According to its directional distribution, solar radiation incident on the earth’s surface may be divided into two components. The direct component consists of parallel rays incident at a fixed zenith angle θ , while the diffuse component consists of radiation that may be approximated as being diffusely distributed with θ . Consider clear sky conditions for which the direct radiation is incident at 0 = 30 ∘ , with a total flux (based on an area that is normal to the rays) of q d i r " = 1000 W / m 2 , and the total intensity of the diffuse radiation is I d i f = 70 W / m 2 · s r . What is the total solar irradiation at the earth’ s surface?
According to its directional distribution, solar radiation incident on the earth’s surface may be divided into two components. The direct component consists of parallel rays incident at a fixed zenith angle θ , while the diffuse component consists of radiation that may be approximated as being diffusely distributed with θ . Consider clear sky conditions for which the direct radiation is incident at 0 = 30 ∘ , with a total flux (based on an area that is normal to the rays) of q d i r " = 1000 W / m 2 , and the total intensity of the diffuse radiation is I d i f = 70 W / m 2 · s r . What is the total solar irradiation at the earth’ s surface?
Solution Summary: The author calculates the total solar irradiation at the surface of earth as the sum of direct radiation and diffuse radiation.
According to its directional distribution, solar radiation incident on the earth’s surface may be divided into two components. The direct component consists of parallel rays incident at a fixed zenith angle
θ
, while the diffuse component consists of radiation that may be approximated as being diffusely distributed with
θ
. Consider clear sky conditions for which the direct radiation is incident at
0
=
30
∘
, with a total flux (based on an area that is normal to the rays) of
q
d
i
r
"
=
1000
W
/
m
2
, and the total intensity of the diffuse radiation is
I
d
i
f
=
70
W
/
m
2
·
s
r
. What is the total solar irradiation at the earth’ s surface?
Experiment
تكنولوجيا السيارات
- Internal Forced convenction Heat transfer Air Flow through Rectangular Duct.
objective: Study the convection heat transfer of
air
flow through rectangular duct.
Valve Th
Top Dead Centre
Exhaust Valve Class
CP.
N; ~
RIVavg Ti
K
2.11
Te To
18.8 21.3 45.8
Nath Ne
Pre
Calculations:.
Q = m cp (Te-Ti)
m: Varg Ac Acca*b
Q=hexp As (Ts-Tm)
2
2.61
18.5 20.846.3
Tm = Te-Ti =
25
AS-PL
= (a+b)*2*L
Nu exp=
Re-Vavy D
heep Dh
k
2ab
a+b
Nu
Dh
the-
(TS-Tm)
Ts. Tmy Name / Nu exp
Naxe
بب ارتدان
العشري
Procedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to find
Procedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of the
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