A 1-in (internal diameter) pipe is used to supply water to a low-pressure tank with an MAWP of 20 psig. The water is supplied through a regulator from a source with a maximum pressure of 100 psig. Determine the diameter of a conventional spring-operated relief required to protect the vessel from a regulator failure. Be sure to state clearly any assumptions and your justifications for them.

Answer:

mmmmmm

Explanation:

ffmfmfmmfmfmmfmfmfmfmfmffmfmfmfmfmfmfmfmfmfmfmfmfffmmfmffmmmfmfmfmfmfmfmfmfffmfmfmfmfmfmmfmfmmfmfmfmfmfmfffmfmfffmfmfmfmffmmfmfmffmfmfmfmfffmfmfmfmfmmmmfmfmfmfmfmfmfmmmfmfmfmfmfmfmfmfmmfmfmfmfmfmfmfmfmfmfmffmfmfmfmfmfmmfmfmmmmmfmfmfmffmfmfmfmffmffmfm

Solution :

Given :

Height at which the metal is poured, h = 10 in

Diameter of the runner , d = 0.4 in

Assume viscosity, μ = 0.004 Pa-s

Now considering Bernoulli's equation to find velocity,

As there is no loss in energy, Δ Pressure energy ≅ 0

So,

[tex]$\frac{v^2_1}{2g}+z_1=\frac{v^2_2}{2g}+z_2$[/tex]

Here 1 and 2 represents top and bottom section of the sprue.

[tex]$\frac{v^2_1}{2g}+z_1=\frac{v^2_2}{2g}+z_2 \ \ \ (v_1=0)$[/tex]

[tex]$v_2=\sqrt{2g \Delta z}$[/tex]

Now substituting [tex]$32.20 \ ft/s^2 = 386.4 \ in/s^2$[/tex] for g and 10 in for Δz in velocity equation,

[tex]$v_2=\sqrt{2 \times 386.4 \times 10}$[/tex]

[tex]$v_2= 87.91 \ in\s = 7.32 \ ft/s$[/tex]

Calculating the area of the basin

[tex]$A=\frac{\pi}{4}d^2$[/tex]

Substitute .04 in  for d in the above equation

[tex]$A =\frac{\pi}{4} \times (0.4)^2$[/tex]

[tex]$A= 0.1256 \ in^2$[/tex]

Calculating the flow rate

Q = 0.1256 x 87.91

[tex]$ Q= 11 .04 \ in^3/s $[/tex]

Hence the viscosity is [tex]$v_2 = 87.91 \ in/s$[/tex] and the flow rate is [tex]$Q=11.04 \ in^3/s$[/tex]

Calculating the Reynolds number of the flow,

[tex]$Re = \frac{\rho v d}{\mu}$[/tex]

[tex]$Re = \frac{0.097544 \times 87.91 \times 0.40}{5.8 \times 10^{-7}}$[/tex]

[tex]$Re=5.9 \times 10^6$[/tex]

Therefore, the flow is turbulent.

Now considering the solidification time,

[tex]$t=c \times \left(\frac{V}{A}\right)^2$[/tex]

[tex]$t=c \times \left(\frac{\frac{\pi}{4}d^2h}{2\left(\frac{\pi}{4}d^2\right)+ \pi dh}\right)^2$[/tex]

[tex]$t=c\left(\frac{dh}{2d+4h}\right)^2$[/tex]

Substituting 1.5 for d and 3 for h and 3 min for t to calculate the value of c is

[tex]$3=c\left(\frac{1.5 \times 3}{2 \times 1.5 + 4 \times 3} \right)^2$[/tex]

c = 33.33

For case when height is double i.e. h = 6 in

[tex]$t_h = 33.33 \times \left(\frac{1.5 \times 6}{2 \times 1.5 + 4 \times 6} \right)^2$[/tex]

[tex]$t_h= 3.70 \ min$[/tex]

