“District Cooling System”

By
PM. IR. DR. MUHAMMAD IRWANTO
Faculty of Electrical Engineering Technology
 Concept of District Cooling System
District cooling system is thermal energy networks that
distribute chilled water through insulated pipes to serve
commercial, residential, institutional, and industrial energy
needs for space cooling, and industrial purposes.

The district cooling system has three basic components:

❑ The cooling resource (Central cooling Plant)
❑ Transmission and distribution systems
❑ User or customer

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Concept of District Cooling System

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 Concept of District Cooling System
Central cooling Plant
Chilled water is typically generated at the central chiller plant by
compressor driven by chillers, absorption chillers or other sources
like ambient cooling or “free cooling” from deep lakes, rivers,
aquifers or oceans.

Sea water condensers or fresh water cooling towers can be utilized

to reject waste heat from the central chillers.

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 Concept of District Cooling System
Distribution Network
District chilled water is
distributed from the cooling
source(s) to the user
stations through supply
pipes and is returned after
extracting heat from the
building’s secondary chilled
water systems. Pumps
distribute the chilled water
by creating a pressure
differential between the
supply and return lines.

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 Concept of District Cooling System
User
The interface between the district cooling system and the building
cooling system is commonly referred to as user. The user would
usually comprise of air handling units (AHU) and chilled water
piping in the building. A user is required in each user's building to
connect the DCS distributed chilled water pipe to the building

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 Benefits of District Cooling System (DCS)
Energy Saving
DCS is a very energy-efficient cooling solution, a typical saving of
around 35% and 20% can be achieved when compared with
traditional air-cooled air-conditioning systems and individual water-
cooled air-conditioning systems using cooling towers respectively.

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 Benefits of District Cooling System (DCS)
Reduce Green House Gas Emission
The reduction of energy consumption by using DCS, due to
improved energy efficiency, helps to reduce fossil fuel consumption
for power generation. Consequently, greenhouse gas emission such
as carbon dioxide which causes the global warming can also be
reduced.

Reduce Noise Pollution

With the removal of chiller plants and heat rejecting condensers
from individual buildings, noise, vibration, thermal plume and
waste heat pollution can be significantly reduced. Equipment that
are located at central DCS plant can be kept under better acoustic,
vibration and waste heat control.
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 Benefits of District Cooling System (DCS)
Reduce Refrigerant Usage
The refrigerant used in air-conditioning equipment can be another source
of environmental pollutant. In DCS, the quantity of refrigerant required to
serve a district is less than the total amount of refrigerant required for
individual centralized air-conditioning systems due to smaller overall plant
size as a result of the load diversity and reduction of number of standby
chillers.
Costs Saving
A district cooling system allows the building owner to eliminate their on-
site chiller operation and maintenance. By doing this, the building owner
no longer needs to operate and maintain chillers, and replace chillers at
the end of their life cycles. Therefore, building owners using DCS can
minimize the capital investment and the subsequent maintenance costs
for their air-conditioning plants.
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Chiller

Chiller is a center air

conditioning chilled water
system.

Operation principle:
❑ The chilled water is generated
by the evaporator of chiller.
❑ The chilled water leaves the
evaporator at around 6°C and
is pushed around the building
by the chilled water pump.

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 Chiller
Operation principle (Cont’):

❑ The chilled water flows

up the height of the
building to each floor
in pipes known as
“risers”. These pipes
are known as risers no
matter if the water is
flowing upwards or
downwards within
them.

CHWS: 6.0oC (44oF) CHWR: 12.0oC (54oF)

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 Chiller
Operation principle (Cont’):

❑ The chilled water enters the Air Handling Unit (AHU) and passes
through the cooling coil (a series of thin pipes) where it will
absorb the heat of the air blowing across.
❑ When the chilled water leaves the cooling coil it will now be
warmer at around 12°C.
❑ The warm chilled water then heads back to the evaporator, via
the return riser, and once it enter the evaporator a refrigerant
will absorb the unwanted heat and move this over to the
condenser.
❑ The chilled water will then leave cool again, ready to circulate
around the building and collect more unwanted heat.

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 Components of Chiller
The components of
chiller consist of:
1. Compressor
2. Evaporator
3. Air handling unit
(AHU)
4. Chilled water
pump (CHWP)
5. Condenser
6. Cooling tower
7. Condenser
water pump

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Chiller

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 Performance of Chiller
The cooling effect of chiller is commonly expressed by
refrigeration ton (RT). One RT of cooling effect is defined as
the rate of heat input required to melt one ton of ice in 24
hours period, which is equal to about 3.517 kW.

