University of Perpetual Help System - DALTA

Paciano Rizal,Calamba City Laguna

College of Engineering

MECHANICAL ENGINEERING DEPARTMENT

AIRCONDITIONING AND VENTILATION SYSTEM:

PROPOSED DESIGN OF A CENTRALIZED

AIRCONDITIONING SYSTEM

Barosa, Romart B.
BSME – V

ENGR. RIZAL M. MOSQUERA

INSTRUCTOR

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INTRODUCTION:

      Air-Conditioning is a process which conditions air in a confined space into the

desired level of comfort compared from outside condition. Most of Air-
Conditioning System in service comfort cooling especially in places where the
outside temperature is very high for human comfort. Air-Conditioning does not
only cool a confined space but it also dehumidifies or filter air supplied to the
space. Thus it provides comfort cooling for occupants of the space.
Basically there are 4 components of an Air-Conditioning Unit: Evaporator,
which receives heat from the space; Compressor, which is the heart of the
system it compresses and pumps the refrigerant; Condenser, which rejects heat;
Expansion Valve, used to meter the flow of liquid refrigerant entering the
evaporator at a rate that matches the amount of refrigerant being boiled off in the
evaporator.
There are also 2 Units that play an important part in Air-Conditioning System.
The Air Handling Unit (AHU) and the Cooling Tower. Air Handling Unit, it is the
unit used with the system to minimize the cost especially when using ducts. The
Air Handling Unit controls, mix, condition, and may filter the air flow. It supplies
conditioned air to each room via ducts. On the other hand, Cooling Tower, is
another unit which helps in raising the efficiency of the system. It helps the
condenser to reject heat faster by heat exchanger. The Cooling Tower is a tower
like structure, where the hot water from the heat exchanger of the condenser is
sprinkle down to the raisin of the tower. It is pumped again to the condenser. Air
is blown toward the sprinkled water to removed heat. Make-up water is added
due to evaporation.
Air conditioning systems using centrally located equipment provided only
heated (Tempered) air for comfort and ventilation using relatively simple
ductwork and control. The addition of cooling, dehumidification, and

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humidification equipment allowed year-round comfort to be more attainable in all
climates. By dividing the condition spaces into zones with individual thermostat
controls, better comfort was possible, even where heating and cooling
requirements were not uniform from one part of a building to another. This led to
the need for more sophisticated equipment and controls. Building owners and
occupants have also become more sophisticated and more demanding of the
HVAC systems. In recent years design has been strongly influenced by
increasing emphasis on Indoor Air Quality (IAQ), Energy Conversion,
Environmental Effects, Safety, and Economics.
      The use of digital computers has facilitated many of the advances made in
the HVAC industry. Computers have allowed the design of more complex and
intricate but more reliable components; programs have reduced the time required
for determining building requirements and have permitted better and faster
design of duct and piping systems. Complex HVAC systems can be controlled
using Direct Digital Control (DDC) systems which can be integrated into building
management system to provide monitoring and almost any control sequence by
the owners.
Cooling Load Calculation:
      Cooling load calculation for air conditioning system design are mainly used to
determine the volume flow rate of the air system as well as the coil and
refrigeration load of the equipment to size the HVAC & R equipment and to
provide the inputs to the system for the energy use calculations in order to select
optimal design alternatives. Cooling load usually can be classified into two
categories; external and internal.
External Cooling Load – The loads are formed because of heat gain in the
conditioned space from external sources through the building envelope or
building shell and the partition walls. Sources of external loads include the
following cooling loads;

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 1. Heat gain entering from the exterior walls and roofs.
2. Solar heat gain transmitted through the fenestrations.
3. Conductive heat gain coming through the fenestrations.
4. Heat gain entering from the partition walls and interior doors.
5. Infiltration of outdoor air into the conditioned space.
Internal Cooling Loads – These loads are formed by the release of sensible and
latent heat sources inside the conditioned space. These sources contribute
internal cooling loads;

1. People
2. Electric lights
3. Equipment and appliances
   If moisture transfers from the building structures and the finishing are excluded,
only infiltrated air, occupants, equipment, and appliances have both sensible and
latent cooling loads. The remaining components have only sensible cooling
loads.

