Answer:
<em>minimum required diameter of the steel linkage is 3.57 mm</em>
<em></em>
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 = 
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/(
) = 
m^2
recall that area = 
= 
= 
= 
= 
m = 3.57 mm
<em>maximum diameter of the steel linkage d = 3.57 mm</em>
Answer:
......................................................
Concentrating solar power (CSP) plants use mirrors to concentrate the sun's energy to drive traditional steam turbines or engines that create electricity. The thermal energy concentrated in a CSP plant can be stored and used to produce electricity when it is needed, day or night. Today, roughly 1,815 megawatts (MWac) of CSP plants are in operation in the United States.
Parabolic Trough
Parabolic trough systems use curved mirrors to focus the sun’s energy onto a receiver tube that runs down the center of a trough. In the receiver tube, a high-temperature heat transfer fluid (such as a synthetic oil) absorbs the sun’s energy, reaching temperatures of 750°F or higher, and passes through a heat exchanger to heat water and produce steam. The steam drives a conventional steam turbine power system to generate electricity. A typical solar collector field contains hundreds of parallel rows of troughs connected as a series of loops, which are placed on a north-south axis so the troughs can track the sun from east to west. Individual collector modules are typically 15-20 feet tall and 300-450 feet long.
Compact Linear Fresnel Reflector
CLFR uses the principles of curved-mirror trough systems, but with long parallel rows of lower-cost flat mirrors. These modular reflectors focus the sun's energy onto elevated receivers, which consist of a system of tubes through which water flows. The concentrated sunlight boils the water, generating high-pressure steam for direct use in power generation and industrial steam applications.
Answer:
Solution:
Note: Refer the diagram
Drag coefficient data for selected objects table at
Hemisphere (open end facing flow), 
Substituting all parameters,
Then,
![\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3DV_%7Bw%7D-%5Cleft%5B%5Cfrac%7B2%20F_%7BR%7D%7D%7B%5Crho%5Cleft%28C_%7BD%2C%20w%7D%20A_%7Bw%7D%2BC_%7BD%2C%20B%7D%20A_%7Bb%7D%5Cright%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cdots%5C%5C%26V_%7Bw%7D%3D24%20%5Ctimes%201000%20%5Ctimes%20%5Cfrac%7B1%7D%7B3600%7D%5C%5C%26V_%7Bw%7D%3D6.67%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
And the equation becomes,
![\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3D6.67-%5Cleft%5B%5Cfrac%7B2%20%5Ctimes%205.52%7D%7B1.23%281.42%20%5Ctimes%201.17%2B1.2%20%5Ctimes%200.3%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%26V_%7Bb%7D%3D6.67-2.11%5C%5C%26V_%7Bb%7D%3D4.56%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
Thus the floyds travels at 
wind speed.
