This document provides a summary of the cooling load calculations for a hostel building located at the University of Engineering & Technology Lahore-Narowal Campus. It includes an introduction, purpose, assumptions, zoning of the building, calculations of various space heat gains including external and internal loads, and manual calculations of the cooling load for each zone. The four zones are residential rooms, toilets, main dining hall and kitchen, and TV lounge. Detailed cooling load calculations are presented for each room and zone based on the CLTD/SCL/CLF method. The document concludes with the total cooling load calculated for the entire hostel building.
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COOLING LOAD CALCULATIONS FOR
HOSTEL BUILDING UET LAHORE
NAROWAL CAMPUS
A Project Submitted to fulfil the requirement of B.Sc. Mechanical Engineering degree at
Department of Mechanical Engineering
University of Engineering & Technology Lahore-Narowal Campus
Under supervision and kind guidance of
Engr. Aqib Hussain
By
JABIR ALI SIDDIQUE 2012-ME-519
RIZWAN MINHAS 2012-ME-508
MUHAMMAD QAMAR NAEEM 2012-ME-524
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Internal Examiner
NAME: ______________________________
SIGNATURE: _________________________
DATED: __________________________________
External Examiner
NAME: ______________________________
SIGNATURE: _________________________
DATED: __________________________________
PROJECT SUPERVISOR SIGNATURE: ______________________
Department of Mechanical Engineering- Narowal Campus
University of Engineering & Technology Lahore-PAKISTAN
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Abstract:
This document represents the Final Year Project report for the “Cooling load Calculations of
Hostel Building”, located in Narowal Campus of University of engineering & Technology Lahore.
A well designed and adequate HVAC system is essential to maintaining the comfortable,
productive and health living environment. The system is being designed to meet the minimum
ASHRAE and building code standards. Narowal campus was inaugurated in 2012 and the campus
building is under construction consisting of total area of 200 acres. The hostel building will host
more than 450 students and is being built with total area of about 112,222 sq. feet. The building is
to be built on concrete slab with masonry walls. The building comprises of three floors, named as
Ground Floor, First Floor and Second Floor.
The baseline load calculations were manipulated for:
Outdoor/Indoor design conditions
Building Components
Ductwork conditions
Ventilation/Infiltration conditions
Worst Case Scenario (Combining All the safety Factors)
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ACKNOWLEDGMETS:
We would like to express our sincere gratitude to Engr. Aqib Hussain for his solid efforts in
initiation of this project and his extended approach to give us familiarity with basics of HVAC.
Perhaps this motivation was really necessary for starting this Project. We would also like to extend
our gratitude to Engr. Ijaz-Ul-Haq for his continuous support in the completion of our B.Sc.
Mechanical Engineering degree. Perhaps it was not possible to successfully complete this task
without his kind advice that was available at every moment.
Finally, we would like to honor our parents for their continuous support throughout our academic
career. Perhaps they are a true source of inspiration for us.
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Table of Contents
Abstract:.................................................................................................................................................3
ACKNOWLEDGMETS: ..............................................................................................................................4
List of Figures:.......................................................................................................................................10
List of Tables:........................................................................................................................................10
1.1 Introduction:...................................................................................................................................11
1.2 Overview:........................................................................................................................................11
1.2.1 Individual systems:...................................................................................................................11
1.2.2 District networks:.....................................................................................................................12
1.3 Heating: ..........................................................................................................................................12
1.3.1 Generation:..............................................................................................................................12
1.3.2 Distribution:.............................................................................................................................12
1.3.2(a) Water/steam:....................................................................................................................12
1.3.2(b) Air:....................................................................................................................................12
1.4 Ventilation:.....................................................................................................................................13
1.4.1 Mechanical or forced ventilation: ............................................................................................13
1.4.2 Natural ventilation:..................................................................................................................13
1.5 Air conditioning: .............................................................................................................................14
1.5.1 Refrigeration cycle: ..................................................................................................................15
1.5.2 Free cooling:.............................................................................................................................15
1.5.3 Central vs. split system:............................................................................................................15
1.5.4 Dehumidification: ....................................................................................................................16
2 Purpose:.............................................................................................................................................16
2.1 Problem Definition:.....................................................................................................................16
2.2 Scope: .........................................................................................................................................16
3 Assumptions: .....................................................................................................................................17
4 Zoning the Office: ..............................................................................................................................17
4.1 Introduction to Zoning: ...............................................................................................................17
4.2 Need of Zoning:...........................................................................................................................17
4.3 Final Zoning Choice: ....................................................................................................................18
5 Space heat gain:.................................................................................................................................18
5.1 The method of how the heat enters the space:...........................................................................18
5.2 Sensible heat:..............................................................................................................................18
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5.3 The factors involve the sensible heat load:.................................................................................18
5.4 Latent Heat Loads: ......................................................................................................................19
5.4.1 The factors involve the sensible heat load:.............................................................................19
5.5 Space heat gain and cooling load (heat storage effect):..............................................................19
5.6 Space cooling and cooling load (Coil): .........................................................................................20
5.7 Components of cooling load:.......................................................................................................20
6 CLTD/SCL/CLF METHOD OF LOAD CALCULATION: ..............................................................................22
6.1 External Cooling Load:.................................................................................................................22
6.1.1 Roof: .....................................................................................................................................22
6.1.2 Walls:....................................................................................................................................23
6.1.3 Solar load through glass:........................................................................................................23
6.1.4 Partition, ceilings and floors:..................................................................................................25
6.2 Internal Cooling Loads:................................................................................................................25
6.2.1 People:..................................................................................................................................26
6.2.2 Lights: ...................................................................................................................................27
6.2.3 Appliances:............................................................................................................................28
6.2.4 Infiltration Air:.......................................................................................................................28
Excel Calculations .................................................................................................................................29
Zone 1: Residential Rooms....................................................................................................................29
Zone 2: Toilets ......................................................................................................................................33
Zone 3: Main Dining Hall & Kitchen ......................................................................................................35
Zone 4: TV Lounge ................................................................................................................................35
Manual Calculations .............................................................................................................................36
7 Zone1(Residential Rooms): ................................................................................................................36
7.1 Specifications:.............................................................................................................................36
7.2 Conduction through exterior surfaces:........................................................................................36
7.3 Conduction through Walls:..........................................................................................................36
Room 1 & 27 .........................................................................................................................................37
7.4 Solar radiation through Windows Glass:.....................................................................................37
7.5 Conduction through interior surfaces: ........................................................................................38
7.6 Conduction through Lightings: ....................................................................................................38
7.7 Conduction through Task Lightings and Bracket Fans: ................................................................38
7.8 Conduction through electrical equipment:..................................................................................38
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7.9 Conduction through People: .......................................................................................................39
7.10 Infiltration Load: .......................................................................................................................39
Total Cooling load of Each Room 1 & 27: ..............................................................................................39
Room 2, 3, 17, 18, 24, 25 & 26 ..............................................................................................................40
Total Cooling load of each Room 2, 3, 17, 18, 24, 25 & 26:....................................................................42
Room 4 .................................................................................................................................................42
Total Cooling load of Room 4:...............................................................................................................45
Room 5 .................................................................................................................................................45
Total Cooling load of Room 5:...............................................................................................................47
Room 6, 7, 37, 38 & 39..........................................................................................................................47
Total Cooling load of Room 6, 7, 37, 38 & 39: .......................................................................................50
Room 8 & 36 .........................................................................................................................................50
Total Cooling load of Room 8 & 36:.......................................................................................................52
Room 9 & 35 .........................................................................................................................................53
Total Cooling load of each Room 9 & 35: ..............................................................................................55
