# Solar Energy Hot Water and Calculation

| January 14, 2016

Solar Energy Hot Water and Calculation A four story building located in Hammond Indiana has 100 rooms and is used as a dormitory for students. A sketch of the building is shown in Figure 1. There are 25 rooms per floor and each room is occupied by three students. The school operates for fall (15th of August to 15th of December), spring (15th of January to15th of May) summer (15th of May to 5th of August) semesters. Each room has a 3ft x 6ft window. The specifications for walls and windows are given below. The hot water load by month is determined and is shown in Table 1. The average hot water demand is 13.1 Gal/student per day. Due to the relatively large static head, approximately 50ft, and the freeze potential at the site, the conceptual system selected for this site should be a closed loop antifreeze system. Assume \$40/ft2 for evacuated and \$30/ ft2 for flat collector and an average of \$20/ ft2 for installation for evacuated and flat collectors. The heating for each room is done by electric and 70ºF is preferred. The sun exposure schedule is given below. a. Select the system, either flat plate or evacuated tube collector b. Select the dimensions for the collector c. Calculate the total collector size d. Calculate the storage size e. Select the flow rate and pump size f. Determine the energy required to maintain water load per month as shown in Table 1. g. Calculate the solar heat contribution through the widows per room for each month. h. Calculate the total kWh needed per year to heat each room. i. Assuming \$0.1/kWh for electricity, calculate the total cost of heating per year. j. Investigate the installation of solar heating k. Calculate the total cost of such system, if 1000 BTU of heat from a solar panel cost \$100. l. Calculate total cost m. Calculate the pay pack (simple pay back method 5% interest rate) Figure 1.Building layout Wall and Window Characteristics For the wall, the components 1 and 3 are 10cm brick and the center one is 5 cm stagnant air. The wall is plastered 25 mm on both sides. Wall outside resistance = 0.044 Plaster 25 mm = 0.025/1.39 = 0.018 Brick 10 cm = 0.10/.25 = 0.4 Stagnant Air 50 mm = 0.18 Brick 10 cm = 0.10/0.25 = 0.40 Plaster 25 mm = 0.025/1.39 = 0.018 Inside Surface Resistance = 0.12 Table 1 Hot water requirement per month Month Gallons Cold Water Temp(ºF) Hot Water Temp(ºF) January 122,000 48 140 February 110,000 48 140 March 122,000 50 140 April 118,000 52 140 May 61,000 57 140 June 59,000 59 140 July 61,000 61 140 August 61,000 62 140 September 118,000 63 140 October 122,000 60 140 November 118,000 55 140 December 112,000 50 140 Date Sunrise Sunset Length Change Dawn Dusk Length Change Today 07:18 17:49 10:31 06:49 18:17 11:28 +1 day 07:19 17:48 10:29 00:02 shorter 06:50 18:16 11:26 00:02 shorter +1 week 06:26 16:40 10:14 00:17 shorter 05:57 17:09 11:12 00:16 shorter +2 weeks 06:35 16:33 9:58 00:33 shorter 06:05 17:02 10:57 00:31 shorter +1 month 06:54 16:22 9:28 01:03 shorter 06:23 16:53 10:30 00:58 shorter +2 months 07:16 16:27 9:11 01:20 shorter 06:45 16:59 10:14 01:14 shorter +3 months 07:07 16:59 9:52 00:39 shorter 06:37 17:29 10:52 00:36 shorter +6 months 05:50 19:45 13:55 03:24 longer 05:20 20:15 14:55 03:27 longer Notes: Daylight saving time, * = Next day. Darkness Dawn Sunshine Dusk Notes: How to read this graph? Change preferences. Variable I II III IV V VI VII VIII IX X XI XII Insolation, kWh/m²/day 1.73 2.56 3.61 4.60 5.38 6.00 6.04 5.25 4.29 2.94 1.77 1.50 Clearness, 0 – 1 0.46 0.48 0.49 0.49 0.49 0.52 0.54 0.53 0.53 0.49 0.42 0.44 Temperature, °C – – 2.10 8.72 14.81 19.82 21.84 20.56 16.53 10.58 3.96 – 6.19 3.60 3.36 Wind speed, m/s 6.76 6.49 6.72 6.72 6.01 5.60 5.20 5.04 5.46 6.09 6.42 6.43 Precipitation, mm 43 37 64 96 92 102 98 92 94 70 77 61 Wet days, d 11.7 9.9 12.6 12.6 10.8 9.8 9.6 9.2 9.2 9.2 11.0 12.4 These data were obtained from the NASA Langley Research Center Atmospheric Science Data Center; New et al. 2002 Hammond, Indiana, United States – Basic information Latitude: +41.58333 (41°34’59.988″N) Longitude: -87.5 (87°30’00″W) Time zone: UTC-6 hours Local time: 19:01:28 Country: Indiana, United States Continent: Americas Sub-region: Northern America Distance: ~3.8 km (from your IP) Altitude: ~180 m

