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3 years ago

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In general, this of the following methods yields the most conservative fatigue strength proof (a) Saderberg method (b)-Goodman m

ethod (c)-Gerber line (d)-The ASME elliptic curve.

1 answer:

photoshop1234 [79]3 years ago

7 0

Answer:

a). Soderberg method

Explanation:

A straight line joining the endurance limit, $S_{e}$ on the ordinate and to the yield strength, $S_{yt}$ on the abscissa is know as Soderberg line.

The Soderberg line is the most conservative failure criteria and in this there is no need to consider yielding point in this case.

The equation for Soderberg is given by

$\frac{\sigma _{m}}{S_{yt}}+\frac{\sigma _{a}}{S_{e}}=1$

where $\sigma _{m}$ is mean stress

$\sigma _{a}$ is amplitude stress

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Choose the person who best completes each statement.

victus00 [196]

The exercise is about finding the missing medical terms from the list below.. See the sentences below for the missing words.

<h3>What are the completed sentences?</h3>

The outbreak and spread of epidemic diseases, such as swine flu, are tracked by a <u>Public Health Surveillance Team.</u>
The person who would use equipment to show an image of a baby before birth is a <u>Diagnostic Medical Sonographer.</u>
The education to become a <u>Medical Practitioner</u> requires more than four years of college.
When a water source becomes contaminated with an illness-causing bacteria, a <u>Microbiologist</u> could be brought in to study how to get rid of the microorganisms.

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7 0

2 years ago

Read 2 more answers

What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin

DochEvi [55]

Answer:

$m_{LP}=0.45\,kg$

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

$Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]$

$Q_{water} = 3599.435\,kJ$

The heat liberated by the LP gas is:

$Q_{LP} = \frac{3599.435\,kJ}{0.16}$

$Q_{LP} = 22496.469\,kJ$

A kilogram of LP gas has a minimum combustion power of $50028\,kJ$ . Then, the required mass is:

$m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }$

$m_{LP}=0.45\,kg$

6 0

2 years ago

A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m² of water vapor in equilibrium at 800 kPa. Heat is transferre

docker41 [41]

Answer:

Initial temperature = 170. 414 °C

Total mass = 94.478 Kg

Final volumen = 33.1181 m^3

Diagram = see picture.

Explanation:

We can consider this system as a close system, because there is not information about any output or input of water, so the mass in the system is constant.

The information tells us that the system is in equilibrium with two phases: liquid and steam. When a system is a two phases region (equilibrium) the temperature and pressure keep constant until the change is completed (either condensation or evaporation). Since we know that we are in a two-phase region and we know the pressure of the system, we can check the thermodynamics tables to know the temperature, because there is a unique temperature in which with this pressure (800 kPa) the system can be in two-phases region (reach the equilibrium condition).

For water in equilibrium at 800 kPa the temperature of saturation is 170.414 °C which is the initial temperature of the system.

to calculate the total mass of the system, we need to estimate the mass of steam and liquid water and add them. To get these values we use the specific volume for both, liquid and steam for the initial condition. We can get them from the thermodynamics tables.

For the condition of 800 kPa and 170.414 °C using the thermodynamics tables we get:

Vg (Specific Volume of Saturated Steam) = 0.240328 m^3/kg

Vf (Specific Volume of Saturated Liquid) = 0.00111479 m^3/kg

if you divide the volume of liquid and steam provided in the statement by the specific volume of saturated liquid and steam, we can obtain the value of mass of vapor and liquid in the system.

Steam mass = *0.9 m^3 / 0.240328 m^3/kg = 3.74488 Kg

Liquid mass = 0.1 m^3 /0.00111479 m^3/kg = 89.70299 Kg

Total mass of the system = 3.74488 Kg + 89.70299 Kg = 93,4478 Kg

If we keep the pressure constant increasing the temperature the system will experience a phase-change (see the diagram) going from two-phase region to superheated steam. When we check for properties for the condition of P= 800 kPa and T= 350°C we see that is in the region of superheated steam, so we don’t have liquid water in this condition.

If we want to get the final volume of the water (steam) in the system, we need to get the specific volume for this condition from the thermodynamics tables.

Specific Volume of Superheated Steam at 800 kPa and 350°C = 0.354411 m^3/kg

We already know that this a close system so the mass in it keeps constant during the process.

If we multiply the mass of the system by the specific volume in the final condition, we can get the final volume for the system.

Final volume = 93.4478 Kg * 0.354411 m^3/kg = 33.1189 m^3

You can the P-v diagram for this system in the picture.

For the initial condition you can calculate the quality of the steam (measure of the proportion of steam on the mixture) to see how far the point is from for the condition on all the mix is steam. Is a value between 0 and 1, where 0 is saturated liquid and 1 is saturated steam.

Quality of steam = mass of steam / total mass of the system

Quality of steam = 3.74488 Kg /93.4478 Kg = 0,040 this value is usually present as a percentage so is 4%.

Since this a low value we can say that we are very close the saturated liquid point in the diagram.

6 0

3 years ago

Given the strings s1 and s2, not necessarily of the same length, create a new string consisting of alternating characters of s1

andrew11 [14]

Answer:

THE ANSA IZ A

Explanation:

4 0

3 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

2 years ago