final report thermal

7/29/2019 Final Report Thermal

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MEKELLE UNIVERSITY

ETHIOPIAN INSTITUTE OF TECHNOLOGY, EiT-M

DEPARTMENT OF MECHANICAL ENGINEERING

M.Sc. Program in Energy Technology

June, 2012

Course Title: Thermal System Design

Air Conditioning and Ventilation Assignment

Submitted to: Demiss A. (PhD) Prepared by: Alemnew Ebabu

Ashenafi K. (MSc.) Akatew Haile

Binyam Gebray


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Assignment -1-

Design of ventilation system .

The Restaurant shown below is to be maintained at a constant environmental temperature of

22o

C for a plant operation of 12 hours per day.The Restaurant area is on the ground floor of anSingle storey building located at 51.7oNThe internal construction is lightweight partitions,

concrete hollow slab floors and suspended ceilings.it was desired to design a ventilation system

for the restaurant.

the following ste[ps are neccesary to be carried out the plan view of the restaurant is shown

belaow

Calculate Ventilation rates.

Decide on number of fans and grilles/diffusers.

Draw scale layout drawing and Position fan(s).

Lay out ductwork. Lay out grilles and diffusers.

Indicate flow rates on drawing. Size ductwork

Size fan Size grilles and diffusers

14.0 m

7.7

9.5

FemaleToilet

MaleToilet

Receptio

Entranc

10.0Lobby

Restaurant

3.5

Kitchen

2.7m

Cold

Store

Prep.

Room

PLAN

South

Height of wall to

eaves

Height of

ceiling at


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Figure : the elevation plan view

Sn.

Design data for the restaurant

1 Occupants 70

2 Infiltration 1.0 ACH

3 Outside air temperature 28oC.

4 Area of window 2.8 m2

5 Total area of glass 28.0 m2

6 Area of glass facing South 28.0 m2

7 Area of wall facing South 28.0 m2

8 Floor area 140 m2

9 Room volume 623 m3

10 Building classification lightweight, fast response building

South


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Table: ventilation system design data for occupancy, building category and exterior structure

dimensions

Ventilation requirements;

In this design it is proposed to ventilate the kitchen, the preparation room and the restaurant

and toilets separately using 4 exhaust fans the position of the fans its to be decided after the

longest path for which the maximum pressure drop occurs is determined the supply fan should be

as far as possible from the restaurant to avoid noise creation

All rooms the cheaper alternative is an extract system with replacement air coming in through

doors and operable windows. Fans should be located in accessible positions; in the corridor,

above ceilings. In-duct axial flow fans take up less space than centrifugal fans. Possible use ofwall mounted fan in Food Prep. Room. Upward with diffusers installed in suspended ceilings.

Use separate extract systems so that smoke will not spread from room to room through ducts.

Consider fire dampers in ducts. Fans are positioned as far away from Conference room and

restaurant as possible. Choose fans from catalogue with decibel levels less than 85dB if possible.

Table: restaurant rooms ventilation requirements as recommended by the Table B2.3 (CIBSE

1986)

ventilation requirements for restaurant rooms

Room or building type Recommended fresh air

supply rate

Recommended total air

supply rate(ACH)

Restaurant

Kitchen 10

Female toilet 8

Male toilet 8


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Table: calculated ventilation rates for the rooms suggested requiring ventilation

The next is to calculate the fresh rate this was done by assuming that the outdoor air supply per

person should suit for a heavy smoking area this is because of the possibility that heavy smokers

may arrive as consumers at restaurants. And since we are dealing with restaurants it is fair to

assume a heavy smoking area with a recommended air supply rate per person of24 liters per

second. Therefore similarly like the ventilation requirements and assuming a peak of70 persons

the outdoor fresh air supply was found to be 1680

Table: recommended outdoor air supply rates

ROOMS volume(m3) Recommended (ACH) ventilation rate (m3/hr) ventilation rate(m3/s)

kitchen 217 35 7595 2.109722222

preparationroom 47.25 10 472.5 0.13125

male toilet 84 8 672 0.186666667

female

toilet 140 8 1120 0.311111111

restaurant 980 10 9800 2.722222222

12064.5 3.35125

Level of SmokingOutdoor air supply rate

(liter/s per person)

No smoking 8

Some smoking 16

Heavy smoking 24

Very heavy smoking 36


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Table: calculated outdoor air supply rate

CALCULATING THE FRESH AIR RATE

occupants outdoor supply air(heavy smoking l/s/person) fresh air rate(l/s)

70 24 1680

Duct sizing

The recommended velocity for main duct and branch ducts for restaurants are 7.5m/s and 5m/s

respectively therefore by choosing a square duct with aspect ratio of 1 we can determine the

nearest standard sizes of the ducts and the following standard sizes were obtained for the main

duct and the branching ducts to the rooms which require ventilation.