Wall Gain Load:

      The wall gain load, sometimes called the wall leakage load, is a measure of
the heat flow rate by conduction through the walls of the refrigerated space from
outside to inside. The wall gain load is a measure of the heat flow rate by the
conduction through the walls of the refrigerated space from the outside to the
inside. Since there is no perfect insulation, there is always a certain amount of
heat passing from the outside to the inside whenever the inside temperature is
below that of the outside. The wall gain load is common to all refrigeration

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application and is ordinarily a considerable part of total cooling load. Also include
the flooring, roofing and doors.

FACTORS DETERMINING THE WALL GAIN LOAD:

      The quantity of heat transmitted through the walls of a refrigerated space/unit

of time is the function of three factors whose relationship is expressed in the
following equation;
Q = UATD

Where:  Q = the quantity of heat transferred, Btu/Hr.

            A = the outside surface area of the wall, ft².
            U = the overall coefficient of heat transmission, Btu/hr-ft²-°F
            TD = the temperature differential across the wall, °F

DESIGN CONDITIONS:
FORMULA:
Heat transfer throughout the design building using formula:
Q = UATD
Where:
Q = heat transferred / connected across a surface through a wall
thickness.
A = area which heat flows.
TD = temperature difference passing through the wall.
U = overall heat transfer coefficient.

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 1
U=
R

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 Chapter 1

Lot Area= 340 sq. m

Roof Area= 340 sq. m
Wall Area 1 & 3= 20m x 3m =60m2
Wall Area 2 & 4= 17m x 3m = 51m2
Door Area= 2m x 2.5m = 5m2 4m x 2.5m = 10m2
Single Door Double Door
Glass Window Area= 1.25m x 2m =2.5m2
I. Design Consideration
Outside Condition
To =38° =311k
RH =87%
Inside Condition

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Ti =21° =294k
RH= 53%
Ho=22.6979
H1=9.3629

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 Chapter 2
II. Structural Cooling Load
 Wall
Materials K(thermal Conductivity) T(Thickness)
Cement Portland 0.29 0.127
Hollow Block 0.463 0.2
Steel 45 0.015
Cement Portland 0.29 0.127

Paint 1.38 6.35x10-6

I
U=
R
1
¿
1 x 1
+Σ +
ho k h1

1
¿
1 .127 0.2 0.015 0.127 6.35 x 10−6 1
22.6979
+ (
+
.29 0.463
+
45
+)(
0.29
+ )(
1.38
+ )
9.3629

1
¿
0.0440569391+ ( 0.437931035+ 0.4319654428 )+ ( 3.33 x 10− 4+ 0.4379310345 ) +4.601449275 x 10−6 +0.10
¿

1
¿
0.0440569391+ 0.8698964778+ 0.4382640345+ 0.1068091171
w
U =0.5389269582 .K
m2
Q 1&3¿ UATD

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¿ 0.5389269582(60)(311−294)
Q1∧3=549.7054973 W
Q 2∧4 =UATD

¿ 0.5389269582 (51 ) ( 311−294 )

Q2∧4 =467.2496727 W

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  Ground
Materials Thickness (x) Thermal Conductivity
(k)
Tiles 0.012 0.25
Portland Cement 0.1524 0.29
Gravel 0.1016 0.7
Soil 3 1.5
I
U=
R
1
¿
1 x 1
+Σ +
ho k h1
1
¿
1 0.012 0.1524 0.1016 3 1
22.6979
+( .25
+
0.29 )(
+
0.7
+ + )
1.5 9.3629
1
¿
0.04405693919+ ( 0.048+0.5255172414 ) + ( 0.1451428571+2 ) +0.1068045157
w
U =0.3484901512 .k
m2
Q=UATD
¿ 0.3484901512 (340 )( 311−294 )
Q=2014.273074W

