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43 Heat Transfer of
Human Body A verage Metabo l i c r a t e Read
44 Heat Exchanger H E - - Read
45 Heat Exchanger H E - - Read
46 Heat Exchanger Hea t Exchange r Ma in tenance Read
47 Health Care Facility Hosp i ta l Ope ra t i on Room Read
48 Global Warming
Greenhouse E f f ec t and G loba l
Warm ing Read
49 Fans Fan Curve Read
50 Fan Fan Law Read
51 Fan Mu l t i p l e Fan Sys tems - Fans i n
Se r i e s and Pa ra l l e l Read
52 ETTV and RETV ETTV RETV Read
53 ETTV and RETV ETTV V ) Read
54 ETTV and RETV ETTV V ) Read
55 Equipment Control HVAC Con t ro l - Ch i l l ed Wate r
Fan Co i l Un i t s Read
56 Engineered Smoke
Sys tem - Ove rv i ew Read
57 Engineered Smoke
Eng inee red Smoke Con t ro l
Sys tem - Examp le Ca l cu l a t i on Read
58 Energy Saving Guide
Energy Smar t O f f i c e Read
59 Energy Saving Guide
Line Energy Smar t Ho te l Read
60 Energy Saving Bill Unders tand ing on Ene rgy B i l l Read
61 Energy Saving Hea t P ipe Read
62 Energy Efficiency Energy E f f i c i ency Index (EEI ) Read
63 Energy Efficiency Energy E f f i c i ency - kW/ton , COP
and EER Read
Read
65
Re f r i ge ra t i on (Vapor -
compress io ) Read
66 Decision Making
Process Dec i s i on Mak ing Cha r t Read
67 Cooling Tower C T ) Read
68 Chiller Plant Room Over v i ew o f Ch i l l e r P l an t Room Read
69 Chiller Plant Room Ch i l l ed Wate r Sys tem Des ign ( I ) Read
70 Chiller Plant Room Ch i l l ed Wate r Sys tem Des ign
( I I ) Read
71 Chiller Plant Room Ch i l l ed Wate r Sys tem Des ign
( I I I ) Read
72 Chiller Plant Room
S ta r t i ng Sequence and
Shu tdown Sequece o f Ch i l l e r
P l an t Room
Read
Read
74 Chiller Plant Room Ch i l l e r Se lec t i on Read
75 Chiller Plant Room C S -C )
Read
and Efficiency
Ch i l l e r P l an t Des ing and
E f f i c i ency (Fu l l Load ) Read
77 Chiller P R) N P mach ine (R 123 ch i l l e r )
Read
78
Water Pump Power
Pump ing Power Ca l cu l a t i on Read
79 Chapter-7 ( Part 3
of 3) V en t i l a t i on ( Par t 3 o f 3 ) Read
82 Chapter-2 (Part 2 of
3)
Unders tand ing P sych romet r i c s
(Pa r t 2 o f 3 ) Read
3)
3)
85 Chapter-1 Fundamenta l and Bas i c Concep t Read
86 Chapter – 4 (Part 2
of 2) A i r Compresso r s ( Par t 2 o f 2 ) Read
of 2) A i r Compresso r s ( Par t 1 o f 2 ) Read
of 2)
Dis t r i bu t i on o f Compres sed A i r
of 2)
Dis t r i bu t i on o f Compres sed A i r
90 Chapter 1 Fundamenta l Concep t
Units
Read
91 Chapter 1 In t roduc t i on to P rog rammab le
Log i c Con t ro l l e r s D Read
92 Chapter - 6 (Part 2
of 3)
Fans and B lowers (Pa r t 2 o f 3 ) Read
of 3) Fans and B lowers (Pa r t 1 o f 3 ) Read
of 3) Fans and B lowers (Pa r t 3 o f 3 ) Read
of 3)
A i r D i s t r i bu t i on Sys tems (Par t 3
o f 3 ) Read
