why is excessive moisture in steam undesirable in steam turbines what is the moisture content allowed

A saturated soil with a mass of 43.2g. When dried in the oven its mass was 30g, the limit of contraction is 0.4.

For the limit of contraction, we know that:

Limit of contraction (Lc) = (Vw - Vd) / Vw

Lc = (20 - 12) / 20

Lc = 8 / 20

Lc = 0.4

Therefore, the limit of contraction is 0.4.

Now for the dry weight,

Dry unit weight (γd) = γw / (1 + e)

e = (Gs - 1) / Gs

n = e / (1 + e)

w = (wet weight - dry weight) / dry weight

In this case,

γw = 9.81 kN/\(m^3\)

Gs = 2.74.

So,

γd = 2010 / (1 + ((2.74 - 1) / 2.74))

γd = 2010 / (1 + 0.727)

γd = 2010 / 1.727

γd = 1163.54 kg/m^3

e = (2.74 - 1) / 2.74

e = 1.74 / 2.74

e ≈ 0.635

n = 0.635 / (1 + 0.635)

n = 0.635 / 1.635

n ≈ 0.388

w = (43.2 - 30) / 30

w = 13.2 / 30

w ≈ 0.44

Therefore, the dry unit weight (γd) is approximately 1163.54 kg/m^3, the void ratio (e) is approximately 0.635, the porosity (n) is approximately 0.388, and the moisture content (w) is approximately 0.44.

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Answer:

See explanation.

Explanation:

Since no figure was given I solved a problem that was similar to the one you described that I worked in my mechanics of materials class. The method should be very similar for your figure. See attached image for my work.

If it is subjected to the couple determine the maximum shear stress in regions AC and CB of the shaft. G st = 75 GPa. Than the answer will be 52Mpa.

We need to perform a two step process to obtain the maximum shear stress on the shaft. For the solid shaft,

P=2×pi×N×T/60 or T=60×p/2×pi×N

Where P=power transmitted by the shaft=50×10³W

N=rotation speed of the shaft in rpm=730rpm

Pi=3.142

T is the twisting moment

By substituting the values for pi, N and P, we get

T=654Nm or 654×10³Nmm

Also, T=pi×rho×d³/16 or rho=16×T/pi×d³

Where rho=maximum shear stress

T = twisting moment=654×10³Nmm

d= diameter of shaft= 40mm

By substituting T, pi and d

Rho=52Mpa

b. For a hollow shaft, the value for rho is unknown

T=pi×rho(do⁴-di⁴/do)/16

Rho=T×16×do/pi×(do⁴-di⁴)

Where

T= twisting moment=654×10³Nmm gotten above

do=outside shaft diawter=40mm

di= inside shaft diameter =30mm

Pi=3.142

Substituting values for pi, do, di and T.

Rho=76Mpa

Therefore, If it is subjected to the couple determine the maximum shear stress in regions AC and CB of the shaft. G st = 75 GPa. Than the answer will be 52Mpa.

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Tonya, as a network engineer, recognizes that performing a risk assessment is crucial for developing a new security policy for her company's IT infrastructure. In this process, she needs to consider and understand the assets, likelihoods, threats, and vulnerabilities associated with the organization's systems and data.

A risk assessment involves identifying and evaluating the assets, such as hardware, software, and sensitive information, that need protection. Tonya will assess the likelihood of various threats or potential incidents occurring, such as unauthorized access, data breaches, or system failures. Additionally, she will analyze the existing vulnerabilities within the IT infrastructure that could be exploited by threats.

By thoroughly understanding these components of a risk assessment, Tonya will be able to assess the level of risk associated with each asset, prioritize security measures, and develop an effective security policy that addresses the identified risks.

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Answer:

hello your question has some missing information attached below is the missing information

answer :

a) 9.17431 kg/s

b) 2598.4374 kW

Explanation:

a) Calculate mass flow rate at inlet 1

Given data :

AV1 = 0.01 m^3/s

pressure ( P1 ) = 5 bar , temp ( T1 ) = 150°C

therefore : m1 = AV1 / v1 ------ ( 1 )

where : v1 = 0.00109 m^3/kg  ( value gotten from compressed table )

back to equation 1

m1 =  ( 0.01 m^3/s  / 0.00109 m^3/kg )  = 9.17431 kg/s

b) Calculate the rate of heat transfer

first determine mass balance

m3 = m1 + m2  

     = 9.17431 + 19 = 28.17431 kg/s

apply the energy balance equation to determine the rate of heat transfer

m1*h1 + m2*h2 = m3*h3 + Q

∴ Q = m1*h1 + m2*h2 - m3*h3 ------------ ( 1 )

where :

h1 = 632.266 kJ/kg  value gotten from compressed water table

h2 = 2812.45 kJ/kg  value gotten from superheated steam table

h3 = 2010.2965 kJ/kg value gotten from saturated pressure table

input values into equation 1 above

Q = 2598.4374 kW  . the energy moves from the chamber to the surroundings

Answer:

see and make me brainlist

Explanation:

What is oxy fuel cutting used for?

