Total Cooling Required
Space cooling load = 18630 Btu/hr
CFM Ventilation = 100 CFM Ventilation Load = 14000 Btu/hr
Total cooling load = 32,630 Btu/hr
Sensible heat gain (Qs) = 4320 Btu/hr

Latent heat gain (QL) = 9680 Btu/hr

Total Ventilation Load = 14000 Btu/hr

Comparison with direct calculation –Refer Grand total PRACTICE 1 :
Total cooling load Hi (80btu/hr.ft2) = 60000 Btu/hr
Total cooling load Avg (60btu/hr.ft2) = 45000 Btu/hr Calculate for cooling load for office interior. Total
Total cooling load Lo (45btu/hr.ft2) = 33750 Btu/hr occupancy is 75 person

The design conditions

a) Indoor design condition; 21C

Total Cooling load calculated base parameter DR, OT, Coil b) Outdoor design condition; 33C
design
Total cooling load = 32630 Btu/hr c) Outdoor design humidity; 75 %

d) Saturated coil ; 12C

Load difference = + 27370 Btu/hr Hi e) Use AVG value 50 m

+ 12370 Btu/hr Avg
+ 1120 Btu/hr Lo

45 m

7

JAWAPAN PRACTICE 1

SUMMARY Total CFM Air Required

Space cooling load (CFM) = ………………. CFM

Ventilation Load (CFM) = ………………. CFM

Total cooling load CFM = ……………….. CFM

SUMMARY Total Cooling Required

Space cooling load = ………………. Btu/hr

Ventilation Load = ………………. Btu/hr

Total cooling load = ………………. Btu/hr

ASSIGNMENT 1 : Calculate the Heat load for MAIDIN MALL. ASSIGNMENT 1 – MAIDIN MALL

Estimate occupancy load 120 m

The design conditions 1. SHOP 1 =5 SURAU
a) Indoor design condition; 21C
b) Outdoor design condition; 33C 2. SHOP 2 = 20 OFFICE 1 OFFICE 2 STORE 18 m
c) Outdoor design humidity; 70% 86 m 85 m
d) Saturated coil ; 10 C 3. RESTAURENT = 70 6m MEETING
7m ROOM
4. SHOPPING AREA = 200
7m

5. FOOD COURT= 60 FOOD
COURT
6. OFFICE 1 =7 35 m
20 m
7. OFFICE 2 =1 SHOPPING AREA

8. SURAU = 10

9. MEETING ROOM = 10

10. STORE = 35 25 m 5 m 45 m 5m 40 m
SHOP 2 RESTAURENT
WALK WALK
WAY 12 m
SHOP 1 WAY

SUMMARY TABLE FOR ASSIGNMENT 1

LOCATION AREA Cooling load
(ft2) (BTU/hr)
1. SHOP 1
2. SHOP 2
3. RESTAURENT
4. SHOPPING AREA
5. FOOD COURT
6. OFFICE 1
7. OFFICE 2
8. SURAU
9. MEETING ROOM
10. STORE

8

PERTUKARAN UNIT PENGENALAN

• 2 Jenis unit yang biasa digunakan di dunia. Pertimbangan terhadap persekitaran yang perlu ada dalam
• Sistem metric (International System Unit -SI) rekabentuk biasanya adalah:
• Sistem Imperial (British Imperial-BI) • Dry bulb temperature
• Moisture content of the air (humidity, both relative)
1 RT = 12000 Btu/hr = 3.516 kW 1 foot (ft) = 12 inchi • Air movement
1MBH = 1000 Btu/hr • Air quality
1 CMS = 2119 CFM 1 ft = 0.3048 m
1 CMH = 0.588 CFM Tugas air conditioning system ialah mengawal suhu dan
1 CMS = 1000 L/s = 3600 CMH 1 Pa = 0.004 inWG kelembapan udara bagi keadaan berikut:
1 Bar = 14.7 Psi = 101325 Pa 1 inWG = 0.083 ftWG • Heating
1 CMH = 4.403 USGPM
• Cooling
1 hr = 60 m = 3600 s • Dehumidification
• Humidification
1 kw = 3412 Btu/hr • Melibatkan chiller plant room yang besar

REFRESHMENT PENGENALAN
PERTUKARAN UNIT-Conversion Table
Apakah Sistem Chiller dan Chiller Plant
Flow rate = room?
=
1. 500 USGPM kepada L/s = • Komponen utama (system penyejukan - RAC)
2. 10 CMH kepada USGPM = • Vapor Compression cycle
3. 55 L/s kepada CFM • Proses penyejukan/pemanasan
4. 875 CFM kepada CMS & CMH = • Cth Water cooled atau air cooled
= • Komponen sokongan seperti pump, cooling tower,
Pressure
= piping system, dan lain-lain lagi.
1. 800 Pa kepada ft.W.G = • Sistem yang besar
2. 235 Psi kepada ft.W.G =

Cooling capasity

1. 6500000 Btu/hr kepada MBH
2. 1060 kW kepada BTU/hr & RT
3. 850 RT kepada MBH & kW

PENGENALAN DESIGN ACRMV Mengapa chilled water sebagai pilihan?

