Assignment ME 5044 Assignment 2

8/10/2019 Assignment ME 5044 Assignment 2

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Familiarizing with modeling & simulation ofBuilding Energy


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Design Data:

1.Place : Lecture Hall A at Colombo

2. Design Condition

Outdoor design condition

Dry Bulb temperature : 30Relative Humidity : 80 % RH

Indoor design condition

Dry Bulb temperature : 24.0 Relative Humidity : 55% RH

3. Glass Window

Type and Thickness : Single Clear Glazing with 3mm thickness

Frame : 30 mm Aluminum

Size : 4000mm1500mm

4. Glass Doors

Type and Thickness : Double glazing 13mm air space

Frame : 40 mm Aluminum

Size : 2000mm2500mm

5. Heat transfer co-efficient of air

Still Air : 10 W/m2K

Moving Air : 15 W/m2K


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Required Preliminary Data and Assumptions

1.

Design month Assume :July

2.

Design solar Time Assume :14.00 hr

3. Location :Colombo, Sri Lanka (Latitude-70N)

4. Material of Construction of the building

External Wall

1. U value of the wall

Thermal resistance (R) of the outer wall can be calculated as

= 11 +22 +

11 =

0.02

0.5+

0.225

0.85+

0.02

0.5 = 0.345 m2K/w

Here,

1:Wall plaster thickness2:Brick thickness

1:Thermal conductivity of wall plaster2:Thermal conductivity of Brick

= 1 =1

0.345 = 2.9 w/m2K

2.

Wall construction Group is Group G

3. Area separated from partition wall is conditioned Area

4. Partition wall construction is same as external wall : U = 2.9 W/m2Floor and ceiling

1. U value of the Floor

Thermal resistance (R) of the floor can be calculated as

= 11 +22 +

33 =

0.01

1.2+

0.02

0.85+

0.2

1.7 = 0.15 m2K/w

Here,

1:Ceramic Tiles2:Ceramic Plaster3:Reinforced Concrete

1:Thermal conductivity of Ceramic Tiles2:Thermal conductivity of Ceramic Plaster3:Thermal conductivity of Reinforced Concrete


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= 1 =1

0.15 = 6.67 w/m2K

2. Ground floor space is conditioned : Ground Floor Conditioned

3. Since the second floor is well insulated, heat transferred to the Lecture halls can be taken as

negligible.

4. U value of the ceiling

Thermal resistance (R) of the ceiling can be calculated as

= =0.01

0.08 = 0.125 m2K/w

Here,

:Ceiling sheet thickness :Thermal conductivity ceiling sheet

= 1 = 10.125 = 8.0 w/m2KWindows and Doors

1.

U value of the Window is selected as 7.37 w/m2Kas per the given criteria according to table6.1.6

2. No external fins and overhangs for windows : No external shade

3. No interior shade (like curtains) for windows : No internal shade

4. Window made up of : Aluminum frame without

Thermal Break

5.

Convective heat transfer coefficient near the window : ho= 10.0 W/m2

6.

U value of the Door is selected as 4.58 w/m2Kas per the given criteria according to table6.1.6

Occupancy

1. Considering maximum people in the lecture hall A be 100 persons

2.

Activity level of occupancy is seated at rest ( according to table 6.1.5 , Sensible Heat gain = 70

W, Latent Heat gain = 30 W)

Lighting

1. Building is used for day operation, so assume need all light for the same time (Ful= 1.0)

Power

1. Equipments with motors are not used in the office space.

Appliance

1. Computers with CRT monitor. ( Sensible Heat Gain = 155 W)

Ventilation and Infiltration

1. Outdoor air requirement is 2.5 l/s-person


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Cooling Load Calculation

1. Outdoor Loads

1.1.Roof

Since the Lecture room A is located in the 1stfloor of the 3 storied building, no need to

accommodate outdoor load from the roof.

