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MT518 SolarEnergyTheoryandApplication MidTermTestSolutions
(10 points)
1.
Explain the advantages and limitations of solar energy. List five (5) different factors in each category to support your argument.
Advantages: Renewable
Plentiful No lethal radiation Not centralized Non- pollutant
Disadvantages: Ignorance of the American public about what solar energy is able to do.
If people want solar equipment they do not know where to find it. Most buildings that use solar energy for space heating have been custom built at a relatively high cost.
Financing the initial cost is particularly difficult when building costs are high, and when mortgage money is difficult to obtain. Shortage of contractors and architectural engineering designers who are equipped to handle the present complexity of solar energy designs.
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(10 points)
2.
Calculate the declination, the zenith angle, and the azimuth angle of the sun for New York City (Lat. 40.8N) on October 1 at 2:00 pm solar time.
Solar Declination, S: S = 23.45 sin[360(284+n)/365] n = 274 (October 1) Therefore, S = 23.45 sin[360(284+274)/365] S = - 4.22 (Ref. Slide #3.3)
Zenith Angle, z: Solar Time = 14.00 hours (2:00 pm) Solar Hour Angle, hs = 15 x Hours from local solar noon
= 15(14 12) = 30 sin = sin L sin s + cos L cos s cos hs (Ref. Slide #3.22) = (sin 40.8)(sin -4.22) + (cos 40.8)(cos -4.22)(cos 30) = 0.606 = 37.3 z = 90 = 90 37.3 z = 52.7 (Ref. Slide #3.20)
Azimuth Angle, s: sin s = cos s sin hs / cos (Ref. Slide #3.22)
sin s = (cos -4.22)( sin 30) / cos 37.3 = 0.627 s = 38.8
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(20 points)
3.
Two rows of flat plate collectors are mounted on a flat roof at a 40 angle with the horizontal (see sketch) and are facing true
south. What should the minimum distance D between collectors be such that they do not shade each other on the shortest day of the year when shadows are longest, December 21, between 10:00 am and 2:00 pm solar time. The installation is at 40N Lat.
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D2 = 6.0Cos 40 D1 = H/tan D = D1 + D2 = (H/ tan ) + 6.0Cos 40 H = 6.0sin 40 = 3.86 ft sin = sin L sin s + cos L cos s cos hs (Ref. Slide #3.27)
S = 23.45 sin[360(284+n)/365] = 23.45 sin[360(284+355)/365] = - 23.5
(Ref. Slide #3.3) n = 355 (Dec. 21)
Solar Time = 10.00 am and 2:00 pm (14.00 hours) Solar Hour Angle, hs = 15 x Hours from local solar noon At 10:00 am, hs = 15(12 10) = 30 At 2:00 pm, hs = 15(14 12) = 30 Use the hs for 10:00 am and 2:00 pm because they will produce the longest shadow.
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Therefore, sin = (sin 40)( sin -23.5) + (cos 40)(cos - 23.5)( cos 30) = 0.353 or = 20.7
D = D1 + D2 = (H/ tan ) + 6.0Cos 40 = (3.86/tan 20.7) +6.0Cos 40 = 10.2 + 4.60 D = 14.8 ft
(10 points)
4. Explain the need for temperature stratification in water and rock heat storage.
Stratification is important. Temperature stratification is when one end of a rock bin is hotter than the other end. The advantage of stratification is that room temp air is returned to collectors therefore resulting in higher collector efficiencies. A solar hot air heating system, with a temperature stratified rock bin, tends to perform 5 10% more efficiently than one that has a thermally mixed rock storage. Stratification occurs easily in a vertical rock storage bin because the solar heated air enters the rock bin at the top plenum flowing downward. Thus the rock bed is hot at the top and relatively cool at the bottom where the air exits the bottom plenum. Stratification is easy to maintain because the rocks are stationary. Similarly, stratification in hot water tank storage is also important to the efficient performance of liquid type collectors. But, unlike rock bed storage, stratification in hot water tank storage is harder to maintain because any pumping action within the tank and natural convection tend to disturb stratification.
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(20 points)
5.
Design a vertical storage rock bed for a hybrid solar heating system in Newark, NJ. The rock bed is to be built in the basement of the house, which has an 8 ft. high ceiling. Allow 1.0 ft. for each air-in and air-out plenum. To maintain a comfortable air temperature in the sunspace, heat removed is at a rate of 10,000 Btu/hr for 3 hours and stored in the rock bed. The air-in from the sunspace is at 95 F and leaves the rock bed at 70F.
Determine the size of the rock bed and the pressure drop through it.