As shown on the table below the main duct area is 0.45m

2

and the branching duct are also shownby considering a square duct the nearest standard sizes were determined for the rooms kitchen,

preparation room ,restaurant and toilets respectively

Table: standard size duct selection

Branching ducts(m2) Main duct area(m2) Square duct size(m)

Nearest

std.size(mm)

0.421944444 0.446833333 0.649572509 300*300mm

0.02625 0.162018517 150*150mm

0.037333333 0.193218357 150*150mm

0.062222222 0.249443826 150*150mm

0.668455932

500mm*500mm(std.)


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Table : Recommended Duct velocities

Design tool duct sizer

BuildingAir Velocity (m/s)

Main Duct Branch

Domestic 3 2

Auditoria 4 3

Hotel bedroom, Conference hall 5 3

Private office, Library, Hospital ward 6 4

General office, Restaurant, Dept. store 7.5 5

Cafeteria, Supermarket, Machine room 9 6

Factory, Workshop 10-12 7.5


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As t can be seen from the results of the duct sizer air at 370c and 23% relative humidity at 1

atm is assumed the results were found to be consistent with spreadsheet results moreover the

velocity pressure and the static pressure losses were found to be 0.749pa/m and 31.5pa for the

dynamic losses therefore the total head loss is given by

Total head loss= velocity pressure loss+ static head loss .

To compute the total pressure loss it is first necessary to identify the index run that is the longest

path of the duct work layout a suitable duct layout system is proposed next which roughly

identifies the longest path for which the highest pressure drop occurs and which is late used as a

selection criteria for the fan sizing.

Figure: rough duct layout system for the main and branching for the whole building to estimate

the index run

The duct layout can be laid out as shown above to roughly estimate the index run from the

supply side of the fan therefore the maximum run could be the duct system from the fan position

at the reception site to the branch duct for the preparation room which runs half the perimeter of

the whole building so it is estimated as 22m.


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the next step is therefore to estimate the total static head loss for the index run which enables to

select the appropriate fan type by using the flow rate already obtained for their and the total

pressure earlier the pressure loss length was found as 0.79pa/m which gives

Total static pressure loss across the index run=22*0.79=31.5pa

Total pressure loss=static head loss dynamic head loss=49pa

from the calculations and from the table above showing the standard duct size selection

unequal size of branching ducts with duct dimensions ranging from150mm*150mm-

300*300mm were selected and 500mm*500mm standards size square duct were chosen .the

total pressure loss that the supply fan must overcome is found to be 49pa

Fan selection

Given the total pressure drop along the index run and the air flow rate for the main supply duct it

is possible to choose the appropriate fan from available catalogs from the above designcalculations the operation criteria was found to be 12064m

3/hr and 49pa.

Figure: fan performance curves for various models obtained from catalog


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the fan performance curve for the model 634-b model is suitable for the air flow rate and total as

as shown by the red lines corresponding to the flow rate and total pressure losses obtained from

calculations 12604m3/hr and 49pa.

Table; product specification of fans type cc 630

For more comprehensive design one can look at the power requirements and noise conditions byreferring to the specifications provided in the above table the model we have chosen has a noise

generation of71db which is acceptable for the restaurant compared to 85db limit.


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AC System Design for the Assembly Hall

Introduction

An air conditioning system is used to give comfort for humans by creating favorable conditions

in residence places, working environments, assembly halls etc. The system does this byhumidifying or dehumidifying the air and maintaining the temperature level of the air. During

the process heat is either supplied or extracted from an area. The cooling is done using a simple

refrigeration cycle.

Air conditioners and refrigerators work on the same principle. Instead of cooling just the small,

insulated space inside of a refrigerator, an air conditioner cools a room, a whole house, or anentire business. Air conditioners use refrigerants that can be easily converted from a gas to a

liquid and back again. This chemical is used to transfer heat from the air inside of a home to the

outside air.

The air conditioning unit has three main parts. These are a compressor, a condenser and an

evaporator. The compressor and condenser are usually located on the outside air portion of theair conditioner. The evaporator is located inside the house.

The working fluid arrives at the compressor as a cool, low-pressure gas. The compressorsqueezes the fluid. This packs the molecule of the fluid closer together. The closer the molecules

are together, the higher its energy and its temperature.

