Objectives

Objectives

• Review the cooling load calculation example

• Learn about Heating & Cooling Systems

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.

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

Example solutionFor 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 the roof 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 61 F = 6.6 kBtu/hWindow 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 30 F = 3.2 kBtu/hWindow solar gains = A x SC x SHGF =80 ft2 x 0.71 x 109 Btu/hft2 = 6.2 kBtu/h total= 9.4 kBtu/h

For the example critical hour is July 12 PM.

Heating systems

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?

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

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

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

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

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

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

Furnace

• Load demand, load profile• Amount of heat• Response time

• Efficiency• 80 – 90 %• Electricity is ~100 %

• Combustion air supply• Flue gas discharge (stack height)

Choosing a Boiler

• Fuel source

• Transfer medium

• Operating temperatures/pressures

• Equipment• Type• Space requirements• Auxiliary systems

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

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

Location

• Depends on type• Aesthetics• Stack height• Integration with cooling systems

Cooling

Why should architectural engineers know about cooling machine?

Equipment

Selection

example

Need 1.2 ton

Of water cooling

1 ton = 12000 Btu/h

Capacity is 1.35 ton

only for:

115 F air condenser temp

50 F of water temperature

What is the COP?

A. Congressional Observer Publications

B. California Offset Printers, Inc

C. Coefficient of Performance

D. Slang for a policeman

What is the efficiency of a typical residential air conditioner?

A. 10%

B. 50%

C. 80%

D. 100%

E. 300%

COPCoefficient of Performance

[W]energy electric Used

[W]energy cooling ProvidedCOP

Vapor Compression Cycle

Expansion valve

Outdoor 105°FIndoor 75°F

Thermodynamics - review

Thermodynamics - review

Thermodynamics - review

• 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

Refrigeration Cycle

Reading Assignment

Tao and Janis Chapter 5

Heating systems

Tao and Janis Chapter 6

Cooling systems