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THERMAL LOSSES

Thermal Losses Calculations

Employer : 4M SA

Project : ASHRAE Office Room

: Example from ASHRAE 2013 Handbook - Fundamentals

: Chapter 18, Single Room Example Peak Heating Load (p. 18.45)

Location : Atlanta, Georgia

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1. INTRODUCTION

This study is based upon the ASHRAE methodology. Furthermore, the following literature was also used:

i) ASHRAE Handbook of Fundamentals 2013

ii) ASHRAE Cooling and Heating Load Calculations Principles

2. ASSUMPTIONS & RULES OF CALCULATION

The general procedure for calculating the design heat losses of a structure is the following: 1. Select outdoor design conditions.

2. Select indoor design conditions to be maintained. 3. Estimate temperature in any adjacent unheated spaces.

4. Select transmission coefficients and compute heat losses for walls, floors, ceilings, windows and doors. 5. Compute heat load through infiltration and any other outdoor air introduced directly to the space.

6. Sum the losses caused by transmission and infiltration.

2.1 Heat losses due to transmission

2.3.1) The heat transferred through walls, ceiling, roof, window glass, floors and doors is all sensible heat transfer, referred to as transmission heat loss and computed from:

q = U * A * (ti – to)

where

U : Overall heat transfer coefficient or U-factor, (W/m²K) A : Surface area, normal to heat flow, (m²)

ti :Inside design temperature, (°C) to : Outside design temperature, (°C)

A separate calculation is made for each different surface in all rooms of the structure.

2.3.2) The heat loss through below-grade walls and floors is given by:

q = Uavg * A * (ti – tgr)

where,

Uavg : Average U-factor for below-grade surface, (W/m²K) tgr : Ground surface temperature (°C)

2.3.3) The heat loss from at-grade floor slabs is given by:

q = p * Fp * (ti – to)

where,

p : Perimeter (exposed edge) of floor, (m)

Fp : Heat loss coefficient per metre of perimeter, (W/mK)

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where,

U : Average U-factor for below-grade surface, (W/m²K)

tb : Partition temperature (°C)

2.2 Heat losses due to infiltration

Infiltration is treated as a room load and during winter has a sensible component. The sensible infiltration heating load assuming standard air conditions is given by:

qs = 1.23 * Qs * (ti – to)

where

qs : Sensible heat load due to infiltration, (W)

Qs : Infiltration airflow at standard air conditions, (m³/s)

to : Outdoor air temperature, (°C)

ti : Indoor air temperature, (°C)

1.23: Air sensible heat factor at standard air conditions, (W/(m³s°C)) temperature, (°C)

3. PRESENTATION OF RESULTS

The computed results are presented in a table form as follows:

i) In the upper part of the table the building elements that have heat losses due to thermal heat conductivity are presented with their characteristics. The table columns correspond to the following data:

• Surface Type (e.g. W=wall, O=opening, C=ceiling F=floor)

• Orientation • Adjacent room • Thickness • Length • Height or Width • Surface area

• Number of equal surfaces

• Total surface area

• Subtracted surface area

• Calculated surface area

• U-factor coefficient

• U equivalent coefficient

• Temperature difference

• Net Thermal Heat Losses

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Building Parameters

City Atlanta

Design External Temperature (°C) -5.8 Desired Indoor Temperature (°C) 22.2 Not Heated Spaces Temperature (°C) 10 Soil Temperature (°C) 10 Number of Levels (Floors) 2 Floor on the Ground Level 1

Calculation Method ASHRAE HB 2013

Energy Units W

Structural Elements

Structural Elements - Outer Walls

Outer Walls Description Outer Walls

U Factor (W/m²K)

W1 Brick wall 0.45

W2 Spandrel wall 0.44

Structural Elements - Ceilings

Ceilings Description Ceilings

U Factor (W/m²K)

C1 Flat metal deck 0.18

Structural Elements - Openings

Openings Description Width

(m) Height (m) Openings U Factor (W/m²K)

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Calculations

Level : Second floor Space : 1

Space Name : Office room

Surface Type Adjacent room Length (m) Height or Width (m) Surface (m²) Equal Surface Number Total Surface (m²) Calculat. Surface (m²) U-Factor (W/m²K) Temperat. Difference (°C) Thermal Losses (W) W1 Outer Space 1 9.29 9.29 1 9.29 9.29 0.45 28.00 117.1 W2 Outer Space 1 16.72 16.72 1 16.72 16.72 0.44 28.00 206.0 O1 Outer Space 1.91 1.95 3.72 2 7.44 7.44 3.18 28.00 662.5 C1 Outer Space 1 12.7 12.70 1 12.70 12.70 0.18 28.00 64.01

Losses due to Building Elements QT (W) : 1050

Losses due to infiltration (W): qs = 1,23*Qs*∆t = 316.6

Space Volume (m³): V = 3.96*3.05*2.74 = 33

Air Changes Number per hour n = 1

Total Increment Z due to losses in the air distribution system = 0 % 0

TOTAL THERMAL LOSSES (W): Qtot = (QT + qs) * (1+Z) = 1366

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SPACES TOTAL THERMAL LOSSES (W)

Level : First floor

Total Level Thermal Losses : 0

Level : Second floor

1. Office room : 1366

Total Level Thermal Losses : 1366

References

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