Room 10, 11, 12, 33 & 34......................................................................................................................55
Total Cooling load of each Room 10, 11, 12, 33 & 34: ...........................................................................58
Room 13, 14, 15, 21, 22, 23,29 & 30......................................................................................................58
Total Cooling load of each Room 13, 14, 15, 21, 22, 23,29 & 30:...........................................................60
Room 16, 28 & 20(Electric Room) .........................................................................................................60
Total Cooling load of each Room 16 28 & 20:........................................................................................63
Room 19 (Warden Room) .....................................................................................................................63
Total Cooling load of Room 19:.............................................................................................................65
Room 31 ...............................................................................................................................................66
Total Cooling load of Room 31:.............................................................................................................68
Room 32 ...............................................................................................................................................68
Total Cooling load of Room 32:.............................................................................................................71
Room 41, 42, 56, 57, 63, 64 & 65...........................................................................................................71
Total Cooling load of each Room 41, 42, 56, 57, 63, 64 & 65.................................................................73
Room 58 (Warden Room) .....................................................................................................................73
Total Cooling load of Room 58:.............................................................................................................76
8 Zone2(Toilets):...................................................................................................................................77
8.1 Specifications:.............................................................................................................................77
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8.2 Conduction through exterior surfaces:........................................................................................77
8.3 Conduction through Walls:..........................................................................................................77
Toilet 1, 2, 9, 10, 13, 14, 22, 23, 30, 31, 34 & 35 ....................................................................................78
8.4 Solar radiation through Windows Glass:.....................................................................................78
8.5 Conduction through interior surfaces: ........................................................................................79
8.6 Conduction through Lightings: ....................................................................................................79
8.7 Conduction through Task Lightings and Bracket Fans: ................................................................79
8.8 Conduction through electrical equipment:..................................................................................79
8.9 Conduction through People: .......................................................................................................79
Total Cooling load of each Toilet 1, 2, 9, 10, 13, 14, 22, 23, 30, 31, 34 & 35: .........................................79
Toilet 3, 4, 15, 16, 24, 25, 36 & 37.........................................................................................................80
Total Cooling load of each Toilet 3, 4, 15, 16, 24, 25, 36 & 37: ..............................................................81
Toilet 5, 6, 17, 18, 26, 27, 38 & 39.........................................................................................................82
Total Cooling load of each Toilet 5, 6, 17, 18, 26, 27, 38 & 39: ..............................................................83
Toilet 7, 8, 11, 12, 19, 20, 28, 29, 32, 33, 40 & 41 ..................................................................................84
Total Cooling load of each Toilet 7, 8, 11, 12, 19, 20, 28, 29, 32, 33, 40 & 41:........................................85
9 Zone3(Main Dining Hall & Kitchen)....................................................................................................86
9.1 Specifications:.............................................................................................................................86
9.2 Conduction through exterior surfaces:........................................................................................86
9.3 Conduction through Walls:..........................................................................................................86
9.4 Solar radiation through Windows Glass:.....................................................................................87
9.5 Conduction through interior surfaces: ........................................................................................88
9.6 Conduction through Lightings: ....................................................................................................88
9.7 Conduction through Task Lightings and Bracket Fans: ................................................................88
9.8 Conduction through electrical equipment:..................................................................................88
9.9 Conduction through People: .......................................................................................................88
Total Cooling load of Main Kitchen:......................................................................................................89
Main Dining Hall ...................................................................................................................................89
Total Cooling load of Main Dining Hall:.................................................................................................92
10 Zone4(TV Lounge) ............................................................................................................................93
10.1 Specifications: ...........................................................................................................................93
10.2 Conduction through exterior surfaces:......................................................................................93
10.3 Conduction through Walls:........................................................................................................93
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10.4 Solar radiation through Windows Glass: ...................................................................................94
10.5 Conduction through interior surfaces: ......................................................................................95
10.6 Conduction through Lightings: ..................................................................................................95
10.7 Conduction through Task Lightings and Bracket Fans: ..............................................................95
10.8 Conduction through electrical equipment:................................................................................95
10.9 Conduction through People: .....................................................................................................96
Total Cooling load of TV Lounge: ..........................................................................................................96
Total Cooling load of Hostel Building....................................................................................................97
Conclusion: ...........................................................................................................................................98
References:...........................................................................................................................................99
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List of Figures:
Figure 1: Residential HVAC Design Process……………………………………………………………………………………10
Figure 2: Difference b/w space heat and space cooling load………………………………………………………….15
Figure 3: Solar Heat Gain for Lights……………………………………………………………………………………………….16
Figure 4: External Loads and internal Loads…………………………………………………………………………..….….16
Figure 5: Heat Gain Locations………………………………………………………………………………………………….…...17
Figure 6: Winter and Summer Comfort Zone………………………………………………………………………………….18
List of Tables:
Table 1: Cooling Load Temperature Difference for Roof and External Walls (Dark)……………………….19
Table 2: Cooling Load Factor for Window Glass with Indoor Shading Devices………………………………..20
Table 3: Maximum Solar Heat Gain Factor for Sunlit Glass on Average Cloudiness Days………………..21
Table 4: Heat Gain from Occupants at Various Activities……………………………………………………………….22
Table 5: Cooling Load Factors (CLF) for Lights………………………………………………………………………………..23
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1.1 Introduction:
It was natural that the first HVAC controllers would be pneumatic since engineers understood fluid
control. Thus, mechanical engineers could use their experience with the properties of steam and air
to control the flow of heated or cooled air. After the control of air flow and temperature was
standardized, the use of electromechanical relays in ladder logic to switch dampers became
standardized. Eventually, the relays became electronic switches, as transistors eventually could
handle greater current loads [1]. By 1985, pneumatic controls could no longer compete with this
new technology although pneumatic control systems (sometimes decades old) are still common in
many older buildings. By the year 2000, computerized controllers were common. Today, some of
these controllers can even be accessed by web browsers, which need no longer be in the same
building as the HVAC equipment. This allows some economies of scale, as a single operations center
can easily monitor multiple buildings. [2]
HVAC (heating, ventilating, and air conditioning) is the technology of indoor and vehicular
environmental comfort. Its goal is to provide thermal comfort and acceptable indoor air quality.
HVAC system design is a sub discipline of mechanical engineering, based on the principles of
thermodynamics, fluid mechanics, and heat transfer. HVAC is important in the design of medium
to a large industrial and office buildings such as skyscrapers, onboard vessels, and in marine
environments such as aquariums, where safe and healthy building conditions are regulated with
respect to temperature and humidity, using fresh air from outdoors. Ventilating or ventilation is the
process of exchanging or replacing air in any space to provide high indoor air quality which involves
temperature control, oxygen replenishment, and removal of moisture, odors, smoke, heat, dust,
airborne bacteria, and carbon dioxide. Ventilation removes unpleasant smells and excessive
moisture, introduces outside air, keeps interior building air circulating, and prevents stagnation of
the interior air. Ventilation includes both the exchange of air to the outside as well as circulation of
air within the building. It is one of the most important factors for maintaining acceptable indoor air
quality in buildings. Methods for ventilating a building may be divided into mechanical/forced and
natural types. [3]
1.2 Overview:
The three central functions of heating, ventilation, and air-conditioning are interrelated, especially
with the need to provide thermal comfort and acceptable indoor air quality within reasonable
installation, operation, and maintenance costs. HVAC systems can provide ventilation, reduce air
infiltration, and maintain pressure relationships between spaces. The means of air delivery and
removal from spaces is known as room air distribution. [4]
1.2.1 Individual systems:
In modern buildings the design, installation, and control systems of these functions are integrated
into one or more HVAC systems. For very small buildings, contractors normally estimate the
capacity and select HVAC systems and equipment. For larger buildings, building service designers,
mechanical engineers, or building services engineers analyze, design, and specify the HVAC systems.
Specialty mechanical contractors then fabricate and commission the systems. Building permits and
code-compliance inspections of the installations are normally required for all sizes of building.
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1.2.2 District networks:
Although HVAC is executed in individual buildings or other enclosed spaces (like NORAD's
underground headquarters), the equipment involved is in some cases an extension of a larger district
heating (DH) or district cooling (DC) network, or a combined DHC network. In such cases, the
operating and maintenance aspects are simplified and metering becomes necessary to bill for the
energy that is consumed, and in some cases energy that is returned to the larger system. For example,
at a given time one building may be utilizing chilled water for air conditioning and the warm water it
returns may be used in another building for heating, or for the overall heating-portion of the DHC
network (likely with energy added to boost the temperature). [5]
1.3 Heating:
Heaters are appliances whose purpose is to generate heat for the building. This can be done via
central heating. Such a system contains a boiler, furnace or heat pump to heat water, steam or air in
a central location such as a furnace room in a home or a mechanical room in a large building. The
heat can be transferred by convection, conduction or radiation.
1.3.1 Generation:
Heaters exist for various types of fuel, including solid fuels, liquids, and gases. Another type of heat
source is electricity, typically heating ribbons made of high resistance wire. This principle is also used
for baseboard heaters and portable heaters. Electrical heaters are often used as backup or
supplemental heat for heat pump systems.
The heat pump gained popularity in the 1950s in the US and Japan [6]. Heat pumps can extract
heat from various sources, such as environmental air, exhaust air from a building, or from the
ground. Initially, heat pump HVAC systems were used in moderate climates, but with improvements
in low temperature operation and reduced loads due to more efficient homes, they are increasing in
popularity in cooler climates.