Calculating Heat Loss of Windows

Example 1

A house in State College, PA has 380 ft2 of windows (R = 1.1), 2750 ft2 of walls and 1920 ft2 of roof (R = 30). The composite R-Value of the walls is 19. Calculate the heating requirement for the house for the heating season. What is the percentage of heat that is lost through the windows?

Solution:

Heat loss in a heating season is given by

Heat Loss=Area×HDD×24Rvalue

Heat Loss through windows =

380 ft 2 ×6000   0 F   days×24 h  /day1.1ft 2    o F   h  Btus  =49,745,455 Btus

Heat loss through walls =

2750 ft 2 ×6000   0 F   days×24 h  /day19ft 2    o F   h  Btus  =20,842,105 Btus

Heat loss through roof =

1920 ft 2 ×6000   0 F   days×24 h  /day30ft 2    o F   h  Btus  =9,216,000 Btus

Total heat loss = 79,803,560 BTUS

Percentage of heat loss through the windows =

Heat Loss=49.74 MMBtus79.8 MMBtus ×100=62.3%

Example 2

Windows in the house described in Example 1 are upgraded at a cost of \$1,550. The upgraded windows have an R-value of 4.0.

• What is the percent savings in the energy and the heating bill if the energy cost is 11.15/MMBTUs.
• What is the pay back period for this modification?

Solution:

1. a) New heat loss for the same window size with the new R-value is

380 ft 2 ×6000   0 F   days×24  h  /day4.0ft 2    0 F    h  Btus  =13,680,000 Btus

Annual energy savings = 49.745 MMBTUs -13.680 MMBTUs = 36.06 MMBTUs

The percent savings is

36.06MMBTU79.84MMBTU ×100=45.1%

 Cost effectiveness of using improved windows Performance Base Model Recommended Level Best Available Window Description Double-paned, clear glass, aluminum frame Double-paned, low-e coating, wood or vinyl frame Triple-paned, tinted, two spectrally selective low-e coatings, krypton-filled, wood or vinyl frame SHGCa 0.61 0.55 0.20 U-factor b 0.87 0.40 0.15 Annual Heating Energy Use 547 therms 429 therms 426 therms Annual Cooling Energy Use 1,134 kWh 1,103 kWh 588 kWh Annual Energy Cost \$290 \$240 \$210 Lifetime Energy Cost c \$4,700 \$3,900 \$3,400 Lifetime Energy Cost Savings – \$800 \$1,300

The old heating bill would be

79.803MMBtu*\$11.15MMBtu =\$889.80

The new heating bill would be

43.743MMBtu*\$11.15MMBtu =\$487.73

The monetary savings = \$402.06 per year.

The Pay Back Period =

Additional InvestmentSavings per year =\$1550.00\$402.06 =3.85 years

The table shows the cost effectiveness of replacing old windows with new and improved windows. The costs are calculated using a computer program called RESFEN developed by US Department of Energy.
aSHGC, or Solar Heat Gain Coefficient, is a measure of the solar radiation admitted through a window. SHGC ranges between 0 and 1; the lower the number, the lower the transmission of solar heat. SHGC has replaced shading coefficient (SC) as the standard indicator of a window’s shading ability. SHGC is approximately equal to the SC multiplied by 0.87.

b U-factor is a measure of the rate of heat flow through a window. The U-factor is the inverse of the R-value, or resistance, the common measure of insulation.

c Lifetime energy cost savings is the sum of the discounted value of annual energy cost savings, based on average usage and an assumed window life of 25 years. Future energy price trends and a discount rate of 3.4 percent are based on Federal guidelines (effective from April 2000 to March 2001).Assumed electricity price: \$0.06/kWh, the Federal average electricity price in the U.S.Assumed gas price: \$0.40/therm, the Federal average gas price in the U.S.

Cost-Effectiveness Assumptions: The model shown above is the result of a simulation using a residential windows modeling program called RESFEN. Calculations are based on a prototype house: 1,540 sq. ft., two stories, a standard efficiency gas furnace and central air conditioner, and window area covering 15 percent of the exterior wall surface area.

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