of 3)
o f 3 ) Read
of 3)
o f 3 ) Read
of 3) Coo l i ng Towers (Pa r t 1 o f 3 ) Read
101
Chapter - 3 Air
Handling Units (Part
2 of 2)
Chap te r - 3 A i r Hand l i ng Un i t s
102
Chapter - 3 Air
Handling Units (Part
1 of 2)
Chap te r - 3 A i r Hand l i ng Un i t s
103 Chapter - 2
C - C A S (App l i c a t i on )
Read
104 Chapter - 1 Fundamenta l and Bas i c Concep t Read
105 Building M&E
Systems Scope o f Bu i l d i ng M&E Sys tems Read
106
Building Automation
Systems (BAS) -
Direct Digital
Controllers (DDC)
Di rec t D ig i t a l Con t ro l l e r s (DDC) Read
107 Building Automation
System
A u tomat i c Con t ro l Sys tems in
Bu i l d ings Read
108 Building Automation
Management Sys tem ( IBMS) Read
109 Building Automation Bu i l d ing Au tomat ion Sys tem Introduction to Building Read
Monox ide Mon i to r i ng and
Ven t i l a t i on Fan Con t ro l Sys tem
Read
111 ASHRAE Pocket
Guide Load and A i r F l ow Es t ima te Read
112 Alarm A la rm Pro cess ing Read
113 Air Side AHU Coo l i ng Co i l Read
114 Air Side A -H U Read
115 Air Side V a r i ab l e A i r Vo lume (VAV)
Sys tem- In t roduc t i on Read
116 Air Side Duc t S i z i ng Me thods Read
117 Air Side F r i c t i on Los ses i n Duc t 1 Read
118 Air Side F r i c t i on Los ses i n Duc t 2 Read
119 Air Side Dynamic Los ses Read
120 Air Side D S P L E )
Read
121 Air Handling Unit Componen ts o f A i r Hand l i ng
Un i t Read
122 Air Handling Unit AHU B lower Spec i f i c a t i on Read
123 Air Handling Unit A i r Hand l i ng Un i t and F i re Mode Read
124 Air Conditioning
129 Air Conditioning C T ) - Read
Equipment
130 Air Conditioning How to a i r - cond i t i on ou tdoo r
spaces Read
131 AHU Cooling Coils AHU Coo l i ng Co i l Spec i f i ca t i on Read
132 AC Induction Motor Moto r Speed and S l i p Read
133 AC Induction Motor Unders tand ing moto r
namep la tes Read
134 AC Induction Motor Energy Sav ing f rom Moto r
E f f i c i ency Read
135 AC Induction Motor Es t ima t i on o f Mo to r Load ing Read
136 AC Induction Motor Power Fac to r Read
137 AC Induction Motor Supp ly Vo l t age to Moto r Read
138 AC Induction Motor Insu la t i on , Se rv i ce Faco t r and
Enc losu res Read
R ead
Read
141
Code o f P rac t i c e fo r A i r -
cond i t i on ing and mechan i ca l
ven t i l a t i on i n bu i l d i ngs
Read
B) C A C S
C A C S R T C A C S A C C W C C ) H T AHU FCU C E C W C T C E C W C W C L C W
O L C W C L C W C T O L C L O L ACMV S HVAC
T
C L C W S C W
C W - C L C W S P AHU FCU C) P S
H T H T
8/15/2019 PLC ,AirConditioning and Mechanical Ventilation ,M&E, Automation Lecture( )
C BTU F)
C BTU F) B 92 B B
C L C W S ) C L
C L C W Conductivity (micro S) S C L C W H H A
A C L C W H
--
C L C W T FTU) FAU T FTU) T FTU)
- T FTU)
)
-
S F
Inside Corrosion
C

