THERMAL LOSSES Thermal Losses Calculations

Transcription

1 Calculations -1- THERMAL LOSSES Thermal Losses Calculations Employer : 4M SA Project Location : ASHRAE Office Room : Example from ASHRAE 2013 Handbook - Fundamentals : Chapter 18, Single Room Example Peak Heating Load (p ) : Atlanta, Georgia

2 Calculations 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: where q = U * A * (ti to) U A ti to : Overall heat transfer coefficient or U-factor, : area, normal to heat flow, (m²) : Inside design temperature, ( C) : Outside design temperature, ( C) A separate calculation is made for each different surface in all rooms of the structure ) The heat loss through below-grade walls and floors is given by: where, q = Uavg * A * (ti tgr) Uavg tgr : Average U-factor for below-grade surface, : Ground surface temperature ( C) 2.3.3) The heat loss from at-grade floor slabs is given by: where, q = p * Fp * (ti to) p Fp : Perimeter (exposed edge) of floor, : Heat loss coefficient per metre of perimeter, (W/mK) 2.3.4) The heat loss to adjacent unconditioned or semiconditioned spaces is given by: q = U * A * (ti tb)

3 Calculations -3- where, FINE HVAC 14 Thermal Losses U tb : Average U-factor for below-grade surface, : 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: where q s Q s t o t i q s = 1.23 * Q s * (ti to) : Sensible heat load due to infiltration, (W) : Infiltration airflow at standard air conditions, (m³/s) : Outdoor air temperature, ( C) : 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: Type (e.g. W=wall, O=opening, C=ceiling F=floor) Orientation Adjacent room Thickness Length Height or Width 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 ii) In the lowest part of the table the increments as well as the losses due to ventilation are filled in, in detail.

4 Calculations -4- Building Parameters City Atlanta Design External Temperature ( C) -5.8 Desired Indoor Temperature ( C) 22.2 Not Heated s 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 - Walls Walls Description Walls U Factor W1 Brick wall 0.45 W2 Spandrel wall 0.44 Structural Elements - Ceilings Ceilings Description Ceilings U Factor C1 Flat metal deck 0.18 Structural Elements - Openings Openings Description Width Height O1 Double glazed window Openings U Factor

5 Calculations -5- Calculations Level : Second floor : 1 Name : Office room Type Adjacent room Length Height or Width (m²) Equal Number Total (m²) Calculat. (m²) U-Factor Temperat. Difference ( C) Thermal Losses (W) W1 W2 O1 C Losses due to Building Elements Q T (W) : 1050 Losses due to infiltration (W): q s = 1,23*Qs* t = 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 = (Q T + q s ) * (1+Z) = 1366

6 Calculations -6- Level : First floor SPACES TOTAL THERMAL LOSSES (W) Total Level Thermal Losses : 0 Level : Second floor 1. Office room : 1366 Total Level Thermal Losses : 1366 Total Building Thermal Losses : 1366