Therefore, cooling load of chiller or the rate of heat gain in

evaporator in RT= (Rate of heat gain in evaporator in
kW)/3.517.

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 Performance of Chiller
Motor and compressor coupling
The motor and compressor of chiller can be connected using
direct drive concept.

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 Performance of Chiller
Chiller Efficiency
❑ Chiller efficiency is defined as the magnitude of electrical
power consumed by the motor of compressor or chiller
power consumption to produce one unit of cooling effect.
❑ Chiller efficiency in kW/RT = Electric power consumed by
the motor of compressor in kW/ cooling effect produced in
RT. =
Chiller power consum ption in k W
 ch iller
Chiller cooling load in RT

❑ Chiller efficiency is also expressed by coefficient of

performance (COP) with dimensionless quantity.

Chiller cooling load in k W

COP =
Chiller power consumption in k W

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 Performance of Chiller
Example 1:
A chiller operates 10 hours per day and 350 days of a year. The
average cooling load of chiller is 1000 RT. Present chilled water
supply temperature is 7 °C and the chiller efficiency is 0.51 kW/RT.
If the chilled water supply temperature is increased to 8. 5 °C,
calculate annual energy and electrical cost saving. Assume that
the chiller efficiency is improved by 2% for every 1 °C increasing of
the chilled water supply temperature and electrical tariff is RM
0.43/kWh.

Solution 1:
The chiller cooling load = 1000 RT
Present chiller supply temperature = 7 °C
Present chiller efficiency = 0.51 kW/RT

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 Performance of Chiller
Solution 1:
Proposed chilled water supply temperature = 8.5 °C
Proposed increment of chilled water supply temperature=8.5–7 °C
= 1.5 °C
The improvement of chiller efficiency due to the increasing of
chilled water supply temperature = 1.5 °C x 2%/°C = 3 %
The new chiller efficiency = 0.51 kW/RT x (1-0.03) = 0.495 kW/RT

The annual electrical energy saving= 10 hrs/day x 350 days/year

x 1000 RT x(0.51- 0.495 kW/RT)
= 52,500 kWh/year
The annual electrical cost saving=52,500 kWh/year x RM 0.43/kWh
= RM 22,575/year

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 Pumping System
Chilled water and condenser water pumping system are used in
water cooled central air conditioning chilled water systems.

The function of chilled water pump is to provide the primary force to

overcome the pressure losses caused by the different components
of chilled loops and circulates the required amount of chilled water
through the evaporator of chiller and the cooling coil of AHUs.

The function of condenser water pump is to overcome the

associated pressure losses and circulates the required amount of
condenser water through the condenser of chiller and the cooling
towers.

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 Pumping System

The friction loss in the pipe can be Average flow velocity

calculated as: L V2 of fluid in a pipe can
Pfriction = f
D 2g be calculated as:
Q
V=
A

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 Pumping System
The pressure losses for the The summation of friction loss
fitting (also known as dynamic of the pipe and the fitting
loss) such as elbow, tee-joint, represents the total pressure
valve, strainer of the piping drop of a piping system. It is
system can be calculated: given by

Pt o t a l = (P f r i c tio n + P f i t i n g )  g

V2
 =
Pfiting Co
2g

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 Pumping System
If the vertical height of the
open section of the cooling
tower is HCT , the static
pressure head can be
calculated:
The input power to motor is
Pstatic = H C T  g
given by
QP total
W m o to r, in p u t =
 m o to r co u p lin g p u m p

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 Pumping System

2 2 0

1 3 9

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 Pumping System
Example 2:
Cooling water is supplied at a rate of 500 m^3/h from a cooling tower
to a heat exchanger as shown in Figure below. Pressure loss across
the heat exchanger is 7 m of water. The diameter and total length of
pipe are 250 mm and 150 m, respectively. The piping system
consists of 6 fully open gate valves, 2 strainers and 20 90o standard
elbows. The loss coefficient of strainer is 8. The efficiencies of pump,
coupling and motor are 0.6, 0.98 and 0.8, respectively. Estimate the
total head and input power to the motor of pump to maintain the flow.

