It is desired to enrich the partial pressure of hydrogen in a hydrogen–nitrogen gas mixture for which the partial pressures of both gases are 0.1013 MPa (1 atm). It has been proposed to accomplish this by passing both gases through a thin sheet of some metal at an elevated temperature; in as much as hydrogen diffuses through the plate at a higher rate than does nitrogen, the partial pressure of hydrogen will be higher on the exit side of the sheet. The design calls for partial pressures of 0.051 MPa (0.5 atm) and 0.01013 MPa (0.1 atm), respectively, for hydrogen and nitrogen. The concentrations of hydrogen and nitrogen (CHC
H and CNC N , in mol/m3mol/m
3 ) in this metal are functions of gas partial pressures (pH2 and pN2p
H 2 and p N , in MPa) and absolute temperature and are given by the following expressions:
CH=2.5×103√pH2exp(−27,800J/mol/RT)
CN=2.75×103√pN2exp(−37,600J/mol/RT )
​Furthermore, the diffusion coefficients for the diffusion of these gases in this metal are functions of the absolute temperature, as follows:
DH(m2/s)=1.4×10−7exp(−13,400J/mol/RT)
DN(m2/s)=3.0×10−7exp(−76,150J/mol/RT)
Is it possible to purify hydrogen gas in this manner? If so, specify a temperature at which the process may be carried out, and also the thickness of metal sheet that would be required. If this procedure is not possible, then state the reason(s) why.

Answer:

Option B: An MSDS

Explanation:

A dictionary is used to check up the meaning of general words and not for checking if a substance being used is hazardous. Option A is wrong.

MSDS means "Material Safety Data Sheet" and it contains documents with information that relates to occupational health & safety for checking various substances and products. Thus, option B is correct.

SAE stands for Society of Automotive Engineering and their standards pertain to mainly Automobiles. Thus option C is wrong.

EPA guidelines are mainly for checking facility and environmental health and safety compliance. Thus, option D is wrong.

Answer:

Explanation:

7. False

8. True

9. True

10.True

11. True

Answer:

72 MPa

Explanation:

Critical resolved shear stress = 36 MPa

calculate the maximum possible yield strength for a single crystal of the metal

first we have to express the critical resolved shear stress as

бy = 2т[tex]_{critical}[/tex]

given that the minimum stress needed to introduce yielding curve will occur  at Ψ = λ = 45

where : бy = yielding stress

             t[tex]_{critical}[/tex] = 36MPa

therefore the maximum possible yield strength

= (2) * (36)

= 72 MPa

Answer:

It's equalization of the system

Explanation: If there was a leak it would likely leak all the time even if the car was turned off. Plus, a system leak bad enough to hear would drain the system of refrigerant very quickly and would no longer cool.

Answer:

Speed V1 at the beginning of the skid  = 57.075 mph

Explanation:

Calculate the V1 speed at the beginning of the skid

V1 ( speed before brake was applied ) = ?

distance travelled by Vehicle 1 = 195 ft

curve tangent for vehicle 1 =  - 3% = - 0.03

coefficient of friction = 0.48

speed at Impact = 25 mph

To determine the V1 speed at the beginning of the skid we have to apply AASTHO

d1 = [tex]( \frac{VA^{2} - VA^{12} }{30(F+N)} )[/tex]

195 =  [tex](\frac{VA^2- 25^2}{30(0.48-0.03)} )[/tex]

Hence : VA = 57.075 mph

Answer:

116,090 Btus

Explanation:

Answer:

The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.

The rotor inertia may differ from these assumption due to differences in the shape of cross section.

Explanation:

We assume that rotor can be represented as a solid cylinder of radius [tex]r[/tex], length [tex]l[/tex], made of cooper ([tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex]) and whose axis of rotation passes through its center of mass and is parallel to its cross section. By definition of Moment of Inertia and Theorem of Parallel Axes, the moment of inertia of the rotot is:

[tex]I = \frac{1}{4}\cdot \rho \cdot \left(\frac{\pi}{4} \right) \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex]

[tex]I = \frac{\pi}{16}\cdot \rho \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex] (Eq. 1)

Where:

[tex]\rho[/tex] - Density of copper, measured in kilograms per cubic meter.

[tex]R[/tex] - Radius of the rotor, measured in meters.

[tex]L[/tex] - Length of the rotor, measured in meters.

[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.

If we know that [tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex], [tex]L = 25\times 10^{-3}\,m[/tex] and [tex]R = 8.98\times 10^{-3}\,m[/tex], the estimated moment of inertia of the rotor is:

[tex]I = \frac{\pi}{16}\cdot \left(8960\,\frac{kg}{m^{3}} \right)\cdot (8.98\times 10^{-3}\,m)^{3}\cdot [3\cdot (8.98\times 10^{-3}\,m)^{2}+(25\times 10^{-3}\,m)^{2}][/tex]

[tex]I \approx 1.105\times 10^{-6}\,kg\cdot m^{2}[/tex]

The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.

From D'Alembert's Formula we know that net force of rigid bodies experimenting rotation equals the product of moment of inertia and angular acceleration. In this case, the purpose is minimizing moment of inertia and it is done by modifying the shape of the cross section so that rotor could be aerodynamically more efficient.