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  Second and Third Floor
Materials Thickness (x) Thermal Conductivity
(k)
Tiles 0.012 0.25
Portland Cement 0.1524 0.29
Steel Beams 0.25 50.2
Concrete Dense 0.1016 1.5
I
U=
R
1
¿
1 x 1
+Σ +
ho k h1
1
¿
1 0.012 0.1524 0.25 0.1016 1
22.6979 (
+
0.25
+
0.29
+ )(
50.2
+
1.5
+ )
9.3629

¿ 0.0440569319+ ( 0.048+0.5255172414 ) + ( 4.980079681 x 10−3 + 0.06777333 )+ 0.1068045157

w
U =0.83990966 .k
m2
Q=UATD
¿ 0.83990966 ( 340 ) ( 311−294 )
Q=4854.677869W

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  Roof
Thermal Conductivity
Materials Thickness (x)
(k)
Portland Cement 0.1524 0.29
Steel Beams 0.25 50.2
Concrete Dense 0.1016 1.5
I
U=
R
1
¿
1 x 1
+Σ +
ho k h1
1
¿
0.1524 0.25 0.1016
0.4405693919+ ( +
0.29 50.2
+
1.5 )
+ 0.1068045157

w
U =0.872870586 .k
m2
Q=UATD
¿ 0.872870586 ( 340 ) ( 311−294 )
Q=5045.191987 W

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  Single Door
Thermal Conductivity
Materials Thickness (x)
(k)
Oak wood 0.05715 0.17
I
U=
R
1
¿
1 x 1
+Σ +
ho k h1
1
¿
0.05715
0.4405693919+( 0.17 )
+0.1068045157

w
U =1.131800518 .k
m2
Q=UATD

¿ 1.131800518(5 m2)(311−294)
Q=96.20304403W

 Double Door
Materials Thickness (x) Thermal Conductivity
Oak Wood 0.05081 0.17
I
U=
R
1
¿
1 x 1
+Σ +
ho k h1
1
¿
0.05081
0.4405693919+( 0.17 )
+( 0.1068045157)

w
U =1.811675158 .k
m2

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 Q=UATD

¿ 1.811675158 ( 10 m 2 ) ( 311−294 )
Q=200.8847768W

 Fire Exit
Materials Thickness (x) Thermal Conductivity
(k)
Steel 0.05 50.2
I
U=
R
1
¿
1 x 1
+Σ +
ho k h1
1
¿
0.05
0.4405693919+ ( )
50.2
+0.1068043157

w
U =1.823586519 .k
m2
Q=UATD
¿ 1.823586519 (5 )( 311−294 )
Q=155.0048541 W

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  Glass
Area 1.5 m x 1.5 m = 2.25m 2

6 windows (2.5 m)
North Side
15m2

6 windows (2.5 m)
South Side
15m 2

Materials Thickness (x) Thermal Conductivity

(k)
Glass 0.00635 0.96
I
U=
R
1
¿
1 x 1
+Σ +
ho k h1
1
¿
0.00635
0.4405693919+ (0.96 )
+0.1068043157

w
U =6.350172454 .k
m2
Q=UATD
¿ 6.350172454 ( 15 ) ( 311−294 )
Q=1619.293976 W

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Heat Gain per Occupant
To know the no. of people use the formula:
lot area
× no . of floors
area per person

Area per person is based on The Engineering ToolBox. Since my design is a library I will
use 10 persons per area.

340 m2
= ×3
10
¿ 102 person
Heat gain

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Mechanical Engineering |
 metabollic
QS= Number of people x human heat gain x
100
340 82
=34 ( 3 ) x 215 x
10 100
QS=17982.6 W

Sensible Heat Latent Heat Sensible Heat Latent Heat

78 52 86 44

48
Ql = 34 (3) X 185 x
100
Ql =9057.6 W
Chapter III
III. Infiltration

QV = 0.172 x (Area of Doors + Area of Windows) √ 1+ ( ¿−ti )

Qv = 0.172 x (45m 2 +30 m2 ) √ 1+ ( 311−294 )

m3
QV = 54.73006486
s
o For Wind Action
Qv
C=
A

54.7300648m 3 /s
C=
75 m2
1m 3600 s
¿ 0.7292341973 x x
1000 m 1 hr
C=2.627043113 km/hr