Microbiological Controls, Corrosion Rate Controls and Water
Quality Paramete C Microbiological Controls
Microbiological Analysis Frequency Control Criteria
1 Standard Plate Count Monthly < 100,000 CFU/mL
2 Legionalla Pneumophilla Once every 3 months Negative detection
3 Sulfate Reducing Bacteria Once every 6 months Negative detection
4 Denitrifying Bacteria Once every 6 months Negative detection
5 Pseudomonas Once every 6 months < 500 CFU/mL
The above test should follow international standards
Standard Plate Count
CFU - ) C L CFU Legionalla Pneumophilla L L P Sulfate Reducing Bacteria C
C L Pseudomonas Bacteria CFU

1 Mild Steel < 1 mpy
2 Copper < 0.1 mpy
B Corrosion R )
1 pH Value Field Test Kit & Laboratory method APHA 3120B
Between 7 to 8.8
2 Total Dissolved Solid (TDS) Field Test Kit & Laboratory method APHA 3120B
< 3000 ppm
3 Iron (Fe) Field Test Kit & Laboratory method APHA 3120B
< 1.00 mg/L
4 Copper (Cu) Laboratory method APHA 3120B <0.50 mg/L
5 Zinc (Zn) Laboratory method APHA 3120B <2 mg/L
6 Chloride Laboratory method APHA 3120B < 500 ppm Cl
7 Total Hardness Laboratory method APHA 3120B <800 ppm
Service Provider can provide an alternative equivalent method but subject to Owner approval.
C H V C H V C C C
C Total Dissolved Solid (TDS) T D S C C T D S TDS) C I F) C
I F)
C Z Z) C
Z Z)
S C C T C W R) W T S C W C
Air Conditioning > Water Side > Friction Losses in Pipes >
Content
Chapter - 2 Understanding Psychrometrics
Chapter - 4 Cooling Towers
Chapter - 7 Ventilation
1 Air Side Friction Losses in Duct 1 Read
2 Water Side Friction Losses in Pipes Read
3 Air Side Friction Losses in Duct 2 Read
4 Air Side Dynamic Losses Read
5 Air Side D S P L E ) Read
6 Pumping System Pressuer Losses for Pump or Pump Head Read
To download all ACMV lecuters in PDF format
V VAV) S - VAV B
V VAV) VAV B S VAV B ) ACMV
VAV B )
VAV B
(2) ability of the VAV box controller to measure and control the desired minimum and maximum
airflow set points;
VAV
(3) first costs of the VAV box, its installation, and controls;
VAV
VAV B
M) D P) P D) VAV B D P)
A F R VAV B P D) VAV B P D)
VAV B VAV B VAV B VAV B Pressure Dr) P D) VAV B
A) T P

VAV B P D D
Air Flow Rate 1000CFM B S P D D A F R CFM)
) VAV B A F R CFM Pressu D D ) VAV B P D D
A F R CFM

(2) ability of the VAV box controller to measure and control the desired minimum and maximum
airflow set points;
VAV
VAV M M VAV I D) VAV M M M M

(3) first costs of the VAV box, its installation, and controls;
VAV VAV B VAV )
(4) noise generation;
VAV B
VAV B C) C) VAV B VAV B
VAV B Pressure Independent VAV Box P VAV B P I VAV B P I VAV B P I
Pressure Independent control P T
Variable Air Volume (VAV) System ACMV L
Maintenance > Type of Maintenance > Type of Maintenance -
>
S S P E S T S S B

S
Predict Maintenance
P P
P V A C E P T C I T S T I P M P D
Performance based Maintenance
C P E WRT

1 Type of Maintenance T M -
Cooling Process > The evaporative cooling process > The evaporative cooling process and Web bulb Temperature >
Air Conditioning and Mechanical Ventilation System (Vol. 1)
Content
Air conditioning > SS553:2009 (CP13) > Outdoor Air Supply >
SS CP) O A S
SS CP ) S S ACMV C CP
C P C P ME C ACMV SSCP )
SSCP )
A C
O A S) O A S) S S N C
O A )
O A )

(How to use table)
T C C W
- O A S) LS) O A) CMH) O A)
LS)

www.bca.gov.sg/publications/others/handbook_for_solar _ pv _ systems.pdf
1.1 Introduction
PV PV
PV PV DC) e DC DC AC ACDB)
Types of Solar PV System
S PV - - -) - ) PV
‘ AC ACDB)
PV size
of the
PV
)
PV PV
C - - - )
Cadmium Telluride (CdTe) 9-12%
Amorphous Silicon (a-Si) 5-7%
PV PV PV C E PV C E
(for the same nominal capacity under Standard Test C STC) )
PV
C

CIGS -0.32 to -0.36
R H C C R M ) C R M Refri
H P) C C R
P) C Refrig R R R R R R R C P T P E E
A C R G R R R
G)
) HFC R HFC ) HFC Hydrofluorocarbons HFC HFC

C F
HFC- CFC- CFC-
CFC- CFC- HFC-134
(optimize the
) ) C R )
) C R

condenser c C W P C
W C C T - C W C Opened
System
ACMV C
End Suction Pump
H H V
)
P
- C W ) AHU FCU - B C L C W F P
P
P
C W O )
H = hd - hs
The total discharge head is made from three separate heads:
hd = hsd + hpd + hfd
hd = total discharge head
The total suction head also consists of three separate heads
hs = hss + hps - hfs
hs = total suction head
= P )
C W T
T ) The liquid level in the suction tank