For case when the diameter is doubled i.e. 3 in for d and 3 in  for h,

[tex]$t_d = 33.33 \times \left(\frac{3 \times 3}{2 \times 3 + 4 \times 3} \right)^2$[/tex]

[tex]$t_d= 8.3325 \ min$[/tex]

Answer:

[tex]1.67\times 10^{9}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

[tex]3.33\times 10^{19}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

[tex]3.33\times 10^{19}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

Explanation:

[tex]p_0[/tex] = Hole concentration = [tex]2\times 10^{16}\ \text{cm}^{-3}[/tex]

[tex]n_i[/tex] = Intrinsic concentration = [tex]10^{10}\ \text{cm}^{-3}[/tex]

[tex]\tau_{n0}[/tex] = Minority carrier life time = [tex]3\times 10^{-6}\ \text{s}[/tex]

[tex]\delta n[/tex] = Excess concentration of electrons = [tex]10^{13}\ \text{cm}^{-3}[/tex]

Majority carrier electron concentration is given by

[tex]n_0=\dfrac{n_i^2}{p_0}\\\Rightarrow n_0=\dfrac{(10^{10})^2}{2\times 10^{16}}\\\Rightarrow n_0=5000\ \text{cm}^{-3}[/tex]

Recombination rate is given by

[tex]R_{n0}=\dfrac{n_0}{\tau_{n0}}\\\Rightarrow R_{n0}=\dfrac{5000}{3\times 10^{-6}}\\\Rightarrow R_{n0}=1.67\times 10^{9}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

The recombination rate is [tex]1.67\times 10^{9}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

Recombination rate is given by

[tex]R_n=\dfrac{\delta_n}{\tau_{n0}}\\\Rightarrow R_n=\dfrac{10^{13}}{3\times 10^{-7}}\\\Rightarrow R_n=3.33\times 10^{19}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

The recombination rate is [tex]3.33\times 10^{19}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

Change in the recombination rate is

[tex]\Delta R_n=3.33\times 10^{19}-1.67\times 10^{9}\\\Rightarrow \Delta R_n=3.33\times 10^{19}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

The change in the recombination rate is [tex]3.33\times 10^{19}\ \text{cm}^{-3}\text{s}^{-1}[/tex]

Answer:

Following are the solution to this question:

Explanation:

In point a:

This takes me six weeks for both the took ideas that I was searching for but it continued for 3 years (12 weeks) as it's an intern.  

In point b:

Finding job:

[tex]\to f = \frac{1}{6} = 0.166[/tex] jobs weekly  

Separation of jobs:

[tex]\to \frac{ 1}{12}=0.083[/tex] employment per week.  

In point c:

Its natural rate of unemployment is:  [tex]\frac{U}{L} = s+(s \times f)[/tex].  

The normal level of employment for that community I represent, once we add up from that preceding section, is as follows:

[tex]\to \frac{U}{L} = 0.083+ (0.083\times 0.166) = 0.096[/tex]

If on average, it requires six weeks to find another job or the work lasted 12 weeks, the group's unemployment level is [tex]0.096 \ \%[/tex].

Answer:

B

Explanation:

Answer:

add them to a project you are doing. You know the purpose of robotics is to be creative

Explanation:

Answer:

True

Explanation:

Answer:A

Explanation:

B - it never talked about a steep path or not.

C- it never stated the speed.

d- more experienced driver makes a little bit of sence since she is new but she is new at the farm not driving.  

thats my guess.