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 C
Vratio=
wind velocity
2.627043113 km/hr
¿
19.8
Vratio=0.1326789451

QDoor= Areadoor (Vratio)

= 45 (0.132678945)
Qdoor=5.970552531
1000 L 1 m
=5.970552531 m 2 x x
1 m2 60 s
l
Qdoor=99.50920884
s
o Infiltration Sesible

Qsi =1.23(Qdoor)( ∆T )
Qsi =1.23 (99.50920884) (311-294)
Qsi = 2080.737557 W
Qli = 3000( QDoor )(Wo – Wi)
Psat = P (RH)
Psat
WO¿
Patm−psat

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Interpolate the saturation pressure between 20°C and 22°C

22° 2.642
21° X

20° 2.337

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22−21 2.642−x
=
22−20 2.642−2.337
x=2.4895 KPa

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 Saturation pressure @38℃ =6.624 KPa
Psat
WO¿
Patm−psat
0.85(6.624 KPa)
Wo=0.622
101.325−(0.85)(6.624)
kj
Wo−0.03659672333
kg
(0.5)(2.4895)
Wi=0.622
101.325−( 0.85 )( 6.624 )
kj
Wi=0.00773613675
kg
Infiltration Latent Heat
Qli=3000 ( Qdoor ) ( Wo−Wi )
¿ 3000 ( 99.50920884 ) ( 0.0369672333−0.00773613675)
kg
Qli=8725.394291
m3
Ventilation

kg
Air Density ¿ 1.4048
m3
Total Occupancy = 102 person
l
Occupancy Air Required ¿ 10
s

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Vf =( total occupancy ) ( occupancy air required )

( LS )
Vf =( 102 ) 10

l
Vf =1020
S

m3
Vf =1.02
s
Interpolate the 𝜌air between 20°C and 30°C
Dair Pair
20°C 1.324 kg/m3
21°C x
30°C 2.297 kg/m3

20−21 1.324−x
x= =
20−30 1.324−1.297
kg
x=1.3213
m3
Computing the m
m= pv

kg m3
m=(1.3213 )(1.02 )
m3 s
kg
m=1.347726
s

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SH Load
Qs=¿ ṁCp ∆ T
kg kj
(
Qs= 1.347726
s )(
1.0067
kg ℃ )
(311 ℃−294 ℃ )

Qs=23.06484799kW
Qs=23064.8799W
Mass water Removal
Mwr=m ( Wo−Wi )
¿ 1.347726 ( 0.03659672333−0.0077361365 )
kg
Mwr=0.038896116291
s
Interpolate the enthalpy of liquid and gas between 20℃ and 22℃ (from table 10)

Temperature Liquid Gas

20℃ 83.96 2538.2
21℃ x x
22℃ 92.23 2541.8

22−21 92.23−x
x= =
22−20 92.93 .83 .86
x=88.045
22−21 2541.8−x
y= =
22−21 2541.8−2538.2
y=2540

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Computing the Hfg
Hfg=Hg−Hf
Hfg=2540−88.045
kj
Hg=2541.935
kg
Latent Heat Computation
Ql=masswaterremoval ( Hfg )
Ql=0.038896116291 ( 2541.955 )
Ql=95.37152682 kW
Chapter IV
IV. Fenestration Cooling Load

North South
10 123 41
14 x y
20 88 38

Interpolate
20−14 88−x
x= =
20−10 88−123
x=109
20−14 38− y
y= =
20−10 38−41
y=39.8

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North South
=109 w /m2 (15 m2) =39.8 w/m2 (15 m 2)
North =1635 w South=597 w