The working fluid leaves the compressor as a hot, high pressure gas and flows into the

condenser. When the working fluid leaves the condenser, its temperature is much cooler and ithas changed from a gas to a liquid under high pressure. The liquid goes into the evaporator

through a very tiny, narrow hole. On the other side, the liquid's pressure drops. When it does it

begins to evaporate into a gas.

As the liquid changes to gas and evaporates, it extracts heat from the air around it. The heat in

the air is needed to separate the molecules of the fluid from a liquid to a gas. The evaporator has

metal fins to help in exchange the thermal energy with the surrounding air. By the time theworking fluid leaves the evaporator, it is a cool, low pressure gas. It then returns to the

compressor to begin its trip all over again. Connected to the evaporator is a fan that circulates the

air inside the house to blow across the evaporator fins. Hot air is lighter than cold air, so the hotair in the room rises to the top of a room.

There is a vent there where air is sucked into the air conditioner and goes down ducts. The hot air

is used to cool the gas in the evaporator. As the heat is removed from the air, the air is cooled. Itis then blown into the house through other ducts usually at the floor level.


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In selecting a suitable air conditioning system for a particular application, consideration is given

to the following:

System constraints : Cooling load, Zoning requirements, Heating and ventilation

Architectural Constraints : Size and appearance of terminal devices, acceptable noise

level, Space available to house equipment and its location relative to the conditioned

space, acceptability of components in the conditioned space

Financial Constraints : Capital cost, Operating cost, Maintenance cost

Figure: Basic refrigeration cycle used in air conditioning systems

Air conditioning systems are categorized in three main parts. These are: central plants, room air

conditioning units and fan coil units. In central plant air conditioning units, there is one source of

conditioned air which is distributed in different parts through a network of ducts. Room air

conditioning units use refrigerant to give cooling effects for a room. Fan coil air conditioning

types use chilled water instead of refrigerant to produce conditioned air.

Figure: Central air conditioning system


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Room air conditioning units fall into two main categories: These are split type with indoor and

outdoor unit and window/wall units with evaporator at inside and condenser at the outside face.

Split Room air conditioning units: Split air conditioners have two main parts, the outdoor unit

is the section which generates the cold refrigerant gas and the indoor unit uses this cold

refrigerant to cool the air in a space.

The outdoor unit uses a compressor and air cooled condenser to provide cold refrigerant to a

cooling coil in the indoor unit. A fan then blows air across the cooling coil and into the room.

The indoor unit can either be ceiling mounted (cassette unit), floor mounted or duct type.

Figure: Split air conditioning units

Window/Wall Units: Window or wall units are more compact than split units since all the plant

items are contained in one box. Window units are installed into an appropriate hole in the

window and supported from a metal frame.


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Wall units like the one shown below are built into an external wall and contain all the necessary

items of equipment to provide cool air in summer and some may even provide heating in winter.

Figure: wall air conditioning units


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Objective

The main objective of this project work is to design air conditioning system for an assembly hall.

Given Parameters

Number of people: 375

Building height: 6 meters

Number of windows: 6

Window size: 2x4 meters

Orientation of windows: south east and south west

Basic Steps for Design

Air conditioning is required in buildings which have a high heat gain and as a result a high

internal temperature. The heat gain may be from solar radiation and/or internal gains such as

people, lights, business machines etc.

If the inside temperature of a space rises to above 25 oC then air conditioning is necessary to

maintain comfort levels. This internal temperature (around 25oC) may change depending on a

number of variables such as:

type of building

location of building

duration of high internal temperature

Expected comfort conditions.

degree of air movement

percentage saturation

Cooling load or heating load determination is based on annual climate data, building data and

occupancy data. The climate data should include monthly dry bulb temperature, wet bulb

temperature, average solar radiation, relative humidity etc.

The building data includes floor area, building height to determine the total volume of the space.

In addition to this, a number of building parameters are required for heating or cooling load

calculation.

The occupancy data are important for heating/cooling load determination. The flow rate

requirement of conditioned air at different parts of a room also depends on this parameter.


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Figure: Air conditioning system design flow process chart

Cooling Load Determination

The load on an air-conditioning system can be divided into the following sections:

1. Sensible Transmission through glass.

2. Solar Gain through glass.

3. Internal Heat gains


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4. Heat gain through walls.

5. Heat gain through roof.

6. Ventilation and/ or infiltration gains.

The heat gain through the glass windows is divided into two parts since there is a heat gain due totemperature difference between outside and inside and another gain due to solar radiation shining

through windows.

Heat gains through solid ground floors are minimal and can be neglected.

Sensible Transmission Through Glass

This is the Solar Gain due to differences between inside and outside temperatures. In very warmcountries this can be quite significant. This gain only applies to materials of negligible thermal

capacity i.e. glass.

