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Murrr4er [49]
3 years ago
7

How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.

Engineering
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

N_3=\dfrac{T_1}{T_3}N_1

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

Speed\ of\ gear 1=N_1

Number\ of\ teeth\ of\ gear 1=T_1

Speed\ of\ gear 3=N_3

Number\ of\ teeth\ of\ gear 3=T_3

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}

N_3=\dfrac{T_1}{T_3}N_1

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The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto
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Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

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3 years ago
Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.
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Answer: Eye injury

Explanation: small material such as dust, dirt, and metal shards can harm your eyes with potential blindness or infection.

7 0
2 years ago
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The answer is answered! Explanation:

4 0
3 years ago
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A company has a stack that emits a hazardous air pollutant. The ground mass concentration directly downwind of the plume sometim
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do the wam wam

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The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s
Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

lA=\sqrt{0.2^{2}+0.2^{2}  }=0.2\sqrt{2}m

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B} (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+      \frac{1}{2}k(l_{A}-l_{ul})  ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul})  ^{2}

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

\frac{m_{C}v_{B}^{2}   }{R}-N_{B}-k(l_{B}-l_{ul})=0

Replacing values and clearing NB:

NB = 694 N

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3 years ago
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