Oxy-fuel cutting is used for the cutting of mild steel. Only metals whose oxides have a lower melting point than the base metal itself can be cut with this process. Otherwise as soon as the metal oxidises it terminates the oxidation by forming a protective crust.

True.

According to article 320-15 of the National Electrical Code (NEC), exposed Type AC cable must closely follow the surface of the building. However, there is an exception to this rule for the use of Type AC cable on the underside of floor joists in basements. In this case, the cable can be used if it is supported at each joist and is not subject to damage. This means that if the cable is in an area where it could be damaged, such as a location where heavy objects may fall on it, it cannot be used in that location.

Type AC cable is a versatile and durable wiring solution often utilized in residential, commercial, and industrial settings. It consists of insulated conductors surrounded by a flexible metal sheath, which provides protection against external damage, as well as grounding capabilities. This design ensures the cable's longevity and reduces the risk of electrical hazards. In basements, where the risk of damage from moisture, debris, or other external factors may be higher, it is essential to support the cable at each joist. Proper support not only maintains the cable's structural integrity but also prevents sagging or strain, which can lead to potential damage and safety risks.

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Solving these equations simultaneously will give you the pressure forces P1 and P2 and their respective x and y components. Remember that we have neglected the shear force and weight in this problem.

To analyze the forces in a 90-degree bend in a circular pipe with a uniform diameter, we need to consider the forces acting on the fluid in both the x and y directions.
Let's denote the pressure forces acting on the fluid as \(P_{1}\)and  \(P_{2}\), the velocity of the fluid entering the pipe as V1, and the velocity of the fluid leaving the pipe as\(V_{2}\). The radius of the circular bend is R.
First, we'll write the equation for the x-direction (horizontal) forces. The net force in the x-direction can be written as:
\(F_{x}\) = P1 * A - P2 * A * cos(90)
Since cos(90) = 0:
\(F_{x}\) = P1 * A
Now, let's write the equation for the y-direction (vertical) forces. The net force in the y-direction can be written as:
\(F_{y}\) = P2 * A * sin(90)
Since sin(90) = 1:
\(F_{y}\) = P2 * A
Now, we can use the conservation of momentum to relate the forces to the change in momentum of the fluid. For the x-direction, we have:
\(F_{x}\) = m * (\(V_{2}\)- \(V_{1}\))
For the y-direction, we have:
\(F_{y}\) = m * (0 - (-\(V_{1}\)))

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The crystallinity of cellulose as well as its polar qualities makes it less difficult to dissolve. It has a three-dimensional hydrogen bond (H-bond) arrangement with the hydroxyl group connected to each polymer chain. On the other hand, Hydrocarbon polymers are nonpolar, and nonpolar solvents are used to dissolve them.

Cellulose is an organic molecule with the formula n that is a polysaccharide made up of a linear chain of hundreds to thousands of β  linked D-glucose units. Cellulose is a structural component of the major cell wall of green plants, many algae, and oomycetes.

Cellulose is the principal component present in plant cell walls and aids in the plant's stiffness and strength. Although cellulose cannot be digested by humans, it is an essential source of fiber in the diet. Cellulose is utilized in the production of clothing and paper.

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The inclusion of a slight slope on flat roofs, typically from front to rear, is essential for effective drainage, preventing water accumulation, and maintaining the integrity of the roof structure. It is a key design feature that ensures the long-term performance and durability of flat roofs.

The primary reason for incorporating this slope is to prevent water from pooling or accumulating on the roof's surface. If a flat roof were entirely level, water could collect in low-lying areas and cause structural issues, leaks, or damage to the roofing material over time. By introducing a gentle slope, the roof encourages water to flow towards drainage points, such as gutters or scuppers, where it can be efficiently directed away from the building.

In addition to enhancing drainage, a slight slope can also contribute to the overall longevity and durability of the roof. By minimizing the amount of time water remains in contact with the roof surface, the potential for water-related problems, such as leaks or deterioration, is reduced. Proper drainage helps prevent excessive moisture buildup and supports the longevity of the roofing system.

Overall, the inclusion of a slight slope on flat roofs, typically from front to rear, is essential for effective drainage, preventing water accumulation, and maintaining the integrity of the roof structure. It is a key design feature that ensures the long-term performance and durability of flat roofs.