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 Pumping System
Solution 2:
Water flow rate, Q = 500 m^3/h
= 500 m^3/3600 s
= 0.139 m^3/s
= 0.139 x 1000 l/s
= 139 l/s
D 2 3.14(0.25)2
Cross section area of pipe, A = = = 0.049 m 2
4 4

Diameter of pipe, D = 250 mm

= 0.25 m
Water flow velocity, V = Q/A = 0.139/0.049 = 2.837 m/s
From friction loss chart, for Q = 139 l/m and D = 250 mm, thus the
friction loss per unit length of pipe ∆P = 220 Pa/m

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 Pumping System
Solution 2:

Calculate the friction loss of pipe:

 P f r i ct i o n =  P x L = 2 2 0 P a / m x 1 5 0 m = 3 3 0 0 0 P a = 3 3 k P a

Calculate the head loss of 6 fully open gate valves:

From the table of loss coefficient, for D = 250 mm, Co=0.06, thus
V 2 1000(2.837)2
Pvalve = 6 x C o = 6 x0.06 x = 1.45 kPa
2 2

Calculate the head loss of 2 strainers:

V 2 1000(2.837)2
Pstrainer = 2 x Co = 2x8 x = 64.4kPa
2 2

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 Pumping System
Solution 2:
Calculate the head loss of 20 90o standard elbows:
From the table of loss coefficient, for 90o standard elbow with D=250
mm, thus Co= 0.25
V 2 = 1000(2.837) 2
= 20.1kPa
Pelbow = 20 x C o 20x0.25 x
2 2

Calculate the head loss across the heat exchanger:

P exchanger = H  g = 7 x 1000 x 9.81 = 68.7 k P a

Calculate the head loss of the cooling tower:

PCT = H g = 4 x 1000 x 9.81 = 39.2 kPa

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 Pumping System
Solution 2:
Calculate the total head loss of pump:
P t o t a l = P f r i c t i o n + P v a l v e + P s t r a i n e r + P e l b o w + P e x c h a n g e r + P CT
= 3 3 + 1 .4 5 + 6 4 .4 + 2 0 .1 + 6 8 .7 + 3 9 .2
= 226.85kPa

The input power of pump motor:

QP total
W m oto r, in put =
 m oto r co up ling pu m p
0.139 x 226850
=
0.6 x 0.98 x 0.8
= 67,03kW

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 Air Handling Unit (AHU)
In the building with central air conditioning system, air is treated
(cooled and dehumidified) in AHU and then distributed to the
various parts of building to control the relative humidity and
temperature of spaces.

Main components of AHU:

❑ Fan
❑ Cooling coil
❑ Filter
❑ Damper
❑ Duct system

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 Air Handling Unit (AHU)
Example 3:
A AHU fan consumes 16 kW of power and delivers air at a rate of
15 m^3/s. Presently, media filter of 70 Pa pressure drop is used to
in the system. The total pressure drop (including the media filter) of
AHU system is 550 Pa. Calculate the achievable fan power saving
if the media filter is replaced with an electronic air cleaner which
has a negligible pressure drop. Present air flow rate and efficiency
required to be maintained after replacing the media filter with the
electronic air cleaner.

Solution 3:
The initial fan power consumption, W1 = 16 kW
The air flow rate, Q1 = Q2 = 15 m^3/s
The pressure drop across media filter, ∆Pfilter = 70 Pa
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 Air Handling Unit (AHU)
Solution 3:
The initial total pressure drop (including the media filter), ∆P1 =
550 Pa
The total pressure drop after replacing media filter, ∆P2
∆P2 = ∆P1 - ∆Pfilter
= 550-70
= 480 Pa
The fan power consumption after replacing media filter can be
calculated based on comparison of fan power consumption before
and after replacing.
Q1P1
W1 1 P1 P2 480
= =  W2 = W1 x = 16000 x = 13963.64W
W2 Q 1P 2 P2 P1 550
2

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 Air Handling Unit (AHU)
Solution 3:
The achievable fan power saving = W1-W2
= 16000 -13963.64
= 2036.36 W

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 Cooling Tower System
The cooling tower systems are used to reject heat from the water
cooled central air conditioning systems, water cooled package units
and process cooling system to the atmosphere.

Main components of
cooling tower:
❑ Water spray system
❑ Packing material (fill
packing)
❑ Fan

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 Cooling Tower System
The warm water is sprayed over the fill packing material of the
cooling tower. Heat is transferred from the exposed surface of water
film to the following air as sensible heat, Qs and latent heat, QL .
Qs = Ahc (Tw − Ta )

Q L = k m (m v, f − m v,a )Ah fg

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 Cooling Tower System
The mass friction of water The mass friction of water
pavour at the exposed surface pavour at the air stream is
of water film is given by. given by.

