What forced induction device is more efficient?

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

4 years ago

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of

Orlov [11]

Answer: 133.88 MPa approximately 134 MPa

Explanation:

Given

Plane strains fracture toughness, k = 26 MPa

Stress at which fracture occurs, σ = 112 MPa

Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m

Critical internal crack length, l' = 6 mm = 6*10^-3 m

We know that

σ = K/(Y.√πa), where

112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]

112 MPa = 26 MPa / Y.√(3.142 * 0.043)

112 = 26 / Y.√1.35*10^-2

112 = 26 / Y * 0.116

Y = 26 / 112 * 0.116

Y = 26 / 13

Y = 2

σ = K/(Y.√πa), using l'instead of l and, using Y as 2

σ = 26 / 2 * [√3.142 * (6*10^-3/2)]

σ = 26 / 2 * √(3.142 *3*10^-3)

σ = 26 / 2 * √0.009426

σ = 26 / 2 * 0.0971

σ = 26 / 0.1942

σ = 133.88 MPa

8 0

3 years ago

What are the main causes of injuries when using forklifts?

Julli [10]

Answer:

1, 3, 4, 5

Explanation:

just did it

4 0

3 years ago

1. What should a technician look for during a visual inspection?

USPshnik [31]

: B (damaged or broken components)

8 0

3 years ago

Describe in your own words the three strengthening mechanisms discussed in this chapter (i.e., grain size reduction, solid-solut

Kobotan [32]

Explanation:

Strengthening by grain size reduction

It is based on the fact that dislocations will experience hindrances while trying to move from a grain into the next because of abrupt change in orientation of planes.

Hindrances can be two types: forcible change of slip direction, and discontinuous slip plane.
Smaller the grain size, often a dislocation encounters a hindrance. Yield strength of material will be increased.
Yield strength is related to grain size (diameter, d ) as Hall Petch relation:

$\sigma_{y}=\sigma_{i}+k d^{-1 / 2}$

Strengthening by Grain size reduction (contd..)

Grain size reduction improves not only strength, but also the toughness of many alloys.
If d is average grain diameter, $S_{v}$ is grain boundary area per unit volume, $N_{L}$ is mean number of intercepts of grain boundaries per unit length of test line, $N_{A}$ is number of grains per unit area on a polished surface:

$S_{v}=2 N_{L} \quad d=\frac{3}{S_{v}}=\frac{3}{2 N_{L}} \quad d=\sqrt{\frac{6}{\pi V_{A}}}$

Grain size can also be measured by comparing the grains at a fixed magnification with standard grain size charts.

Other method: Use of ASTM grain size number (Z). It is related to grain diameter, (in mm) as follows:

$D=\frac{1}{100} \sqrt{\frac{645}{2^{6-1}}}$

Solid solution strengthening

Impure foreign atoms in a single phase material produces lattice strains which can anchor the dislocations.

Effectiveness of this strengthening depends on two factors size difference and volume fraction of solute. Solute atoms interact with dislocations in many ways:

- elastic interaction

- modulus interaction

- stacking-fault interaction

- electrical interaction

- short-range order interaction

- long-range order interaction

3 0

4 years ago