Qg = Ag . Ug (To- Tr) .

Where;

Qg = Sensible heat gain through glass (W)

Ag = Surface area of glass (m2)

Ug = 'U' value for glass (W/m2 oC)

To = outside air temperature (oC)..

Tr = room air temperature (oC)

Solar Gain Through Windows

This gain is when the sun shines though windows. The cooling loads per metre squared window

area have been tabulated in stanadrds. for various; locations, times, dates and orientations. Thesefigures are then multiplied by correction factors for; shading and air node correction factor.

Heat load is determined as follows;

Qsg = Fc . Fs . qsg . Ag

Where: Qsg = Actual cooling load (W)

Qsg = Tabulated cooling load (W/m2)

Fc = Air node correction factor from Table

Fs = Shading factor.

Ag = Area of glass (m2)


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Internal Heat Gains

Internal gains can account for most heat gain in buildings. These gains are from occupants,

lights, equipment and machinery, as detailed in the figure below. Occupants Sensible and latent

heat gains can be obtained from tables.

Figure: Internal heat gains for a building

Table: Internal heat gains from occupants at different activity levels

Conditions Typical building

Sensible

Heat Gain

(Watts)

Latent Heat Gain

(Watts)

Seated very light workOffices, hotels,

apartments70 45

Moderate office workOffices, hotels,

apartments75 55

Standing, light work;

walking

Department store, retail

store 75 55

Walking standing Bank 75 70

Sedentary work Restaurant 80 80

Light bench work Factory 80 140


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Heat Gain Through Walls

This is the unsteady-state heat flow through a wall due to the varying intensity of solar radiation

on the outer surface.

4.1 Sol-Air Temperature

In the calculation of this heat flow use is made of the concept of sol-air temperature, which is

defined as; the value of the outside air temperature which would, in the absence of all radiation

exchanges, give the same rate of heat flow into the outer surface of the wall as the actual

combination of temperature difference and radiation exchanges.

Teo = Ta + ( )

Where:

Teo = sol-air Temperature (oC)

Ta = outside air temperature (oC)

= absorption coefficient of surface

I = intensity of direct solar radiation on a surface at right angles (W/m2)

A = solar altitude (degrees)

n = wall-solar azimuth angle (degrees)

Is = intensity of scattered radiation normal to a surface (W/m2

)

hso = external surface heat transfer coefficient (W/m2oC)

4.2 Thermal Capacity

The heat flow through a wall is complicated by the presence of thermal capacity, so that some of

the heat passing through it is stored, being released at a later time. Thick heavy walls with a high

thermal capacity will damp temperature swings considerably, whereas thin light walls with a

small thermal capacity will have little damping effect, and fluctuations in outside surface

temperature will be apparent almost immediately.

Heat Gain Through Roof

The heat gain through a roof uses the same equation as for a wall as shown below.

Q+Roof = A U [( Tem - Tr) + f ( Teo - Tem)]

. I . cos a . cos n + Is

hso


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Ventilation and/or Infiltration Gains

Heat load is found from;

Qsi = n . V (To- Tr) / 3

where Qsi = Sensible heat gain (W)

n = number of air changes per hour (h-1)

V = volume of room (m3)

To = outside air temperature (oC)

Tr = room air temperature (oC)

Infiltration gains should be added to the room heat gains. Recommended infiltration rates are 1/2

air change per hour for most air-conditioning cases. Ventilation or fresh air supply loads can be

added to either the room or central plant loads but should only be accounted for once.

Total Room Load From Heat Gains

Q total = Qg + Qsg + Qint. + QWall + Q Roof + Qsi

Cooling Load for the Assembly Hall

For determination of the cooling load of the assembly hall, we have taken the following data and

assumptions:

Data:

Number of People = 375

Number of windows = 6

Window size = 2x4 m

Orientation of windows = South east and south west

Average building height = 6 meters


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Assumptions:

Infiltration = 0.8 air changes per hour

Building classification = medium weight

Building response = fast

External wall 'U' value = 0.45 W/m2o

C

External wall colour = light.

External wall decrement factor f = 0.7

Glass type & 'U' value = clear 5mm, double glazing, U = 2.80 W/m2oC

Lighting = 20 Watts / m2 floor area

Heat gain from equipments = 1000 Watts

Maximum outside temperature = 280C

Room maintained temperature = 220C

Area calculation:

Area of window = 2 x 4 = 8 m2.

Total area of glass = 8 x 6 No. windows = 48 m2.

Area of glass facing South = 48 m2

Area of wall facing South = 24 m x 6 m high = 96 m2

less glass

= 144 - 48 = 96 m2

Floor area = (92 + 20) m2.

