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Fig 106 How Refrigerators Work
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Refrigeration Savings
Refrigeration Savings
Calculating the Cooling Load
First, two de nitions: Btu = heat that changes the temperature of 1 pound of water by 1F Heat of fusion of water (heat absorbed by 1 pound of ice when it melts) = 144 Btu Next, Figure 107 shows the basic equation of conductive heat loss: H = A T/R where: H = heat loss through A, in Btu/hour A = area of surface in square feet T = temperature difference, Toutside TinsideF R = thermal resistance To this conductive heat loss through the refrigerator walls we have to add the losses of in ltration (air leakage through the lid and drain) and cooldown (heat that must be removed from warm objects added to the box)
Fig 107 Conductive Heat Loss
T Tinside A H R H Toutside
Fig 108 Typical Refrigerator Box
Before Retrofit
The Refrigerator Box
Figure 108 shows the dimensions of a typical 8-cubicfoot refrigerator The walls are constructed of 2-inch urethane foam, lined with berglass Plugging the numbers into the heat-loss equation above: A = 28 square feet (insulation mid-plane) T = 75 F outside 40 F inside = 35F R = 2 inches R-6/inch = 12 H = 28 35/12 = 82 Btu/hour If this box had a perfect lid and we never opened it, it would melt 82 Btu/hour/144 Btu/pound = 057 pound of ice per hour or 137 pounds of ice per day If we add another third (27 Btu/hour) to the heat loss for in ltration and cooldown, our totals become 109 Btu/hour and 18 pounds of ice per day How much electricity will it take to replace the ice The answer is it all depends It depends primarily on the ef ciency of the refrigeration system, the refrigerator s coef cient of performance (COP), de ned as: COP = watts removed/watts used We ll also need a conversion factor: Ah/day = 0586 Btu/hour
190 Conser vation for Liveaboar ds
24" 28" 24" 24"
After Retrofit
20" 28" 20" 20"
COP depends on how hard the compressor has to work pumping heat from the evaporator (cold) side to the condenser (hot) side When the temperatures of evaporator and condenser are the same, the compressor can remove about 3 Btu of heat energy, using the equivalent of 1 Btu of electrical energy, yielding a COP of 30 At the opposite extreme, the typical refrigerator is capable of moving heat across a maximum difference of about 150F , at which point its COP becomes zero Figure 109 shows the relationship between COP and condenser temperature for a typical small refrigerator, assuming the evaporator is at 20 F The temperature of the condenser depends on its type and location Table 101 shows typical condenser operating temperatures and resulting COPs for different condenser types and locations in the boat Assume the condenser is located in free air at 75 F but has no fan According to Table 101, its COP will be about 13 Our example refrigerator will then draw 190 Btu/hr 0586/13 = 49 Ah/day
Fig 109 Refrigerator COP vs Temperature
Coefficient of Performance (COP)
0 20
60 80 100 120 Condenser Temperature, F
Table 101 Refrigerator COP vs Location
Condenser/Location Condenser Temp, F COP
A Typical Refrigerator Retro t
To reduce the refrigeration load we can take three steps:
1 Increase wall insulation 2 Decrease in ltration with a better lid gasket and a drain trap 3 Increase condenser COP with a fan
Air-cooled in 110 F engine compartment, no fan Air-cooled in 110 F engine compartment with fan Air-cooled in 75 F free air with no fan Air-cooled in 75 F free air with fan Water-cooled, intake water 75 F Water-cooled, intake water 60 F
140 120 105 85 85 70
06 10 13 17 17 20
Wall Insulation At the cost of 34 cubic feet of interior space, we can double the insulation of all surfaces and increase their R-values to R-24 Because we also have decreased the heat loss area (measured at the midpoint of the insulation) to 24 square feet, conductive heat loss is decreased by 47 Btu/hour to 35 Btu/hour In ltration By installing a better gasket on the lid and a better meltwater drain trap, we might decrease the losses due to air leaks by half, or 13 Btu/hour, and the total heat loss by 60 Btu/hour to 49 Btu/hour Condensor COP By installing a thermostatically controlled 12-volt DC fan to move air through the condenser coils, we can increase the condenser COP from about 13 to 17 at the cost of about 2 Ah per day to power the fan Most
marine and RV refrigeration manufacturers offer such fans as an option If not, small fans that run on 12 VDC are readily available at electronics stores, such as RadioShack DIY Project 13 ( 14) is a circuit to control such a fan Manufacturers also sometimes offer water cooling, where the condenser is cooled by water circulated to a heat exchanger mounted on the bottom of the hull While more ef cient in cooler waters, such systems are susceptible to becoming fouled with growth
Net Refrigerator Savings With the above improvements, our new refrigeration load becomes 60 Btu/hour 0586/17 = 207 Ah/day a decrease of more than 50%
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