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Anarel [89]
4 years ago
13

You are designing a geartrain with three spur gears: one input gear, one idler gear,and one output gear. The diametral pitch for

the geartrain is 16. The diameter of the input gear needs to be twice the diameter of the idler gear and three times the diameter of the output gear. The entire geartrain needs to fit into a rectangular footprint of no larger than 22 in. high and 15 in. long. Determine an appropriate number of teeth of the smallest gear.
Engineering
1 answer:
Mrac [35]4 years ago
4 0

Answer:

The appropriate number of teeth of the smallest gear should is 58 teeth.

Explanation:

The given parameters include;

The diametral pitch = 16

Number of gears = 3

Diameter of the input gear = 2 × diameter of the idler gear

Diameter of the input gear = 3 × diameter of the output gear

Height of footprint = 22 in.  

Length of footprint = 15 in.  

Let the size of the output gear = X

Therefore, the input gear = 3·X

The diameter of the idle gear = 2·X

Therefore, total width of the gear train = X + 2·X + 3·X = 6·X

Where 6·X = 22, X = 22/6 = 11/3 in.

Since the diametral pitch = 16 then we have;

Diametral \, Pitch = \frac{Number \, of \, teeth}{Pitch \, diameter}

\therefore Diametral \, Pitch \, of  \, the \, smallest \, gear =16 =\frac{Number \, of \, teeth \, of  \, the \, smallest \, gear}{\frac{11}{3} }

Hence, number of teeth of the smallest gear = 16 × 11/3 = 176/3 = 58\tfrac{2}{3}

The appropriate number of teeth of the smallest gear should be 58

From which we have the diameter of the smallest gear = 58/16 = 3.625

The diameter of the input gear is then 3 × 3.625 = 10.875 in.

The diameter of the idler = 2 × 3.625 = 7.25 in.

The appropriate number of teeth of the smallest gear should = 58 teeth.

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Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

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Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

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h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

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4 0
3 years ago
A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 27 . It has b
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Answer:

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

Explanation:

Given data;

Let,

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surface length of the crack = a

dimensionless parameter = Y.

Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m

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Cc = Kic/(Y*√pia*a)

Cc = 27/(2.05*√pia*3.1*10-³)

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The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

For more understanding, I have provided an attachment to the solution.

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4 years ago
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Answer:

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__

If the heating element is open-circuit, no heating will occur. A gasket leak may cause a puddle, but may have nothing to do with the end of the brewing cycle. (Loss of water can be expected to end boiling, rather than prolong it.)

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