Answer:

carried courage buried

Explanation:

its Filipino

Answer:

The estimated batching production is 210 cy/hr

Explanation:

Given;

capacity of the batching chamber of the plant, Q = 12.5 cy

average batching cycle time, t = 3 min

plant efficiency, n = 84 %

The estimated batching production is calculated as efficiency of the machine multiplied by the production rate in hour. This is given by the solution below;

Batching production = (nQ )/ t

[tex]Batching \ production = 0.84(\frac{12.5 \ cy}{3 \ min} *\frac{60 \ min}{1 \ hr})\\\\ Batching \ production = 210 \ cy/hr[/tex]

Therefore, the estimated batching production is 210 cy/hr

Answer:

Your mobile phone has more computing power than the computers used for the Apollo 11 moon landing.

Mobile phones have to “work harder” to get a signal if you are in a moving vehicle.

The first mobile phone was made in 1973.

The first mobile phones that went on sale in 1983 cost nearly $4,000 each.

In 2012 Apple sold 340,000 phones per day.

4 out of 10 Brits admit to snooping on their partners phone.

Out of the 53% of snoopers that found incriminating evidence on their partner’s phone, 5% went on to terminate their relationship.

Waterproof mobile phones came to market because Japanese youngsters like to use them in the shower.

Apparently mobile phones have 18 times more bacteria on them than toilet handles!

Phubbing describes the act of snubbing someone by using your mobile phone in their company.

In 2015 more people died from taking selfies than shark attacks.

Teenagers that use a phone more than 2 hours a day increase their risk of depression and anxiety.

Nomobophobia is severe anxiety caused by the thought or act of losing your phone or running out of battery.

Explanation:

Answer:

  no

Explanation:

No. More power is dissipated when the transistor is in its active region. In general, transistors in switching circuits are biased either "on" or "off". Time spent in the active region is minimized.

_____

On the other hand, speed can be enhanced if the transistors are active. So, it's a speed/power trade-off. Usually power is of more interest, particularly when there are millions of switching circuits. However, in certain applications, speed may be the priority, so the transistor will be biased in its active region.

Answer:

Resolution of Fourier transform = 2Hz

Explanation:

Given:

Sample signals = 16,384 samples/second

Number of data points = 8,192 oints

Find:

Resolution of Fourier transform

Computation:

Resolution of Fourier transform = fs / N

Resolution of Fourier transform = 16,384 / 8,192

Resolution of Fourier transform = 2Hz

Answer:

a Network Engineer (architect)

Explanation:

Indeed, as a Network Engineer, we would expect Nicole to be in charge of planning, setting up, and managing the software and hardware components of the computer networks so that they function as intended.

A Network Engineer therefore would be responsible for setting up the networking capabilities and policies that will govern the workstations when connected to the company network.

Answer:

a) According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.

b) According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.

Explanation:

First, we need to determine the torque experimented by the shaft ([tex]T[/tex]), measured in kilonewton-meters, whose formula is described:

[tex]T = \frac{\dot W}{\omega}[/tex] (Eq. 1)

Where:

[tex]\dot W[/tex] - Power, measured in kilowatts.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

If we know that [tex]\dot W = 10\,kW[/tex] and [tex]\omega = 20.944\,\frac{rad}{s}[/tex], then the torque experimented by the shaft:

[tex]T = \frac{10\,kW}{20.944\,\frac{rad}{s} }[/tex]

[tex]T =0.478\,kN\cdot m[/tex]

Let consider that shaft has a circular form, such that shear stress is determined by the following formula:

[tex]\tau = \frac{16\cdot T}{\pi\cdot D^{3}}[/tex] (Eq. 2)

Where:

[tex]D[/tex] - Diameter of the shaft, measured in meters.

[tex]\tau[/tex] - Torsional shear stress, measured in kilopascals.

If we know that [tex]D = 0.03\,m[/tex] and [tex]T =0.478\,kN\cdot m[/tex], the torsional shear stress is:

[tex]\tau = \frac{16\cdot (0.478\,kN\cdot m)}{\pi\cdot (0.03\,m)^{3}}[/tex]

[tex]\tau \approx 90164.223\,kPa[/tex]

a) According to the maximum-shear-stress failure theory, we get that maximum shear stress limit is:

[tex]S_{ys} = 0.5\cdot S_{ut}[/tex] (Eq. 3)

Where:

[tex]S_{ys}[/tex] - Ultimate shear stress, measured in kilopascals.

[tex]S_{ut}[/tex] - Ultimate tensile stress, measured in kilopascals.

If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:

[tex]S_{ys} = 0.5\cdot (440\times 10^{3}\,kPa)[/tex]

[tex]S_{ys} = 220\times 10^{3}\,kPa[/tex]

Lastly, the static factor of safety of the shaft ([tex]n[/tex]), dimensionless, is:

[tex]n = \frac{S_{ys}}{\tau}[/tex] (Eq. 4)

If we know that [tex]S_{ys} = 220\times 10^{3}\,kPa[/tex] and [tex]\tau \approx 90164.223\,kPa[/tex], the static factor of safety of the shaft is:

[tex]n = \frac{220\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]

[tex]n = 2.440[/tex]

According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.

b) According to the distortion-energy failure theory, we get that maximum shear stress limit is:

[tex]S_{ys} = 0.577\cdot S_{ut}[/tex] (Eq. 5)

If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:

[tex]S_{ys} = 0.577\cdot (440\times 10^{3}\,kPa)[/tex]

[tex]S_{ys} = 253.88\times 10^{3}\,kPa[/tex]

Lastly, the static factor of safety of the shaft is:

[tex]n = \frac{253.88\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]

[tex]n = 2.816[/tex]

According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.