Air conditioning system & refrigeration design amat meluas • Kaedah mudah menggerakkan tenaga di dalam bangunan
dan kadangkala amat rumit kerana memerlukan • Chilled Water Coil tanpa valve mempunyai kawalan
pertimbangan serta pilihan yang ada untuk dinilai bagi sempurna
setiap projek. • Nyahkelembapan
Antaranya termasuklah • Efisen
• Rekabentuk dan kesannya • Lokasi berjauhan daripada penghuni
• Aspek teknikal ACRMV yang dipilih • Servis
• Kos, teknikal, personality dan kemahiran personel • Bunyi
• Keselamatan
Setiap jurutera/estimator yang membuat rekabentuk
adalah mengikut pengalaman/kemahiran masing2.

Nilai anggaran mungkin lebih / kurang??

1

Asas Sistem Chiller Tunggal Langkah merekabentuk sistem ACMV

Building drawing Load Estimation

VAV box Diffuser

System selection

Cooling tower Ducting system

VAV system
AHU

Warm Water Make-up Chilled water
tank 7C

35 C 12C

29 C Piping system
close

Piping system Chillers
open

Condenser water pump Chilled water pump

ASAS KOMPONEN WATER COOLED CHILLER Asas sistem paip Chilled Water (CHW)

cooling tower / 7C
condenser
Option
cooling coil Unit chiller
(AHU / FCU) 13 C

pumps Peralatan termasuklah.
• Gegelung yang menghasilkan cooling/heating
• 2Way /3W valve yang mengawal aliran ke gegelung
• Piping dan pump untuk mengalirkan dan memacu air
• Expansion tank

KOMPONEN UNIT CHILLER RUMUS BAGI KAPASITI SISTEM
(AIR SEBAGAI MEDIUM)
Vapor compression cycle Unit
reject heat chiller QBtu/hr = 500  flow rate  T

D condenser C compressor
expansion energy in

device

evaporator B
A

absorb heat

2

Soalan 1 : Imperial Unit Asas system pipe terbuka

Satu system ACRMV = 60TR = Q 35 C
Suhu air masuk chiller = 55 F
Suhu air keluar chiller = 45 F 29.4 C
Berapakah nilai kadar aliran water dalam paip?

RUMUS BAGI KAPASITI SISTEM Sistem Piping water cooled chiller
(AIR SEBAGAI MEDIUM)

QW = 4,184  flow rate  T

Di mana,
 Q = load, Watt
 flow rate = aliran air melalui chiller, L/s , sekiranya CMS (kW)
 T = perbezaan suhu (air masuk & keluar chiller), ºC

Soalan 2 : Metrix Unit Contoh chiller system dan aksesori

4.4 liter per second aliran air dengan suhu 37 °C memasuki
condenser unit water cooled dan keluar pada 42°C.
Berapakah jumlah beban haba Q?

3

Rekabentuk kadar aliran air dan suhu JENIS BAHAN PAIP

95°F 44°F

[35°C] [6.7°C]

85°F ARI conditions

[29.4°C] evaporator
flow rate
54°F
condenser
[12.2°C] flow rate

Design Delta T (rekabentuk beza suhu) Kawalan injap aliran water dalam paip
Heating systems

Cooling systems

Cooling systems

Rekabentuk aliran pipe–Konsep 1 Sistem sokongan aliran water dalam paip

Coil 4

Main supply Main return

Chiller

Two pipe direct return system

Sistem sokongan aliran water dalam paip Layout building plan – Factory base

Block 1 – Factory –heavy activity Block 2 – Office general perimeter
180 ft 45 ft

60 ft

75 ft

AHU AHU AHU
Room 1 Room 2 Room 3

100 ft

Chiller room CT

180 ft

MARI MENCUBA DALAM MEREKABENTUK
SISTEM ACRMV

Langkah kerja- sekiranya memilih WC Chiller system Layout building plan – Factory base

1. Pilih saiz WC chiller system Block 1 – Factory –heavy activity Block 2 – Office general perimeter
180 ft 45 ft
2. Pilih saiz AHU yang bertepatan
60 ft
3. Rekabentuk ducting system and diffusers
75 ft
4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR).
5. Pemilihan pump air- CHWP AHU AHU AHU
Room 1 Room 2 Room 3
6. Rekabentuk cooling tower
100 ft
7. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
8. Pemilihan pump- CWP Chiller room CT
9. Rekabentuk Mark-up tank
180 ft
10. Drawing AutoCAD
11. B.O.M

5

Determine Heat load of buildings 1 Air-Cooled or Water-Cooled??

Heat load = Area x Coefficient
=
= 145 RT

Tukar ke RT 10 water-cooled
1 RT =
= air-cooled
=
= 7.5

0 tons 500 tons 1,000 tons 1,500 tons 2,000 tons 2,500 tons 3,000 tons
[0 kW] [1,759 kW] [3,517 kW] [5,276 kW] [7,034 kW] [8,793 kW] [10,551 kW]

chiller capacity

Determine Heat load of buildings 2 Menentukan Chiller capacity

Heat load = Area x Coefficient Reference = 145 RT
=
= catalogue Daikin = Water cooled Single Screw Chiller
= 153.6 RT / 540 kW
Model selected = CUWD 160 B5Y

Tukar ke RT Catalog DB = Water cooled rotary Screw Water Chiller
Model selected = 156 RT / 538 kW
1 RT = = WCFXE-23S

=

= Dapatkan semua data sistem– Model ??