1.3.2 Distribution:
1.3.2(a) Water/steam:
In the case of heated water or steam, piping is used to transport the heat to the rooms. Most modern
hot water boiler heating systems have a circulator, which is a pump, to move hot water through the
distribution system (as opposed to older gravity-fed systems). The heat can be transferred to the
surrounding air using radiators, hot water coils (hydro-air), or other heat exchangers. The radiators
may be mounted on walls or installed within the floor to give floor heat. The use of water as the heat
transfer medium is known as hydronic. The heated water can also supply an auxiliary heat exchanger
to supply hot water for bathing and washing. [7]
1.3.2(b) Air:
Warm air systems distribute heated air through duct work systems of supply and return air through
metal or fiberglass ducts. Many systems use the same ducts to distribute air cooled by an evaporator
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coil for air conditioning. The air supply is typically filtered through air cleaners to remove dust and
pollen particles. [8]
1.4 Ventilation:
Ventilation is the process of changing or replacing air in any space to control temperature or remove
any combination of moisture, odors, smoke, heat, dust, airborne bacteria, or carbon dioxide, and to
replenish oxygen. Ventilation includes both the exchange of air with the outside as well as circulation
of air within the building. It is one of the most important factors for maintaining acceptable indoor
air quality in buildings. Methods for ventilating a building may be divided into mechanical/forced
and natural types.
1.4.1 Mechanical or forced ventilation:
Mechanical or forced ventilation is provided by an air handler and used to control indoor air quality.
Excess humidity, odors, and contaminants can often be controlled via dilution or replacement with
outside air. However, in humid climates much energy is required to remove excess moisture from
ventilation air. Kitchens and bathrooms typically have mechanical exhausts to control odors and
sometimes humidity. Factors in the design of such systems include the flow rate (which is a function
of the fan speed and exhaust vent size) and noise level. Direct drive fans are available for many
applications, and can reduce maintenance needs. Ceiling fans and table/floor fans circulate air within
a room for the purpose of reducing the perceived temperature by increasing evaporation of
perspiration on the skin of the occupants. Because hot air rises, ceiling fans may be used to keep a
room warmer in the winter by circulating the warm stratified air from the ceiling to the floor. [9]
1.4.2 Natural ventilation:
Natural ventilation is the ventilation of a building with outside air without using fans or other
mechanical systems. It can be via operable windows, louvers, or trickle vents when spaces are small
and the architecture permits. In more complex schemes, warm air is allowed to rise and flow out
high building openings to the outside (stack effect), causing cool outside air to be drawn into low
building openings. Natural ventilation schemes can use very little energy, but care must be taken to
ensure comfort. In warm or humid climates, maintaining thermal comfort solely via natural
ventilation may not be possible. Air conditioning systems are used, either as backups or supplements.
Air-side economizers also use outside air to condition spaces, but do so using fans, ducts, dampers,
and control systems to introduce and distribute cool outdoor air when appropriate.
An important component of natural ventilation is air change rate or air changes per hour: the hourly
rate of ventilation divided by the volume of the space. For example, six air changes per hour means
an amount of new air, equal to the volume of the space, is added every ten minutes. For human
comfort, a minimum of four air changes per hour is typical, though warehouses might have only two.
Too high of an air change rate may be uncomfortable, akin to a wind tunnel which have thousands
of changes per hour. The highest air change rates are for crowded spaces, bars, night clubs,
commercial kitchens at around 30 to 50 air changes per hour. Room pressure can be either positive
or negative with respect to outside the room. Positive pressure occurs when there is more air being
supplied than exhausted, and is common to reduce the infiltration of outside contaminants. [10]
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[11]
1.5 Air conditioning:
An air conditioning system, or a standalone air conditioner, provides cooling and humidity control
for all or part of a building. Air conditioned buildings often have sealed windows, because open
windows would work against the system intended to maintain constant indoor air conditions.
Outside, fresh air is generally drawn into the system by a vent into the indoor heat exchanger section,
creating positive air pressure. The percentage of return air made up of fresh air can usually be
manipulated by adjusting the opening of this vent. Typical fresh air intake is about 10%. Air
conditioning and refrigeration are provided through the removal of heat. Heat can be removed
through radiation, convection, or conduction. Refrigeration conduction media such as water, air, ice,
and chemicals are referred to as refrigerants. A refrigerant is employed either in a heat pump system
in which a compressor is used to drive thermodynamic refrigeration cycle, or in a free cooling system
which uses pumps to circulate a cool refrigerant (typically water or a glycol mix). [12]
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1.5.1 Refrigeration cycle:
The refrigeration cycle uses four essential elements to cool.
The system refrigerant starts its cycle in a gaseous state. The compressor pumps the
refrigerant gas up to a high pressure and temperature.
From there it enters a heat exchanger (sometimes called a condensing coil or condenser)
where it loses energy (heat) to the outside, cools, and condenses into its liquid phase.
An expansion valve (also called metering device) regulates the refrigerant liquid to flow at the
proper rate.
The liquid refrigerant is returned to another heat exchanger where it is allowed to evaporate,
hence the heat exchanger is often called an evaporating coil or evaporator. As the liquid
refrigerant evaporates it absorbs energy (heat) from the inside air, returns to the compressor,
and repeats the cycle. In the process, heat is absorbed from indoors and transferred
outdoors, resulting in cooling of the building.
In variable climates, the system may include a reversing valve that switches from heating in winter to
cooling in summer. By reversing the flow of refrigerant, the heat pump refrigeration cycle is changed
from cooling to heating or vice versa. This allows a facility to be heated and cooled by a single piece
of equipment by the same means, and with the same hardware. [13]
1.5.2 Free cooling:
Free cooling systems can have very high efficiencies, and are sometimes combined with seasonal
thermal energy storage so the cold of winter can be used for summer air conditioning. Common
storage mediums are deep aquifers or a natural underground rock mass accessed via a cluster of
small-diameter, heat exchanger equipped boreholes. Some systems with small storages are hybrids,
using free cooling early in the cooling season, and later employing a heat pump to chill the circulation
coming from the storage. The heat pump is added-in because the storage acts as a heat sink when
the system is in cooling (as opposed to charging) mode, causing the temperature to gradually increase
during the cooling season. [14]
Some systems include an "economizer mode", which is sometimes called a "free cooling mode".
When economizing, the control system will open (fully or partially) the outside air damper and close
(fully or partially) the return air damper. This will cause fresh, outside air to be supplied to the
system. When the outside air is cooler than the demanded cool air, this will allow the demand to be
met without using the mechanical supply of cooling (typically chilled water or a direct expansion
"DX" unit), thus saving energy. The control system can compare the temperature of the outside air
vs. return air, or it can compare the enthalpy of the air, as is frequently done in climates where
humidity is more of an issue. In both cases, the outside air must be less energetic than the return air
for the system to enter the economizer mode.
1.5.3 Central vs. split system:
Central air conditioning systems (or package systems) with a combined outdoor
condenser/evaporator unit are often installed in modern residences, offices, and public buildings,
but are difficult to retrofit (install in a building that was not designed to receive it) because of the
bulky air ducts required [15]. An alternative to central systems is the use of separate indoor and
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outdoor coils in split systems. These systems, although most often seen in residential applications,
are gaining popularity in small commercial buildings. The evaporator coil is connected to a remote
condenser unit using refrigerant piping between an indoor and outdoor unit instead of ducting air
directly from the outdoor unit. Indoor units with directional vents mount onto walls, suspended from
ceilings, or fit into the ceiling. Other indoor units mount inside the ceiling cavity, so that short lengths
of duct handle air from the indoor unit to vents or diffusers around the rooms.
1.5.4 Dehumidification:
Dehumidification (air drying) in an air conditioning system is provided by the evaporator. Since the
evaporator operates at a temperature below the dew point, moisture in the air condenses on the
evaporator coil tubes. This moisture is collected at the bottom of the evaporator in a pan and
removed by piping to a central drain or onto the ground outside.
A dehumidifier is an air-conditioner-like device that controls the humidity of a room or building. It
is often employed in basements which have a higher relative humidity because of their lower
temperature (and propensity for damp floors and walls). In food retailing establishments, large open
chiller cabinets are highly effective at dehumidifying the internal air. Conversely, a humidifier
increases the humidity of a building. [16]
2 Purpose:
The aim of the project is to find the cooling and heating load of under construction Boys Hostel of
UET Lahore’s Narowal campus. A well designed and adequate HVAC system is essential to
maintaining the comfortable, productive and health living environment. The system is being
designed to meet the minimum ASHRAE and building code standards.
2.1 Problem Definition:
University of Engineering and Technology Lahore is pioneer engineering institution in Pakistan
providing engineering serving the cause of education since 1923. The university is currently
managing 4 sub campuses in addition to main campus. Narowal campus was inaugurated in 2012
and the campus building is under construction consisting of total area of 200 acres.
The hostel building will host more than 450 students and is being built with total area of about
112,222 Sft. The building will be oriented so that the front entrance will be pointing south. The
building is to be built on concrete slab with masonry walls.
2.2 Scope:
The scope of the design will involve the following considerations.
The zones of the complete hostel building
The building insulation, doors and windows for consideration of heat transfer
The internal heat generation of the building from equipment and lighting
Heating and cooling load
Ventilation requirements for each zone
Humidification requirements
Dehumidification requirements
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Ducting layout and specifications
Air distribution means
Annual cooling cost
Overall energy uses
3 Assumptions:
Refer to ASHRAE Standard 62 for ventilation requirements.
