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Rasek [7]
4 years ago
12

A particular motor rotates at 3000 revolutions per minute (rpm). What is its speed in rad/sec, and how many seconds does it take

to make one revolution?
Engineering
2 answers:
raketka [301]4 years ago
5 0

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

Now by putting the values in the above equation

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

\dfrac{1}{50}=0.02\ s

Murljashka [212]4 years ago
4 0

Answer:

6284 rad/sec

Explanation:

1 revolution per minute = 2pi/60 rad per second

3000rpm = 2pi/60 * 3000

= 6284rad/sec

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What is the effect of the workpiece specific cutting energy on the cutting forces, and why?
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Explanation:

Specific cutting energy:

   It the ratio of power required to cut the material to metal removal rate of material.If we take the force required to cut the material is F and velocity of cutting tool is V then cutting power will be the product of force and the cutting tool velocity.

Power P = F x V

Lets take the metal removal rate =MRR

Then the specific energy will be

    sp=\dfrac{F\times V}{MRR}

If we consider that metal removal rate and cutting tool velocity is constant then when we increases the cutting force then specific energy will also increase.

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4 years ago
A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at
Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, T_{s}=280^{o}C

Temperature of air, T_{\infinity}=20^{o}C

Velocity of air flow V=5.5m/s

The film temperature is determined as:

T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C

from the table, properties of air at 1 atm pressure

At T_{f}=150^{o}C

Thermal conductivity, K = 0.03443 W/m^oC; kinematic viscosity v=2.860 \times 10^{-5} m^2/s; Prandtl number Pr=0.70275

The reynolds number for the flow is determined as:

Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92

sice the obtained reynolds number is less than 2\times10^5, the flow is said to be laminar.

The nusselt number is determined from the relation given by:

Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}

Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11

The covective heat transfer coefficient is given by:

Nu_{cyl}=\frac{hD}{k}

Rewrite and solve for h

h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m

The rate of heat transfer from the wire to the air per meter length is Q=340.42W/m

6 0
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Answer:

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Explanation:

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The Answer Is C. A collection of computer programs or applications along with its related data
Hope this helps
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