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Answer:

option A. Inferring

Explanation:

inferring/ inference as reading strategy simply is the process by which one uses what he/she knows to make a guess about what you don't know or reading between the lines. Readers in making  inferences uses  clues found inside text along with their own  views or experiences to help them figure out what is not directly said,thereby causing a  personal and memorable text. for one to draw an inference from the passage via reading, Identify if its an Inference Question.inferring involves Trusting the Passage or what you are seeing,  then you start Hunting for Clues thereafter you Narrow Down the Choices. and then come to a conclusion or Practice.

Answer:

C) Skull and crossbones

Explanation:

Its on the OSHA website under HCS Pictograms and Hazards

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:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

The values of h at the two internal boundaries are :

Given data :

Z₁ = Z₂ = Z₃ = 25 m

h top = 120 m

h bottom = 100 m

K₁ = 0.0001 m/s

K₂ = 0.0005 m/s

K₃ = 0.0010 m/s

we will apply the formula below since flow is perpendicular to the bedding plane

Keq = \(\frac{Z1 + Z2 + Z3 }{\frac{Z1}{K1}+\frac{Z2}{K2} + \frac{Z3}{K3} }\)  ----- ( 1 )

Insert values given above into equation 1

Therefore ; Keq = 2.307 * 10⁻⁴ m/s

Hydraulic gradient ( Ieq ) = head loss / length

                                          = ( 120 - 100 ) / 3 * 25

                                   Ieq  = 0.266

Given that the flow is perpendicular to bedding plane

q1 = q2 = q3

V₁ = V₂ = V₃ = V

K₁i₁ = K₂i₂ = K₃i₃ = Keq * ieq

Hence :

V = Keq* Ieq

   = 2.307 * 10⁻⁴ * 0.266

   = 6.15 * 10⁻⁵ m/s .

Also;

K₁i₁ =  Keq * ieq = K₂i₂ = K₃i₃

therefore :

  i₁ = 0.615

  i₂ =  0.123

  i₃ = 0.0615

Pressure at point 1 ( i.e. pressure between first two formations )

h₁ = h top - i₁L₁

   = 120 - 0.615 * 25

   = 104.625 m

Pressure at point 2 ( i.e. pressure between the 2nd and 3rd formation )

h₂ = h₁ - i₂L₂

    = 104.625 - 0.123 * 25

    = 101.55 m

Therefore we can conclude that The values of h at the two internal boundaries are :  h₁ = 104.625 m , h₂ = 101.55 m

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The 16-input multiplexer will have one output line, and it will have four select lines.

A multiplexer is used to select one of several input signals and forward the selected input to a single output line.

The number of select lines for a multiplexer is determined by the number of input lines, where the select lines are equal to the base-2 logarithm of the number of input lines. For example, a 16-input multiplexer requires four select lines because 2^4 = 16.

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Answer:

Terminal 2 on a potential relay typically goes to the start winding of a motor.

Explanation:

The potential relay is an electromechanical device used in some types of motors to provide a temporary boost to the starting torque. Terminal 2 is connected to the start winding to assist in starting the motor.

Answer:

steep grading

Explanation:

the farthest

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\(_{critical}\)

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

where : бy = yielding stress

             t\(_{critical}\) = 36MPa

therefore the maximum possible yield strength

= (2) * (36)

= 72 MPa

The full description of the vessel includes details like as the vessel's speed and consumption rates, which are crucial for the Charterer to be aware of in order to precisely and efficiently arrange the voyages they must make during the Time Charter and plan the cargo operations.

This is a hotly debated subject and one of the Owners' and Charterers' most frequent strategic choices. The market's turbulence, the availability of cargo, the vessel's repositioning, and financial considerations are some of the major variables influencing this decision.

Both parties have expectations as a result of these, and these expectations are effectively represented in market swings and freight prices. Low vessel supply will cause freight rates to increase, and low cargo demand will cause freight costs to decrease. This is often referred to as the spot market.

Therefore, The full description of the vessel includes details like as the vessel's speed and consumption rates, which are crucial for the Charterer to be aware of in order to precisely and efficiently arrange the voyages they must make during the Time Charter and plan the cargo operations.

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Answer:

The answer is "shear".

Explanation:

Every transformation could be shown by some kind of conventional matrix.

Its standard matrices of shape k here could be any real value enabling shear transformation to parallel to the y axis.

\(\left[\begin{array}{cc}1&0\\k&1\\\end{array}\right]\)

This coefficient matrix A is the standard matrix during transformation. With the use of a the transformation could be entered into T(x)=Ax

And standard transfer function is standard.

