Download presentation

Presentation is loading. Please wait.

Published bySheila Norrington Modified over 2 years ago

1
HEAT LOSS & HEAT GAIN HEAT FLOW

2
Heat flows from high temperature to low temperature There are 3 types/methods of heat flow/transfer: 1.Conduction- through solids 2.Convection- through liquids & through convection currents in air 3.Radiation- through space (w/ or w/o air) Heat flow is measured in BTUs (amt. of heat needed to create a 1°F change in 1 pound of water. Flow is usually in BTU/hr. or BTUH.

3
HEAT FLOW

4
Insulation slows heat flow from conduction, convection &/or radiation. Most insulation uses trapped air to minimize conduction & convection (foam, fiberglass, dual pane glass). Radiation barriers are reflective and must face an air space to function.

5
Resistance to heat flow is R Higher is better R values can be for a specific thickness or expressed in R/inch. We usually look at assemblies of materials (construction) for walls, ceilings, roofs, floors or etc. This requires us to add the R values of all the materials in the assembly plus allowances for air films on the inside and outside surfaces.

6
CONDUCTANCE or U We typically look at the reciprocal value of this summation of Rs (1/R). This is the opposite of resistance which is conductance or U. It is measured by U=(BTUH) (1SF) (1°F).

7
TYPICAL U VALUES

8
Resistance Calculation- Example The first example in your textbook is for a concrete wall w/ brick veneer. Notice that the wood furring is not used in the calculations because it is not continuous. We look at the continuous layers of materials. In this case: the outside air film (0.2), 4 brick (0.4), air space w/ foil (3.0), 2 polyurethane (12.0), 5 concrete (0.5), air space w/o foil (1.0), ½ gyp. bd. (0.5) and the inside air film (0.7) for a total R of 18.3.

9
The internal temperature of the wall assembly can be determined at each layer of material by using the R values of the different layers in comparison to the total R value of the assembly times the Temperature Differential (TD) between the inside and the outside Why do we care about the temperature of any of these layers? Two reasons: 1.Comfort of people; low surface temperatures will increase the loss of body heat by radiation (you feel cold). 2.For many structures if there is humid air passing through the assembly there can be condensation inside the assembly causing problems w/ mold, mildew & deterioration of the materials (such as rotting/rusting framing).

10
HEAT FLOW Infiltration & Ventilation We need fresh air to flow into buildings in order to maintain a healthy environment. This air can come from infiltration- air that leaks into buildings through cracks and openings in the exterior shell or from ventilation- air purposely brought into the building using fans.

11
We usually allow 15 cubic feet per minute (CFM) per person. We also consider the number of air changes per hour. Both the recommended CFM/person and the air changes/hour can vary w/ conditions. There are tables and charts that give these recommendations or averages (15.2).

12
Infiltration is typically calculated by: OA= (Volume) (ACH) 60 (mins.) Where OA is the outdoor air in CFM; Volume is the CF of the buildings interior; ACH is the rate of air change per hour and 60 minutes gives you the air flow in minutes-CFM.

13
Ventilation Ventilation is typically calculated using the recommended cfm/person times the number of occupants.

14
In order to minimize the cost of heating and cooling: 1.Use outdoor air when it can warm (in winter) or cool (in summer) the inside. 2.Keep out outdoor air when it will impose additional heating or cooling loads. 3.When ventilation is needed, reduce energy loss by using a heat exchanger to temper the incoming air (warm it or cool it as appropriate).

15
Estimating Building Heat Losses Use the largest or highest heat loss. Calculate heat loss for: 1.Conduction through walls, roof, ceilings, floor & etc. Q= U x A x TD. 2.Infiltration of cold, outside air Q = CFM x 1.08 x TD.

16
Conduction: Q = U x A x TD Q= quant. Of heat flow in BTUH U= conductance in BTUH per sf per°F of the assembly A= area of the assembly in sf TD= Temperature Difference in °F between indoor & outdoor

17
You will need to calculate Q for each assembly (wall, roof, floor etc.) and total all the Qs.

18
Infiltration (outside air) Q= CFM x 1.08 x TD Q= quant. of heat flow in BTUH 1.08 is a constant; the # of BTUH needed to increase the temperature of 1 CFM of air 1°F. TD= the temperature difference in °F between indoor and outdoor air

19
INFO & CALCS REQUIRED For most assemblies you will need to calculate U by adding the R values of the components of the assembly. 1/Total R=U. Will need to calculate the SF of the assembly. (may need to deduct doors, windows & etc.) Will need to know the outside temperature & usually assume an indoor temperature of 70°F. Will also need the ventilation rate, if there is one, and use the higher of infiltration or ventilation CFM.

20
INFO & CALCS REQUIRED (cont.) You will also need to consider: *Ducts outside the insulated space (attic). *Basement walls & floors. *Unheated spaces adjacent to heated space (garage or storage). *Slab edge if it is exposed on the outside.

21
Use the form shown on p. 206 for calculating Building Heat Loss (or Gain) What you will need to know (for winter heat loss): 1.The winter design temperature (Fig. 15.8). 2.Infiltration & ventilation rates; pick the highest CFM (usually infiltration). 3.U values for various assemblies: glass (windows), roof, walls, floor, slab edge, doors & etc. 4.SF of each of these assemblies. 5.Add 10% if ductwork is in noninsulated space.

22
ON THE FORM Fill in factors and quantities, calculate each BTUH loss and total all the losses to get the Total BTUH Loss.

23
Office Building

24
Office Building Heat Loss 1. Location- 10°F winter temp. & 70°F indoor temp= 60°F TD. 2. Floor area= (180x60)2 floors = 21,600 sf. 3. Infiltration=0.75 air changes/hr x 21,600sf x 11clgs= 178,200CF /60 min= 2,970 CFM. 4. Given other top: glass 0.6 U, people 160, ceiling/roof 0.05 U, walls 0.07 U, floor factor of 2, door 1.1 U, slab edge 0.8 U.

25
Office Building Heat Loss (cont.) 5. Glass total given 2,400 sf. 6. Ceiling-roof & floor is ½ of 21,600 sf= 10,800 sf. 7. Wall area given 10,880 sf. 8. Slab edge (67) = 494 lf. 9. Doors 5x7= 70 sf. 10. Fill in info, do calcs, sum the numbers & get Total of 406,850 BTUH.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google