Answer:
Rated power = 1345.66 W/m²
Mechanical power developed = 3169035.1875 W
Explanation:
Wind speed, V = 13 m/s
Coefficient of performance of turbine, = 0.3
Rotor diameter, d = 100 m
or
Radius = 50 m
Air density, ρ = 1.225 kg/m³
Now,
Rated power =
or
Rated power =
or
Rated power = 1345.66 W/m²
b) Mechanical power developed =
Here, A is the area of the rotor
or
A = π × 50²
thus,
Mechanical power developed =
or
Mechanical power developed = 3169035.1875 W
Answer:
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
Explanation:
The peak voltage after the 6 to 1 step down is . Then, the peak voltage of the rectified output is
V_{d}[/tex] and according to the statement, the diodes can be modeled to be
. Then, the peak voltage in the load is
.
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
The average output voltage is calculated as:
The average current in the load is calculated as:
[ find the attachment ]
Answer:
a) 0.978
b) 0.9191
c) 1.056
d) 0.849
Explanation:
Given data :
Stiffness of each bolt = 1.0 MN/mm
Stiffness of the members = 2.6 MN/mm per bolt
Bolts are preloaded to 75% of proof strength
The bolts are M6 × 1 class 5.8 with rolled threads
Pmax =60 kN, Pmin = 20kN
<u>a) Determine the yielding factor of safety</u>
------ ( 1 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
Input the given values into the equation above
equation 1 becomes ( np ) = = 0.978
note : values above are derived values whose solution are not basically part of the required solution hence they are not included
<u>b) Determine the overload factor of safety</u>
------- ( 2 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
input values into equation 2 above
hence : = 0.9191
<u>C) Determine the factor of safety based on joint separation</u>
Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,
input values into equation above
Hence = 1.056
<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>
nf = 0.849
attached below is the detailed solution .
