Lecture 16.75– Heat Pumps, AC HeatPump,%Operang%Principles ...
Transcript of Lecture 16.75– Heat Pumps, AC HeatPump,%Operang%Principles ...
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Lecture 16.75– Heat Pumps, AC Pete Schwartz Cal Poly Physics
Heat Pump, Opera-ng Principles • Overall transfer of heat from cold to warm (against the macro temperature gradient)
• At each point in the system, heat flow is from warm to cold
• Relies on the fact that a gas cools when it expands, and is heated when it is compressed (work is done on it), to create local temperature gradients contrary to the macro-‐gradient
• You can MOVE a lot more than one Joule with one Joule of energy, depending on the temperature difference you have to move it across
Qhigh = Wnet + Qlow
HOT (reactor)
COLD (Ocean)
Work or Electricity
QH
QC
Heat Engines
€
η =WQH
=QH −QC
QH
η <TH −TCTH
Qhigh = Wnet + Qlow
Work or Electricity
€
COP =QH
W=
QH
QH −QC
COP <TH
TH −TC
Refrigerator Air Conditioner
(Freezer, Summer House)
(Hot Outside World)
Heat Pump
(Cold Outside)
(Winter House)
€
COP =QC
W=
QC
QH −QC
COP <TC
TH −TC
HOT
COLD
QH
QC
COP (Coefficient of Performance) >>1, depends Often more than 5, and decrease with larger ΔT
From your reading on Wikipedia
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Freezer
http://www.truehvac.com
Changing the direction of flow will turn the A/C into a heater
Tcondenser
Tevaporator
Tindoor
Toutdoor
Apparent T lift
Real T lift
ΔTH
ΔTL
If Tcondenser= 40°C Toutdoor= 30°C Tindoor= 16°C Tevaporator = 6°C, then Apparent Carnot COP = 20.6 Real Carnot COP= 8.5 Actual COP= 5.53 if machine efficiency = 0.65
Heat Flow
Heat Flow
Optimize Actual efficiency by minimizing total temperature drop (Real T lift)
Freezer
House
Compressor
Hot Coils
Cold Coils
Heat Pump to cool house (Freezer)
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Thus, to reduce heat pump energy use, • Distribute heat at the lowest possible temperature (e.g., at 30°C instead of 60°C – using radiant floor hea-ng or radiant ceiling)
• Distribute coldness at the warmest possible temperature (e.g., at 20°C instead of 6°C – using chilled ceiling or chilled floor slab)
• Minimize ΔTH and ΔTL by -‐ minimizing the required heat flows (which must balance heat loss or heat gain)
-‐ using as large a radiator surface as possible
http://www.truehvac.com
Geothermal Reservoirs: very high COP for winter heating.
Climate master. Geothermal heat pump systems
Geothermal Heat Pumps Energy Required to Move Air or Water through Ducts or Pipes
• Power imparted to fluid Pfluid=ΔP x Flow
• Electric power required Pelectric = (ΔP x Q)/(η) but ΔP α Flow2 for turbulent flow, so Pfluid α Flow3
Radiant ceiling cooling As it turns out, ven-la-on air flow
requirements can be reduced by a factor of two by using displacement ven1la1on
rather than ceiling-‐based mixing ven-la-on, while improving air quality and reducing total hea-ng loads on the
chillers
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Energy required to deliver heat by circula-ng warm water vs warm air
• Rate of heat supply to a room is equal to the rate of heat loss from the circula-ng air or water, which is given by QH=ρcpQ (Tsupply-‐Treturn) = ρcpQ ΔT
• The ra-o of energy supplied to move the fluid to heat delivered is given by ΔP/ ρcp ΔT
• Given the large ρ and cp for water compared to air, and given typical ΔPs and ΔTs, moving heat with water requires less energy than with moving air (by a factor of 25).