=

Layout building plan – Factory base Dapatkan semua data sistem– Model ??

Block 1 – Factory –heavy activity Block 2 – Office general perimeter Dunham Bush Daikin
180 ft 45 ft

60 ft 24 RT 75 ft Model WCFX 23S CUWD160 B5Y
Cooling capacity 156 RT 153.6 RT
108 RT Power 400V/3P/50Hz 380-415V/3P/50Hz
Cooling capacity 538 kW 540 kW

AHU AHU AHU Compressor Screw semi Semi hermetic Screw
Room 1 Room 2 Room 3 85.7 kW
Power input 88.7 kW Star delta
143.5A
Starter Star delta
329.7
108 RT + 24 RT RLA 146

132 RT x 10% D.safety factor 100 ft LRA 711

145 RT Compressor inrush

Bergantung untuk tambah. Catatan: Dunham bush – warranty part 1 year

Chiller room 180 ft

6

Sambungan…… Perhatian: Kesesuaian memilih sistem Chiller

Model Dunham Bush Daikin Jurutera perlu membuat pertimbangan yang sewajarnya serta cadangan
yang sesuai :
Evaporator WCFX 23S CUWD160 B5Y
Qty • Performance requirements – keselesaan, kebisingan, kawalan,flexible,
Water flow rate Flooded Shell & tube kod amalan
Pressure drop 1 1
Pipe size 367 USGPM 1548 LPM(408.9USGPM) • Capacity requirements – julat kapasiti, unit pelbagai, zon/seksyen.
10.4 ftWG 6.8 ftWG
• Spatial requirement – plant room space, ruangan ducting, piping (vertical
shafts), terminal unit, diffuser

• Costs – kos awal, kos operasi, kos penyelenggaraan

• Energy consumption – ekonomik dan mesra pengguna/persekitaran

• System qualities – e.g. jangkahayat, ketahanan, penyelenggraan mudah,
spareparts.

Sambungan….. PELARASAN TR bagi setiap bangunan

Dunham Bush Daikin Block 1 – Factory –heavy activity Block 2 – Office general perimeter
180 ft 45 ft
Condenser Flooded Shell Flooded shell
Model
Water flow rate 429 USGPM 1935 LPM (511 USGPM) 60 ft 28 RT 75 ft
Pressure drop
Pipe size 5.1 Psi (11.78 ftWG) 8.7 ftWG 128 RT

General information AHU AHU AHU
Room 1 Room 2 Room 3

Length 4.08 m 3.34 m

Width 1.2 m 1.84 m

Height 2.2 m 2.1 m

Weight - 128 + 28 RT 100 ft
= 156 RT
Operating weight CT
Chiller room
Refrigerant R134a R134a 180 ft

Charging Catatan: Dunham bush – warranty part 1 year

Estimate charging per 100 TR –Dunham bush Dimanakan kita berada???

Dunham Bush DAIKIN 1. Pilih saiz WC chiller system

156 RT = 220 kg 145 RT = 215 kg 2. Pilih saiz AHU yang bertepatan
100 RT = X kg 100 RT = X kg
3. Rekabentuk ducting system and diffusers
X (156 RT) = 220 kg ( 100 RT) X (145 RT) = 215 kg ( 100 RT)
4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR).
X = 220 (100) / 156 X = 215 (100) / 145 5. Pemilihan pump air- CHWP

= 141 kg = 148 kg 6. Rekabentuk cooling tower
100 RT = 141 kg 100 TR = 148 kg
7. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
Peratusan beza 2 model 8. Pemilihan pump- CWP
9. Rekabentuk Mark-up tank
(148-141) / 148 x100
10. Drawing Auto CAD
= 4.7 % 11. B.O.M

7

Accessories at AHU system – Standard Malaysia JKR Jenis –Jenis Fan
Additional HEPA Filter (Clean room)

Kegunaan kelengkapan

perubatan, automotif,

kapal terbang dan

HEPA : High efficiency particulate arrestance kediaman. HEPA perlu

memenuhi piawai

kecekapan.

Reka bentuk 1 chiller dan 3 AHU Jenis –Jenis Fan merujuk katalog AHU

Coil AHU room 3
AHU room 2
Main supply AHU room 1 FS BF Axial curve
Forward curve Backward curve
Main return
centrifugal centrifugal
Chiller

Two pipe direct return system

Permudahkan reka bentuk 1 chiller dan 2 AHU PELARASAN AHU bagi setiap bangunan

Coil 2 Block 1 – Factory –heavy activity Block 2 – Office general perimeter
Coil 1 180 ft 45 ft

AHU room 2 60 ft 28 TR 75 ft
AHU room 1
128 TR

Bypass line AHU AHU
Room 1 Room 2

Main supply Main return

Chiller 128 + 28 TR 100 ft
= 156 TR
Two pipe direct return system CT
Chiller room
180 ft

8

Pembahagian haba ruangan Determine AHU 1 – Kes menggunakan 70%

Haba deria = 70 % Menggunakan Delta T = 27 F
(Sensible heat)
Menggunakan Delta T = 25 F
Haba pendam = 30 %
(Latent heat)

Haba keseluruhan = 100 %
(Total heat)

