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3 years ago

What are the characteristic features of stress corrosion cracks?

Engineering

1 answer:

Oduvanchick [21]3 years ago

5 0

Answer and Explanation:

The crack formation growth that takes place in an environment corrosive.

Stress corrosion cracks can be defined as the spontaneous failures of the metal alloy as a result of the combined action of corrosion and high tensile stress.

Some of the characteristic features of stress corrosion cracks are:

These occur at high temperatures.
Occurrence of failures in metals mechanically.
Occurrence of sudden and unexpected failures under tensile stress.
The rate of work hardening of the metal alloy is high.
Time
An environment that is specific for stress corrosion cracking.

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The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325 P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0

3 years ago

The first thing you should do is develop a ____________________ to determine what vehicle you can afford.

swat32

The first thing you should do is develop a <u>budget</u> to determine what vehicle you can afford.

<h3>What is an automobile?</h3>

An automobile is also referred to as a vehicle, car or motorcar and it can be defined as a four-wheeled vehicle that is designed and developed to be propelled by an internal-combustion (gasoline) engine, especially for the purpose of transportation from one location to another.

<h3>What is a budget?</h3>

A budget can be defined as a financial plan that is typically used for the estimation of revenue and expenditures of an individual, business organization or government for a specified period of time, often one year.

In this context, we can reasonably infer and logically deduce that the first thing anyone should do is to develop a <u>budget</u> in order to determine what vehicle they can afford.

Read more on budget here: brainly.com/question/13964173

#SPJ1

7 0

1 year ago

The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si

alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

<u>A) Classifying the soil according to USCS system</u>

( using 2nd image attached below )

<em>description of sand</em> :

The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0

3 years ago

A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion

kati45 [8]

Answer:

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

Explanation:

Given data:

Diffusion constant for nitrogen is $= 1.85\times 10^{-10} m^2/s$

Diffusion flux $= 1.0\times 10^{-7} kg/m^2-s$

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen concentration is 0.5 kg/m^3 ......?

from fick's first law

$J = D \frac{C_A C_B}{X_A X_B}$

Take C_A as point on which nitrogen concentration is 2 kg/m^3

$x_B = X_A + D\frac{C_A -C_B}{J}$

Assume X_A is zero at the surface

$X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}$

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

4 0

3 years ago

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

nikitadnepr [17]

Answer:

(a) dynamic viscosity = $1.812\times 10^{-5}Pa-sec$

(b) kinematic viscosity = $1.4732\times 10^{-5}m^2/sec$

Explanation:

We have given temperature T = 288.15 K

Density $d=1.23kg/m^3$

According to Sutherland's Formula dynamic viscosity is given by

${\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}$ , here

μ = dynamic viscosity in (Pa·s) at input temperature T,

$\mu _0$ = reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

$T_0$ = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

$\mu _0=4\pi \times 10^{-7}$

$T_0$ = 291.15

$\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s$ when T = 288.15 K

For kinematic viscosity :

$\nu = \frac {\mu} {\rho}$

$kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec$

3 0

3 years ago