The typical values below:
Auditoriums, theaters - 15 cfm/person
Sleeping rooms - 15 cfm/person
Bedroom - 30 cfm/room
Classroom - 15 cfm/person
Communication centers - 20 cfm/person
Conference rooms - 20 cfm/person
Corridors - 0.1 cfm/sq ft
Dining - 20 cfm/person
Lobbies - 15 cfm/person
Locker, dressing rooms - 0.5 cfm/sq ft
Lounges, bars - 30 cfm/person
Offices - 20 cfm/person
Toilet, bath (private) - 35 cfm/room
Toilet (public) - 50 cfm/water closet or urinal [17]
4 Zoning the Office:
4.1 Introduction to Zoning:
A temperature zoning system is a system that allows to:
Support a consistent temperature in a given part of a house, regardless of external
conditions, such as sun, rain, snow, clouds, wind or phase of moon.
Support different temperatures in different parts of the house at different times as desired
by the owner. In an essence, each zone will have its own climate. It is possible, for
example, to have some rooms be cooler or warmer than others (individual preferences),
and it is also possible to keep the same rooms at different temperature depending on,
for example, whether the rooms are [supposed to be] occupied or not.
When using a zone control system, it is important that realistic demands are made on the system
such as keeping all the zones desired temperatures to within 5 to 8c of other zone temperatures or
of the master controlling thermostat. [18]
4.2 Need of Zoning:
We need Zoned Temperature Control in our Building if one or more of these conditions exist:
Family lifestyles dictate different temperatures in different areas of the home.
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Heating and cooling temperature patterns vary at different times of the day.
Our Building has:
More than one level.
Large, open areas such as vaulted ceilings or lofts, an atrium or a solarium.
A room off the back or over the garage.
Finished rooms in the basement or attic.
A room or rooms with expansive glass areas.
A portion built over a concrete slab floor.
A rambling floor plan or wings extending off the main living area. [19]
4.3 Final Zoning Choice:
There are total 4 zones in the hostel building.
Zone1: Residential rooms (129) Zone2: Toilets (64)
Zone3: Main Dining Hall & Main Kitchen Zone4: TV Lounge
5 Space heat gain:
It's the rate of heat gained when heat enters the space or heat generated within a space.
5.1 The method of how the heat enters the space:
Solar radiation through the window or any transparent surfaces.
Heat conduction through walls, roof and windows of the class.
Heat conduction through interior partitions, ceilings and floors.
The generated heat by the occupants such as lights, appliances, equipment and processes.
The loads that are results of ventilation and infiltration of outdoor air.
Other miscellaneous heat gains.
5.2 Sensible heat:
It's about heat at which a substance absorbs. During rising the temperature of the substance, the
substance doesn't change state. Sensible heat gain is directly added to the conditioned space by
conduction, convection, and radiation. Note that the sensible heat gain entering the conditioned
space does not equal the sensible cooling load during the same time interval because of the stored
heat in the building envelope. Only the convective heat becomes cooling load instantaneously. [20]
5.3 The factors involve the sensible heat load:
Heat transmitted through floors, ceilings, walls.
Occupant's body heat.
Appliance & Light heat.
Solar Heat gain through glass.
Infiltration of outside air.
Air introduced by Ventilation.
19. 19 | P a g e
5.4 Latent Heat Loads:
Latent heat gain occurs when moisture is added to the space either from internal sources (e.g.
vapor emitted by occupants and equipment) or from outdoor air as a result of infiltration or
ventilation to maintain proper indoor air quality.
5.4.1 The factors involve the sensible heat load:
Moisture-laden outside air form Infiltration & Ventilation.
Occupant Respiration & Activities.
Moisture from Equipment & Appliances.
5.5 Space heat gain and cooling load (heat storage effect):
The heat that is collected from the heat sources (conduction, convection, solar radiation, lightning,
people, equipment, etc...) doesn't go directly to heating the room. However, only some part of the
heat sources that is absorbed air in the conditioned space (class), leading to a quick change in its
temperature. Most of the radiation heat especially from sun, lighting, people is first absorbed by the
internal surfaces, which include ceiling, floor, internal walls, furniture etc. Due to the large but finite
thermal capacity of the roof, floor, walls etc., their temperature increases slowly due to absorption
of radiant heat. The radiant portion introduces a time lag and also a decrement factor depending
upon the dynamic characteristics of the surfaces. Due to the time lag, the effect of radiation will be
felt even when the source of radiation, in this case the sun is removed. [21]
[22]
The relation between heat gain and cooling load and the effect of the mass of the structure (light,
medium & heavy) is shown below. From figure it is evident that, there is a delay in the peak heat,
especially for heavy construction.
20. 20 | P a g e
[23]
5.6 Space cooling and cooling load (Coil):
Space cooling is the rate at which heat must be removed from the spaces to maintain air temperature
at a constant value. Cooling load, on the other hand, is the rate at which energy is removed at the
cooling coil that serves one or more conditioned spaces in any central air conditioning system. [24]
5.7 Components of cooling load:
[25]
21. 21 | P a g e
The total residential room cooling load consists of heat transferred through the room envelope
(walls, roof, floor, windows, doors etc.) and heat generated by occupants, equipment, and lights. The
load due to heat transfer through the envelope is called as external load, while all other loads are
The total cooling load on any building consists of both sensible as well as latent load components.
The sensible load affects the dry bulb temperature, while the latent load affects the moisture content
of the conditioned space. [26]
Residential room may be classified as externally loaded and internally loaded as you can see from
figure. In externally loaded room, the cooling load on the room is mainly due to heat transfer
between the surroundings and the internal conditioned space. Since the surrounding conditions are
highly variable in any given day, the cooling load of an externally loaded building varies widely.
[27]
22. 22 | P a g e
6 CLTD/SCL/CLF METHOD OF LOAD CALCULATION:
CLTD is a theoretical temperature difference that accounts for the combined effects of inside and
outside air temp difference, daily temp range, solar radiation and heat storage in the construction
assembly/building mass. It is affected by orientation, tilt, month, day, hour, latitude, etc. CLTD
factors are used for adjustment to conductive heat gains from walls, roof, floor and glass. CLF
accounts for the fact that all the radiant energy that enters the conditioned space at a particular time
does not become a part of the cooling load instantly. The CLF values for various surfaces have been
calculated as functions of solar time and orientation and are available in the form of tables in
ASHRAE Handbooks. CLF factors are used for adjustment to heat gains from internal loads such
as lights, occupancy, power appliances. [28]
6.1 External Cooling Load:
6.1.1 Roof:
If the roof is exposed directly to the sun, it absorbs maximum heat. If there is other room above the
air-conditioned room, then the amount of heat gained by the roof reduces. The heat gained by the
partitions of the room depends upon the type of partition. Roof calculation formula is given below:
Q = U * A * (CLTD)
Q = cooling load.
U = Coefficient of heat transfer roof or wall or glass.
A = area of roof.
CLTD = cooling load temperature difference.
Since the ASHRAE tables provide hourly CLTD values for one typical set of conditions i.e. outdoor
maximum temperature of 95°F with mean temperature of 85°C and daily range of 21°F, the equation
is further adjusted to apply correction factors for conditions other than the mentioned base case.
Thus,
Q Roof = U * A * CLTD Roof Corrected
[29]
23. 23 | P a g e
6.1.2 Walls:
The walls of the room gain heat from the sun by way of conduction. The amount of heat depends
on the wall material and its alignment with respect to sun. If the wall of the room is exposed to the
west direction, it will gain maximum heat between 2 to 5 pm. The southern wall will gain maximum
heat in the mid-day between 12 to 2 pm. The heat gained by the wall facing north direction is the
least. The heat gained by the walls in day-time gets stored in them, and it is released into the rooms
at the night time thus causing excessive heating of the room. If the walls of the room are insulated
the amount of heat gained by them reduces drastically.
The cooling load from walls is treated in a similar way as roof:
Q Wall = U * A * CLTD Wall Corrected
Where
Q Wall = Load through the walls.
U = Thermal Transmittance for walls.
A = area of walls.
CLTD = Cooling Load Temperature Difference for walls.