Soalan berkaitan CFM udara – Q sensible (Qs) Kes sekiranya Determine AHU 1 dan 1a

Contoh satu system ACRMV = 60TR = Q Untuk memasang 2 unit AHU

Suhu udara masuk AHU = 80 F Unit AHU - 1
Suhu udara keluar AHU = 53 F
Model =
Berapakah nilai kadar aliran udara dalam duct? Row =
GPM =
Pd =
SC =

Unit AHU – 1a

Model =
Row =
GPM =
Pd =
SC =

Determine AHU 1 Determine AHU 2

Rujuk katalog AHU – Senaraikan jumlah MBH pada + - (10%) Qs = 1.08 x CFM x delta t

Contoh = 1536 + 153.6 (10%) = 1689 MBH CFM = Q SC
1.08 x delta t
Contoh = 1536 – 153.6 (10%) = 1382 MBH
= 257.8 MBH x 1000
Pertimbangan memilih AHU 1.08 x (27) F
1. MBH terhampir
2. Tekanan jatuh seminimum yang mungkin = 8840.9 CFM
3. Coil sekecil yang mungkin (row)
4. Keperluan penggunaan /melibatkan kos

9

Sketching AHU No. & CFM at building plan

Block 1 – Factory –heavy activity Block 2 – Office general perimeter DUCTING Design

180 ft 45 ft Kriteria pemilihan

60 ft – Face Air velocity (kelajuan angin bekal-diffusers)
– Static pressure (Tekanan static)
75 ft – Type round / rectangular (Jenis duct)
– Appearance (Rupa bentuk duct)
AHU AHU – Length (Panjang duct)
Room 1 Room 2 – Materials selection (Bahan digunakan)

Dimanakan kita berada??? Komponen pengagihan

1. Pilih saiz WC chiller system – Supply fan (Kipas udara bekal)
– Supply ductwork
2. Pilih saiz AHU yang bertepatan – Transition fitting
– Discharge grills
3. Rekabentuk ducting system and diffusers – Return grills
– Return ductwork
4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR).
5. Pemilihan pump air- CHWP
6. Rekabentuk expansion tank

7. Rekabentuk cooling tower

8. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
9. Pemilihan pump- CWP
10. Rekabentuk Mark-up tank

Kaedah asas merekabentuk system ducting Jenis-jenis ducting

Equal velocity method – jurutera memilih kelajuan 10
angin yang malar di dalam keseluruhan system.

Velocity reduction method – jurutera memilih kelajuan
yang berubah-rubah mengikut setiap seksyen atau
cawangan duct

Equal friction method – Saiz dan friction loss ditentukan
melalui Ductulator. Panduan CFM dan FPM yang diambil
daripada design chart.

Friction yang malar digunakan disepanjang laluan ducting
tersebut. Perlu membuat balance damper pada setiap
outlet untuk mendapatkan CFM/FPM supply.

Contoh Ducting design– Standard Malaysia JKR Pemilihan Diffuser

• Standad kelajuan angin antara 700 -900 FPM (Neck
velocity = Vp = 4005/ 2 )

• Kadar maksima kelajuan angin keluar diffuser untuk
kegunaan domestik = 800 FPM

• Kadar kelajuan angin balik = 600 FPM
• Sekiranya lebih daripada nilai diatas akan berlaku noise.
• Angin balik yang melebihi 400 FPM akan memberi kesan

terhadap penapis udara.
• Elakkan perselisihan/overlap / hadangan

Contoh Ducting design– Standard Malaysia JKR Pemilihan Diffuser- Aturan aliran udara
With fresh air duct

Pemilihan diffusers Sketching AHU No. & CFM at building plan

Kriteria pemilihan diffusers Block 1 – Factory –heavy activity Block 2 – Office general perimeter

– Aliran udara (CFM) 180 ft 45 ft
– Lontaran
– Tekanan jatuh 60 ft
– Noise Criteria (NC) Level
– Rupa luaran 75 ft
– Kelajuan udara daripada diffuser = contoh 5 m/s
AHU AHU
1 m/s = 196.8 FPM Room 1 Room 2

11

Pemilihan diffusers AHU-1 Contoh 2-Duct calculator

Rujuk - aplikasi factory…??? Tiada • Air flow – 15000 cfm
Rujuk – anggar kpd Office • Duct type : Round
• Friction loss : 0.08 in.wg/100ft
General open office • The round duct size : ________?
• The rectangular duct size : ___________?
Bahagikan jumlah diffuser yg perlu – Rujuk katalog diffuser

= 40370 CFM Pilihan diffusers
60 Size = 15 in x 15 in
Q = 698 CFM
= 672.8 CFM NV = 500 FPM (Neck velocity)
Throw = 19, 27, 37 ft
= 673 CFM NC = 28 dB
Order No. = SCD-4S-15” x 15” - 9010

Relocated 60 unit diffuser – Building 1

0 13.8 27.7 41.5 55.4 69.2 83.1 96.9 110.8 124.6 138.5 152.3 166.2 Cara baca ducting rectangular

180 b
a
10

20

30

40

50

60

60 40370 CFM
AF520 1000X

MEREKABENTUK Contoh 3-Duct calculator
DUCTING AHU -1
• Air flow – 8500 cfm
Refreshment • Duct type : Round
Contoh 1-Duct calculator • Friction loss : 0.05 in.wg/100ft
• The round duct size : ________?
• Air flow – 2000 cfm • The rectangular duct size : ___________?
• Duct type : Round • Velocity in duct : ___________?
• Friction loss : 0.1 in.wg/100ft
• The round duct size : ________?