T P
hs = hss + hps - hfs = -6 + 0 - 4 = -10 feet of liquid gauge at rated flow
The total discharge head calculation
1. The static discharge head is:
= E = )

T F V F
hfd = 25 feet at rated flow
4. The total discharge head is:
hd = hsd + hpd + hfd = 125 + 0 + 25 = 150 feet of liquid gauge at rated flow
The total system head calculation:
H = hd - hs = 150 - (-10)= 160 feet of liquid at rated flow
Note: did you notice that when we subtracted a minus number (-10) from a positive number
(150) we ended up with a positive 160.
chart section
Specifications:
T
P - A S - S
5. Discharge piping rises 40 feet vertically above the pump centerline and then runs 400 feet
horizontally. There is one 90° flanged elbow in this line.
D
6. Suction piping has a square edge inlet, four feet of pipe, one gate valve, and one 90° flanged
elbow all of which are 6" in diameter.
S one 90° f
V
To calculate suction surface pressure use one of the following formulas:
inches of mercury x 1.133/ specific gravity = feet of liquid
pounds per square inch x 2.31/specific gravity = feet of liquid
Millimeters of mercury / (22.4 x specific gravity) = feet of liquid
pu
Total suction head calculation
1. The suction side of the system shows a minimum static head of 5 feet above suction
centerline. Therefore, the static suction head is:
hss = 5 feet
2. Using the first conversion formula, the suction surface pressure is:
hps = -20 Hg x 1.133/ 0.98 = -23.12 feet gauge
3. The suction friction head, hfs, equals the sum of all the friction losses in the suction line.
Friction loss in 6" pipe at 1000 gpm from table 15 of the Hydraulic Institute Engineering Data
Book, is 6.17 feet per 100 feet of pipe.
in 4 feet of pipe friction loss = 4/100 x 6.17 = 0.3 feet
Friction loss coefficients (K factors) for the inlet, elbow and valve can be added together and
multiplied by the velocity head:
6" Gate valve 0.11 32 (b)
Total coefficient, K = 0.90
4. The total suction head then becomes:
hs = hss + hps - hfs = 5 + (-23.12) - 2.0 = -20.12 feet, gauge at 1000 gpm.
Total discharge head calculation
2. Discharge surface pressure = hpd = 0 feet gauge
3. Discharge friction head = hfd = sum of the following losses :
Friction loss in 6" pipe at 1000 gpm. from table 15, is 6.17 feet per hundred feet of pipe.
In 440 feet of pipe the friction loss = 440/100 x 6.17 = 27.2 feet
Friction loss in 6" elbow:
from table 32 (a), K = 0,29
from table 15, V2/2g = 1.92 at 1000 gpm.
Friction loss = K V2/2g = 0.29 x 1.92 = 0.6 feet
The friction loss in the sudden enlargement at the end of the discharge line is called the exit
loss. In systems of this type where the area of the discharge tank is very large in comparison to
the area of the discharge pipe, the loss equals V2/2g, as shown in table 32 (b).
Friction loss at exit = V2/2g = 1.9 feet
The discharge friction head is the sum of the above losses, that is:
hfd = 27.2 + 0.6 + 1.9 = 29.7 feet at 1000 gpm.
4. The total discharge head then becomes:
hd = hsd + hpd + hfd = 40 + 0 + 29.7 = 69.7 feet, gauge at 1000 gpm.
c. Total system head calculation:

ACMV HVAC C H
T C) C) T seconda C S

the primary system produces more chilled or hot water than what the secondary system

On the other hand, if the secondary system requires more water than that produced by the
primary system, the flow of water in the decoupler pipe will be from return to supply.
C C RT US C W )
In such primary-secondary systems, hydraulic isolation allows the secondary pumps to vary the

