A homeowner wants to replace an electric hot-water heater. There are two models in the store. The inexpensive model costs $280 and has no insulation between the inner and outer walls. Due to natural convection, the space between the inner and outer walls has an effective conductivity of 3 times that of air. The more expensive model costs $310 and has fiberglass insulation in the gap between the walls. Both models are 3.0 m tall and have a cylindrical shape with an inner wall diameter of 0.60 m and a 5 cm gap. The surrounding air is at 25°C, and the convective heat transfer coefficient on the outside is 15 W/(m2 K). The hot water inside the tank results in an inside wall temperature of 60°C. If energy costs 6 cents per kilowatt-hour, estimate how long it will take to pay back the extra
investment in the more expensive hot-water heater. State your assumptions.
GIVEN
FIND
- The time it will take to pay back the extra investment in the more expensive hot-water heater
ASSUMPTIONS
- Since the diameter is large compared to the wall thickness, one-dimensional heat transfer is
assumed
- To simplify the analysis, we will assume there is no water drawn from the heater, therefore the
inside wall is always at 60°C
- Steady state conditions prevail
SKETCH
PROPERTIES AND CONSTANTS
fiberglass (ki) = 0.035 W/(m K) at 20°C
dry air (ka) = 0.0279 W/(m K) at 60°C
The areas of the inner and outer walls are
The average area for the air or insulation between the walls (Aa) = 6.8 m2.The thermal circuit for water
heater #1 is
The rate of heat loss for water heater #1 is
Therefore the cost to operate water heater #1 is
Cost1 = q1 (energy cost) = 0.361 kW ($0.06/kWh) (24 h/day) = $0.52/day
The thermal circuit for water heater #2 is
The rate of heat loss from water heater #2 is
Therefore the cost of operating water heater #2 is
Cost2 = q2 (energy cost) = 0.16 kW ($0.06/kWh) (24 h/day) = $0.23/day
The time to pay back the additional investment is the additional investment divided by the difference
in operating costs
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