Answer:

 b is the answer

Explanation:

Answer:

hello some part of your question is missing below is the complete question

answer :

A) 162750 Ib.ft

B) - 64950 Ib.ft

Explanation:

Applying Muller-Breslau's law

we will make assumptions which include assuming an imaginary hinge at G

therefore the height of I.LD for B.M at G = ( 12 * 8 ) / 20 = 4.8

height of I.L.D at C = 2.4 ( calculated )

height of I.L.D at F = 1.5 ( calculated )

A) Determine Maximum positive moment produced at G

[tex]M ^+[/tex] = [ (1/2 * 20 * 4.8 ) ( 600 + 300 ) ] + [ ( 25 * 4.8 * 10^3 ) ] - [ ( 1/2 *2.4*20 ) * 300 ] + [ (1/2 * 1.5 * 10 ) ( 600 + 300 ) ]

      = 162750 Ib.ft

B)   Determine the maximum negative moment produced at G

[tex]M ^-[/tex] = [ ( 1/2 * 20 * 4.8 ) * 300 ] - [ ( 1/2 * 2.4 * 20 ) ( 600 + 300 ) ] - [ (2.5 * 10^3 * 2.4 ) ] + [ ( 1/2 * 1.5 * 10) * 300 ]

     = - 64950 Ib.ft

Answer:

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Explanation:

From Fluid Mechanics, we remember that absolute pressure ([tex]p_{abs}[/tex]), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:

[tex]p_{abs} = p_{atm}+p_{g}[/tex] (1)

Where:

[tex]p_{atm}[/tex] - Atmospheric pressure, measured in pounds per square inch.

[tex]p_{g}[/tex] - Working pressured of the boiler (gauge pressure), measured in pounds per square inch.

If we suppose that [tex]p_{atm} = 14.696\,psi[/tex] and [tex]p_{g} = 10\,psi[/tex], then the absolute pressure is:

[tex]p_{abs} = 14.696\,psi+10\,psi[/tex]

[tex]p_{abs} = 24.696\,psi[/tex]

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Answer:

a) 13 mins

b) 18 mins

Explanation:

solidification time = 10 minutes

height-to-diameter ratio = 1

applying the expression for calculating solidification time

t = B  [tex](\frac{V}{A} ) ^n[/tex]  --------- ( 1 )

n = 1.5

B = ?

t = 10 minutes

V = [tex]\frac{\pi d^2}{4} H[/tex]

A = [tex]\frac{\pi d^2}{2} + \pi dH[/tex]

back to equation 1

making B subject of the formula ; B = [tex]\frac{147}{d^{1.5} }[/tex]

a) Determine the solidification time if the cylinder height is doubled

t = B  [tex](\frac{V}{A} ) ^n[/tex]

substitute :  B = [tex]\frac{147}{d^{1.5} }[/tex]  , H = 2D  , V = [tex]\frac{\pi d^2}{4} H[/tex] , A = [tex]\frac{\pi d^2}{2} + \pi dH[/tex] , n = 1.5

therefore ; t ( solidification time when height is doubled ) ≈ 13 mins

b) Determine the time if the diameter is doubled

t = B  [tex](\frac{V}{A} ) ^n[/tex]

substitute ; B = [tex]\frac{147}{h^{1.5} }[/tex] , D = 2H , n = 1.5 , V = [tex]\frac{\pi d^2}{4} H[/tex], A = [tex]\frac{\pi d^2}{2} + \pi dH[/tex]

therefore ; t ( solidification time when diameter is doubled ) ≈ 18 mins

Answer:

Engineers spend up to 25 percent of their design process formulating and analyzing

the team that will work on the project.

Explanation:

During the design engineering stage of a project, the team is a very important decision item because determining the required team also determines the cost and the technical skill requirements of the project.  The cost depends, to a large extent, on the technical skills of those that will work on the project.  When the technical skill requirements are not met, the project will be massively jeopardized.

Answer:

the team that will work on the project.

Explanation:

Answer:

zero off your workpiece so you can work from a datum, usually the centre of your work on a lathe. change your tool to a drill and drill a hole to a size smaller than your thread diameter, change out your tool for a threaded tap and away you go.

I'm not sure which part they want but I'd say ensure your tool is set to the right height, you have the tool lines up where you want to cut and that you have calculated the speed you need to cut at safety. Drill a hole before you tap though.

If you have a CNC lathe you just set the programme to do the processes and tool change for you.

The first step to cutting internal threads on an engine lathe is to make calculations so that the thread will have proper dimensions.