Table 1 Cooling Load Temperature Difference for Roof and External Walls (Dark). [30]
Solar time, hour 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Roof 14 12 10 8 7 5 4 4 6 8 11 15 18 22 25 28 29 30 29 27 24 21 19 16
External wall
North
North-east
East
South-east
South
South-west
West
North-west
8
9
11
11
11
15
17
14
7
8
10
10
10
14
15
12
7
7
8
9
8
12
13
11
6
6
7
7
7
10
12
9
5
5
6
6
6
9
10
8
4
5
5
5
5
8
9
7
3
4
5
5
4
6
7
6
3
4
5
5
4
5
6
5
3
6
7
5
3
5
5
4
3
8
10
7
3
4
5
4
4
10
13
10
4
4
5
4
4
19
15.5
12
20.5
29
35
41
5
12
17
14
7
5
6
5
6
13
18
16
9
7
6
6
6
13
18
17
11
9
8
7
7
13
18
18
13
12
10
8
8
14
18
18
15
15
12
10
9
14
18
18
16
18
17
12
10
14
17
17
16
20
10
15
11
13
17
17
16
21
11
17
11
13
16
16
15
21
12
18
10
12
15
15
14
20
11
17
10
11
13
14
13
19
11
16
9
10
12
12
12
17
19
15
6.1.3 Solar load through glass:
Solar load through glass has two components: 1) Conductive and 2) Solar Transmission The
absorbed and then conductive portion of the radiation through the windows is treated like the roof
& walls where CLTD values for standard glazing are tabulated in ASHARE fundamentals handbook.
24. 24 | P a g e
For solar transmission, the cooling load is calculated by the cooling load SCL factor and shading
coefficient (SC).
The cooling load equations for glass are:
Conductive Q Glass Conductive = U * A * CLTD Glass Corrected
Solar Transmission Q Glass Solar = A * SC * SCL
Where
Q Conductive = Conductive load through the glass.
Q Solar = Solar transmission load through the glass.
U = Thermal Transmittance for glass.
A = area of glass.
CLTD = Cooling Load Temperature Difference for glass.
SC = Shading coefficient.
SCL = Solar Cooling Load Factor.
Table 2 Cooling Load Factor for Window Glass with Indoor Shading Devices. [31]
Solar time,
hour
1 2 3 4 5 6 7 8 9 10 11 12
Orientation:
North
North-east
East
South-east
South
South-west
West
North-west
Horizontal
0.08
0.03
0.03
0.03
0.04
0.05
0.05
0.05
0.06
0.07
0.02
0.02
0.03
0.04
0.05
0.05
0.04
0.05
0.06
0.02
0.02
0.02
0.03
0.04
0.04
0.04
0.04
0.06
0.02
0.02
0.02
0.03
0.04
0.04
0.03
0.04
0.07
0.02
0.02
0.02
0.03
0.03
0.03
0.03
0.03
0.73
0.56
0.47
0.30
0.09
0.07
0.06
0.07
0.12
0.66
0.76
0.72
0.57
0.16
0.11
0.09
0.11
0.27
0.65
0.74
0.80
0.74
0.23
0.14
0.11
0.14
0.44
0.73
0.58
0.76
0.81
0.38
0.16
0.13
0.17
0.59
0.80
0.37
0.62
0.79
0.58
0.19
0.15
0.19
0.72
0.86
0.29
0.41
0.68
0.75
0.22
0.16
0.20
0.81
0.91
0.86
0.81
0.82
0.83
0.81
0.82
0.865
0.85
25. 25 | P a g e
Table 3 Maximum Solar Heat Gain Factor for Sunlit Glass on Average Cloudiness Days [32]
Month Maximum solar heat gain factor for 22-degree north latitude, W/m2
North North-east /
north-west
East / west South-east /
south-west-
South Horizontal
January.
February.
March.
April
May
June
July
August
September
October
November
December
88
97
107
119
142
180
44
123
112
100
88
84
140
265
404
513
572
589
565
502
388
262
142
101
617
704
743
719
687
666
149
694
705
676
606
579
789
759
663
516
404
355
391
496
639
735
786
790
696
578
398
210
139
134
171
223
392
563
686
730
704
808
882
899
892
880
877
879
854
792
699
657
6.1.4 Partition, ceilings and floors:
The various internal loads consist of sensible and latent heat transfers due to occupants, products,
processes appliances and lighting. The lighting load is only sensible. The conversion of sensible heat
gains (from lighting, people, appliances, etc.) to space cooling load is affected by the thermal storage
characteristics of that space and is thus subject to appropriate cooling load factors (CLF) to account
for the time lag of the cooling load caused by the building mass. The weighting factors equation
determines the CLF factors.
CLF = Q cooling load / Q internal gains [33]
6.2 Internal Cooling Loads:
The various internal loads consist of sensible and latent heat transfers due to occupants, products,
processes appliances and lighting. The lighting load is only sensible. The conversion of sensible heat
gains (from lighting, people, appliances, etc.) to space cooling load is affected by the thermal storage
characteristics of that space and is thus subject to appropriate cooling load factors (CLF) to account
for the time lag of the cooling load caused by the building mass. The weighting factors equation
determines the CLF factors.
26. 26 | P a g e
CLF = Q cooling load / Q internal gains
Note that the latent heat gains are considered instantaneous.
6.2.1 People:
Human beings release both sensible heat and latent heat to the conditioned space when they stay in
it as you can see the figures in the table (11). You might have noticed that when a small room is filled
with people, it tends to become warmer. People emit heat primarily through breathing and
perspiration, and, to a lesser extent, through radiation. This heat translates into an increased cooling
load on your cooling systems. The heat gain by the occupants in the building is separated into
sensible and latent heat. The number of people, the type of activity they are performing, and the
CLF determines sensible and latent heat. The CLF is determined by the time the occupants come
into the building and for how long they stay in the building. [34]
The heat gain from the occupancy or people is given be equation:
Q sensible = N (QS) (CLF)
Q latent = N (QL)
N = number of students in space (class room).
QS, QL = Sensible and Latent heat gain from occupancy is given in Table 3.
CLF = Cooling Load Factor, by hour of occupancy in Table 37.
Table 4 Heat Gain from Occupants at Various Activities [35]
Activity Total heat, W Sensible heat, W Latent heat, W
Adult, male Adjusted
Seated at rest
Seated, very light work, writing
Seated, eating
Seated, light work, typing,
Standing, light work or walking slowly,
Light bench work
Light machine work
Heavy work
Moderate dancing
Athletics
115
140
150
185
235
255
305
470
400
585
100
120
170b
150
185
230
305
470
375
525
60
65
75
250
90
100
100
165
120
185
40
55
95
200
95
130
205
305
255
340
27. 27 | P a g e
6.2.2 Lights:
The primary source of heat from lighting comes from light-emitting elements. Table (13), indicate
and explain more about the lights effects on the heat gain when it's off or on. Calculation of this load
component is not straightforward; the rate of heat gain at any given moment can be quite different
from the heat equivalent of power supplied instantaneously to those lights. Only part of the energy
from lights is in the form of convective heat, which is picked up instantaneously by the air-
conditioning apparatus. The remaining portion is in the form of radiation, which affects the
conditioned space only after having been absorbed and re-released by walls, floors, furniture, etc.
This absorbed energy contributes to space cooling load only after a time lag, with some part of such
energy still present and reradiating after the lights have been switched off. [36]
Generally, the instantaneous rate of heat gain from electric lighting may be calculated from:
Q = 3.41 x W x FUT x FSA
Cooling load factors are used to convert instantaneous heat gain from lighting to the sensible cooling
load; thus the equation is modified to:
Q = 3.41 x W x FUT x FSA x (CLF).
Where:
W = Watts input from electrical lighting plan or lighting load data.
FUT = Lighting use factor, as appropriate.
FSA = special ballast allowance factor, as appropriate.
CLF = Cooling Load Factor, by hour of occupancy, Table 38.
Table 5 Cooling Load Factors (CLF) for Lights: [37]
Lights Number of Hours the lights are turned ON
For 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
8 0.85 0.92 0.95 0.96 0.97 0.97 0.97 0.98 0.13 0.06 0.04 0.03 0.02 0.02 0.02 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01
10 0.85 0.93 0.95 0.97 0.97 0.97 0.98 0.98 0.98 0.98 0.14 0.07 0.04 0.03 0.02 0.02 0.02 0.02 0.02 0.02 0.01 0.01 0.01 0.01
12 0.86 0.93 0.96 0.97 0.97 0.98 0.98 0.98 0.98 0.98 0.98 1.0 0.14 0.07 0.04 0.03 0.03 0.02 0.02 0.02 0.02 0.02 0.02 0.02
14 0.86 0.93 0.96 0.97 0.98 0.98 0.98 0.98 0.98 0.98 0.99 0.99 0.99 0.99 0.15 0.07 0.05 0.03 0.03 0.03 0.02 0.02 0.02 0.02
16 0.87 0.94 0.96 0.97 0.98 0.98 0.98 0.99 0.99 0.99 0.99 0.99 0.99 0.99 0.99 0.99 0.15 0.08 0.05 0.04 0.03 0.03 0.03 0.02
28. 28 | P a g e
6.2.3 Appliances:
In a cooling load estimate, heat gain from all appliances-electrical, gas, or steam-should be taken into
account. Because of the variety of appliances, applications, schedules, use, and installations,
estimates can be very subjective. Often, the only information available about heat gain from
equipment is that on its name-plate. [38]
Q Sensible = Qin x Fu x Fr x (CLF)
Q Latent = Qin x Fu
Where
Qin = rated energy input from appliances. See Table 5 through 9 or use manufacturer's data.