Desired Heat Storage = 10,000 Btu/hr x 3 hrs = 30,000 Btu tair in = 95F, tair out = 70F Working air t = tair in (into rock bed) - tair out (return from rock bed) = 95 70 = 25F Working bed t (tair in - tair out) (95 70) 12.5F
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Volumetric Specific Heat of 1 dia. Rocks ( 35% voids), Cp =
= 22.8 Btu/ft3-F
(Ref. Slide #6.32)
Rock Bed Volume (ft3) =
Desired Heat Storage, Btu 3 = (Volumetric Specific Heat of Rock Bed, Btu/ft - F)(t working bed )
= 30,000 / (22.8 x 12.5) = 105 ft3
Heat Transport Rate, Btu/hr Air Flow (CFM) = (Volumetric Specific Heat of Air, Btu/hr/ F - CFM)(t working air )
Volumetric Specific Heat of Air = 1.08 Btu/hr/F-CFM (at sea level) (Ref. Slide #6.36) Air Flow (CFM) = 10,000 / (1.08 x 25) = 370 CFM
Rock Bed Length, L = 8.0 2(1.0) = 6.0 ft Face Area = Volume of Bed /L = 105/6.0 = 17.5 ft2 One side of rock bed = (17.5) = 4.18 ft Rock bed size = 4.2 ft x 4.2 ft x 6.0 ft Face Velocity =CFM/Face Area = 370/17.5 = 21.1 ft/min
For 1.0 Dia. rocks with a face velocity of 21.1 ft/min, Procks/L = 0.021 H2O or by interpolation from Table of Slide #6.34
Procks = 0.021 x 6 ft Procks = 0.126 H2O
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(20 points)
6.
Determine the instantaneous efficiency, c of a collector at 11:00 am solar time on July 21, located in New York City (Lat. 40.8N) and facing south. Collector size = 4.0 x 6.0 ft. Collector tilt = 40 Flow rate in collector = 4.8 lb/min of 50% Ethylene Glycol-50% water Water inlet temperature = 65F Water outlet temperature = 80F
Instantaneous Collector Efficiency,
c =
qu Ac I c
qu = Ac Gc p ( t f ,out t f ,in )
Where, G = m/Ac = mass flow rate per unit surface area, lb/hr-ft2 = 4.8 lb/min / 24 ft2 = 0.2 lb/min-ft2 or 12.0 lb/hr-ft2
Cp (50% Ethylene Glycol/Water) = 0.831 Btu/lb-F qu = AcGCp(t f,out t f,in) = 24.0 x 12.0 x 0.831 (80 65) = 3.59 x 103 Btu/hr
(Ref. Slide #6.6)
Clear sky insolation (striking the collector) on a 40 tilt, south facing surface at Lat = 40, on July 21 @ 11:00 am = 281 Btu/hr-ft2 (Ref. Table A2.6 Textbook)
c = qu / (Ac x Ic) = 3.59 x 103 / (24 x 281) c = 53.2%
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(10
points)
7.
A cylindrical hot water storage tank (see sketch) for a set of collectors is located in the basement of a dwelling. The tank is made of 0.080 thick steel and insulated with 3 thick fiberglass batt insulation. If the basement air temperature is 65F and the liquid in storage tank is at 120F, what are the heat losses through the vertical cylindrical surface of the tank? Assume inner surface of tank is at 120F
Cross Section of Hot Water Tank
Rtotal = Rsteel + Rfiberglass + Rnat.conv.
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Since Rsteel is negligible compared to the fiberglass and the natural convection resistances, ignore it. K fiberglass batt = 0.31Btu/hr-ft2-F/in Area of vertical surface of tank = (2r)(h) = (2 x 1.0)(4.5) = 28.3 ft2 R fiberglass batt = L/KA = 3.0/(0.31)(28.3) = 0.342 F-hr/Btu Rnc = 1/Ahc F-hr/Btu hc = C(K/L)(L3t)d Btu/hr-F-ft2
Assume outer surface temperature of fiberglass = 105 F Therefore, Mean air temperature = (105 + 65)/2 = 85.0 F and t = 40 F = 1.25 x 106 1/ft3-F (From Slide #2.11 for mean air temp. of 85 F) L = 2 ft. (limited height) K = 0.0157 Btu/hr-ft-F (From Slide #2.11 for mean air temp. of 85 F) L3t = (1.25 x 106)(2.0)3(40) = 4.0 x 108 Therefore, d = 0.25 (From Slide #2.9) hc = C(K/L)(L3t)d hc = 0.50(0.0157/2.0)[4.0 x 108]0.25 = 0.553 Btu/hr-ft2-F Rnc = 1/Ahc = 1/(28.3)(0.553) = 0.064 F-hr/Btu Rtotal = Rsteel + Rfiberglass + Rnat.conv. Rtotal = 0 + 0.342 + 0.064 = 0.406 F-hr/Btu q = t /Rtotal = (120 65)/0.406 = 135 Btu/hr and C = 0.50 (From Slide #2.10)
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Verify that the outside temperature of the fiberglass was correctly assumed to be 105 F. Rtotal (between the steel tank liner and the outside surface of the fiberglass) = 0.342 F-hr/Btu Therefore, q = t /Rtotal
135 = (to 65)/0.342 to = (135 x 0.342) + 65 = 111 F The calculation should be done again with an outside surface temperature of 111 F.