The technique of thread cutting on the lathe results in a helical ridge with a consistent section on the workpiece.

To work from a datum, often the center of your work on a lathe, zero off your workpiece. Use a drill to create a hole that is less in diameter than the thread, then switch to a threaded tap and carry on.

One would advise making sure your tool is adjusted to the appropriate height, that it is lined up where you want to cut, and that you have determined the speed you must cut in order to be safe. Before you tap, drill a hole.

With a CNC lathe, you can simply program the machine to perform the processes and tool changes for you.

Therefore,  to do this, make a series of cuts with a threading toolkit that matches the needed thread form.

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Answer:

Explanation:

RC low Pass Filter is an electronic circuit that comprises of a resistor and capacitor and it functions to permit low-frequency signals depending on the design and reject the high-frequency signals above a given frequency known as the cutoff frequency.

From the diagram attached below:

[tex]V_{in[/tex] = the input signal  

[tex]V_o[/tex] = the output signal

Since; [tex]V_o[/tex]  is used across the capacitor C,

By using the potential divider equation we have:

[tex]V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }[/tex]

From above; [tex]X_C[/tex] = capacitive reactance ;

and The total impedance Z is illustrated as [tex]Z = \sqrt{R^2+X_C^2}[/tex]

Thus;

[tex]V_o =V_{in} \times \dfrac{X_C}{Z}[/tex]

Recall that;

[tex]X_C = \dfrac{1}{2 \pi fC}[/tex]

Here; f denotes the frequency of the input signal

Since the cutoff frequency is related to the frequency at which the capacitive reactance and resistance are said to be the same, then:

The Cutoff frequency can be expressed as:

[tex]F_C = \dfrac{1}{2 \pi RC}[/tex]

Also;

the frequency of input signal [tex]f = \dfrac{F_c}{4}[/tex]

[tex]f = \dfrac{1}{8 \pi RC}[/tex]

Hence;

[tex]X_C = \dfrac{1}{2 \pi fC}[/tex]

[tex]X_C = \dfrac{1}{2 \pi \times \dfrac{1}{8 \pi RC} \times C}[/tex]

[tex]X_C = 4R[/tex]

Finally;

From [tex]Z = \sqrt{R^2+X_C^2}[/tex]

[tex]Z = \sqrt{R^2+(4R)^2}[/tex]

[tex]Z = \sqrt{17R^2}[/tex]

Z [tex]\simeq[/tex] 4.12R

As such, the output will be:

[tex]V_o =V_{in} \times \dfrac{X_C}{\sqrt{ R^2+X_C^2} }[/tex]

[tex]V_o =V_{in} \times \dfrac{4R}{4.12R^2} }[/tex]

[tex]V_o =0.97V_{in}[/tex]

So, if we regard [tex]A_{in}[/tex] to be the input amplitude, then [tex]A_{out}[/tex] i.e the output amplitude will also be [tex]A_{out}[/tex] = [tex]0.97 A_{in}[/tex]

Answer:

73.24 K byte

Explanation:

Assuming that

N = total number of samples

N = 10 * 5kHz

N = 50*10^3

Also, the total number of bits, T

T = 12 * N

T = 12 * 50*10^3

T = 600 * 10^3

And then, finally, the total number of byte,

B = 600*10^(3/8)

B = 75*10^3 byte

75*10^3 byte = 75*10^3/1024 kilo byte

And on converting to decimal, we will have

= 73.24 K byte

Therefore, the memory required = 73.24 K byte

Answer:

c

Explanation:

Fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the head loss will double.

The amount of piping head loss is computed by equating the lengths and distances of all fittings to a particular inner pipe diameter. The length of a flow through a conduit before it completely develops is referred to as the entry length in fluid dynamics.

Entrance length is the length of the entry zone, which is the area immediately following the pipe entrance where effects from the inner wall of the pipe disseminate into the flow as an expanding boundary layer.

As the boundary layer completely fills the pipe and the flow characteristics stop changing as the pipe's length rises, the emerging flow becomes fully formed.