Room volume = 112 x 6 = 672 m3.

Gains:

1. Sensible transmission through glass Qg = Ag Ug (To - Tr)

Qg = 48 x 2.8 (28 22)

Qg = 806.4 Watts


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2. Solar Gain through glass Qsg = Fc Fs qsg Ag

Where:

Qsg = Actual cooling load (W)

Fc = Air node correction factor from forinternal blind, fast response - 0.91.

Fs = Shading factor fast response 0.95.

qsg = 260 W/m2

Ag = Area of glass facing South (m2)

Qsg = 0.91 x 0.95 x 260 x 48

Qsg = 10788.96 Watts

3. Internal Qint. = Qint.

Qint. = Lights (20 W/m2 x 112) + 1000 W + People (375 x 100)

Qint. = 2240 + 1,000 + 37500

Qint. = 378,240 Watts

4. External wall Q Wall = A U [( Tem - Tr) + f ( Teo - Tem)]

Where:

Q = heat gain through wall at time q+f (Watts)

A = area of wall facing South (m2)

U = overall thermal transmittance given in question as 0.45 W/m2 o

C

Ttem = 240C

Tr

= constant dry resultant temperature (oC). Room dry bulb of 21oC is given

f = decrement factor for wall is given as 0.65

Tteo = 370C


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Q+Wall = 96 x 0.45 [( 24 22) + 0.65 ( 37 24)]

Q+Wall = 451.44 Watts

5. Roof Q+Roof = Nil for intermediate floor

6. Ventilation Qsi = n V (to - tr) / 3

Qsi = 0.8x672 (28 22) / 3

Qsi = 1075.2 Watts

7. Q total = Qg + Qsg + Qint. + Q+Wall + Q +Roof + Qsi

Q total = 806.4 + 10788.96 + 378,240 + 451.44 + 0 + 1075.2

Q total = 391,362 Watts or 391.362KW

Air Flow Rate Determination

Once we have determined the cooling load of the building, For an Air Conditioning system the

supply air flow rate for cooling is found from the following formulae:

m = H / (Cp x (tr ts))

Where:

H = Sensible heat gain (kW)m = mass flow rate of air (kg/s)

Cp = Specific heat capacity of air (1.005 kJ/kg K)

tr = room temperature (oC)

ts = supply air temperature (oC)

m = H / (Cp x (tr ts))

= 391.362/ (1.005* (22 15)

= 55.63 kg/s

Mass flow rate is converted to a volume flow rate as follows:

Volume flow rate (m3/s) = mass flow rate (kg/s) / density of air (kg/m3)

= 55.63 (kg/s)/1.225 (kg/m3)

= 4.52 m3/s or 16,351 m3/h


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Duct sizing

From the above table the recommended air velocity for a conference hall for the main and branch

ducts are highlighted and are 5m/s and 3m/s respectively. in order to be able to size the duct the

recommended 5m/s is first taken to dimension the duct with a flow rate of16,351 m3/hr.

Building

Air Velocity (m/s)

Main

DuctBranch

Domestic 3 2

Auditoria 4 3

Hotel bedroom,

Conference hall5 3

Private office, Library,

Hospital ward6 4

General office, Restaurant,

Dept. store7.5 5

Cafeteria, Supermarket,

Machine room9 6

Factory, Workshop 10-12 7.5


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The duct characteristics were calculated using the duct sizer as shown in the above for the main

supply duct and it was found out that the static head loss per length was 0.103pa/m and the

dynamic pressure loss of15.2pa.

The next step would be to compute the total pressure loss and in turn find out the supply fan

Index run should be identified to calculate the total static head loss therefore 65m is assumed to

be the longest run to determine the total static head loss.

Hence total static pressure loss=index run*0.103pa/m

=65*0.056=3.64pa

Total pressure loss=6.7+15.2

=21.9pa

The duct sizes for the branching ducts can also be determine similarly using the e duct sizer and

the following results were found for the entrance air conditions assumed.

The equivalent diameter for the branching ducts are with a flow area of by assuming a square

duct and since for uniform air distribution the diffusers must supply equal flow rates the flow

rate pre each diffuser equals 16351/16=1022m3/hr and the recommended air velocity for the

branches for the conference halls is 3m/s this means that a flow area of

1022/3*3600m2=0.104m

2is the minimum duct area for the diffusers .by assuming square duct

which is 323mmduct and the nearest standard size duct cab be assumed with 300mm*300mm

16351 m3/hr

1022m3/hr

300mm*300mm

Square duct


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final report thermal

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