Advanced Thermography Investigations

Transcription

1 Advanced Thermography Investigations Examples of Thermography to Extend our Knowledge 1

2 Infrared Training Center Global Leader in IR Thermography Training. ISO-9001 Registered NETA accredited in IR Thermography Training Level I, II and III Condition Monitoring Infrared Training Certifications Building Investigation Courses R&D Courses Electrical Application Courses Mechanical Application Courses Roofing Application Courses 2

3 3 Your Presentor Jay Bowen Electrician 34 years Thermographer 17 years Level III Teaching Thermography 16 years Master Electrician - Wisconsin Certified Electrical Inspector - Wisconsin Designer of Electrical Systems - Wisconsin Associate Degree Electronics Technology BPI Certified Building Analysis Involved in construction since 1977 Green Bay, Wisconsin Packer Fan (season ticket holder), Coin Collector, Fishing, Metal Detectorist

4 Missing insulation Things we normally look for! 4

5 Covered real well by ASTM C1060 C a 5

6 Covered real well by RESNET 6

7 Air problems Things we normally look for! 7

8 Covered real well by ASTM E1186 8

9 Moisture Things we normally look for! 9

10 Not Covered real well by specific standards ASTM E (2008) Standard Test Method for Field Determination of Water Penetration of Installed Exterior Windows, Skylights, Doors, and Curtain Walls, by Uniform or Cyclic Static Air Pressure Difference ASTM C Standard Practice for Location of Wet Insulation in Roofing Systems Using Infrared Imaging ASTM E331-00(2009) Standard Test Method for Water Penetration of Exterior Windows, Skylights, Doors, and Curtain Walls by Uniform Static Air Pressure Difference 10

11 Lets step up and look at these another way 11

12 What else can we find with these typical images? Outdoor temp is 10F q U A T q =.25 x.5 x 43= Btu / ft 2 /hr Btu/hr x 24hr x 180 days= 23,220 BTU If we use the camera to get some basic temperatures and we surmise the wall cavity R value. We can determine a heat loss value for this deficiency. 23,220 Btu x 100,000 x x $1.11 = $.264 BTU / get to the CCF or 100 cubic feet. There is roughly 1000 BTU for 1 CF for natural gas. Then the therms per CCF which is what you pay for. 12

13 What else can we find with these typical images? Outdoor temp is 10F q A ( T T ) 1 2 R q = (.5 x 43) / 4 = Btu / ft 2 /hr Btu/hr x 24hr x 180 days= 23,220 BTU If we use the camera to get some basic temperatures and we surmise the wall cavity R value. We can determine a heat loss value for this deficiency. 23,220 Btu x 100,000 x x $1.11 = $.264 BTU / get to the CCF or 100 cubic feet. There is roughly 1000 BTU for 1 CF for natural gas. Then the therms per CCF which is what you pay for. 13

14 Fuels and energy & R-Values BTU Comparison of fuels: 1 cubic foot of natural gas 1,000 BTUs 1 gallon of #2 fuel oil 132,000 BTUs 1 gallon of HD5 propane 91,700 BTUs 1 ton of coal 16 million BTUs 1 kw of electricity 3,413 BTUs 1 ton of wood pellets 13.9 million BTUs 1 cord white birch 20.3 million BTUs Infrared Training Center. All rights reserved. 14

15 What else can we find with these typical images? Outdoor temp is 15F q 1.08 ACH V ( volume) ( T 1 T2 ) 60 q = (1.08 x.05 x 250 x 42)/ 60= 9.45 Btu/hr If we use the camera to get some basic temperatures and we surmise the air volume and ACH. We can determine a heat loss value for this deficiency Btu/hr x 24hr x 180 days= BTU/hr 40824BTU/hr / 100,000 x x $1.11 = $.464 BTU / get to the CCF or cubic feet. There is roughly 1000 BTU for 1 CF for natural gas. Then the therms per CCF which is what you pay for. 15

16 What else can we find with these typical images? Q h A ( T a Ts ) q = 14 x 2 x (70-52)= 504 Btu 504 Btu / hr x 24hr x 10 days = BTU BTU / 100,000 x x $1.11 = $1.375 We can determine a heat loss value for this deficiency. Though this scenario is very difficult as the parameters that are involved are not constant. Q 1060 cfm 60 x q = 60 x 1060 x.075 x.5 x.3= Btu / hr Btu / hr x 24hr x 10 days= BTU BTU / 100,000 x x $1.11 = $1.95 BTU / get to the CCF or cubic feet. There is roughly 1000 BTU for 1 CF for natural gas. Then the therms per CCF which is what you pay for. 17