12

Contoh 4-Duct calculator Ducting layout dan susunan– Building 1

• Air flow – 4000 cfm 0 13.8 27.7 41.5 55.4 69.2 83.1 96.9 110.8 124.6 Ø 18 138.5 152.3 166.2
• Duct type : Round duct
• Application duct : Theaters 2019 2019 Ø 14 Ø 12 180
• Duct designation : Main duct
• Duct velocities:________ ? 673 673 673 1346 1346 673
• The duct size : ________?
10

673 673 673 673 673 673
Ø 24 4038

20

673 673 673

Ø 31 8076

30

673 673 673 673
12114 Ø 36
673 673

40

673 673 673 673 673 673 Ø 40
16152
16152 20190
50 Ø 44

20190

60 40380 Ø58
40370 CFM

AF520 1000X 0.1 inWG

Contoh 5 -Duct calculator Design 2 Ducting and Diffusers- AHU 1
150’
• Air flow – 5000 cfm
• Duct type : Round duct 13.6’ 27.2 40.8 54.4 68 81.6 95.2 108.8 122.4 136
• Application duct : Industrial
• Duct designation : Branch duct 10’
• Duct velocities:________ ?
• The duct size : ________? 20’

30’
40’

25575 CFM
FS 240

Refresh layout for duct design

Block 1 – Factory –heavy activity Block 2 – Office general perimeter

180 ft 45 ft MEREKABENTUK
DIFFUSERS DAN DUCTING
60 ft 28 TR 75 ft
AHU -2
128 TR SAMA MACAM AHU-1
FOLLOW STEP SAHAJA
AHU AHU
Room 1 Room 2 13

40370 CFM 9270 CFM
AF520 1000X FS 100

Ceiling height = 10ft

Pemilihan diffusers-AHU2 Pemilihan return ceiling diffusers-AHU2

Rujuk - aplikasi ofis. Rujuk - aplikasi office
Rujuk – aplikasi general office Rujuk – aplikasi general office

NC = 35 - 45 Bahagikan jumlah diffuser yg perlu – Rujuk katalog diffuser

Bahagikan jumlah diffuser yg perlu – Rujuk katalog diffuser = 9270 CFM Pilihan return ceiling diffusers
12 Size = 18 in x 12 in
= 9270 CFM Q = 802 CFM
15 = 772.5 CFM NV = 600 FPM (Neck velocity)
= 773 CFM NC = 36 dB
= 618 CFM Order No. = RCD-4R-18” x 12” - 9010

= 618 CFM

Design Ducting and Diffusers- AHU 2 Return Ducting and Diffusers type 1- AHU 2
12.5’ 25’ 37.5’ 50’ 62.5’

75’

688

11.25

12x10

8250 CFM 22.5
FS 80

33.75

MEREKABENTUK RETURN DUCTING Dimanakan kita berada???
& RETURN DIFFUSER
1. Pilih saiz WC chiller system
Pemilihan Return Ceiling Diffuser
2. Pilih saiz AHU yang bertepatan
Kriteria utama
3. Rekabentuk ducting system and diffusers
– Air flow
– Noise Criteria (NC) Level 4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR).
– Appearance 5. Pemilihan pump air- CHWP
– Kadar angin balik 600 FPM, NV
– Kebisingan NC ikut aplikasi bangunan 6. Rekabentuk cooling tower

7. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
8. Pemilihan pump- CWP
9. Rekabentuk Mark-up tank

10. Drawing Auto CAD
11. B.O.M

14

Route piping Chilled water -anggaran ukuran ft

128 RT 28 RT
308.9 USGPM 71.4 USGPM
PD = 8.1 ft.WG PD = 9 ft.WG

C C’ D D’

40’ 40’

MEREKABENTUK PIPING B 70’
CHWS/CHWR
60’ 80’
Chiller A 70’
156 RT
374.9 USGPM 20’ 30’ 50’ E 80’
PD = 10.9 ft.WG F

G 50’
Water Pump

Rekabentuk kadar aliran air dan suhu Route piping Condenser Water system

10’ Cooling tower

5’
Y

95°F 44°F

[35°C] [6.7°C]

85°F ARI conditions 25’

[29.4°C] evaporator Condenser Y’
flow rate 156 RT 25’
54°F 468 USGPM 40’
condenser PD 11.78 inWG Z 40’
[12.2°C] flow rate
30’
30’
X
Check jumlah RT yang dipilih dan aliran air =??

90’ Mark -up tank

DATA CHILLER YANG TELAH DIPILIH SELEPAS MEMBUAT ASHRAE STANDARD HANDBOOK-FUNDAMENTAL 1993
KERJA-KERJA PENGIRAAN COOLING LOAD BANGUNAN
Pipe friction loss = 1 to 4 ft/100ft
156 RT CHILLER SYSTEM Kawalan kebisingan = kelajuan 4 ft/s dalam 2 inci paip

CHWS/CHWR CWS/CWR dan kebawah
Pressure drop limit = 4 ft/100ft for pipe size 2 in above.

