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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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Explanation:

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4 years ago

A heat engine uses fuel of energy content 43.1 MJ/kg and produces 17.4 kW of useful power. The heat rejection rate (through the

solmaris [256]

Answer:

a)27.9%

b) $\dot{m}=3.06 \frac{Kg}{h}$

Explanation:

Given that

Fuel energy content = 73.1 MJ/kg

Useful power = 17.4 KW

Heat rejection rate = 44.8 KW

From first law of thermodynamics

Heat addition rate =Heat rejection rate + Power out put

Now by putting the values in the above formula

Heat addition rate = 44.8 + 17.4

Heat addition rate =62.2 KW

We know that efficiency is given as follows

$\eta =\dfrac{power\ out\ put}{Heat\ addition\ rate}$

$\eta =\dfrac{17.4}{62.2}$

$\eta =0.279$

So the efficiency is 27.9%.

Now to find usage rate of fuel

Lets take usage rate is $\dot{m}$

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$73100\times \dfrac{\dot{m}}{3600}=62.2$

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3 years ago

The acceleration of a particle is given by a = 4t – 30, where a is in meters per second squared and t is in seconds. Determine t

mestny [16]

Answer:

The velocity of the particle is given by $v(t) = 2\cdot t^{2}-30\cdot t + 3$ , where $v$ is in meters per second and $t$ is in seconds.

The displacement of the particle is given by $s(t) = \frac{2}{3}\cdot t^{3}-15\cdot t^{2}+3\cdot t -5$ , where $s$ is in meters and $t$ is in seconds.

Explanation:

The acceleration of the particle is given by $a(t) = 4\cdot t -30$ , the expression for velocities are obtained by integration:

Velocity

$v(t) = \int {a(t)} \, dt$

$v(t) = \int {4\cdot t-30} \, dt$

$v(t) = 4\int {t} \, dt -30\int \, dt$

$v(t) = 2\cdot t^{2}-30\cdot t + v_{o}$

Where $v_{o}$ is the initial velocity of the particle, measured in meters per second.

If $t = 0$ and $v(0) = 3$ , then:

$3 = 2\cdot (0)^{2}-30\cdot (0)+v_{o}$

$v_{o} = 3$

The velocity of the particle is given by $v(t) = 2\cdot t^{2}-30\cdot t + 3$ , where $v$ is in meters per second and $t$ is in seconds.

Displacement

$s(t) = \int {v(t)} \, dt$

$s(t) = \int {2\cdot t^{2}-30\cdot t+3} \, dt$

$s(t) = 2\int {t^{2}} \, dt - 30\int {t} \, dt +3\int \, dt$

$s(t) = \frac{2}{3}\cdot t^{3}-15\cdot t^{2}+3\cdot t +s_{o}$

If $t = 0$ and $s(0) = -5$ , then:

$-5 = \frac{2}{3}\cdot (0)^{3}-15\cdot (0)^{2}+3\cdot (0)+s_{o}$

$s_{o} = -5$

The displacement of the particle is given by $s(t) = \frac{2}{3}\cdot t^{3}-15\cdot t^{2}+3\cdot t -5$ , where $s$ is in meters and $t$ is in seconds.

7 0

3 years ago