# Question 9.14: A Modest Household Demand. Estimate the monthly energy deman......

A Modest Household Demand. Estimate the monthly energy demand for a cabin with all ac appliances, consisting of a 19-cu. ft refrigerator, six 30-W compact fluorescents (CFLs) used 5 h/day, a 19-in. TV turned on 3 h/day and connected to a satellite, a cordless phone, a 1000-W microwave used 6 min/day, and a 100-ft deep well that supplies 120 gallons/day.

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Using data from Table 9.10, we can put together the following table of power and energy demands. The total is just over 3.11 kWh/day.
Notice how little of the energy for TV/satellite is used during the 3 h that it is actually turned on (255 Wh out of a total of 698 Wh, or 36.5%). This household really needs to provide a power strip to allow manual complete shut off of these electricity vampires.

 Appliance Power (W) Hours Watt-hours/day Percentage Refrigerator, 19  cu. ft 300 1140 37% Lights (6 @ 30  W) 180 5 900 29% TV, 19-in., active mode 68 3 204 7% TV, 19-in., standby mode 5.1 21 107 3% Satellite, active mode 17 3 51 2% Satellite, standby mode 16 21 336 11% Cordless phone 4 24 96 3% Microwave 1000 0.1 100 3% Washing machine 250 0.2 50 2% Well pump, 100  ft, 1.6  gpm 100 1.25 125 4% Total 3109 100%
 TABLE 9.10 Power Requirements of Typical Loads Kitchen Appliances Power Refrigerator: ac EnergyStar, 14 cu. ft 300 W, 1080 Wh/day Refrigerator: ac EnergyStar, 19 cu. ft 300 W, 1140 Wh/day Refrigerator: ac EnergyStar, 22 cu. ft 300 W, 1250 Wh/day Refrigerator: dc Sun Frost, 12 cu. ft 58 W, 560 Wh/day Freezer: ac 7.5 cu. ft 300 W, 540 Wh/day Freezer: dc Sun Frost, 10 cu. ft 88 W, 880 Wh/day Electric range (small burner) 1250 W Electric range (large burner) 2100 W Dishwasher: cool dry 700 W Dishwasher: hot dry 1450 W Microwave oven 750–1100 W Coffeemaker (brewing) 1200 W Coffeemaker (warming) 600 W Toaster 800–1400 W General Household Clothes washer: vertical axis 500 W Clothes washer: horizontal axis 250 W Dryer (gas) 500 W Vacuum cleaner 1000–1400 W Furnace fan: 1/4 hp 600 W Furnace fan: 1/3 hp 700 W Furnace fan: 1/2 hp 875 W Ceiling fan 65–175 W Whole house fan 240–750 W Air conditioner: window, 10,000 Btu 1200 W Heater (portable) 1200–1875 W Compact fluorescent lamp (100-W equivalent) 30 W Compact fluorescent lamp (60-W equivalent) 16 W Electric blanket, single/double 60/100 W Clothes iron 1000–1800 W Electric clock 4 W Consumer Electronics TV: >39-in. (active/standby) 142/3.5 W TV: 25 to 27-in. color (active/standby) 90/4.9 W TV: 19 to 20-in. color (active/standby) 68/5.1 W Analog cable box (active/standby) 12/11 W Satellite receiver (active/standby) 17/16 W VCR (active/standby) 17/5.9 W Component stereo (active/standby) 44/3 W Compact stereo (active/standby) 22/9.8 W Cordless phone 4 W Clock radio (active/standby) 2.0/1.7 W Computer, desktop (active/idle/standby) 125/80/2.2 W Laptop computer 20 W Ink-jet printer 35 W Dot-matrix printer 200 W Laser printer 900 W Shop Circular saw, 7 ${1}/{4^{\prime \prime}}$ 900 W Table saw, 10-in. 1800 W Hand drill, ${1}/{4^{\prime \prime}}$ 250 W Water Pumping Centrifugal pump: 36 Vdc, 50-ft @ 10 gpm 450 W Submersible pump: 24 Vdc, 100-ft @ 1.6 gpm 100 W Submersible pump: 48 Vdc, 300-ft @ 1.5 gpm 180 W DC pump (house pressure system), typical use 1–2 h/day 60 W Source: Rosen and Meier (2000) and others.

Question: 9.22

## Sizing an Array for a 150-ft Well in Santa Maria, California. Suppose that the goal is to pump at least 1200 gallons per day from the 150-ft well described in Example 9.21 using the Jacuzzi SJ1C11 pump. Size the PV array based on Siemens SR100 modules with rated current 5.9 A, mounted at an L + 15º ...

From the solar radiation tables given in Appendix ...
Question: 9.21

## Total Dynamic Head for a Well. What pumping head would be required to deliver 4 gpm from a depth of 150 ft. The well is 80 ft from the storage tank, and the delivery pipe rises another 10 ft. The piping is 3/4-in. diameter plastic, and there are three 90º elbows, one swing-type check valve, and one ...

The total length of pipe is 150 + 80 + 10 = 240 ft...
Question: 9.20

## PVs for the Cabin in Salt Lake City. The cabin from Example 9.18 needs 3000 Wh/day of ac delivered from an 85%-efficient inverter. For a 24-V system voltage, a 90% Coulomb efficiency, and 10% de-rating (de-rating factor = 0.90), size a PV array using Kyocera KC120 modules. ...

From Table 9.3, the Kyocera KC120 is a 120-W modul...
Question: 9.18

## Battery Sizing for an Off-Grid Cabin. A cabin near Salt Lake City, Utah, has an ac demand of 3000 Wh/day in the winter months. A decision has been made to size the batteries such that a 95% system availability will be provided, and a back-up generator will be kept in reserve to cover the other 5%. ...

With an 85% efficient inverter, the dc load is [la...
Question: 9.15

## Accounting for Inverter Losses. Suppose that a dc refrigerator that uses 800 Wh/day is being considered instead of the 1140 Wh/d ac one given in Example 9.14. Estimate the dc load that the batteries must provide if an 85% efficient inverter is used (a) with all loads running on ac and (b) with ...

a. With all 3109 Wh/day running on an 85% efficien...
Question: 9.19

## Impact of a Blocking Diode to Control Nighttime Battery Leakage. A PV module is made up of 36 cells, each having a reverse saturation current I0 of 1 × 10^−10 A and a parallel resistance of 8 Ω. The PVs provide the equivalent of 5 A for 6 h each day. The module is connected without a blocking diode ...

The voltage across each PV cell will be about 12.5...
Question: 9.17

## Losses at High and Low Charging Rates. A 100-Ah, 12-V battery with a rest voltage of 12.5 V (at its current SOC) is charged at a C/5 rate, during which time the applied voltage is 13.2 V. Using a simple Thevenin equivalent: a. Estimate the internal resistance of the battery. b. What fraction of the ...

a. At C/5, the current is 100 Ah/5 h = 20 A. The i...
Question: 9.16