Answer:
d
Explanation:
Beryllium copper combines high strength with non-magnetic and non-sparking qualities. It has excellent metalworking, forming and machining properties. It has many specialized applications in tools for hazardous environments, musical instruments, precision measurement devices, bullets, and aerospace.
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 = ![\frac{1}{2}\rho V^3](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Crho%20V%5E3)
or
Rated power = ![\frac{1}{2}\times1.225\times13^3](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes1.225%5Ctimes13%5E3)
or
Rated power = 1345.66 W/m²
b) Mechanical power developed = ![\frac{1}{2}\rho AV^3C_p](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Crho%20AV%5E3C_p)
Here, A is the area of the rotor
or
A = π × 50²
thus,
Mechanical power developed = ![\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes1.225%5Ctimes%5Cpi%5Ctimes50%5E2%5Ctimes13%5E3%5Ctimes0.3)
or
Mechanical power developed = 3169035.1875 W
Flip-flops are normally used for all of the following applications, except logic gates.
<h3>What are Flip flops?</h3>
Flip flops are known to be tools that are used for counting. They come in different ranges.
Note that Flip flops are one that can be seen on counters, storage registers, and others and as such, Flip-flops are normally used for all of the following applications, except logic gates.
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Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 ×
= 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 ×
/ 3
so d = 0.0133 m
so diameter is 14 mm
The amount of work done by steady flow devices varies with the particular gas volume. The kinetic energy of gas particles decreases during cooling.
When the gas is subjected to intermediate cooling during compression, the gas specific volume is reduced, which lowers the compressor's power consumption. Compression is less adiabatic and more isothermal because the compressed gas must be cooled between stages since compression produces heat. The system's thermodynamic cycle's cold sink temperature is lowered by cooling the compressor coils. By increasing the temperature difference between the heat source and the cold sink, this improves efficiency.
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