SHGT= North + South

SHGT= 1635 w + 597 w
SHGT =2232 w
Chapter V
V. Miscellaneous
Appliance Quantity Watts Total
Computer 60 250 3750
Wifi-Router 5 24 120
Ip Camera 18 30 540
Flourescent Light 300 100 30000
Vacuum Cleaner 15 1440 21600
Printer 30 50 1500
Amplifier 10 800 8000
Speaker 15 200 3000
Shredder 10 200 2000
Coffer maker 15 1500 22500
Flat Screen Tv 15 240 3600
Electric Kettle 10 1500 15000
Dehumidifier 24 785 18840
Laptop 60 80 4800
Projector 30 80 2400
Toaster 10 1400 14000
Water Dispenser 15 600 9000
Microwave oven 10 1100 11000
Scanner 30 50 1500
Espresso Coffee 9 1300 11700

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 Machine
Incandescent
300 100 30000
Lamp
Total Watts 220,850 W

Structural Load Watts (w)

Wall 1&3 549.7554973
Wall 2&4 467.2496727
Ground 2014.273014
2 and 3rd Floor
nd
4854.677869
Roof 5045.191987
Single Door 96.20304403
Double Door 200.8447768
Fire Exit 155.0048541
Glass 1619.293976
Total Watts 15002.49469

Grand Total Cooling Load

Sensible Heat (w)

Heat Gain 17982.6
Infiltration 2080.737557
Ventilation 23064.84799
Fenestration 2232
Miscellaneous 220850
Structural Load 15002.49469
Total of Sensible Heat (w) 281212.6802

Latent Heat (w)

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 Heat Gain 9037.6
Infiltration 8725.394291
Ventilation 95371.52682
Total of Latent Heat (w) 113134.5211

Computing the Sensible Heat Ratior using the Formula:

Qs
SHR=
Qs+Ql
281212.6802
SHR=
281212.6802+113134.5211
SHR=0.713093597

Plot the Computed SHR to the Psychrometric Chart to get the Supply
Temperature:

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T3=11℃

Getting the Grand Total Cooling Load

Subtotal=Qs+Ql

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¿ 281212.6802+113134.5211
Subtotal=39437.2013 w
Subtotal=394.3472013 kW
Safety Factor of 10%
Safety Factor of 10 %=Subtotal ( 10 % )
Safety Factor of 10 %=39437.2013 w ( 0.10 )
Safety Factor of 10 %=39434.72013 w
Safety Factor of 10 %=39.43472013 kW

Grand Total Cooling Load

Grand Total Cooling Load=Subtotal + Safety Factor of 10 %
Grand Total Cooling Load=394347.2013+39434.72013
Grand Total Cooling Load=433781.9214 w
Grand Total Cooling Load=433.7819214 kW
Design Cooling Load
Grand Total Cooling Load
DCL=
Working Hours
Operating Hours
433.7819214 kW
DCL=
8 hours
12 hours
1 TOR
DCL=650.6728821kW ( )
3.5168
DCL=185.0184492 TOR

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(Schematic Diagram of Supply return system)

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Finding the Mass Flowrate
Qs
Ms=
Cp ( t 4−t 3 )
281.2126802
Ms=
kj
(1.0062 )(21 ℃−11 ℃ )
kg . k
kg
Ms=27.94799048
s
Finding the Volume of Air Supply
P 1V 1=mRT
V1 RT
=
M 1 P−Pv
kj
R 1=0.287
kg ℃
T =inside temp .∈k
P=101.235 KPa
Pv=Saturation Pressureof inside Temp .
kj
(0.287 )(294 ℃)
kg ℃
¿
(101.325 KPa−2.4895 )

V1 m3
=0.8537215879
M1 kg
Finding the value of Vs

Vs=Ms ( MV 11 )
kg m3
Vs=27.94779048
s(0.8537215879
kg )

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 m3
Vs=23.85980281
s
Computing the Flowrate on the Outside

m3
Vo=1.02
s
Vr=Vs−Vo
Vr=23.85980281−1.02

m3
Vr=22.83980281
s
Computation for getting the Percentage of Vo and Percentage of Vr
Vo
%Vo= X 100
Vs
1.02
%Vo= x 100
23.85980281
%Vo=4.274972464 %
Vr
%Vr= x 100
Vs
22.83980281
%Vr= x 100
23.85980281
%Vr=95.72502754 %
Desired Temperature of the building =21℃

Computing the Psat of the desired Temperature of the building.