\(\left[\begin{array}{cc}1&0\\6&1\\\end{array}\right]\)

The matrix in the form of the shear matrix.

The best alternative about the red-light shift is It is used to observe the expandable universe and one expression of the Doppler effect.

The red-light shift refers to a phenomenon observed in light waves where the wavelength of the light is stretched, causing a shift towards the red end of the spectrum. Option B correctly states that the red-light shift is used to observe the expanding universe and is an expression of the Doppler effect.

When light waves from distant galaxies or celestial objects travel through space, the expansion of the universe causes the wavelength of the light to stretch. This stretching leads to a shift towards longer wavelengths, which appear as a red shift when observed from Earth. This phenomenon is an essential tool in studying the expansion and evolution of the universe.

The Doppler effect is the change in frequency or wavelength of a wave as perceived by an observer when there is relative motion between the source of the wave and the observer. In the context of the red-light shift, the motion of distant galaxies away from us causes the observed wavelength of their light to increase, resulting in a red shift.

By observing the red-light shift, scientists can gather valuable information about the distance, speed, and behavior of celestial objects. It provides evidence for the expansion of the universe and helps in understanding its dynamics and history.

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Answer:

xcvjchjcnjkkhjmklk heyyyy biiiiiiiittcccchhh

Answer:

Jeanette does better on her exam when she takes it in the same room where she studies

Explanation:

Certain respondents react to that question in a self-administered survey, where respondents are permitted to go back and answer earlier questions after reading later ones. Thus, option D is correct.

The cognitive psychology concept known as the “context effect” examines how environmental factors affect how people react to stimuli in their surroundings.

Context effects happen when the information in the questions that come before or even after one particular question impacts.

According to the notion, how people are exposed to context based on environmental elements impacts their decisions.

The context effect theory holds true for both simpler psychological processes like making judgments about purchases and more complicated life activities like reading a book.

Therefore, Jeanette does better on her exam when she takes it in the same room where she studies.

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Answer:

they are made from petroleum

Answer:

Utilizing flush valves for toilets and urinals and/or low-flow versions of each, as well as installing low-flow faucets, or aerators can significantly reduce water use.

Explanation:

Answer:

the answer is simply 5000250 because the phenomicals add up to the mainstream with beacoups

When adding refrigerant in vapor form to a refrigeration system, the system motor/compressor assembly is running. This statement is false.

You MUST flash or throttle the refrigerant when injecting liquid refrigerant into the suction pipe while the compressor is running. Otherwise, liquid refrigerant could enter the compressor, thus this must be done (slugging). The oil in the compressor may be flushed out during slugging. Charge refrigerant vapor into the low side of the refrigeration system while the system is under high evaporator heat loading and continue doing so until the sight glass stops bubbling. When a system runs out of refrigerant, liquid refrigerant can be passed into the king valve on the liquid line or receiver. It passes via the condenser and evaporator. The liquid refrigerant is let to enter a system that is evacuated or vacuumed up until the flow has almost stopped.

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An average 350-ton R-123 chiller at 0 psig would likely have a small amount of refrigerant vapor remaining once all the R-123 liquid has been removed.

The exact quantity can vary based on various factors, but it is generally expected to be a minimal amount. When all the liquid R-123 has been removed from a chiller, there may still be some refrigerant vapor left behind. The quantity of vapor remaining depends on several factors, including the chiller's design, the efficiency of the liquid removal process, and the time given for the vapor to dissipate. However, since the chiller operates at 0 psig, the pressure is close to atmospheric pressure. At this low pressure, the amount of vapor left would likely be minimal. Chiller systems are designed to hold and circulate a specific amount of refrigerant, and any vapor remaining would typically be a small fraction of the total refrigerant charge.

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If an engine fails during rollout or just before takeoff, immediately shut both throttles and land the aircraft safely. Before reaching a safe single engine speed right away after takeoff, drop your nose to increase velocity.

Closing the throttle, turning off the fuel pump, setting the mixture control to idle cutoff, and simply cranking the engine is the most reliable hot start method I've found.

If an engine fails during rollout or just before takeoff, immediately shut both throttles and land the aircraft safely. Before reaching a safe single engine speed right away after takeoff, drop your nose to increase velocity. If you are unable to climb, close both throttles and land straight ahead.

The typical practice for the majority of aircraft would be to abandon takeoff if an engine failed during takeoff. In small aircraft, the pilot should turn the throttles down to idle, activate the speed brakes (if provided), and apply the brakes as needed if the engine fails before VR (Rotation Speed).

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