For
computers, monitors, printers and miscellaneous office equipment, see 2001 ASHRAE
Fundamentals, Chapter 29, Tables 8, 9, & 10.
Fu = Usage factor. See 1997 ASHRAE Fundamentals, Chapter 28, Table 6 and 7.
Fr = Radiation factor. See 1997 ASHRAE Fundamentals, Chapter 28, Table 6 and 7.
CLF = Cooling Load Factor, by hour of occupancy. See 1997 ASHRAE Fundamentals,
Chapter 28, Table 37 and 39.
6.2.4 Infiltration Air:
Q sensible = 1.08 x CFM x (To - Ti)
Q latent = 4840 x CFM x (Wo - Wi)
Q total = 4.5 x CFM x (ho - hi)
Where
CFM = Infiltration air flow rate. See 1997 ASHRAE Fundamentals, Chapter 25, for
determining infiltration
To, Ti = Outside/Inside dry bulb temperature.
Wo, Wi = Outside/Inside humidity ratio.
ho, hi = Outside/Inside air enthalpy.
35. 35 | P a g e
Zone 3: Main Dining Hall & Kitchen
Serial No. Q Walls Q Glass Q Lights Q Inf Q People Q Total
MDH 2081.55 4586.84 818.88 27000 4491.235 38978.505
Mkitchen 2132.3198 4104.01 511.8 4500 4491.235 15739.3648
Total 4213.8698 8690.85 1330.68 31500 8982.47 54717.8698
Unit of Heat Gain Q is BTU/hr
Total Heat Gain for Zone 3 is Q =54717.8698 BTU/hr
Zone 4: TV Lounge
Serial No. Q Walls Q Glass Q Lights Q Inf Q People Q Total
TV Lounge 2632.84 5950.82 511.8 9000 5188.94 23284.4
Unit of Heat Gain Q is BTU/hr
Total Heat Gain for Zone 4 is Q = 23284.4 BTU/hr
36. 36 | P a g e
Manual Calculations
7 Zone1(Residential Rooms):
7.1 Specifications:
Each room direction is 18x12 Ft
2
Total number of residential rooms are 129
Ceiling height is taken to be 10 ft.
Total occupancy of each room is 4 people
Nature of working is regular office work.
7.2 Conduction through exterior surfaces:
The cooling load caused by the conduction heat gains through the exterior roofs, walls, windows and
glass are each found by the formula
Q = U X A X CLTD
Where Q = Cooling load for roofs, glass or walls in BTU/hr
U = Overall heat co efficient for glass, roofs or walls in BTU/hr-ft2 F
A = Area of roofs, walls or glass in ft2
CLTD = Cooling load temperature difference in F
7.3 Conduction through Walls:
Conduction through walls can be found out using the same formula
Qw = Uv X Aw X CLTDVt.
The wall is made up of the following materials having specific thermal resistances.
So overall heat transfer coefficient is
Layer, Inside to Outside Thickness R
Inside surface resistance 0 0.685
4.5 in face brick 4.5 in 0.495
Air space 1 in 1
4.5 in face brick 4.5 in 0.495
Outside surface resistance 0 0.333
37. 37 | P a g e
Room 1 & 27
Now conduction through,
South-West Wall:
Area of south-west wall = A = 76.5 ft
2
(Excluding window)
Considering the wall of type D, for south-west direction CLTDmax =35
So heat gain through south-west wall is
Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr
East-South Wall:
At ground floor, there is another wall of room is joined with this wall. Therefore, there is no direct
exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e.
Q =0
North-West Wall:
Area of north-west wall = A = 180 ft
2
Considering the wall of type D, for north-west direction CLTDmax =30
So heat gain through north-west wall is
Q = 0.332 X 180 X 30 = 1792.8 BTU/hr
East-North Wall:
Area of east-north wall = A = 120 ft2
Considering the wall of type D, for east-north direction CLTDmax =15.5
So heat gain through east-north wall is
Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr
There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.
So, Q = 308.76 BTU/hr
Total heat gains through walls
Qw = 308.76 + 1792.8+ 888.93+0= 2990.49 BTU/hr
7.4 Solar radiation through Windows Glass:
Radiant energy through sun passes through the transparent material such as glass and became a
heat gain to a room. The solar cooling load can be found by using the following formula
QG = SHGF X A G X S C X CLF
Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2
AG = Area of the glass
38. 38 | P a g e
SC = Shading coefficient
CLF = Cooling load factor for a glass.
Area of the window is
AG = 6 X 7.25 = 43.5 ft2
Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67
Now for North window
Solar heat gain factor for south-west direction = SHGF = 195.5
Cooling load factor for south-west direction = CLF = 0.825
Heat gain for window is
Q = 195.5 X 0.67 X 0.825 X 43.5 = 4700.72 BTU/hr
7.5 Conduction through interior surfaces:
The heat that flows through the interior unconditioned spaces to the conditioned spaces through
partition, floors and ceilings also increase the cooling load.
In designing the air conditioning of hostel building, every portion of the building has to be
conditioned including rooms, toilets, Kitchen, Dining Hall and TV Lounge as well. Therefore,
there is no portion in a floor that remains unconditioned.
Therefore, conduction through interior surfaces is taken to be
QINT = 0 BTU/hr
7.6 Conduction through Lightings:
The equation for determining cooling load due to heat gain from lighting is
QL = 3.412 X W X BF X CLF
Where W = Light capacity in watts
BF = Ballast factor
CLF = Cooling load factor for lightings
The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights
therefore ballast factor i.e. BF = 1.25
Cooling load factor for the light is taken to be 1 in case of operational.
Total number of vented lights in the Room are 2
The wattage of one light is 24 W
So cooling load due to heat gain from lighting is
Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr
7.7 Conduction through Task Lightings and Bracket Fans:
The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also
increase the cooling load of a building.
The heat addition through bracket fans can be calculated as
QBF = 3.412 X W X CLF
There is no bracket fan used in the building so,
QBF = 0
7.8 Conduction through electrical equipment:
The heat gain from the electrical equipment plays a major role in the cooling load increment of the
space. In hostel building, the laptops are widely used and heat rejected from them must be taken
39. 39 | P a g e
into account for the calculation of tonnage of air conditioning.
The heat addition through laptops can be calculated as
Qlaptop = 3.412 X W X CLF
Cooling load factor for the laptop is taken to be 1 in case of operational.
Total number of laptops in Room are 4
The wattage of one laptop doing word processing and mass storage is 65 W
So cooling load due to heat gain from laptops is
Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr
7.9 Conduction through People:
The heat gained from people is composed of two parts, sensible heat and latent heat resulting from
perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat.
The equations from sensible and latent heat gains from people are
Qs = q 3 X N X CLF
Ql = q l X N
Where qs, ql = Sensible and latent heat gains per person
N = Number of people
CLF = Cooling load factor for people
Total number of people in Room are 4
For regular work, Sensible heat gain per person is qs = 250 BTU/hr
Latent heat gain per person is = 200 BTU/hr
Sensible heat gain for a Room is
Qs = 250 X 4 X 1 = 1000 BTU/hr
Latent heat gain for Room1 is
Ql; = 200 X 4 = 800 BTU/hr
Total cooling load required for the working of people is
Qp = Qs + Ql = 1000+800 = 1800 BTU/hr
7.10 Infiltration Load:
Q sensible = 1.08 x CFM x (To - Ti)
Q latent = 4840 x CFM x (Wo - Wi)
Q total = 4.5 x CFM x (ho - hi)
Q = 2995.64
Total Cooling load of Each Room 1 & 27:
Total cooling load of the room is simply the combined cooling load of individual components and
can be calculated by adding all the cooling loads calculated above.
Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp
Q = 2990.49+4700.72+0+204.72+2995.64+887.12+1800 = 13578.69 BTU/hr
40. 40 | P a g e
Room 2, 3, 17, 18, 24, 25 & 26
Now conduction through,
South-West Wall:
Area of south-west wall = A = 76.5 ft
2
(Excluding window)
Considering the wall of type D, for south-west direction CLTDmax =35
So heat gain through south-west wall is
Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr
East-South Wall:
At ground floor, there is another wall of room joined from the east-south wall. Therefore, there is
no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to
be zero i.e. Q =0
North-West Wall:
At ground floor, there is another wall of room joined from the north-west wall. Therefore, there is
no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e.
Q =0
East-North Wall:
Area of east-north wall = A = 120 ft2
Considering the wall of type D, for east-north direction CLTDmax =15.5
So heat gain through east-north wall is
Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr
There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.