C O S) S
C C T C C P S P B S P G) B C R F M CHM C O S) S P
Answer;
Pressure Head & Square of Condenser water Flow Rate
Pressure Head= C x Square of Condenser water Flow Rate ( C is a constant) ------ )
Delta P = P2 - P1 = 0.4 ) - =
D P = P - P = )
2.1 bar = C x 4000( Square of Condenser water Flow Rate)
2.1 = C x (4000X4000)
C P CMH ) S P
P = 0.000000131 x Square of Q + 0.4
P = 0.000000131 x (3500x3500) + 0.4
P = 2.0 bar
Open systems the static pressure difference or independent pressure due to height difference is added to the system curve. The system curve is parabolic in shape since the pressure losses
in the system are proportional to the
square of the flow.
Distribution System ACMV L
P -
C W ) AHU FCU - B C L C W F P
P P
C W O )
P
)
) S
) S fl
) variable VFD) adjustable frequency

-


) A L ) L the pump
W What will the reduction in pump power consumptionbe if the
pump speed is reduced to provide the design water flow of 20 L/s?
) T pressure
drops of 80 kN/m2 and 40 kN/m2, respectively. If the water flow rate
required is 150 L/s, calculate the saving in pump power for the condenser
water pump if the chiller with the lower pressure drop is used instead of
the chiller with the higher pressure drop.
)A L N
P1 _ 55 kW
N1 _ 1400 rpm
N2 _ 1120 rpm Q2 = Q1 x (N2/N1) = 120 x (1120/1400) = 96 L/s
P2 = P1 x (N2/N1)3 = 55 x (1120/1400)3 = 28 kW
W W) P H
P A L E S F Affinity Law F F (power & speed3).

Flow rate 10 L/s F L W F L VSD P
P
From pump affinity laws, the pump speed can be reduced to give the design flow
as follows:
E P M) Pump Ma
p
C C conden pressure
C A N C B N C L Chiller A C B

S H D H S H
D H C ) (rec ) )

) - - ) = H
N S H -)
(Static Head, ft) = (Discharge Head, ft) - (Suction Head, ft) = (125, ft) - (-5, ft) = 130ft

the pump datum poi P S H )
(Static Head, ft) = (Discharge Head, ft) - (Suction Head, ft) = (25, ft) - (+15, ft) = 10ft
C T C W P S H
C W P C B C H B
P S H ) DH
(Static Head, ft) = (Discharge Head, ft) - (Suction Head, ft) = (15, ft) - (+10, ft) = 5ft