Psat of T 4=Saturation Pressure x Relative Humidity

Psat of T 4=2.4895 ( 53 % )
Psat of T 4=1.3139435 kPa

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 Psat
W 4=0.622( )
Pt −Psat
1.3139435
¿ 0.622( )
101.325−1.3139435
kj
W 4=0.008171825052
kg
Computing the Psat of the supply temperature of the building.
Psat of T 3=Saturation Pressure x Relative Humidity
Psat of T 3=1.3142 ( 53 % )
Psat of T 3=0.696526 kPa
Psat
W 3=0.622( )
Pt −Psat
0.696526
W 3=0.622( )
101.325−0.696526
kj
W 3=0.004305333816
kg
Computing the H4
h 4=CpT 4+W 4 hg
h 4= (1.0062 ) ( 21 ) +8.17182052 ( 2540 )
kj
h 4=41.89713563
kg
Computing the Ql
Ql=Ms ( 2500 ) ( W 4−W 3 )
Ql=27.94799048 ( 2500 )( 0.008171825052−0.004305333816 )
Ql=270.1516506 w

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Computing the H3
h 3=Cpt 3+ W 3 hg
¿ ( 1.0067 ) ( 11 ) ( 0.004305333816 )( 2521.75 )
h 3=21.93067555

Get the V3 and H3 using the pyschrometric chart.

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V3=0.818
H3=36.8
Computing the mass of outside air
Mo=0.45(Ms)
Mo=0.45(27.94799048)
Mo=12.57659572kg/s
Computing the Volume Flowrate, V3=0.818
Vo=Mo ( V 3 )
¿ 12.576559572(0.818)

m3
Vo=10.2876553
s
POINT 2
Plot H2 in the psychometric chart
H2=36.2
Mass flow rate of recirculated air
Mr=Ms−Mo
Mr=27.94799048−12.57659572
kg
Mr=15.37139476
S
WO=W1
Kj
W1¿ 0.03659672333
kg

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Computing H1 and H5
h 1=Cpt +WoHg
kj
(
h 1= 1.0062
kg )
. k ( 38 )+ ( 0.003659672333 )( 2570.8 )

kj
h 1=47.64388563 .k
kg
( Mr ) ( h 4 )
h 5=( mo )( h 1 ) +
Ms
( 12.57659572 ) ( 47.64388563 ) +(15.37139776)( 41.89713563)
h 5=
27.94799048
kj
h 5=44.4844095
kg
List of H1 to H5
kj
H1=47.64388563
kg
kj
H2=36.2
kg
kj
H3=36.8
kg
kj
H4=41.89713563
kg
kj
H5=44.4833095
kg

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Refrigerating the Capacity of Air Handling Unit (AHU)

QAHU =Ms ( h 5−h3 ) ( 128 )

QAHU =27.94799048 ( 44.4833095−36.2 ) ( 128 )
QAHU =231.501855 ( 128 )
QAHU =154.33457 kW
Q EVAP=Q AHU + Q AHU ( Leakage )

Q EVAP=154.33457+154.33457 ( 0.2 )
Q EVAP=185.201484 kW

Air handling unit

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Cooling Capacity =154.33457 kW
Air Flow Rate= 25000

SELECTION OF REFRIGERATION SYSTEM COMPONENTS

The Evaporator

The purpose of the evaporator is to remove the unwanted heat from the product,

via the liquid refrigerant. The liquid refrigerant contained within the evaporator is boiling

at a low-pressure. The level of this pressure can be determined by two factors:

 The rate at which the low-pressure vapor is removed from the

evaporator by the compressor.

 The rate at which the heat is absorbed from the product to the

liquid refrigerant in the evaporator.

The Compressor

The purpose of the compressor is to draw the low-temperature, low-pressure

vapor from the evaporator via the suction line. Once drawn, the vapor is compressed.

When vapor is compressed, it rises the temperature. Therefore, the compressor transforms

the vapor from low-temperature vapor to a high-temperature vapor, in turn increasing the

pressure. The vapor is then released from the compressor to the discharge line.