So, Q = 308.76 BTU/hr
Total heat gains through walls
Qw = 308.76 + 0+ 888.93+0= 1197.69 BTU/hr
Solar radiation through Windows Glass:
Radiant energy through sun passes through the transparent material such as glass and became a
heat gain to a room. The solar cooling load can be found by using the following formula
QG = SHGF X A G X S C X CLF
Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2
AG = Area of the glass
SC = Shading coefficient
CLF = Cooling load factor for a glass.
41. 41 | P a g e
Area of the window is
AG = 6 X 7.25 = 43.5 ft2
Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67
Now for North window
Solar heat gain factor for south-west direction = SHGF = 195.5
Cooling load factor for south-west direction = CLF = 0.825
Heat gain for window is
Q = 195.5 X 0.67 X 0.825 X 43.5 = 4700.72 BTU/hr
Conduction through interior surfaces:
The heat that flows through the interior unconditioned spaces to the conditioned spaces through
partition, floors and ceilings also increase the cooling load. There is no portion in a floor that
remains unconditioned.
Therefore, conduction through interior surfaces is taken to be
QINT = 0 BTU/hr
Conduction through Lightings:
The equation for determining cooling load due to heat gain from lighting is
QL = 3.412 X W X BF X CLF
Where W = Light capacity in watts
BF = Ballast factor
CLF = Cooling load factor for lightings
The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights
therefore ballast factor i.e. BF = 1.25
Cooling load factor for the light is taken to be 1 in case of operational.
Total number of vented lights in each Room are 2
The wattage of one light is 24 W
So cooling load due to heat gain from lighting is
Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr
Conduction through Task Lightings and Bracket Fans:
The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also
increase the cooling load of a building.
The heat addition through bracket fans can be calculated as
QBF = 3.412 X W X CLF
There is no bracket fan used in the building so,
QBF = 0
Conduction through electrical equipment:
The heat addition through laptops can be calculated as
Qlaptop = 3.412 X W X CLF
Cooling load factor for the laptop is taken to be 1 in case of operational.
Total number of laptops in each Room are 4
The wattage of one laptop doing word processing and mass storage is 65 W
So cooling load due to heat gain from laptops is
Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr
42. 42 | P a g e
Conduction through People:
The heat gained from people is composed of two parts, sensible heat and latent heat resulting from
perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat.
The equations from sensible and latent heat gains from people are
Qs = q 3 X N X CLF
Ql = q l X N
Where qs, ql = Sensible and latent heat gains per person
N = Number of people
CLF = Cooling load factor for people
Total number of people in Room are 4
For regular work, Sensible heat gain per person is qs = 250 BTU/hr
Latent heat gain per person is = 200 BTU/hr
Sensible heat gain for a Room is
Qs = 250 X 4 X 1 = 1000 BTU/hr
Latent heat gain for Room is
Ql; = 200 X 4 = 800 BTU/hr
Total cooling load required for the working of people is
Qp = Qs + Ql = 1000+800 = 1800 BTU/hr
Total Cooling load of each Room 2, 3, 17, 18, 24, 25 & 26:
Total cooling load of the room is simply the combined cooling load of individual components and
can be calculated by adding all the cooling loads calculated above.
Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp
Q = 1197.69+4700.72+0+204.72+2995.64+887.12+1800 = 11785.89 BTU/hr
Room 4
Now conduction through,
South-West Wall:
Area of south-west wall = A = 76.5 ft
2
(Excluding window)
Considering the wall of type D, for south-west direction CLTDmax =35
So heat gain through south-west wall is
Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr
East-South Wall:
Area of east-south wall = A = 180 ft
2
Considering the wall of type D, for east-south direction CLTDmax =20.5
So heat gain through south-west wall is
43. 43 | P a g e
Q = 0.332 X 180 X 20.5 = 1225.08 BTU/hr
North-West Wall:
At ground floor, there is another wall of room joined with this wall. Therefore, there is no direct
exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0
East-North Wall:
Area of east-north wall = A = 120 ft2
Considering the wall of type D, for east-north direction CLTDmax =15.5
So heat gain through east-north wall is
Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr
There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.
So, Q = 308.76 BTU/hr
Total heat gains through walls
Qw = 308.76 + 1225.08+ 888.93+0= 2422.77 BTU/hr
Solar radiation through Windows Glass:
Radiant energy through sun passes through the transparent material such as glass and became a
heat gain to a room. The solar cooling load can be found by using the following formula
QG = SHGF X A G X S C X CLF
Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2
AG = Area of the glass
SC = Shading coefficient
CLF = Cooling load factor for a glass.
Area of the window is
AG = 6 X 7.25 = 43.5 ft2
Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67
Now for North window
Solar heat gain factor for south-west direction = SHGF = 195.5
Cooling load factor for south-west direction = CLF = 0.825
Heat gain for window is
Q = 195.5 X 0.67 X 0.825 X 43.5 = 4700.72 BTU/hr
Conduction through interior surfaces:
The heat that flows through the interior unconditioned spaces to the conditioned spaces through
partition, floors and ceilings also increase the cooling load. There is no portion in a floor that
remains unconditioned.
Therefore, conduction through interior surfaces is taken to be
QINT = 0 BTU/hr
44. 44 | P a g e
Conduction through Lightings:
The equation for determining cooling load due to heat gain from lighting is
QL = 3.412 X W X BF X CLF
Where W = Light capacity in watts
BF = Ballast factor
CLF = Cooling load factor for lightings
The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights
therefore ballast factor i.e. BF = 1.25
Cooling load factor for the light is taken to be 1 in case of operational.
Total number of vented lights in the Room are 2
The wattage of one light is 24 W
So cooling load due to heat gain from lighting is
Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr
Conduction through Task Lightings and Bracket Fans:
The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also
increase the cooling load of a building.
The heat addition through bracket fans can be calculated as
QBF = 3.412 X W X CLF
There is no bracket fan used in the building so,
QBF = 0
Conduction through electrical equipment:
The heat addition through laptops can be calculated as
Qlaptop = 3.412 X W X CLF
Cooling load factor for the laptop is taken to be 1 in case of operational.
Total number of laptops in Room are 4
The wattage of one laptop doing word processing and mass storage is 65 W
So cooling load due to heat gain from laptops is
Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr
Conduction through People:
The heat gained from people is composed of two parts, sensible heat and latent heat resulting from
perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat.
The equations from sensible and latent heat gains from people are
Qs = q 3 X N X CLF
Ql = q l X N
Where qs, ql = Sensible and latent heat gains per person
N = Number of people
CLF = Cooling load factor for people
Total number of people in Room are 4
For regular work, Sensible heat gain per person is qs = 250 BTU/hr
Latent heat gain per person is = 200 BTU/hr
Sensible heat gain for a Room is
Qs = 250 X 4 X 1 = 1000 BTU/hr
Latent heat gain for Room is
Ql; = 200 X 4 = 800 BTU/hr
Total cooling load required for the working of people is
45. 45 | P a g e
Qp = Qs + Ql = 1000+800 = 1800 BTU/hr
Total Cooling load of Room 4:
Total cooling load of the room is simply the combined cooling load of individual components and
can be calculated by adding all the cooling loads calculated above
Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp
Q = 2422.77+4700.72+0+204.72+2995.64+887.12+1800 = 13010.33 BTU/hr
Room 5
Now conduction through,
South-West Wall:
Area of south-west wall = A = 180 ft
2
Considering the wall of type D, for south-west direction CLTDmax =35
So heat gain through south-west wall is
Q = 0.332 X 180 X 35 = 2091.6 BTU/hr
East-South Wall:
Area of east-south wall = A = 76.5 ft
2
Considering the wall of type D, for east-south direction CLTDmax =20.5
So heat gain through south-west wall is
Q = 0.332 X 76.5 X 20.5 = 520.659 BTU/hr
North-West Wall:
Area of north-west wall = A = 120 ft2
Considering the wall of type D, for east-north direction CLTDmax =15.5
So heat gain through east-north wall is
Q = 0.332 X 120 X 30 = 1195.2 BTU/hr
There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.
So, Q = 597.6 BTU/hr
East-North Wall:
Q=0
Total heat gains through walls
Qw = 2091.6+520.659+597.6= 3209.85 BTU/hr
46. 46 | P a g e
Solar radiation through Windows Glass:
Radiant energy through sun passes through the transparent material such as glass and became a
heat gain to a room. The solar cooling load can be found by using the following formula
QG = SHGF X A G X S C X CLF
Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2
AG = Area of the glass
SC = Shading coefficient
CLF = Cooling load factor for a glass.
