Shielding gases are used to protect the molten metal from what?

An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

A short-circuit experiment is conducted on the high-voltage side of a 500 kVA, 2500 V/250 V, single-phase transformer in its nom

Anarel [89]

Given Information:

Primary secondary voltage ratio = 2500/250 V

Short circuit voltage = Vsc = 100 V

Short circuit current = Isc = 110 A

Short circuit power = Psc = 3200 W

Required Information:

Series impedance = Zeq = ?

Answer:

Series impedance = 0.00264 + j0.00869 Ω

Step-by-step explanation:

Short Circuit Test:

A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.

The series impedance in polar form is given by

Zeq = Vsc/Isc < θ

Where θ is given by

θ = cos⁻¹(Psc/Vsc*Isc)

θ = cos⁻¹(3200/100*110)

θ = 73.08°

Therefore, series impedance in polar form is

Zeq = 100/110 < 73.08°

Zeq = 0.909 < 73.08° Ω

or in rectangular form

Zeq = 0.264 + j0.869 Ω

Where Req is the real part of Zeq and Xeq is the imaginary part of Zeq

Req = 0.264 Ω

Xeq = j0.869 Ω

To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.

Turn ratio = a = Vp/Vs = 2500/250 = 10

Zeq2 = Zeq/a²

Zeq2 = (0.264 + j0.869)/10²

Zeq2 = (0.264 + j0.869)/100

Zeq2 = 0.00264 + j0.00869 Ω

Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.

4 0

3 years ago

HELP ASAP

Illusion [34]

Answer:

You are a historian who is compiling information on the changing materials used in a product in one of the categories listed below. Choose a product that uses a variety of materials that have been discussed in this lesson (for example, a car or a tennis racket). Using online or library resources, research past materials used to create the product. Be sure to discuss the materials currently being used for the design and include information on the sustainability of these resources. Your research must also explain why the materials changes were made (for example, safety, cost, or weight). Keep track of your references so that you can document all of the resources used for your report.

3 0

3 years ago

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

Answer:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

Explanation:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

(i) convert to polar form

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

(i) convert to polar form

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

3 0

3 years ago

3. If nothing can ever be at absolute zero, why does the concept exist?

JulijaS [17]

The reason has to do with the amount of work necessary to remove heat from a substance, which increases substantially the colder you try to go. To reach zero kelvins, you would require an infinite amount of work.

6 0

2 years ago