15

Rumus anggaran saiz pipe GI – Kaedah 1 Gunakan contoh factor geseran = 4 ft/100ft

D = s Q 0.4 Rujuk katalog
Piping
Dimana:
266 USGPM VS 4ft/100ft
D = pipe diameter (in.)
Q = flow rate (ft/s) Plot untuk mendapatkan saiz paip
s = pipe size constant

0.44 = assume 4 ft/100ft pressure drop
atau
0.50 = assume 2 ft/100ft pressure drop

Contoh penggunaan rumus

Chiller saiz = 112 RT, anggarkan kelajuan ….FPS MEREKABENTUK PIPING
CHWP
1 RT = 4.5 FPS  pengalaman estimator
LUKISKAN LAYOUT SISTEM
Jadi, D = s Q 0.4 Cooler – AHU-1
1 RT = 4.5 FPS = 0.44 x (500)0.4 Cooler – AHU-2
= 5.3 inches Expansion tank
= 112 RT x 4.5
= 500 FPS

Rumus anggaran saiz pipe GI – Kaedah 2 Plot dalam Jadual geseran paip – pada 4ft/100ft

Qw = 500 x GPM x delta t Supply A-B = Ø 5” 128 RT 28 RT
Return E-F-G = Ø 5” 308.9 USGPM 71.4 USGPM
D D’
Supply B-C = Ø 4” C C’
Return C’-E = Ø 4” Ø 2.5”
GPM = Qw
Ø 2.5”
500 x delta t Supply B-D = Ø 2.5” Ø 4”
Return D’ -E = Ø 2.5” Ø 5”
Ø 4”
= 111.1 RT x 12000 Btu/hr B F
500 x (55-45)F E
Chiller Ø 5”
= 266.64 USGPM 156 RT A
374.9 USGPM

Tukar kepada CMS, CMH & L/s G
1USGPM = 0.063x10-3m3/s Expansion tank

16

Dimanakan kita berada??? Accessories at AHU system – Standard Malaysia JKR

1. Pilih saiz WC chiller system

2. Pilih saiz AHU yang bertepatan

3. Rekabentuk ducting system and diffusers

4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR).
5. Pemilihan pump air- CHWP

6. Rekabentuk cooling tower

7. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
8. Pemilihan pump- CWP
9. Rekabentuk Mark-up tank

10. Drawing Auto CAD
11. B.O.M

MEREKABENTUK PAM AIR Fast estimate valve and fittings equation,
(CHWP)
pd = k . v2/2g
1) Senarai semua fitting (elbow, tee, valve)
2) Periksa size dimana
3) Dapatkan nilai faktor k pd = pressure drop (ft.)
4) Gunakan RUMUS fitting losses k = resistant coefficient for valve/fitting (Jadual)
v2/2g = velocity head (base on flow GPM of size pipe)
v = velocity of water, ft/s
g = 32.3 ft/s2 gravity constant

Aksesori komponen system - Asas Contoh : Calculate pd for accessories

COOLER / CONDENSER Pd = k . (v2 / 2g) Gate valve, k = 0.17 ,
1) Gate valve – Qty = 1 = (0.17) x (0.227) 2” pipe, USGPM = 40
2) Butterfly valve – Qty = 2 = 0.04 ft head
3) Check valve – Qty = 1 90 degree elbow = 1.0
4) 2 Way modulate valve – Qty = 1 Pd = k . (v2 / 2g)
5) Others (gauge, temperature, air vent) = (4) x (1.0) x (0.227) Tee joint = 1.4
= 0.91 ft head
AHU-1 dan AHU 2
1) Balance valve – QTY = 1 Pd = k . (v2 / 2g)
2) Butterfly valve – Qty = 2 = (2) x (1.4) x (0.227)
3) Strainer – Qty = 1 = 0.64 ft head
4) 3 Way modulate valve – Qty = 1

17

UNTUK MEMUDAHKAN PENGIRAAN-JADUAL 8

MENENTUKAN pressure drop (PD) Supply A-B = Ø 5” 380 USGPM
PD pada cooler
PD pada AHU -1 Return E-F-G = Ø 5” 4ft/100ft 3" Figure
PD pada AHU - 2 Table 7-8

Gunakan jadual yang diberikan Fitting Length pd/100ft Qty k value v2/2g Total pd (ft) Note
Size (ft)

A-B 5 80 0.04 3.68 +15 %
E-f-G 5 130 0.04 5.98 pemanjangan
+15 %
pemanjangan

Tee 5 3 0.32 0.577 0.55 0.34+0.32/2

Elbow 5 3 0.48 0.577 0.83

Gate valve 5 1 0.13 0.577 0.08

Butterfly V 5 2 0.725 0.577 0.84

Check V 5 1 1.6 0.577 0.92

2 way V 5 1 1.44 0.577 0.83 Plug v branc

Others 5 3 1 0.577 1.73 Pipe exit

TOTAL SUPPLY/RETURN BRANCH/ CHILLER COOLER 15.44

Route piping Chilled water - ukuran ft dan saiz paip UNTUK MEMUDAHKAN PENGIRAAN-JADUAL 9

