ENGINEERING COLLEGE

A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 50%, at room temperature, by open-die forging with flat dies. Assume that the coefficient of friction is 0.2. Calculate the forging force at the end of the stroke.

Answers

Answer 1

Answer:

The answer is "45.3 NM".

There at end of the movement, the forging force is given by

$\to F = Y * \pi * r^2 * [1 + ((2 \mu r)/(3h))]$

h is the final height.

$\to h = (100)/(2)= 50 \ mm$

The ultimate radius is determined by following a volume constancy law, which states that volumes before deformation measured amount following distortion.

$\to \pi * 75^2 * 2 * 100 = \pi * r^2 * 2 * 50\n\n\to 75^2 * 2 = r^2\n\n\to r^2 = 11250\n\n\to r = √(11250)\n\n\to r = 106 \ mm\n\n\to E = \In((100)/(50))\n\n\to E = 0.69\n\n$

You may deduce from the graph flow that $Y = 1000\ MPa$ , thus we use the formula.

$= 1000 * 3.14 * 0.106^2 * [1 + (( 2 * 0.2 * 0.106)/(3 * 0.05))]\n\n= 1000 * 3.14 * 0.011236 * [1 + (( 0.0424)/(0.15))]\n\n= 35.3 * 1.2826\n\n = 45.3 \ MN\n\n\n$

Therefore, the answer is "45.3 NM".

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Answer 2

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

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Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are measured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L

Answers

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

How to determine the boundary work done by a gas during an expansion process

A process is a consecution of states of a system. The boundary work (W), in kilojoules, is the work done by the system on surroundings and in a P-Vdiagram this kind of work is equal to the area below the curve, which can be approximated by Riemann sums:

$W = \sum\limits_(i=1)^(n-1) p_(i)\cdot (V_(i+1)-V_(i)) + (1)/(2)\sum\limits_(i=1)^(n-1) (p_(i+1)-p_(i))\cdot (V_(i+1)-V_(i))$ (1)

Where:

p - Pressure, in kilopascals.
V - Volume, in cubic meters.

$W = (1)/(2) \sum\limits_(i=1)^(n-1) (p_(i+1)+p_(i))\cdot (V_(i+1)-V_(i))$

Now we proceed to calculate the boundary work:

W = 0.5 · [(300 kPa + 290 kPa) · (1.1 × 10⁻³ m³ - 1 × 10⁻³ m³) + (270 kPa + 290 kPa) · (1.2 × 10⁻³ m³ - 1.1 × 10⁻³ m³) + (250 kPa + 270 kPa) · (1.4 × 10⁻³ m³ - 1.2 × 10⁻³ m³) + (220 kPa + 250 kPa) · (1.7 × 10⁻³ m³ - 1.4 × 10⁻³ m³) + (200 kPa + 220 kPa) · (2 × 10⁻³ m³ - 1.7 × 10⁻³ m³)]

W = 0.243 kJ

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

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Answer:

attached below

Explanation:

As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine representative power requirements, consider nominal flight conditions for which the plane moves at 100 m/s in air that is at a temperature of -23 degree C. If the characteristic length of the airfoil is L = 2 m and wind tunnel measurements indicate an average friction coefficient of of C_f = 0.0025 for the nominal conditions, what is the average heat flux needed to maintain a surface temperature of T_s = 5 degree C?

Answers

Answer:

Average heat flux=3729.82 W/ $m^(2)$

Explanation:

What is the average distance in microns an electron can travel with a diffusion coefficient of 25 cm^2/s if the electron lifetime is 7.7 microseconds. Three significant digits and fixed point notation.

Answers

Answer: The average distance the electron can travel in microns is 1.387um/s

Explanation: The average distance the electron can travel is the distance an exited electron can travel before it joins together. It is also called the diffusion length of that electron.