and reducing )
Major Head Loss - head loss or pressure loss - due to friction in pipes and ducts.
Minor Head Loss - head loss or pressure loss - due to components as valves, bends, tees and the like in the pipe or duct system
Friction losses
Change in Flow Direction
F H = F M H L D Minor Head Losses
)
Dynamic M H L) TDH
SUCTION HEAD & TDH PROBLEMS
EXAMPLE: The influent pump discharges into a channel where the liquid level is 14 feet above the pump datum line. The pump draws its suction from a wet well, whose water surface is 5 feet above the pump. The friction head is 5.6 ft.
Determine the Static Head, in feet.
Static Head, ft = (Discharge Elev, ft) - (Suction Elev., ft)
Static Head, ft = (14 ft) - (5 ft) = 9 ft Static Head
Calculate the Total Dynamic Head (TDH), in feet.
TDH = (Static Head, ft) + (Friction Head, ft)
TDH = (9 ft) + ( 5.6 ft) = 14.6 ft TDH
PROBLEM: The influent pump discharges into the grit chamber, where the liquid level is 8 feet above the pump datum line. The pump draws its suction from a wet well, whose water surface is 2 feet above the pump. The friction head is estimated at 2.5 ft. Determine the Static Head, in feet. (Ans: 6 ft) Calculate the Total Dynamic Head (TDH), in feet. (Ans: 8.5 ft)
PROBLEM: The polymer makeup pump discharges into the solution tank, where the liquid level is 8 feet above the pump datum line. The pump draws its suction from a sump, whose water surface is 2 feet above the pump. The friction head is 1.5 ft. Determine the Static Head, in feet. (Ans: 6 ft) Calculate the Total Dynamic Head (TDH), in feet. (Ans: 7.5 ft)
SUCTION LIFT & TDH PROBLEMS
EXAMPLE: The influent pump discharges into a channel where the liquid level is 14 feet above the pump datum line. The pump draws its suction from a wet well, whose water surface is 3 feet BELOW the pump. The friction head is 6 ft.
Determine the Static Head, in feet.
Static Head, ft = (Discharge Elev, ft) - (Suction Elev., ft)
Static Head, ft = (14 ft) - (-3 ft) = 17 ft Static Head
Calculate the Total Dynamic Head (TDH), in feet.
TDH = (17 ft) + (6 ft) = 23 ft TDH
PROBLEM: The influent pump discharges into the grit chamber, where the liquid level is 8 feet above the pump datum line. The pump draws its suction from a wet well, whose water surface is 2 feet below the pump. The friction head is estimated at 2.5 ft. Determine the Static Head, in feet. (Ans: 10 ft) Calculate the Total Head (TDH), in feet. (Ans: 12.5 ft)
PROBLEM: The raw water pump discharges into the sand trap, where the liquid level is 18 feet above the pump datum line. The pump draws its suction from a sump in the reservoir, whose water surface is 2 feet below the pump. The friction head is estimated at 4 ft. Determine the Static Head, in feet. (Ans: 20 ft) Calculate the Total Dynamic Head (TDH), in feet. (Ans: 24 ft)
Pump Sizing Example ( Open System)
losses
Friction Losses P S F V) P L P
Dynamic losses changes in flow area (V)
F L ASHRAE H - F F C F
Water Flow rate = 450 CMH = L
P D ) P P D P L = P = P Friction Losses.
L - V F R V C V )
.
Number of Valve = 6 ( Gate Valve, fully open)
Number of Stariner = 2 ( take pressuer loss to be the same as for fully open globe valve)
Number of pipe bends = 15 ( 90 Degree Standard elbow)
Number of Valve = 6 ( Gate Valve, fully open)
K
Pressure drop for fully opened gate valve = K x density of water x Velocity x Velociyt / 2
Pressure drop for fully opened gate valve = 0.05 x 1000 kg/ cu m x 1.7 m/s x 1.7 m/s/ 2 =
72.25Pa/Gate Valve
For 6 Valves x 72.25Pa/Gate Valve = 433.50Pa
Number of Stariner = 2 ( take pressuer loss to be the same as for fully open globe valve)
K
Pressure drop for fully opened globe valve = K x density of water x Velocity x Velociyt / 2
Pressure drop for fully opened globe valve = 5.7 x 1000 kg/ cu m x 1.7 m/s x 1.7 m/s/ 2 =
8236.5Pa
Number of pipe bends = 15 ( 90 Degree Standard elbow)
K
Pressure drop for fully opened globe valve = K x density of water x Velocity x Velociyt / 2
Pressure drop for fully opened globe valve = 5.7 x 1000 kg/ cu m x 1.7 m/s x 1.7 m/s/ 2 =
346.8 Pa/ elbow

Pressuer losses across heat exchanger= High of Water (H) x Desity of Water x G = 5m x 1000 x
9.8m = 49,000 Pa
Cooling Tower Static Head = High of Water (H) x Desity of Water x G = 4m x 1000 x 9.8m =
39,200 Pa

) C D P D F ) P P) P D P D
) P
L C
Content
Content
Content
Content
Content
Content
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Content

Q S
QS Sectio BS Q S ) D C M Q S QS E
QS
QS Engineeri M C ) QS
E )
E E

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QS E
GreenGirl
M C - S-C - S -
N T U NTU ) C ) NTU MOE M E ) C Consultan

C T P P C M C F L F L BCA B C A )
M C M C
S C A C A C S-C
S-C M C N S C C

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M C M C Main Contract C S C C

P C N S-C
M C S C S C S - S C S
QS
) S C QS
GreenGirl
QS S C QS
Q S WK
A quantity surveyor (QS) is a professional person working within the construction industry. The
role of the QS, in general terms, is to manage and control contracts and costs within
construction projects.
Sub Contractor QS
S C M C M C S S C
S C M C QS S C Q
P S C QS P Q Q M C E F
S C Q Q S C
S C
Q R S C QS Q P Q QS
M C Q R S C QS P Q P
R P P M C
P Q R

R S E
S S
S Q S A D ? P S
Q M C P M Q
S M C MC M C Q S C QS M C S
Quo M C P M C L A LA ) ISO

P P M C C C M C
C S C QS C
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M C QS S C QS

C Main C M C C A P ? M C QS
C S C C P C S P M

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M C C QS F A
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