Baixe Cálculos relacionados à energia solar e orientação de janelas em arquitetura e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
20-1. Using Eq. 20-3, compute the hour of sunrise on the shortest day of the year of 40o^ north latitude.
Solution: Eq. 20-
sin β = cos L cos H cos δ + sin L sin δ H = hour angle = 15 degrees per one hour from solar noon. L = 40 o
β = 0 o
- solar altitude at sunrise δ = -23.5o^ - shortest day on winter solstice
sin 0 = cos 40 cos H cos (-23.5) + sin 40 sin (-23.5)
H = 68.6o
or 68.6 / 15 = 4.573333 hrs from noon = 12 - 4. = 7.426667 from midnight = 7:26 A.M. - - - Ans.
20-2. Compute the solar azimuth angle at 32 o north latitude on February 21.
Solution: From Table 4- Solar Time A.M. β φ 7 7 73 8 18 64 9 29 53 10 38 39 11 45 21 12 47 0
β = solar altitude φ = solar azimuth
Eq. 20-
β
δ φ = cos
cossinH sin
for φ≤ 90 o
Eq. 20-3. sin β = cos L cos H cos δ + sin L sin δ
L = latitude = 32 o
H = Hour Angle δ = Declination angle
For February 21 N = 31 + 21 = 52
Eq. 20-2.
360284 N
23.47sin
δ=
23.47sin
δ=
δ = -11.24o sin β = cos L cos H cos δ + sin L sin δ sin β = cos 32 cos H cos (-11.24) + sin 32 sin (-11.24) sin β = 0.83178 cos H - 0.
β
δ φ = cos
cossinH sin
β
φ = cos
cos-11.24sinH sin
β
φ = cos
0.98082sinH Arcsin
Then: H = 15o^ x (No. of hours from Noon)
Ans. Tabulation Solar Time, A.M. H β φ 7 75 6.43 72. 8 60 18.22 63. 9 45 29.00 52. 10 30 38.10 38. 11 15 44.44 20. 12 0 46.76 0.
20-3. (a) What is the angle of incidence of the sun’s rays with a south-facing roof that is sloped at 45 o with the horizontal at 8 A.M. on June 21 at a latitude of 40o^ north? (b) What is the compass direction of the sun at this time?
Solution:
S = 45o L = 40 o At 8 A.M. H = 4 x 15 = 60 o
On June 21, δ = 23.5o
(a) Eq. 20-3. sin β = cos L cos H cos δ + sin L sin δ sin β = cos 40 cos 60 cos 23.5 + sin 40 sin 23. β = 37.
β
δ φ = cos
cossinH sin
cos 37.
cos23.5sin 60 sin φ=
φ = 30. Table 4-13, φ > 90 φ = 180 - 89.04 = 90. γ=φ± ϕ ϕ= 0 γ = 90.96- 0 = 90.
cos29.
cos-20.16sin 0 sin φ=
φ = 0. Then, At 10 A.M., β = 23.66, φ = 30. At 12 N.N., β = 29.84, φ = 0. Eq. 20-8.
cos θ = cos β cos γ sin Σ + sin β cos Σ
Then cos γ = cosφ
Subsitute in Eq. 20-9.
At 10 A.M.
β γ Σ+ β Σ θ = expBsin
Acoscos sin sin cos IDNcos
exp(0.14 sin (23.66 ))
1230 cos23.66cos30.83sin sin23.66cos IDNcos
θ=
1230 0.786513sinΣ 0.401308cosΣ IDNcos
θ=
IDN cosθ=682.502sinΣ+348.238cos Σ
At 12 NN.
β γ Σ+ β Σ θ = expBsin
Acoscos sin sin cos IDNcos
exp( 0.14sin ( 29.84))
1230 cos29.84cos0.0sin sin29.84cos IDNcos
θ=
1230 0.867418sinΣ 0.497580cosΣ IDNcos
θ=
IDN cosθ = 8 05.266sinΣ+461.928cos Σ
Total:
T= ( 682.502+805.26) sinΣ+( 348.238+461.928)cos Σ
T= 1487.77sinΣ+810.166cos Σ Differentiate then equate to zero. T ′=1487.77cosΣ−810.166sinΣ= 0
tan Σ=
ΣΣΣΣ = 61. o
20-5. Plot the efficiency of the collector described in Example 20-3 versus temperature of fluid entering the absorber over the range of 30 to 140 C fluid temperatures. The ambient temperature is 10 C. If the collector is being irradiated at 750 W/m 2 , determine the rate of collection at entering fluid temperatures at (a) 50 C and (b) 100 C.
Solution: Refer to Example 20-3. t (^) ∞=10C tai= 30 to 140 C
Iiθ = 800 W/m Fr = 0. αa = 0. τ (^) c1 =τc2=0.
Eq. 20-12.
Fr I
t t U I
A
q
i
ai c1c i
a
η= = τ τ α − θ
∞ θ
a
t 10 3. 0.87 0.87 0.9 ai
η= −
Tabulation: tai η 30 0. 40 0. 50 0. 60 0. 70 0. 80 0. 90 0. 100 0. 110 0. 120 0. 130 0. 140 0. Plot:
(a) At 50 C, Iiθ = 750 W/m^2
(I U ( t t ))Fr
A
q a (^) = (^) iθ τc1τc2αa− ai− ∞
qa A= ( 750 )(0.87 )( 0.87)( 0.9) −(3.5 )( 50 − 10 )
tao = 127.8 C - - - Ans.
