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iragen [17]
3 years ago
7

Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours

at 1800 rev/min with a load of 2.75 kN with a reliability of 90 percent.
Engineering
1 answer:
svet-max [94.6K]3 years ago
3 0

Answer:

C₁₀ = 6.3 KN

Explanation:

The catalog rating of a bearing can be found by using the following formula:

C₁₀ = F [Ln/L₀n₀]^1/3

where,

C₁₀ = Catalog Rating = ?

F = Design Load = 2.75 KN

L = Design Life = 1800 rev/min

n = No. of Hours Desired = 10000 h

L₀ = Rating Life = 500 rev/min

n₀ = No. of Hours Rated = 3000 h

Therefore,

C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3

C₁₀ = (2.75 KN)(2.289)

<u>C₁₀ = 6.3 KN</u>

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Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original
Liono4ka [1.6K]

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Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:

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Above command is tested in MATLAB command window and is attached in figure below

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3 years ago
For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep
Alisiya [41]

Answer:

MRR = 1.984

Explanation:

Given that                              

Depth of cut ,d=0.105 in

Diameter D= 1 in

Speed V= 105 sfpm

feed f= 0.015 ipr

Now  the metal   removal  rate   given as

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d= depth of cut

V= Speed

f=Feed

MRR= Metal removal rate

By putting the values

MRR= 12 f V d

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1.944

8 0
3 years ago
A plane, opaque, surface M has the following properties: gray, diffuse, absorptivity = 0.7, surface area = 0.5 m2 , temperature
BaLLatris [955]

Answer:

The rate of energy absorbed per unit time is 3500W.

Explanation:

From the question, we were given the following parameters;

Plane, opaque, gray, diffuse surface

â = 0.7

Surface area, A = 0.5m²

Incoming radiant energy, G = 10000w/m²

T = 500°C

Rate of energy absorbed is âAG;

âAG = 0.7 × 0.5 × 10000

âAG = 3500W.

The energy absorbed is measured in watts and denoted by the symbol W.

7 0
3 years ago
(a) If 5 x 10^17 phosphorus atoms per cm3 are add to silicon as a substitutional impurity, determine the percentage of silicon a
Y_Kistochka [10]

Answer:

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice = 0.001 %

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice with boron atoms = 0.4 ×10^{-5} %

Explanation:

No. of phosphorus atoms = 5 × 10^{17} \ cm^{-3}

The volume occupied by a single Si atom

V_{si} = \frac{a^{3} }{8}

V_{si} = \frac{5.43^{3}(10^{-8} )^{3}  }{8}

V_{si} = 2 × 10^{-23} \frac{cm^{3} }{atom}

n_{si} = \frac{1}{V_{si} }

n_{si} = 5 × 10^{22} \frac{atoms}{cm^{3} }

PCT = \frac{N_p}{N_{si}}   100

Put the values in above equation we get

PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100

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These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

(b).

No. of boron atoms = 2 × 10^{15} \ cm^{-3}

The volume occupied by a single Si atom

V_{si} = \frac{a^{3} }{8}

V_{si} = \frac{5.43^{3}(10^{-8} )^{3}  }{8}

V_{si} = 2 × 10^{-23} \frac{cm^{3} }{atom}

n_{si} = \frac{1}{V_{si} }

n_{si} = 5 × 10^{22} \frac{atoms}{cm^{3} }

PCT = \frac{N_p}{N_{si}}   100

Put the values in above equation we get

PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100

PCT = 0.4 ×10^{-5} %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

7 0
3 years ago
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