Objectives

1. Objectives • Review the cooling load calculation example • Learn about Heating & Cooling Systems

2. Example problem • Calculate the cooling load for the building in Pittsburgh PA with the geometry shown on figure. On east north and west sides are buildings which create shade on the whole wall. • Windows: Horizontal slider, Manufacturer:  American Window Alliance, Inc, CDP number AMW-K-3-00028-00003 http://cpd.nfrc.org/pubsearch/psMain.asp • Walls: 4” face brick + 2” insulation + 4” concrete block, Uvalue = 0.1, Dark color • Roof: 2” internal insulation + 4” concrete , Uvalue = 0.120 , Dark color • Below the building is basement wit temperature of 75 F. • Internal design parameters: • air temperature 75 F • Relative humidity 50% • Find the amount of fresh air that needs to be supplied by ventilation system.

3. Example problem • Internal loads: • 10 occupants, who are there from 8:00 A.M. to 5:00 P.M.doing moderately active office work • 1 W/ft2 heat gain from computers and other office equipment from 8:00 A.M. to 5:00 P.M. • 0.2 W/ft2 heat gain from computers and other office equipment from 5:00 P.M. to 8:00 A.M. • 1.5 W/ft2 heat gain from suspended fluorescent lights from 8:00 A.M. to 5:00 P.M. • 0.3 W/ft2 heat gain from suspended fluorescent lights from 5:00 P.M. to 8:00 A.M. • Infiltration: • 0.5 ACH per hour

4. Example solution For which hour to do the calculation when you do manual calculation? • Identify the major single contributor to the cooling load and do the calculation for the hour when the maximum cooling load for this contributor appear. • For example problem major heat gains are through theroof or solar through windows! Roof: maximum TETD=61F at 6 pm (Table 2.12) South windows: max. SHGF=109 Btu/hft2 at 12 pm (July 21st Table 2.15 A) If you are not sure, do the calculation for both hours: at 6 pm Roof gains = A x U x TETD = 900 ft2 x 0.12 Btu/hFft2 x 61F = 6.6 kBtu/h Window solar gains = A x SC x SHGF =80 ft2 x 0.71 x 10 Btu/hft2 = 0.6 kBtu/h total = 7.2 kBtu/h at 12 pm Roof gains = A x U x TETD = 900 ft2 x 0.12 Btu/hFft2 x 30F = 3.2 kBtu/h Window solar gains = A x SC x SHGF =80 ft2 x 0.71 x 109Btu/hft2 = 6.2 kBtu/h total= 9.4 kBtu/h For the example critical hour is July 12 PM.

5. Heating systems

6. Choosing a Heating System • What is it going to burn? • What is it going to heat? • How much is it going to heat it? • What type of equipment? • Where are you going to put it? • What else do you need to make it work?

8. Choosing a Fuel Type • Availability • Emergencies, back-up power, peak demand • Storage • Space requirements, aesthetic impacts, safety • Cost • Capital, operating, maintenance • Code restrictions • Safety, emissions

9. Selecting a Heat Transfer Medium • Air • Not very effective (will see later) • Steam • Necessary for steam loads, little/no pumping • But: lower heat transfer, condensate return, bigger pipes • Water • Better heat transfer, smaller pipes, simpler • But: requires pumps, lower velocities, can require complex systems

10. Choosing Water Temperature • Low temperature water (180 °F – 240 °F) • single buildings, simple • Medium and high temperature (over 350 °F) • Campuses where steam isn’t viable/needed • Requires high temperature and pressure equipment

11. Choosing Steam Pressure • Low pressure (<15 psig) • No pumping for steam • Requires pumping/gravity for condensate • Medium and high-pressure systems • Often used for steam loads

12. Conclusions • Steam needs bigger pipes for same heat transfer • Water is more dense and has better heat transfer properties • You can use steam tables and water properties to calculate heat transfer • Vary design parameters

13. What About Air? • Really bad heat transfer medium • Very low density and specific heat • Requires electricity for fans to move air • Excessive space requirements for ducts • But ! • Can be combined with cooling • Lowest maintenance • Very simple equipment • Still need a heat exchanger

14. Furnace • Load demand, load profile • Amount of heat • Response time • Efficiency • 80 – 90 % • Electricity is ~100 % • Combustion air supply • Flue gas discharge (stack height)

16. Choosing a Boiler • Fuel source • Transfer medium • Operating temperatures/pressures • Equipment • Type • Space requirements • Auxiliary systems

17. Water Boilers Types • Water Tube Boiler • Water in tubes, hot combustion gasses in shell • Quickly respond to changes in loads • Fire Tube Boiler • Hot combustion gasses in tubes, water in shell • Slower to respond to changes in loads

19. Electric Types • Resistance • Resistor gets hot • Typically slow response time (demand issues) • Electrode • Use water as heat conducting medium • Bigger systems • Cheap to buy, very expensive to run • Clean, no local emissions

20. Location • Depends on type • Aesthetics • Stack height • Integration with cooling systems

21. Cooling

22. Equipment Selection example Need 1.2 ton Of water cooling 1 ton = 12000 Btu/h Why should architectural engineers know about cooling machine? Capacity is 1.35 ton only for: 115 F air condenser temp 50 F of water temperature

23. What is the COP? • Congressional Observer Publications • California Offset Printers, Inc • Coefficient of Performance • Slang for a policeman

24. What is the efficiency of a typical residential air conditioner? • 10% • 50% • 80% • 100% • 300%

25. COPCoefficient of Performance

26. Vapor Compression Cycle Expansion valve Indoor 75°F Outdoor 105°F

27. Thermodynamics - review

28. Thermodynamics - review

29. Enthalpy: h [J/kg, Btu/lb] Temperature change ΔT Δh = cp ΔT – only for the same phase (air, water) What if we have change of the phase -evaporation or condensation? Entropy: s [J/kgK, Btu/lb°F] Δh =T Δs for evaporation or condensation Thermodynamics - review

30. Refrigeration Cycle

31. Reading Assignment Tao and Janis Chapter 5 Heating systems Tao and Janis Chapter 6 Cooling systems