The Condenser

The purpose of the condenser is to extract heat from the refrigerant to the

outside air. The condenser is usually installed on the reinforced roof of the building,

Mechanical Engineering |
which enables the transfer of heat. Fans mounted above the condenser unit are used to

draw air through the condenser coils.

The Expansion Valve

Within the refrigeration system, the expansion valve is located at the end of the

liquid line, before the evaporator. The high-pressure liquid reaches the expansion valve,

having come from the condenser. The valve then reduces the pressure, the temperature of

the refrigerant also decreases to a level below the surrounding air. The low-pressure, low-

temperature liquid is then pumped at the evaporator.

Mechanical Engineering |
Heat Supplied=( Ms ) ( h 3−h2 )
kg
¿( 27.94799048 )¿
s
¿ 16.76879429 kW
FanCapacity=msv 3
kg
(
¿ 27.94799048
s )
( 0818 m3 / s )

m3
¿ 22.86145621
s
Chapter VI
VI. Ducting

Mechanical Engineering |
 Size 8
Neck Velocity 1200
Velocity Pressure 0.090
Total Pressure 0.292
Flow Rate 420

Supply Velocity
No .of Ahu=
( Capacity of Ahu air Flow rate )

23.85980281m 3 / s(3600 s)
No .ofAhu=
25000
No .of Ahu=3.435811605
3 Ahu
3 Ahu
¿
3 floor
1 Ahu
No .of Ahu=
floor
Flow Rate (Vs)
No .of Diffuser =
Flow Rate Diffuser
23.8590281
¿
1
420( ) ¿ ¿
60
No .of Diffuser =120.2791475
Safety Factor 11 %=120.2791475 (1.1 )
132.3070623
Safety Factor 11 %=
3 floor
No .of Diffuser =44.10235408 diffuser

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Computation for Duct Loss:

𝐷𝐿 = 𝐷𝑢𝑐𝑡 𝐿𝑒𝑛𝑔𝑡ℎ 𝑥 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐿𝑜𝑠𝑠

Computation for Fittings Loss:

𝐶𝑜𝜌 𝑉𝑜𝑒
𝐹=
2

Computation for Velocity Ratio:

𝑉𝑟𝑎𝑡𝑖𝑜 = 𝑉𝑜𝑢𝑡/𝑉𝑖𝑛

Computation for Area Ratio:

Mechanical Engineering |
 𝐴𝑟𝑎𝑡𝑖𝑜 = 𝐷𝑜𝑢𝑡2 /𝐷𝑖𝑛2

Calculation for Pressure Drop.

A−DB 2=DL A + FL A−B 1+ FL B 1−B 2+ FL B 2− DD 2

¿ 5.6+0.48+1.67 +2.81

¿ 10.54 Pa

A−DD 2=DL A + FL A−C + DLC + FLC−D 1 + FL D 1− D 2+ FLD 2−DD 2

¿ 5.6+3.96+3.85+ 0.46+1.67+2.81

¿ 18.35 Pa

A−DF 2=DL A + FL A −C + DL C + FLC −E + DL E + FL E− F 1 + FL F 1− F 2 + FL F 2− DF 2

¿ 5.6+3.96+3.85+ 3.849+ 3.85+0.46+1.67+ 2.81

¿ 26.049 Pa

¿ 5.6+3.96+3.85+ 3.849+ 3.85+2.92+3.85+0.42+1.67+2.81

¿ 32.779 Pa

Mechanical Engineering |
A−DJ 2=DL A + FL A −C + DL C + FLC− E + DL E + FLE−G + DL G + FLG −I + DL I + FL I−J 1 + FLJ 1−J 2 + FLJ 2−DJ 2
¿ 5.6+3.96+3.85+ 3.849+ 3.85+2.92+3.85+0.42+3.85+3.46+1.67+ 2.81
¿ 40.089 Pa
PRESSURE DROP TOTAL= A−DB2+ A−DD 2+ A−DF 2+ A−DH 2+ A−DJ 2
¿ 10.54+18.35+26.049+ 32.779+ 40.089
¿ 127.807 Pa