Therefore, Fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the head loss will double.

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Answer: False

Explanation:

Answer: 5   6    7

Explanation:

you can run this in python and get this result

The output for the given program is: 5   6    7

This refers to the high-level language that was created and is used for data structures due to its OOP (object-oriented programming).

Hence, this python code asks for an array of numbers in the range of 3 and when that is found, it should make a display of the number and increment it.

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Answer:

9.2 hp

Explanation:

Service factor is the ratio of the gearbox rated horsepower (or torque) to the application's required horsepower (or torque)

SF = Pₒ / Pᵢ

Pᵢ = Pₒ × SF

Pᵢ = 8 hp × 1.15

Pᵢ = 9.2 hp

Answer:

we have zero degree of freedom (F = 0) at the first phase region and one degree of freedom (F = 1) at the second phase region

Explanation:

given the diagram;

C = 2 { Ti and Ni}, N = 1 {Pr is constant}, P = 3 { L and B}

For degree of freedom,

Gibbs phase rule is applied for invariant point (ex.A)

P + F = C + N

3 + F = 2 + 1

F = 3 - 3

F = 0

For the second phase region ( ex.B)

C = 2, N = 1, P = 2

P + F = C + N

2 + F = 2 + 1

F = 3 - 2

F = 1

Therefore we have zero degrees of freedom (F = 0) at the first phase region and one degrees of freedom (F = 1) at the second phase region

Answer:

................................................

Explanation:

plz mark B R A I N L I E S T

lol

Answer:

It is a branch of engineering that deals with the formulation, estimation and evaluation of economic outcomes when there are alternatives to accomplish a defined purpose.

Explanation:

Engineers are able to evaluate the cost and benefits of projects that involve engineering designs and analysis through systematic approach. This process quantifies the costs and benefits of the projects to see enough money is saved to allow capital investments.

Answer:

Following are the solution to this question:

Explanation:

In question 1:

It Cmdlet gets the representatives of items, its features including their methods. Now use a variable (InputObject) and pipe the object to Even getting Member for both the purpose of specifying the object. It static variable utilizes their class members, not the case, to gather information regarding static members. Just use the (MemberType) parameter and receive only some forms of participants, such example NoteProperties.

In question 2:

Its Purchase cmdlet allows us to import components through our process. It can import components via remote sessions through the local client beginning with Window PowerShell 3.0.

In question 3:

The Syntax query to have all cmdlets lists it in the cmdlet title alphabetical order, but instead shows them in noun clauses.

Command:

[tex]\text{Type Cmdlet } | \text{Sort-Object -Property Noun} | \text{Format-Table -GroupBy Noun}[/tex]

In question 4:

Syntax:

[tex]\text{Get node} | \text{format list} | \text{Selection process, device forms, route group title all}[/tex]

[tex]| \text{Layout Table} | \text{Selection process}[/tex]

[tex]\$ s[/tex] = New-PSSession -ComputerName Server

Get-Module -PSSession [tex]\$ s[/tex] -ListAvailable

Answer: The movement of tectonic plates

Explanation:

Tectonic plates are the part of the earth's crust that both the ocean and land rest on. These plates are constantly moving as a result of currents in the mantle.

These movements cause stress on the surface which has the effect of fracturing rocks and thereby creating/ forming faults in the earth's crust. Sometimes faults form when these plates move away from each other and sometimes they are formed when they push into each other.