17 Water evaporating takes quite a lot of heat away calories per gram or 970 Btu/pound -- when it evaporates. That's enough to cool down 1 pound of water by a degree. Latent Heat of Product Evaporation *) (kj/kg) (Btu/lb) Ammonia Aniline Benzene Bromine Carbon bisulphide 160 Carbon dioxide Glycerine Helium 21 9 Hydrogen Iodine Kerosene Methyl chloride 406 Nitrogen Oxygen Propane Propylene Propylene glycol Water

18 Here are 3 other applications not typically encountered by thermographers; Solar collectors and radiant systems efficiency troubleshooting Photo voltaic Collectors troubleshooting R value determination 20

19 Solar collectors Opportunities 21

20 Evacuated Tube 22

21 Evacuated Tube vapor condenses to a liquid (water) vapor rapidly rises to the top boiling point of only around 30 o C (86 o F) space inside evacuated 23

22 Flat panel Copyright by Gary Reysa Reprinted with permission 24

23 Flat Plate Fluid in tubes absorb energy Should never get to the boiling point of the fluids 25

24 Collector Efficiency qout Qout t m t c T 60F c( Toutlet Tinlet ) Qout qout is heat flow in BTU / hr t m 60F is mass flow in lb/hr t F is flow rate in gallons / minute ρ is fluid density in lb / gallon c is specific heat in BTU / (lb-ft) T outlet is outlet temperature in F T inlet is inlet temperature in F q out = 60*F*8.3*1.0*(T outlet - T inlet ) BTU/hr for water 26

25 Density of water vs. temperature Temperature - t - ( o F) Specific Volume - v - (ft 3 /lb) Weight Density - ρ - (lb/ft 3 ) (lb/gallon)

26 Specific heat of water Temperature ( o F) Density (lb m /ft 3 ) Specific Heat (Btu/lb m degr) Viscosity (10-7 lb f sec/ft 2 )

27 Specific Heat Capacity of Ethylene Glycol mixed Water Solutions c p - of ethylene glycol based water solutions at various temperatures Specific Heat - c p - (Btu/lb. o F) Temperature Ethylene Glycol Solution (% by volume) ( o F) ( o C) ) 1) 1) 1) ) ) 1) ) 2) 2) 2) 2) ) 2) 2) 2) 2) 2) below freezing point 2.above boiling point 29

28 Specific Gravity- SG - Temperature Ethylene Glycol Solution (% by volume) ( o F) ( o C) ) 1) 1) 1) ) ) 1) ) 2) 2) 2) 2) 2) ) 2) 2) 2) 2) 2) below freezing point 2.above boiling point 30

29 Specific Heat of Propylene Glycol Solutions Specific Heat - Propylene Glycol Solution(%) by volume c p - (Btu/lb. o F) Specific heat capacity of propylene glycol 1 Btu/(lb mo F) = 4,186.8 J/(kg K) = 1 kcal/(kg o C) 31

30 Propylene Glycol Solution (%) by volume Specific Gravity - SG - 1) Specific Gravity - SG

31 Water based - Then the heat output is Tinlet = 80F Toutlet = 100F Flow Rate = 1.2 gpm q out = 60*Flow*fluid density*heat capacity*(t outlet - T inlet ) BTU/hr q out = 60*1.2*8.3*1.0*(100-80) BTU/hr q out = BTU/hr Assuming clean water 34

32 Glycol based - Then the heat output is Tinlet = 80F Toutlet = 100F Flow Rate = 1.2 gpm q out = 60*Flow*fluid density*heat capacity*(t outlet - T inlet ) BTU/hr q out = 60*1.2*8.82*.86*(100-80) BTU/hr q out = 10,923 BTU/hr Assuming 40% Ethylene glycol and water.86 is the specific heat and the weight is the of water x the specific gravity this mix at