Thus, to minimize the energy use in supplying and delivering heat/coldness and fresh air
• Separate hea-ng/cooling and ven-la-on func-ons • Use chilled water at the warmest possible T for cooling (best: 20°C)
• Use hot water for hea-ng at the coolest possible temperature (best: 30°C)
• Circulate only the amount of air required for ven-la-on purposes using displacement ven-la-on
In many systems,
• Several -mes more air is circulated than is needed for ven-la-on alone, so as to provide adequate cooling through airflow alone
• In “efficient” systems, 80% of the air might be recirculated and mixed with 20% fresh outside air on each circuit, rather than replacing and having to cool and dehumidify 100% outside air
• However, 80% of the internal heat gains picked up by the air will have to be removed by the chillers
In a Dedicated Outdoor Air Supply system (100% outside airflow but only what is needed for ven-la-on) with displacement ven-la-on,
• Heat gains from the ceiling (from ligh-ng) or heat rising to the ceiling is directly vented to the outside – reducing the cooling load on the chillers by up to one third
• Ven-la-on rates can be reduced to near zero when the building is not occupied (because ven-la-on is not used for temperature control) – can save 20-‐30% in total hea-ng+cooling+ven-la-on energy use
Another advantage of using chilled ceilings for cooling is that the required chilled-‐water temperature (18-‐20°C) is
cool enough that it can olen be supplied through evapora-ve cooling
using the chiller cooling tower AIR IN
CLOSED-CIRCUITHEAT-EXCHANGE COIL
HEAT AND HUMIDIFIED AIR OUT
DRIFTELIMINATORS
EXTERNAL WATER
HOTWATER
CLOSEDCIRCUIT
COLDWATER
AIR IN
AIRWATER
PUMP
Evapora:ve Cooling: Electricity is only used to circulate air and water
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Solar Energy in Buildings
• Passive solar hea-ng • Passive ven-la-on • Ac-ve solar thermal collectors, used for – domes-c hot water -‐ space hea-ng -‐desiccant dehumidifica-on systems • PV panels
To maximize passive solar hea-ng requires
• Aoen-on to building form and orienta-on • Use of high-‐performance windows • Use of thermal mass to avoid overhea-ng by day and to release stored heat by night
• High levels of insula-on to retain heat that is released from thermal mass at night
Solar Chimney to induce ven-la-on, Building Research Establishment, Garston, UK
Savings and Costs • With nothing fancy and without requiring detailed
computer simula-ons, this approach will frequently give a 50% savings in annual energy use compared to current prac-ce
• Use of computer simula1on models run by simula-on experts to fully op-mize the design of the building and mechanical systems and use of more advanced designs can push the savings to 60-‐70%
• Savings can be pushed to 75-‐80% with enlightened occupant behaviour
• Buildings achieving such high energy savings some-mes cost no more than conven-onal buildings, due to the downsizing of mechanical equipment, and are superior in other respects
• Some-mes saving more energy costs less
Amory Lovins, Rocky Mountain Institute (Efficiency)
Increased Efficiency
Pro
ject
Cos
ts
Energy savings > Cost of insulation
Diminishing Returns as energy savings are less
Marginal Energy Savings < Marginal Cost of Insulation
Reduce cost by downsizing or eliminating heater or AC
Qhigh = Wnet + Qlow
HOT (reactor)
COLD (Ocean)
Work or Electricity
QH
QC
Heat Engines
€
η =WQH
=QH −QC
QH
η <TH −TCTH
Qhigh = Wnet + Qlow
Work or Electricity
€
COP =QH
W=
QH
QH −QC
COP <TH
TH −TC
Refrigerator Air Conditioner
(Freezer, Summer House)
(Hot Outside World)
Heat Pump
(Cold Outside)
(Winter House)
€
COP =QC
W=
QC
QH −QC
COP <TC
TH −TC
HOT
COLD
QH
QC
COP (Coefficient of Performance) >>1, depends Often more than 5, and decrease with larger ΔT
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http://www.scientificamerican.com/