It is gotten, using the formula below

Ld = √DLt

Ld = diffusion length

D = Diffusion coefficient

Lt = life time

Where

D = 25cm2/s

Lt = 7.7

CONVERT cm2/s to um2/s

1cm2/s = 100000000um2/s

Therefore D is

25cm2/s = 2500000000um2/s = 2.5e9um2/s

Ld = √(2.5e9 × 7.7) = 138744.37um/s

Ld = 1.387e5um/s

This is the average distance the excited electron can travel before it recombine

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.

Answers

Answer:

critical stress required is 18.92 MPa

Explanation:

given data

specific surface energy = 1.0 J/m²

modulus of elasticity = 225 GPa

internal crack of length = 0.8 mm

solution

we get here one half length of internal crack that is

2a = 0.8 mm

so a = 0.4 mm = 0.4 × $10^(-3)$ m

so we get here critical stress that is

$\sigma _c = \sqrt{(2E \gamma )/(\pi a)}$ ...............1

put here value we get

$\sigma _c$ = $\sqrt{(2* 225* 10^9 * 1 )/(\pi * 0.4* 10^(-3))}$

$\sigma _c$ = 18923493.9151 N/m²

$\sigma _c$ = 18.92 MPa

Some organizations prohibit workers from bringing certain kinds of devices into the workplace, such as cameras, cell phones, and USB drives. Some businesses require employees to use clear or see-through backpacks when carrying personal items. What other devices might not be allowed in certain facilities, and why would they be restricted? The video on Google’s Data Center may give you some ideas to write about for this assignment.

Answers

Answer:

Personal computers:

Personal computers may be useful and lead to productivity as using a computer an employee familiars with is a good thing. However, the disadvantages in some facilities especially ones dealing with customer and information security can include data theft, unauthorized data sharing, uses of internet connection for personal purposes, as this can slow down internet connection at the facility, distraction at work place etc.

Hard drive:

Due to large amount of data that can be stored on a hard drive, it might not be allowed in some facilities to avoid data theft and unauthorized transfer.

Music players:

This might be restricted to avoid distraction at work. Noice in places such as libraries would cause abnormality and poor service delivery. An employee with loud speaker at work would not only distracts himself but also other staffs and customers.

PSP Game Device and other game devices:

Playing games during working hour may jeopardize the productivity and therefore might be resctrited in some facilities and working places.

Electronic digital notepad:

Carrying a handheld electronic digital notepad to the work place can cause lack of concentration and division of attention on work and other personal activities. These can harm working harmony and business productivity.

Video recorder:

In some facilities, this device might not be allowed due to facility privacy and protection from unwanted navigation.

Explanation:

A cylindrical bar of metal having a diameter of 19.2 mm and a length of 207 mm is deformed elastically in tension with a force of 52900 N. Given that the elastic modulus and Poisson's ratio of the metal are 61.4 GPa and 0.34, respectively, determine the following: a. The amount by which this specimen will elongate in the direction of the applied stress.
b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answers

Answer:

1)ΔL = 0.616 mm

2)Δd = 0.00194 mm

Explanation:

We are given;

Force; F = 52900 N

Initial length; L_o = 207 mm = 0.207 m

Diameter; d_o = 19.2 mm = 0.0192 m

Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²

Now, from Hooke's law;

E = σ/ε

Where; σ is stress = force/area = F/A

A = πd²/4 = π × 0.0192²/4

A = 0.00009216π

σ = 52900/0.00009216π

ε = ΔL/L_o

ε = ΔL/0.207

Thus,from E = σ/ε, we have;

61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)

Making ΔL the subject, we have;

ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)

ΔL = 0.616 × 10^(-3) m

ΔL = 0.616 mm

B) Poisson's ratio is given as;

υ = ε_x/ε_z

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

υ = (Δd/d_o) ÷ (ΔL/L_o)

Making Δd the subject gives;

Δd = (υ × d_o × ΔL)/L_o

We are given Poisson's ratio to be 0.34.

Thus;

Δd = (0.34 × 19.2 × 0.616)/207

Δd = 0.00194 mm

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