Supply A-B = Ø 5” 128 RT 28 RT Supply B-C = Ø 4” 320 USGPM
Return E-F-G = Ø 5” 308.9 USGPM 71.4 USGPM Return C’ - E = Ø 4” 3" Figure
PD = 8.1 ft.WG PD = 9 ft.WG 7-8
Supply B-C = Ø 4” Table
Return C’-E = Ø 4” C C’ D D’

40’ Fitting Length pd/100ft Qty k value v2/2g Total pd (ft) Note
Size (ft)

Supply B-D = Ø 2.5” 40’
Return D’ -E = Ø 2.5”

B 70’

Chiller 60’ 80’
156 RT A 70’
374.9 USGPM
PD = 10.9 ft.WG 20’ 30’ 50’ E 80’
F

G 50’ TOTAL SUPPLY/RETURN BRANCH AHU-1
+ From chiller (A-B, EFG)
Expansion tank JUMLAH BESAR PD

UNTUK MEMUDAHKAN PENGIRAAN-JADUAL 8 UNTUK MEMUDAHKAN PENGIRAAN-JADUAL 9

Supply A-B = Ø 5” 4ft/100ft 380 USGPM Supply B-C = Ø 4” 320 USGPM
Return E-F-G = Ø 5” pd/100ft Qty 3" Figure Return C’ - E = Ø 4” 3" Figure
7-8 7-8
Table Table

Fitting Length k value v2/2g Total pd (ft) Note Fitting Size Length pd/100ft Qty k value v2/2g Total pd (ft) Note
Size (ft) (ft)
0.04 0.51 1.01
B-C 4 50 1.7 1.01 2.30 +15 %
80 0.04 0.77 1.01 3.68 pemanjangan
C’-E 4 7.1 1.01 +15 %
2 0.51 1.01 pemanjangan
1 1 1.01
Elbow 4 2 1.03
1
Check V 4 1 1.72
Butterfly V 4 3
Strainer 4 1.55
3 Way V 4
Others 4 7.17 Strainer disc

0.52

3.03

TOTAL SUPPLY/RETURN BRANCH/ CHILLER COOLER TOTAL SUPPLY/RETURN BRANCH AHU-1 21.00
+ From chiller (A-B, EFG) 15.44
JUMLAH BESAR PD 36.44

18

UNTUK MEMUDAHKAN PENGIRAAN- JADUAL 10 Penjumlahan tekanan jatuh (pd)
keseluruhan sistem

Supply B-D = Ø 2.5” 70 USGPM Estimate Items pd (ft)
Return D’- E = Ø 2.5” Figure 3- 3" Figure 0.696+0.143/2 AHU 1 8.1
7 7-8 AHU 2 9.0
Chiller unit (Evap) 10.9 From unit catalog
Fitting Length pd/100ft Qty k value v2/2g Total pd (ft) Note Fitting pipe (ABCC’EFG)
Size (ft) Fitting pipe (ABDD’EFG) 36.44 From unit catalog
Total 35.80 From unit catalog
Case others, Total 100.24 Jadual 9
64.44 Jadual 10

TOTAL SUPPLY/RETURN BRANCH AHU-2 Pertukaran unit
+ From chiller (A-B, EFG) 1CMH = 4.403 USGPM
JUMLAH BESAR 1 ft = 0.3048 m

UNTUK MEMUDAHKAN PENGIRAAN- JADUAL 10 Determine Pump Model + Capacity

Supply B-D = Ø 2.5” 70 USGPM Estimate Tukar nilai flowrate condenser Tukar nilai ft kepada meter
Return D’- E = Ø 2.5” Figure 3- 3" Figure 0.696+0.143/2 = 468.6 USGPM
7 7-8 1 ft = 0.3048m
1 CMH = 4.403 USGPM
Fitting Length pd/100ft Qty k value v2/2g Total pd (ft) Note Jadi,
B-D Size (ft) 0.04 5.98 15% Jadi,
100 ft x 0.3048 m
2.5 130 380 USGPM x 1 CMH 1 ft
4.403 USGPM
D’-E 2.5 160 0.04 7.36 15% = 30.6 m
= 86 CMH
Elbow 2.5 4 0.54 0.42 0.91

Check V 2.5 1 1.8 0.42 0.76

Butterfly V 2.5 2 0.81 0.42 0.68

Strainer 2.5 1 7.6 0.42 3.19 Strainer disc

3 Way V 2.5 1 0.54 0.42 0.22

Others 2.5 3 1 0.42 1.26

TOTAL SUPPLY/RETURN BRANCH AHU-2 20.36

+ From chiller (A-B, EFG) 15.44 Rujuk catalog pump untuk
proses memplot
JUMLAH BESAR 35.80

Penjumlahan tekanan jatuh (pd) Perbezaan memilih antara
keseluruhan sistem NK 1450 RPM & NK 2900 RPM

Items pd (ft) 1. Secara umum, 4-pole (1450 @50Hz) motors lebih
AHU 1 sesuai dalam system Chilled water kerana ia boleh
AHU 2 From unit catalog mengurangkan kebisingan.
Chiller unit (Evap)
Fitting pipe (ABCC’EFG) From unit catalog 2. Seterusnya, chilled water memang memerlukan head
Fitting pipe (ABDD’EFG) From unit catalog yang rendah.
Total Jadual 9
Case others, Total Jadual 10 3. Pemilihan pump motor 2900 RPM akan
menghasilkan lebih head berbanding pump motor
Pertukaran unit 1450 RPM, walaupun jumlah HP hampir sama.
1CMH = 4.403 USGPM
1 ft = 0.3048 m 4. Kos awal tinggi.