Area of the window is
AG = 6 X 7.25 = 43.5 ft2
Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67
Now for North window
Solar heat gain factor for east-south direction = SHGF = 124.5
Cooling load factor for east-south direction = CLF = 0.825
Heat gain for window is
Q = 195.5 X 0.67 X 0.825 X 43.5 = 2975.41 BTU/hr
Conduction through interior surfaces:
The heat that flows through the interior unconditioned spaces to the conditioned spaces through
partition, floors and ceilings also increase the cooling load. There is no portion in a floor that
remains unconditioned.
Therefore, conduction through interior surfaces is taken to be
QINT = 0 BTU/hr
Conduction through Lightings:
The equation for determining cooling load due to heat gain from lighting is
QL = 3.412 X W X BF X CLF
Where W = Light capacity in watts
BF = Ballast factor
CLF = Cooling load factor for lightings
The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights
therefore ballast factor i.e. BF = 1.25
Cooling load factor for the light is taken to be 1 in case of operational.
Total number of vented lights in the Room are 2
The wattage of one light is 24 W
So cooling load due to heat gain from lighting is
Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr
Conduction through Task Lightings and Bracket Fans:
The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also
increase the cooling load of a building.
The heat addition through bracket fans can be calculated as
QBF = 3.412 X W X CLF
47. 47 | P a g e
There is no bracket fan used in the building so,
QBF = 0
Conduction through electrical equipment:
The heat addition through laptops can be calculated as
Qlaptop = 3.412 X W X CLF
Cooling load factor for the laptop is taken to be 1 in case of operational.
Total number of laptops in Room5 are 4
The wattage of one laptop doing word processing and mass storage is 65 W
So cooling load due to heat gain from laptops is
Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr
Conduction through People:
The heat gained from people is composed of two parts, sensible heat and latent heat resulting from
perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat.
The equations from sensible and latent heat gains from people are
Qs = q 3 X N X CLF
Ql = q l X N
Where qs, ql = Sensible and latent heat gains per person
N = Number of people
CLF = Cooling load factor for people
Total number of people in Room are 4
For regular work, Sensible heat gain per person is qs = 250 BTU/hr
Latent heat gain per person is = 200 BTU/hr
Sensible heat gain for a Room is
Qs = 250 X 4 X 1 = 1000 BTU/hr
Latent heat gain for Room is
Ql; = 200 X 4 = 800 BTU/hr
Total cooling load required for the working of people is
Qp = Qs + Ql = 1000+800 = 1800 BTU/hr
Total Cooling load of Room 5:
Total cooling load of the room is simply the combined cooling load of individual components and
can be calculated by adding all the cooling loads calculated above
Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp
Q = 3209.85+2975.41+0+204.72+2995.64+887.12+1800 = 12072.1 BTU/hr
Room 6, 7, 37, 38 & 39
Now conduction through,
South-West Wall:
Q = 0 BTU/hr
East-South Wall:
48. 48 | P a g e
Area of east-south wall = A = 76.5 ft
2
Considering the wall of type D, for east-south direction CLTDmax =20.5
So heat gain through south-west wall is
Q = 0.332 X 76.5 X 20.5 = 520.659 BTU/hr
North-West Wall:
Area of north-west wall = A = 120 ft2
Considering the wall of type D, for east-north direction CLTDmax =15.5
So heat gain through east-north wall is
Q = 0.332 X 120 X 30 = 1195.2 BTU/hr
There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.
So, Q = 597.6 BTU/hr
East-North Wall:
Q=0
Total heat gains through walls
Qw = 0+520.659+597.6= 1118.259 BTU/hr
Solar radiation through Windows Glass:
Radiant energy through sun passes through the transparent material such as glass and became a
heat gain to a room. The solar cooling load can be found by using the following formula
QG = SHGF X A G X S C X CLF
Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2
AG = Area of the glass
SC = Shading coefficient
CLF = Cooling load factor for a glass.
Area of the window is
AG = 6 X 7.25 = 43.5 ft2
Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67
Now for window
Solar heat gain factor for east-south direction = SHGF = 124.5
Cooling load factor for east-south direction = CLF = 0.825
Heat gain for window is
Q = 124.5 X 0.67 X 0.825 X 43.5 = 2975.41 BTU/hr
Conduction through interior surfaces:
The heat that flows through the interior unconditioned spaces to the conditioned spaces through
partition, floors and ceilings also increase the cooling load. There is no portion in a floor that
remains unconditioned.
49. 49 | P a g e
Therefore, conduction through interior surfaces is taken to be
QINT = 0 BTU/hr
Conduction through Lightings:
The equation for determining cooling load due to heat gain from lighting is
QL = 3.412 X W X BF X CLF
Where W = Light capacity in watts
BF = Ballast factor
CLF = Cooling load factor for lightings
The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights
therefore ballast factor i.e. BF = 1.25
Cooling load factor for the light is taken to be 1 in case of operational.
Total number of vented lights in the Room are 2
The wattage of one light is 24 W
So cooling load due to heat gain from lighting is
Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr
Conduction through Task Lightings and Bracket Fans:
The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also
increase the cooling load of a building.
The heat addition through bracket fans can be calculated as
QBF = 3.412 X W X CLF
There is no bracket fan used in the building so,
QBF = 0
Conduction through electrical equipment:
The heat addition through laptops can be calculated as
Qlaptop = 3.412 X W X CLF
Cooling load factor for the laptop is taken to be 1 in case of operational.
Total number of laptops in Room are 4
The wattage of one laptop doing word processing and mass storage is 65 W
So cooling load due to heat gain from laptops is
Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr
Conduction through People:
The heat gained from people is composed of two parts, sensible heat and latent heat resulting from
perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat.
The equations from sensible and latent heat gains from people are
Qs = q 3 X N X CLF
Ql = q l X N
Where qs, ql = Sensible and latent heat gains per person
N = Number of people
CLF = Cooling load factor for people
Total number of people in Room are 4
For regular work, Sensible heat gain per person is qs = 250 BTU/hr
Latent heat gain per person is = 200 BTU/hr
Sensible heat gain for a Room is
50. 50 | P a g e
Qs = 250 X 4 X 1 = 1000 BTU/hr
Latent heat gain for Room is
Ql; = 200 X 4 = 800 BTU/hr
Total cooling load required for the working of people is
Qp = Qs + Ql = 1000+800 = 1800 BTU/hr
Total Cooling load of Room 6, 7, 37, 38 & 39:
Total cooling load of the room is simply the combined cooling load of individual components and
can be calculated by adding all the cooling loads calculated above
Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp
Q = 1118.259+2975.41+0+204.72+0+887.12+1800 = 6985.509 BTU/hr
Room 8 & 36
Now conduction through,
South-West Wall:
Q = 0 BTU/hr
East-South Wall:
Area of east-south wall = A = 76.5 ft
2
Considering the wall of type D, for east-south direction CLTDmax =20.5
So heat gain through south-west wall is
Q = 0.332 X 76.5 X 20.5 = 520.659 BTU/hr
North-West Wall:
Area of north-west wall = A = 120 ft2
Considering the wall of type D, for east-north direction CLTDmax =15.5
So heat gain through east-north wall is
Q = 0.332 X 120 X 30 = 1195.2 BTU/hr
There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.
So, Q = 597.6 BTU/hr
East-North Wall:
Area of east-south wall = A = 180 ft
2
Considering the wall of type D, for east-north direction CLTDmax =20.5
51. 51 | P a g e
So heat gain through east-north wall is
Q = 0.332 X 180 X 15.5 = 926.28 BTU/hr
Total heat gains through walls
Qw = 926.28+520.659+597.6= 2343.34 BTU/hr
Solar radiation through Windows Glass:
Radiant energy through sun passes through the transparent material such as glass and became a
heat gain to a room. The solar cooling load can be found by using the following formula
QG = SHGF X A G X S C X CLF
Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2
AG = Area of the glass
SC = Shading coefficient
CLF = Cooling load factor for a glass.
Area of the window is
AG = 6 X 7.25 = 43.5 ft2
Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67
Now for North window
Solar heat gain factor for east-south direction = SHGF = 124.5
Cooling load factor for east-south direction = CLF = 0.825
Heat gain for window is
Q = 195.5 X 0.67 X 0.825 X 43.5 = 2975.41 BTU/hr
Conduction through interior surfaces:
The heat that flows through the interior unconditioned spaces to the conditioned spaces through
partition, floors and ceilings also increase the cooling load. There is no portion in a floor that
remains unconditioned.
Therefore, conduction through interior surfaces is taken to be
QINT = 0 BTU/hr
Conduction through Lightings:
The equation for determining cooling load due to heat gain from lighting is
QL = 3.412 X W X BF X CLF
Where W = Light capacity in watts
BF = Ballast factor
CLF = Cooling load factor for lightings
The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights
therefore ballast factor i.e. BF = 1.25
Cooling load factor for the light is taken to be 1 in case of operational.
Total number of vented lights in the Room are 2
The wattage of one light is 24 W
So cooling load due to heat gain from lighting is
Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr