Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e

r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Answer: hope it helps
Explanation:Moving air has a force that will lift kites and balloons up and down. Air is a mixture ... Here is a simple computer simulation that you can use to explore how wings make lift. ... All these dimensions together combine to control the flight of the plane. A pilot ... When the rudder is turned to one side, the airplane moves left or right.
The HR department should install wheelchair ramps and widen the doorways as they don't want to loose an exceptional systems engineer who hopes to work at the techtonic group.
What is a wheelchair ramps?
A wheelchair ramp is an inclined plane that can be installed in addition to or instead of stairs. Ramps make it easier for wheelchair users, as well as people pushing strollers, carts, or other wheeled objects, to enter a building.
The Americans with Disabilities Act requires wheelchair ramps (or other ways for wheelchair users to access a building, such as a wheelchair lift) in new construction for public accommodations in the United States.
A wheelchair ramp can be fixed, semi-fixed, or portable. Permanent ramps are intended to be bolted or otherwise attached to the ground. Semi-permanent ramps are commonly used in the short term and rest on top of the ground or concrete pad.
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Answer:
1. Make sure the regulator adjustment screws are completely backed out and loose.
2. Open the main cylinder valves slowly.
3. Set the Oxygen and Acetylene working pressures.
4. Light and adjust the torch.
Explanation:
Answer:
1) q=18414.93 W
2) C=12920$
Explanation:
Given data:
pipe length L=25m
pipe diameter D=100mm =0.1 m
air temperature
=
=25
°C.....= 298.15k
pipe surface temp
=150
°C.....=423.15k
surface emissivity e= 0.8
boiler efficiency η=0.90
natural gas price Cg=$0.02 per MJ
1) Total heat loss and radiation heat loss combined
q=
q=
б(
^4-
^4)]....... (1)
б=5.67×10^-8 W/m^2K^4 (boltzmann constant)
area A =L.Dπ=25×0.1π=7.85 m^2
putting all these values in eq (1)
q=18414.93 W
2) suppose boiler is operating non stop annual energy loss will be
E=q.t
=18414.93.3600.24.365
=5.81×10^11 J
to find furnace energy consumption
Ef =E/η
=6.46×10^5 MJ
annual cost
C=Cg. Ef
=12920$