Where, Where,

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 Cooling Tower System
Example4:
A process water enters a water cooling at 40 °C and leaves 30 °C.
The ambient air flows in cross flow configuration through the packing
materials of cooling tower at an average temperature of 32 °C and
relative humidity of 70%. The saturated vapour pressure and latent
heat of vapourisation of water at different temperature are shown in
Table below. Assume that the convective heat and mass transfer
coefficients of circulating air are 5 W/mK and 0.007 kg/m^2s,
respectively. For the average temperature of cooling tower, calculate
the sensible, latent and total heat flux from water to the circulating
air. The relative humidity
of surrounding air is
increase 100%
(saturation condition).

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 Cooling Tower System
Solution 4:
The average temperature in the cooling tower= (40+30 °C)/2= 35 °C.
The saturated pavour pressure at exposed surface water film for
T 35 °C= 5.628 kPa

The partial density of pavour at exposed surface water film:

RH (Psat atT )M v 100%x 5.628x10 3 x18
v, f = 35oC
= = 0.0396 kg / m 3
RuTw 8314.3x (273 + 35)

The density of air pavour mixture at exposed surface water film:

RH(Psat atT )Mv Pa, f M a 100%x 5.628x10 3 x18 (100−100%x5.628)x10 3 x 29
f = 35oC
+ = + =1.1086 kg/ m 3
RuTw RuTw 8314.3x (273+35) 8314.3x (273+35)

The mass friction of water pavour at the exposed surface of water

film: v, f 0.0396
mv, f = = = 0.0357
f 1.1086

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 Cooling Tower System
Solution 4:
The ambient temperature = 32 °C, and RH=70%.
The saturated pavour pressure at T 32 °C= 4.799 kPa (using
interpolation formulation and refer Table of saturated pavour
pressure). (y − y )
y = 2 1 x (x− x1 ) + y1
(x2 − x1 )

The partial density of pavour at air stream:

RH (Psat at T )M v 70% x 4.799 x10 3 x 18
 v,a = 32oC
= = 0.0238 kg / m 3
R u Ta 8314.3 x (273 + 32)

The density of air pavour mixture at air stream:

RH (PsatatT )Mv Pa,bulk M a 70%x 4.799x10 3 x18 (100 − 70%x4.7899) x10 3 x 29
a = 32oC
+ = + =1.1291kg / m 3
RuTa RuTa 8314.3x (273+ 32) 8314.3x (273+ 32)

The mass friction of water pavour in the bulk air:

v,a 0.0238
mv,a = = = 0.0211
a 1.1291
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 Cooling Tower System
Solution 4:
The sensible heat flux per unit area from water film to bulk air:
Q = Ah (T − T ) = 1 x 5(35 − 32) = 15 W / m 2
s c w a

The latent heat flux per unit area from water film to bulk air:
Q = k (m −m )Ah = 0.007 x (0.0357 − 0.0211)1x 2418.6x103 = 247.18 W / m 2
L m v, f v,a fg

The total heat flux per unit area from water film to bulk air:
Q total = Q s + Q L = 15 + 247.18 = 262.18W / m 2

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 Performance of Chiller
Example 5:
Consider three 600 RT capacity chiller options to support the
constant process cooling load of 600 RT. The first cost and
efficiency of the three 600 RT chiller options are presented in the
table below.

Compute the life cycle cost for a 10 year period based on the
following assumption:
- Chiller operates 10 hs/day and 300 days a year
- Electricity tariff = RM 0.43/kWh
- Electricity cost escalation is 2% per year
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 Performance of Chiller
Solution 5:
Consider the chiller efficiency of 0.5 kW/RT, the electrical energy
cost for year-1:
= 10 h/day x 300 day/year x 600 RT x 0.5 kw/RT x RM 0.43 /kWh
= RM 387,000.00

As the escalation of electricity tariff is 2% per year, the energy cost

for year-2 will be
= RM 387,000.00 x (1+2%)= RM 394,740.00

Similarly, energy cost for year-3 and the following years can be
calculated. The life cycle cost comparison for the three chiller
options is given in table below. The maintenance cost for the three
chiller options are assumed to be same and not included here.
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 Performance of Chiller
Solution 2:

From the comparison of three electrical costs, can be analyzed:

1. The electrical cost incurred to operate the option 1 is the
lowest from year-1 onward.
2. The capital cost of option 1 is the highest, however the life
cycle cost (which consists of capital and operating cost for
implementing option 1 is the lowest.
3. Hence, the option 1 should be selected.
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