Transcript 346N_No08_HVAC_Heating_Cooling_Systems

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 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 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/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 30 F = 3.2 kBtu/h
Window solar gains = A x SC x SHGF =80 ft2 x 0.71 x 109 Btu/hft2 = 6.2 kBtu/h
For the example critical hour is July 12 PM.
total= 9.4 kBtu/h
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.
B.
C.
D.
Congressional Observer Publications
California Offset Printers, Inc
Coefficient of Performance
Slang for a policeman
What is the efficiency of a typical
residential air conditioner?
A.
B.
C.
D.
E.
10%
50%
80%
100%
300%
COP
Coefficient of Performance
Provided cooling energy [W]
COP 
Used electric energy [W]
Vapor Compression Cycle
Expansion valve
Indoor 75°F
Outdoor 105°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:
Δh = T Δs
s [J/kgK, Btu/lb°F]
for evaporation or condensation
Refrigeration Cycle
Reading Assignment
Tao and Janis Chapter 5
Heating systems
Tao and Janis Chapter 6
Cooling systems