1.2.Walls

There are only three sunlit walls which are North, South and West faced walls

=(C)Table a: External Wall Details

Direction Gross wall

area/m2(a)

Window

area/m2(b)

Door

area/m2(c)

Net wall area/m

[a-(b+c)]

N 52.5 12.0 0.0 40.5S 52.5 0.0 5.0 47.5

W 42.0 12.0 0.0 30.0

Table b: CLTD Values

Direction CLTD ( Figure 6.1.16)

Wall group G - Time 14:00 h

N 12.5

S 25.0

W 22.5

Table c: Cooling Load for heat gain from walls

DirectionU /

(W/m2.oC)

A / ( m2) CLTD q / (W)

Total Cooling

Load Walls

N North 2.9 40.5 12.5 1468.1

6869.4 WS South 2.9 47.5 25.0 3443.8

W-West 2.9 30.0 22.5 1957.5

1.3.Windows and Doors

1.3.1. Conduction

=()Where;

:Cooling Load in watts:Window, Door Area: Overall Heat Transfer Co-efficient of Glass (6.0 W/m2oC):Cooling Load Temperature Difference


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Table d: Cooling Load for Conduction Heat Gain from windows and doors

Direction

Area of a

Window /

Door

( m2)

U /

(W/m2.oC)

CLTD q / (W)Total Load -

Conduction

N 12.0 7.3712.5

1105.53667.9 WS 5.0 4.58 25.0 572.5

W 12.0 7.37 22.5 1989.9

1.3.2.

Solar

=()()()Where;

:Window, Door Area:Shading Co-efficient:Solar Heat Gain Co-efficient: Cooling Load Factor

Table e: Cooling Load for Heat Gain from Glass windows (Solar)

Direction

Area of a

Window /

Door

( m2)

SC SHGF CLF q / (W)Total Load -

Glass Solar

N 12.0 1.0 243 0.75 2187.00

5230.05 WS 5.0 1.0 123 0.55 338.25

W 12.0 1.0 644 0.35 2704.80

1.4.Partitions

=

Since the Lecture hall B is a conditioned space, the heat transfer through partitions is zero.

1.5.

Floor

=Since the Ground floor is a conditioned space, the heat transfer through partitions is zero.

1.6.Ceiling

=


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Where;

:Cooling Load in watts = 180 2 = 8.0 W/m2(Overall Heat Transfer Co-efficient of Concrete Ceiling) =

= 24.0 (Room Temperature) = 30.0 (Assuming the temperature of the air inside the ceiling is equal to outdoorTemperature)

By substituting values,

= = 8.0 180 6.0 =


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2. Internal Loads

2.1. People

2.1.1. Sensible

=()()Where;

:Sensible Cooling Load in watts = 100(No. of people) = (100 W/Person): Sensible Heat Gain from the Occupancy = 1(Cooling Load Factor, Cooling Off at Night)By substituting values,

=

(

)(

)

= 100 70 1 =

2.1.2. Latent

=()()Where;

:Latent Cooling Load in watts

= 100(No. of people)

= ( 30 /): Latent Heat Gain from the Occupancy = 1(Cooling Load Factor, Cooling Off at Night)By substituting values,

=()() = 100 30 1 =

2.2. Lights

= ()() =Where;

:Heat Gain from Lighting in watts = 40 10 = 400 w: 10 Nos. of 40W incandescent Lamps available in the room = 1: Lighting use factor = 1.25: Lighting Special allowance factor

= 1(Cooling Load Factor, Cooling Off at Night)


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By substituting values,

= = 400 1 1.25 = 500 W

= (

)(

)

= 500 1 =

2.3. Appliance

Following list of appliances used in the office as assumed.

Table f: Contribution for the Cooling Load from Electrical Appliances

ApplianceNo of

Appliance

Recommended Rate

of Heat Gain / (W)

Total Heat Gain /

(W)

Sensible Latent Sensible Latent

Computers

(with CRT monitor)1 150 - 150 -

Projector 1 350 - 350 -

Total Heat Gain from all Appliances / (W) 500 -

2.3.1. Sensible

= ()()Where;

= 500 W = 1(Cooling Load Factor, Cooling Off at Night)By substituting values,

= 500 1 =2.3.2.

Latent

=.


