A photovoltaic panel of dimension 2 m × 4 m isinstalled on the root of a home. The panel is irradiatedwith a solar flux of G s = 700 W/m 2 , oriented normal Wthe top panel surface. The absorptivity of the panel to thesolar irradiation is α s = 0.83 , and the efficiency of conversion of the absorbed flux to electrical power is η = P / α s G s A = 0.553 − 0.001 K -1 T p where T P is thepanel temperature expressed in kelvins and A is the solarpanel area. Determine the electrical power generated for(a) a still summer day, in which T s u r = T ∞ = 35 ° C . h = 10 W/m 2 ⋅ K , and (b) a breezy winter day,for which T s u r = T ∞ = − 15 ° C , h = 30 W/m 2 ⋅ K . The panel emissivity is ∈ = 0.90 .
A photovoltaic panel of dimension 2 m × 4 m isinstalled on the root of a home. The panel is irradiatedwith a solar flux of G s = 700 W/m 2 , oriented normal Wthe top panel surface. The absorptivity of the panel to thesolar irradiation is α s = 0.83 , and the efficiency of conversion of the absorbed flux to electrical power is η = P / α s G s A = 0.553 − 0.001 K -1 T p where T P is thepanel temperature expressed in kelvins and A is the solarpanel area. Determine the electrical power generated for(a) a still summer day, in which T s u r = T ∞ = 35 ° C . h = 10 W/m 2 ⋅ K , and (b) a breezy winter day,for which T s u r = T ∞ = − 15 ° C , h = 30 W/m 2 ⋅ K . The panel emissivity is ∈ = 0.90 .
Solution Summary: The author explains the electrical power generated for a still summer day and the solar flux of irradiation.
A photovoltaic panel of dimension
2
m
×
4
m
isinstalled on the root of a home. The panel is irradiatedwith a solar flux of
G
s
=
700
W/m
2
, oriented normal Wthe top panel surface. The absorptivity of the panel to thesolar irradiation is
α
s
=
0.83
,
and the efficiency of conversion of the absorbed flux to electrical power is
η
=
P
/
α
s
G
s
A
=
0.553
−
0.001
K
-1
T
p
where
T
P
is thepanel temperature expressed in kelvins and A is the solarpanel area. Determine the electrical power generated for(a) a still summer day, in which
T
s
u
r
=
T
∞
=
35
°
C
.
h
=
10
W/m
2
⋅
K
,
and (b) a breezy winter day,for which
T
s
u
r
=
T
∞
=
−
15
°
C
,
h
=
30
W/m
2
⋅
K
. The panel emissivity is
∈
=
0.90
.
Q4/ A compressor is driven motor by mean of a flat belt of thickness 10 mm and a width of
250 mm. The motor pulley is 300 mm diameter and run at 900 rpm and the compressor
pulley is 1500 mm diameter. The shaft center distance is 1.5 m. The angle of contact of
the smaller pulley is 220° and on the larger pulley is 270°. The coefficient of friction
between the belt and the small pulley is 0.3, and between the belt and the large pulley is
0.25. The maximum allowable belt stress is 2 MPa and the belt density is 970 kg/m³.
(a) What is the power capacity of the drive and (b) If the small pulley replaced by
V-grooved pulley of diameter 300 mm, grooved angle of 34° and the coefficient of
friction between belt and grooved pulley is 0.35. What will be the power capacity in this
case, assuming that the diameter of the large pulley remain the same of 1500 mm.
You are tasked with designing a power drive system to transmit power between a motor and a conveyor belt in a manufacturing facility as illustrated in figure.
The design must ensure efficient power transmission, reliability, and safety. Given the following specifications and constraints, design drive system for this application:
Specifications:
Motor Power: The electric motor provides 10 kW of power at 1,500 RPM.
Output Speed: The output shaft should rotate at 150 rpm.
Design Decisions:
Transmission ratio: Determine the necessary drive ratio for the system.
Shaft Diameter: Design the shafts for both the motor and the conveyor end.
Material Selection: Choose appropriate materials for the gears, shafts.
Bearings: Select suitable rolling element bearings.
Constraints:
Space Limitation:
The available space for the gear drive system is limited to a 1-meter-long section.
Attribute 4 of CEP
Depth of knowledge required
Fundamentals-based, first principles analytical approach…
- |
العنوان
In non-continuous dieless drawing process for copper tube as shown in Fig. (1), take the
following data: Do-20mm, to=3mm, D=12mm, ti/to=0.6 and v.-15mm/s. Calculate: (1)
area reduction RA, (2) drawing velocity v. Knowing that: ti: final thickness
V.
Fig. (1)
ofthre
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.