19

Dimanakan kita berada??? Determine Cooling Tower

1. Pilih saiz WC chiller system Tukar nilai flowrate condenser = 468.6 USGPM
1 CMH = 4.403 USGPM
2. Pilih saiz AHU yang bertepatan Jadi,
468.6 USGPM x 1 CMH
3. Rekabentuk ducting system and diffusers 4.403 USGPM
= 106.4 CMH
4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR). Rujuk mukasurat catalog CT – plot
5. Pemilihan pump air- CHWP MODEL CT = 200 RT

6. Rekabentuk cooling tower

7. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
8. Pemilihan pump- CWP
9. Rekabentuk Mark-up tank

10. Drawing Auto CAD
11. B.O.M

MEREKABENTUK COOLING Determine Cooling Tower
TOWER
Ringkasan data cooling tower (Jenis bulat)
1CMH = 4.403 USGPM
1 ft = 0.3048 m Model : BND-200t
Motor : 3P 415V 50 Hz, 5.5 HP
Ukuran : 3500 mm
Tinggi : 3400 mm

Fan : 1890 mm

Airflow : 1250 CMM

Water PD = 40 kPa

Noise level = 62 dB

Rekabentuk dan Pemilihan COOLING TOWER Dimanakan kita berada???

Condenser ADmBb=ie3n2t OWCB 1. Pilih saiz WC chiller system
468.6 USGPMChiller hot water 2W7BC= 270C
PD = 11.78 ftQ.Wv =G69.12 m3/hr 2. Pilih saiz AHU yang bertepatan
156 RT CONPD.dErNopS=E1R8.8S3Y14SmTEWMG
3. Rekabentuk ducting system and diffusers
Temp. =19056oFF oorr 4305oCC
4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR).
Temp. = 8950oFF oorr3320CoC 5. Pemilihan pump air- CHWP

o 6. Rekabentuk cooling tower

7. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
8. Pemilihan pump- CWP
9. Rekabentuk Mark-up tank

10. Drawing Auto CAD
11. B.O.M

20

Complete drawing system sketching- AUTO CAD

Block 1 – Factory –heavy activity Block 2 – Office general perimeter

MEREKABENTUK PIPING AHU AHU
CWP Room 1 Room 2

Pilihan – Buat kerja rumah Chiller room Lukisan Pelan yang tepat dan berskala adalah
diperlukan untuk pemasangan, pembinaan,
Sama seperti Chilled water system menganggarkan dan pengujian pengesahan.

Lukisan juga turut 1. Flow rates of air.; 2.
Ductwork to scale with sizes indicated. 3. Air flow
direction 4. Items of plant

Rekabentuk Mark up Tank Bills Of Materials (B.O.M)

Vt = Vs. [(v2/v1)-1 ] - 3α delta t 1. Senarai Bahan keperluan yang telah dihasilkan
1 – P1/P2 2. Komponen yang telah dipilih
3. Termasuk pemasangan dan bahan2 lainnya
Vs –rujuk drawing P1
5” = 0.938 gal/ft Minimum temperature =10Psig + 14.7Patm
4” = 0.661 gal/ft
3” = 0.384 gal/ft P2
2” = 0.174 gal/ft Maximum temperature =20Psig + 14.7Patm

v1 α = steel, 6.5 x 10-6 in/inF
0.01602 ft/lb at 40F P2
Maximum temperature =20Psig + 14.7Patm
v2
0.01655 ft/lb at 95F Dipermudahkan

Vt = Vs. x 2 x 0.03

Rekabentuk Mark-up Tank Dimanakan kita berada???

Vt = Vs. x 2 x 0.03 1. Pilih saiz WC chiller system
= 375 x 2 x 0.03
= 23 Gal 2. Pilih saiz AHU yang bertepatan

Vs –rujuk drawing Pada paip 5” 3. Rekabentuk ducting system and diffusers
5” = 0.938 gal/ft = 210 ft x 0.938 gal/ft = 197 Gal
4” = 0.661 gal/ft Pada paip 4” 4. Rekabentuk piping system –Chiller to AHU (CHWS/CHWR).
3” = 0.384 gal/ft = 110 ft x 0.661 gal/ft = 73 Gal 5. Pemilihan pump air- CHWP
2” = 0.174 gal/ft Pada paip 2.5” @ 3”
= 270 ft x 0.384 gal/ft = 104 gal 6. Rekabentuk cooling tower
1 Gallon = 3.79 Litre
7. Rekabentuk piping system – Chiller to cooling tower (CWS/CWR)
= 23 gallon x 3.79 Litre 8. Pemilihan pump- CWP
1 gallon 9. Rekabentuk Mark-up tank

= 87.17 Litre Total Vs = 197+73+104 =375 Gal 10. Drawing Auto CAD
11. B.O.M

21

SOALAN KUIZ / UJIAN /
PEPERIKSAAN

SOALAN DMC 3333 –
ACMV TESTING AND

COMMISIONING

















































SOALAN DMC 3113 –
ACMV ACMV COOLING
LOAD CALCULATION

AND EQUIPMENT
SELECTION