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3. Ventilation and Infiltration

= 1.23 = 3010 Where;

= 2.5 100 = 250/: Outdoor air flow rate in l/s, According to ASHRAE 62.1:2007 itis required 2.5 l/s outdoor air per person

= ( ) = ( )kg water/ kg dry air

= 24.0 Room Temperature = 30.0 Outdoor Temperature

= 0.0106kgkg: Room Humidity Ratio (According to pshycrometry) = 0.0218kgkg : Outdoor Humidity Ratio (According to pshycrometry)Sensible Heat Load form Outdoor air to room

= 1.23 = 1.23 250 (30.0 24.0) =Effective Sensible Heat

= 1.23

= 1.23 250 (30.0

24.0) 0.1 =

.

Latent Heat Load form Outdoor air to room

= 3010 = 3010 250 (0.0218 0.0106) =Effective Latent Heat

= 3010 = 3010 250 (0.0218 0.0106) 0.1 =.


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Type of Load Cooling Load Component Sensible Load Latent Load

External Loads

Conduction and

Radiation

Walls 6869.4 W -

Glass Windows (3667.9 + 5230.5) W -

Partitions 0.0 W -

Floor 0.0 W -

Ceiling 8640 W -Internal Loads People 7000 W 3000.0 W

Lighting 500 W -

Appliances 500 W

Ventilation Load Ventilation 1845.0 W 8428.0 W

Total Cooling Load 34252.8 W 11428.0 W

Grand Total Cooling Load 45.68 kW

Total Conditioned Area 180 m

Cooling Load per unit area 253 W/ m

SHR =Room total sensible load

Room total cooling load =34.25

45.68 = 0.75

Required Cooling Coil Capacity = 46 kW with 0.75 SHRBy assuming The Refrigerant is R-22 and Evaporator temperature is -4.Enthalpy of Sub-cooled liquid() = 195 kJ/kgEnthalpy of Super-heated vapour = 403 kJ/kgEnthalpy change at evaporator () = 403 195 = 208 kJ/kgRefrigerant mass flow rate =

45.68208 = 0.22kg/s

COP of the Unit = 3.2

Power consumption of Unit = 45.683.2 = 14.28 kWOperating hours of the building = 17.00 08.00 = 09.00hEnergy consumption of the unit per day = 14.289 = 128.52 kWh

Possible Energy Saving Opportunities

Energy Saving Opportunity Strategy

Reduce heat gain from external walls Introduce Thermal Insulation for the wall

Reduce heat gain from windows Introduce double glazing with a air gap

between the glass surfaces

Introduce external shading

Introduce internal shading, blinds

Reduce lighting Load Introduce high efficient CFL or LED

lighting


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Building Energy Modeling

The Building Energy Use is then Modeled by Using eQUEST Software. Weather file of FLORIDA is

used for the simulation since it is belong to Hot - Humid category which is specified in eQUEST as 2-

A. The weather file of the Colombo is also belongs to the Hot-Humid category. It is performed Whole

Building Energy Simulation by specifying all the parameters relevant for the building.

Figure a: The University Building Modeled by using eQUEST

Lecture

Hall"A"


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Building Energy Simulation

Once modeled the building it is simulated for its energy performance

According to the simulation, the required electricity for space cool on month of July is 5490 kWh. So

assuming number of working days per month as 20, the daily electricity requirement for space cooling

can be calculated as 274.5 kWh549020 = 274.5. Electricity need to space cooling which is calculatedmanually shows 128.5 kWh which is only for lecture hall "A". Since eQUEST is used for whole

building energy simulation, it is difficult to compare these two results. But electricity requirement of

274.5 kWh for the total building is acceptable when compared to 128.5 kWh only for the lecture hall"A".

Figure b: Monthly Electricity Consumption of the Building

Figure c: Monthly electricity consumption of the building - Sector wise comparison


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It is further analyzed the University Building by changing the following measures with respect to the

Baseline Building.

Vertical fenestration of the building envelope

Increase the thermal insulation of vertical fenestration walls

Use of daylight with electrical lights

Use day light effectively in all stories according to weather file of FLORIDA

Use more efficient DX coil

Use DX coil with EER value of 13 instead of previously used coil with EER value of 11

As a conclusion it can be identified that the significant energy savings is only possible for efficient day

light use among all the energy efficiency improvement strtegies.

Figure d: Comparison of different Energy Efficiency Measures applied to University building

Assignment ME 5044 Assignment 2

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