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3 years ago

List two reasons why machined parts often require a high degree of precision.

Engineering

1 answer:

Gekata [30.6K]3 years ago

3 0

Explanation:

Precision machining is a subtractive process used in cases where material needs to be removed from a raw product to create the finished product. Precision machining can be used to create a wide variety of products, items, and parts for any number of different objects and materials. These parts usually require tight tolerances variation from nominal dimensions and from part to part, which means that there is not much room for error in the production of the piece. Repeatability and well-controlled tolerances are hallmarks of precision machining. Components, parts and finished durable products that are designed to maintain extremely tight tolerance margins and a high degree of durability are essential and common drivers for utilization of precision machining. For example, parts that need to work together as part of a machine may need to always align within a certain margin of 0.01mm to 0.05mm. Precision engineering and machining help to ensure these parts can not only be made precisely but can be produced with this level of accuracy over and over again.

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2 years ago

Read 2 more answers

(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1

olya-2409 [2.1K]

Answer:

The answer is " $4.35 \times 10^{-3}\ mm$ and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

$y = yo + \frac{k}{\sqrt{x}}$

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.

Replacing the equation for each condition:

$y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k$

People can get yo = 275 MPa with both equations and k= 15.5 Mpa $mm^{\frac{1}{2}}$ .

To substitute the answer,

$310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm$

In point b:

The equation is $\sigma y = \sigma 0 + k y d^{\frac{1}{2}}$

equation is:

$75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\$

by putting the above value in the formula we get the $\sigma y$ value that is= 157.5 MPa

5 0

2 years ago

A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K

Mashcka [7]

Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

$\Delta S=\dfrac{Q}{T}$

Now by putting the values

$20.3=\dfrac{Q}{400}$

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

8 0

3 years ago