33 Radiant Heating 36

34 Radiant Heating Copyright by Gary Reysa Reprinted with permission 37

35 Glycol based - Then the heat output is Tinlet = 88F Toutlet = 68F Flow Rate =.5 gpm q out = 60*Flow*fluid density*heat capacity*(t outlet - T inlet ) BTU/hr q out = 60*.5*9.06*.768*(88-68) BTU/hr q out = 4175 BTU/hr Assuming 60% Ethylene glycol and water.768 is the specific heat and the weight is the of water x the specific gravity of this mix at

36 Flow Rate = 1 gpm Toutlet = 67F Tinlet = 87F q out = 60*F*fluid density*heat capacity*(t outlet - T inlet ) BTU/hr q out = 60*1*8.595*.895*(87-67) BTU/hr q out = 9231 BTU/hr Assuming 40% propylene glycol and water.895 is the specific heat and the weight is the of water x the specific gravity this mix at

37 Summary We can take very accurate measurements with the cameras, get some data from the system and then make good energy use determinations on the installation. Evaluate system efficiency or distribution of that energy in a uniform and most useful manner. Not many standards would apply to this inspection as the installers and manufacturers would fall under those. 40

38 Solar electric 41

39 Solar electric There is enormous potential to diagnose problems on solar installation sites. High-performance thermal cameras, capable of detecting very small temperature differences and of creating clear images or video of problem areas have hit a price point that promotes the usage of these tools to the solar panel industry 42

40 Solar electric Thermal imagery can identify soiled or shaded individual cells within the systems as hotter areas of resistance. In a photovoltaic installation, efficiency is a function, in part, of temperature; cooler panels run more efficiently. As individual cells fail, they can begin to heat up, rather than collect energy, and this heat increase is easy to see with a thermal imaging device. 43

41 Solar electric Array Panel 44

42 Solar electric 45

43 Solar electric 46

44 Cells Very obvious cell problem Not so obvious cell problems 47

45 Cells A cell problem? No connection box 48

46 Cells The patterning will be particular to the cell or panel construction method Thin film arranged in strips 49

47 Cells Cell issue from the back as the front is impractical to image. InfraMation 2011 Harley Denio Oregon Infrared LLC 50

48 Cells Having a wider perspective either from flyover or adjoining structures. This approach puts the entire collector in view as the array isn't approachable from the same roof InfraMation 2011 Harley Denio Oregon Infrared LLC 51

49 Solar Cells Much study is ongoing for research on designs, materials, efficiency, overall application capability. 12/19/ Infrared Training Center. All rights reserved. 52

50 R value of stud framed walls 53

51 An R Value Calculator Many Caveats to understand. Developed by Dr. Bob Madding of Infrared Training Center 54

52 An R Value Calculator The whole concept is based on these premises and the formula. The largest condition is that the surfaces and temperatures all must be non changing for a significant amount of time. 55

53 Example Published R24 SIP Many caveats and understanding. 56

54 Example drywall Fiberglass batt 4 stud 1 sheathing Vinyl siding Calculates to R21 57

55 How much does that cost me? No attic insulation ~R-3 Good attic insulation ~R-40 Assume 33X33 ft = 1000 sq ft Massachusetts weather Heating and AC Oil Heat 58

56 Energy savings estimate Enter Values Degree days found at: 59

57 Energy savings estimate $1400 View Results 60

58 Summary The calculator works well in controlled circumstances. I advise reading the paper presentations from the InfraMation conferences to understand the situations. A full understanding of cameras for proper parameters and the heat transfer of a wall must also be understood. 61

59 Thanks for attending! 62

60 Thank you for attending Jay Bowen Infrared Training Center Senior Moderator and Course Developer Office: 9 Townsend West Nashua, NH TRAIN-IR 63

61 ASTM E881-92(2009) Standard Practice for Exposure of Solar Collector Cover Materials to Natural Weathering Under Conditions Simulating Stagnation Mode Test Methods and Minimum Standards for Certifying Solar Thermal Collectors FSEC Standard CSA F378 Series, Solar collectors F378.1, Glazed and unglazed liquid heating solar collectors F378.2, Air heating solar collectors Photovoltaic and Solar-Thermal Technologies in Residential Building Codes - National Renewable Energy Laboratory ASTM C a Standard Practice for Thermographic Inspection of Insulation Installations in Envelope Cavities of Frame Buildings ASTM C (2007) Standard Practice for Determining Thermal Resistance of Building Envelope Components from the In-Situ Data SRCC Standard 100 Test Methods And Minimum Standards For Certifying Solar Collectors 64

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