Steam table

T e =22℃

T ahu =16 ℃

Mechanical Engineering |
T refrigerant=10 ℃

T chiller =19℃

kJ
h1 =408.03
kgK

kJ
h2 =426
kgK

kJ
h3 =250
kgK

h 4=h 3

Evaporator

Assume:

T e =0 ℃

T l=−7 ℃

T r=−17 ℃

( T e −T r )−( T l−T r )
LMTD=
( T e −T r )
ln
( T l−T r )

LMTD= ( 0 ℃−(−17 ℃) )−¿ ¿

LMTD=13.19 ℃

LMTD=55.74 ℉

Mechanical Engineering |
 Fig 3.1 T-S Diagram of Evaporator

Calculating the Mass flow rate of the Refrigerant , m:

Mechanical Engineering |
Q Evaporator =1.1 ( DCL )

¿ 1.1 (150.1024 )

¿ 165.1126 k

BTU 3600 s
Q Evaporator =165.1126 kW × ×
1.055 kJ 1 hr

BTU
Q Evaporator =563417.4028
hr

Q Evaporator
Evaporator Capacity=
LMTD

BTU
563417.4028
hr
Evaporator Capacity=
55.74 ℉

BTU
Evaporator Capacity=10107.9548
hr −℉

QEvaporator
m=
(h1−h 4)

165.011 kW
m=
kJ kJ
408.03 −250
kg kg

kg
Mass flow rate of the Refrigerant , m=1.04
s

Condenser Design

Mechanical Engineering |
QCondenser=m(h2−h3 )

kg kJ kJ
Q Condenser =01.04 (426 −250 )
s kg kg

Q Condenser =183.04 kW

Assume:

T e =23 ℃

T l=47 ℃

T r=55 ℃

( T r −T e )−( T r −T l )
LMTD=
( T r −T e )
ln
( T r −T l )

( 55 ℃−23 ℃ )−( 55 ℃−47 ℃ )

LMTD=
( 55 ℃−21 ℃ )
ln
( 55 ℃−47 ℃ )

LMTD=17.31 ℃

LMTD=63.16℉

Mechanical Engineering |
 T-S Diagram of Condenser

BTU 3600 s
Q Condenser=183.04 × ×
1.055 kJ 1 hr

BTU
Q Condenser=624591.4692
hr

Q Condenser
Condenser Capacity=
LMTD

BTU
624591.4692
hr
Condenser Capacity=
63.16 ℉

BTU
Condenser Capacity=9889.0353
hr−℉

Compressor Design

Findingthe Work of Compressor :

W Compressor =m(h2−h1)

kg kJ kJ
W Compressor =1.04 (426 −408.03 )
s kg kg

W Compressor =18.69 kW

W Compressor =25.06 hp

W Compressor =5.35TOR

Mechanical Engineering |
Findingthe Coefficient of Performance(COP) of the Compressor :

QEvaporator
COP=
W Compressor

165.1126 kW
COP=
18.69 kW

COP=8.83

Design Temperature:

Evaporator Temperature ,T e =−18 ℃

Condenser Temperature , T c =48 ℃

Expansion Valve Design

Condenser Temperature , T c =48 ℃

Refrigerant ¿ be used =R−717

Pressure at Condenser , Pc =1932.2 KPa

14.7 psi
1932.2 KPa ×
101.325 KPa

Pressure at Condenser , Pc =280.32 psi

Evaporator Temperature ,T e =−18 ℃

Refrigerant ¿ be used =R−717

Mechanical Engineering |
Pressure at Evaporator , Pe =208.26 KPa

14.7 psi
208.26 KPa ×
101.325 KPa

Pressure at Evaporator , Pe =30.21 psi

Pressure Difference acrossValves :

∆ PValves =P c −P e

∆ PValves =280.32 psi−30.21 psi

∆ PValves =250.11 psi

Chapter VII
VII. Pipings And Calculation

LAYOUT OF THE BUILDING INCLUDING DUCTING AND PIPING

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