(c) If qa = 0
Eq. 20-
q a A= 0 =(I iθ τc1τc2αa−U (t ai−t∞))Fr
0 =( ( 900 )( 0.9)( 0.9) −(6.5 )( tai − 32 ))0.
tai = 144.2 C - - - Ans.
20-7. Two architects have different notions of how to orient windows on the west side of a building in order to be most effective from a solar standpoint-summer and winter. The windows are double-glazed. The two design are shown in Fig. 20-15. Compute at 40o^ north latitude the values of IT from Eq. (20-14) for June 21 at 2 and 6 P.M. and January 21 at 2 P.M. and then evaluate the pros and cons of the two orientations. See Fig. 20-15.
Solution: Eq. 20-
I T= IDN ( cosφ)τ
(a) For notion (a).
At 40o^ north latitude, June 21 at 2 P.M. δ = 23. o
H = 2 x 15 = 30o L = 40 o
Eq. 20- sin β = cos L cos H cos δ + sin L sin δ sin β = cos 40 cos 30 cos 23.5 + sin 40 sin 23. β = 59.
Eq. 20-
β
δ φ = cos
cossinH sin
cos59.
cos23.5sin 30 sin φ=
φ= 65.
Eq. 20-8.
cos θ = cos β cos γ sin Σ + sin β cos Σ Σ = tilt angle = 30 For facing west, Eq. 20-6. γ=φ± ϕ ϕ= 60 γ = 65.91- 60 = 5. cos θ = cos 59.85 cos 5.91 sin 30 + sin 59.85 cos 30 cos θ = 0. θ = 87
Fig. 20-6, Double Glazing τ = 0.
I T= IDN ( cosφ)τ
β
sin exp B
A
IDN
A = 1080 W/m^2 in Mid-summer B = 0.21 in summer
exp (^ 0.21 sin59.85)
IT =
IT = 93.06 W/m 2
o north latitude, June 21, 6 P.M.
δ = 23.5o
H = 6 x 15 = 90 o
L = 40o
Eq. 20- sin β = cos L cos H cos δ + sin L sin δ sin β = cos 40 cos 90 cos 23.5 + sin 40 sin 23. β = 14.
Eq. 20-
β
δ φ = cos
cossinH sin
cos14.
cos23.5sin 90 sin φ=
φ= 71.
But Table 4-13, φ> 90 φ= 180 - 71.58= 108.
Eq. 20-8.
cos θ = cos β cos γ sin Σ + sin β cos Σ Σ = tilt angle = 30 For facing west, Eq. 20-6. γ=φ± ϕ ϕ= 60
γ = 108.42- 60 = 48. cos θ = cos 14.85 cos 48.42 sin 30 + sin 14.85 cos 30 cos θ = 0. θ = 57.
Fig. 20-6, Double Glazing τ = 0.
I T= IDN ( cosφ)τ
β
sin exp B
A
IDN
A = 1230 W/m^2 in December and January B = 0.14 in summer
exp (^ 0.14 sin23.66)
IT =
IT = 493 W/m 2
Then:
June 21, 2 P.M. IT = 93.06 W/m 2
June 21, 6 P.M. IT = 175.65 W/m^2
January 21, 2 P.M. IT = 493 W/m 2
(b) For notion (b).
At 40 o north latitude, June 21 at 2 P.M. δ = 23.5o H = 2 x 15 = 30 o
L = 40o
Eq. 20- sin β = cos L cos H cos δ + sin L sin δ sin β = cos 40 cos 30 cos 23.5 + sin 40 sin 23. β = 59.
Eq. 20-
β
δ φ = cos
cossinH sin
cos59.
cos23.5sin 30 sin φ=
φ= 65.
Eq. 20-8.
cos θ = cos β cos γ sin Σ + sin β cos Σ Σ = tilt angle = - For facing west, Eq. 20-6. γ=φ± ϕ ϕ= 60 γ = 65.91- 60 = 5. cos θ = cos 59.85 cos 5.91 sin (-30) + sin 59.85 cos (-30) cos θ = 0. θ = 60.
Fig. 20-6, Double Glazing τ = 0.
I T= IDN ( cosφ)τ
β
sin exp B
A
IDN
A = 1080 W/m^2 in Mid-summer B = 0.21 in summer
exp (^ 0.21 sin59.85)
IT =
IT = 274.8 W/m 2
o north latitude, June 21, 6 P.M. δ = 23.5o H = 6 x 15 = 90 o
L = 40 o
Eq. 20- sin β = cos L cos H cos δ + sin L sin δ sin β = cos 40 cos 90 cos 23.5 + sin 40 sin 23. β = 14.
Eq. 20-
β
δ φ = cos
cossinH sin
cos14.
cos23.5sin 90 sin φ=
φ= 71. But Table 4-13, φ> 90 φ= 180 - 71.58= 108. Eq. 20-8.
cos θ = cos β cos γ sin Σ + sin β cos Σ Σ = tilt angle = - For facing west, Eq. 20-6. γ=φ± ϕ ϕ= 60 γ = 108.42- 60 = 48. cos θ = cos 14.85 cos 48.42 sin (-30) + sin 14.85 cos (-30) cos θ = -0. θ = 95.67 > 90 Therefore
IT = 0.00 W/m 2
At 40 o north latitude, January 21 at 2 P.M.
360284 N
23.47sin
δ=
N = 21