Answer:

a) 46.76 Lsun

b) 2187 Lsun

c ) You can easily estimate the luminosity of a 3-solar massed giant star because its luminosity is close to the luminosity of the sun

Explanation:

To estimate the luminosity of a celestial body in terms of solar luminosity  we use the expression below

= [tex]\frac{L}{Lsun} = (\frac{M}{Msun} )^{3.5}[/tex]

a) Estimate the luminosity of a 3 solar mass main sequence star

= [tex]\frac{L}{Lsun} = (\frac{3M}{Msun} )^{3.5}[/tex]

Hence  ; L(luminosity of a 3 solar mass star ) = 46.76  Lsun

b) Estimate the luminosity of a 9-solar-mass main sequence star

= [tex]\frac{L}{Lsun} = (\frac{9M}{Msun} )^{3.5}[/tex]

hence L( luminosity of a 9-solar mass star ) =  2187 Lsun

c ) You can easily estimate the luminosity of a 3-solar massed giant star because its luminosity is close to the luminosity of the sun

A) The luminosity of a 3 -solar-mass main-sequence star is; L = 46.765L_sun

B) The luminosity of a 9 -solar-mass main-sequence star is; L = 2187L_sun

C) Yes, we can easily estimate the luminosity of a 3 -solar-mass main-sequence star.

We want to find luminosity of a solar body which in this case is a main sequence star. The formula in terms of solar luminosity is;

L/L_sun = (M/M_sun)^(3.5)

Where;

L is luminosity of the solar body

L_sun is luminosity of the sun

M is mass of the solar body

M_sun is mass of the sun

A) Thus;

For a 3 -solar-mass main-sequence star;

M = 3M_sun

Therefore;

L/L_sun = (3M_sun/M_sun)^(3.5)

L/L_sun = 3^(3.5)

L/L_sun = 46.765

L = 46.765L_sun

B) For a 9-solar-mass main-sequence star;

L/L_sun = (9M_sun/M_sun)^(3.5)

L/L_sun = 9^(3.5)

L/L_sun = 2187

L = 2187L_sun

C) We can see from answer A above that the luminosity of a 3 solar mass body is 46.765 times of the luminosity of sun. Since it's not so much bigger than the luminosity of sun, then it is easy to estimate.

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The reference source that can be consulted is C. The Professional Engineers Act and Board Rules

The reference source that may be consulted to answer questions regarding the Professional Engineers Act is the Professional Engineers Act and Board Rules.

The Professional Engineers Act and Board Rules is an Act that was established in order to regulate the qualifications for professional engineers, register them and also make sure that their conducts and behavior are looked into.

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Answer:

Fracture.

Explanation:

Fracture describes how a mineral looks when it breaks apart in an irregular way.

Answer:

Fracture.

Explanation:Fracture describes how a mineral looks when it breaks apart in an irregular way.

Answer:

a) 294.34 days

b) Δh = 25.361 cm

Explanation:

Given data:

Thickness of clay layer = 18 ft

initial pressure = 0.75 ton/ft^2

final pressure = 1.5 ton/ft^2

Δp = 0.75 ton/ft^2

eo = 1.12

e1 = 0.98

k = 4.3 * 10^-7 cm/sec

A ) determine how long it will take to reach 70% consolidation

attached below is the detailed solution

T( time in days ) = 294.34 days

B) determine settlement at 70%

attached below is the detailed solution

Δh = 25.361 cm

Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.

Explanation: Dalton's Law of Partial Pressure states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:

[tex]P_{total} = P_{1}+P_{2}+...[/tex]

The proportion of each individual gas in the total pressure is expressed in terms of mole fraction:

[tex]X_{i}[/tex] = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: [tex]n=\frac{2000}{14}[/tex] = 142.85mols

molar mass = 44g/mol

mass in the tank = 4000g

number of moles in the tank: [tex]n=\frac{4000}{44}[/tex] = 90.91mols

Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

[tex]P_{i}=P_{total}.X_{i}[/tex]

For Nitrogen gas:

[tex]P_{N_{2}}=1.\frac{142.85}{233.76}[/tex]

[tex]P_{N_{2}}[/tex] = 0.6

For Carbon Dioxide:

[tex]P_{total}=P_{N_{2}}+P_{CO_{2}}[/tex]

[tex]P_{CO_{2}} = P_{total}-P_{N_{2}}[/tex]

[tex]P_{CO_{2}}=1-0.6[/tex]

[tex]P_{CO_{2}}=[/tex] 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.