a turbine operating at steady state at 500 kPa, 860 K and exists at 100 kPa. A temperature sensor indicates that the exit air temperature is 460 K. Stray heat transfer and kinetic and potential energy effects are negligible. The air can be modeled as an ideal gas. Determine if the exit temperature reading can be correct. If yes, determine the power developed by the turbine for an expansion between these states,in kJ/kg of air flowing. If no, provide an explanation with supporting calculations

Find the complete solution in the given attachments.

Note: The complete Question is attached in the first attachment as the provided question was incomplete

Answer:

See explaination

Explanation:

LM358 is the useful IC which works as buffer. It enables circuit to remove overloading effect on each other. Image is in attachment.

We can define a light-emitting diode (LED) as a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons

See attached file for detailed solution of the given problem.

Answer:

3.586.543veh/hr

Explanation:

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

Answer:

A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:

The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.

Explanation:

First of all centripetal movement would be friendlier to our objective as it would not shake the can or the machine itself with disruptive vibrations. Also, we would have to use materials with a good grade of force absorption to eradicate the transmission of the movement to the rest of the structure. Allowing the reduction of the shaking forces while maintaining it effective in the process of mixing.

Answer:

Check the explanation

Explanation:

Assumptions.

1. The surfaces are diffuse, may and opaque

2. steady operating conditions exist

3. Heat transfer from and to the surfaces is only due to Radiation  

Consider the base surface to be surface 2 the top surface to be surface and the side surfaces to surface 3 1. cubical furnace can be considered to be three-surface enclosure. the areas and black body emissive powers of surfaces can be calculated as seen in the attached images below.

Answer:

(a)10.20 mm/s² (b) 403200 s⁴

Explanation:

Solution

Recall that,

The tangible acceleration is a₁ = 10mm/s

The speed = a₁t

Normal acceleration =  aₙ = v₂ /р = a₁²t₂/ р

where р = 0.8m = 800 mm

Now,

When t = 4s

v = (10) (4) = 40 mm/s

Thus,

aₙ = (40)² /800 = 2 mm/s²

Then

The acceleration is,

a = √a₁² + aₙ² = √ (10)² + (2)²

a = 10.20 mm/s²

(b) The time for he magnitude of the acceleration to be 80 mm/s^2

a² =  aₙ² +a₁²

(80)² + [ (10)²t²/800]² + 10²

so,

t⁴ = 403200 s⁴

Answer:

R₁ = 32kΩ

Explanation:

See attached image

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, [tex]g_{c}[/tex] = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

[tex]c_{p}[/tex]= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\[/tex]

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence [tex]v_{a} ^{2} = 0[/tex]

[tex]h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }[/tex]

[tex]T_{1}[/tex] = 20+460 = 480°R

[tex]T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}[/tex]= 533.25°R

Pressure at the inlet of compressor at isentropic condition

[tex]P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}[/tex]

[tex]P_{a}[/tex] = [tex](10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}[/tex]= 14.45 psia

[tex]P_{2}= 8P_{a} = 8(14.45) = 115.6 psia[/tex]

Answer:

a) The temperature and pressure of the gases at every point of the cycle are

T = 38.23 K

P = 2.91 kpa

Respectively

b) The velocity V of the gasses at the nozzle exit = 3590 m/s

Explanation: Please find the attached files for the solutions

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in =  17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

Steam condensing at 120°C

Temperature at saturation = 120°C

hfg of steam at 120°C = 2203 kJ/kg

overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C

U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

temperature of steam going in = temperature of steam out = 120°C

Energy balance:

Heat gained by water = Heat loss by steam

Let specific capacity of steam = 2010kJ/Kg .°C

Find attached the full solution to the question.

The words of No. 138 of the encyclical letter express a profound vision of interconnectedness in the world.

This idea reflects the reality that everything in life is linked in some way. For example, by studying the water cycle, we see how evaporation in one place can lead to precipitation in another, thus affecting the flora and fauna of both places.

Likewise, changes in global temperature impact terrestrial and marine ecosystems, showing how everything is connected in a complex and interdependent system.

The Question in English

A. What is your opinion of the words of No. 138 of the encyclical letter? Is it real that everything is connected? Give some example of this from the texts read.

Answer:

P=8.44 kw

Explanation:

[Find the given attachment for solution]

Answer:

The measurement of the indentation.

Explanation:

Due to disparities in operators making the measurements, the results will vary even under perfect conditions. Less than perfect conditions can cause the variation to increase greatly.

Answer: Liquid

Explanation: This substance is liquid. Liquids are free to move and take the shape of their container, but do not expand to completely fill it.

This substance is MOST likely a liquid. Thus option C is correct.

Liquids can freely move and conform to the structure of their container, but they cannot enlarge to fill it entirely. A beaker is a circular container for holding liquids that is made of glass or plastic.

It is an utilitarian piece of apparatus used to measure liquids, heat them over a Bunsen burner's flame, and contain biochemical processes. It is liquid in nature. Liquids can freely move and conform to the shape of the container, but they cannot enlarge to fill it entirely. Jane fills a beaker with an unidentified chemical.

This material adopts the beaker's form. The chemical doesn't spill out of the open beaker when it's unattended on the lab table. Given that they are flexible and taking on the structure of the container in which they are in volume.  Therefore, option C is the correct option.

Learn more about beaker, Here:

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The question involves calculating the outlet temperature of oil flowing through a tube with known inlet and surface temperatures for two different tube lengths, and comparing the log mean temperature difference to the arithmetic mean temperature difference in an Engineering context.

The subject of the question is Engineering, specifically related to the thermodynamics and fluid mechanics domain. The student is given information about an oil flowing through a tube at a specific rate, with given inlet and surface temperatures, and is asked to find the oil's outlet temperature for tubes of two different lengths. The log mean temperature difference (LMTD) and the arithmetic mean temperature difference (AMTD) should be calculated and compared for both cases. This question involves the principles of heat transfer as well as fluid dynamics, which are typical topics covered in an undergraduate engineering curriculum. Additionally, the student may need to apply the concept of the thermal energy balance to determine the outlet temperature of the oil.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

The break-even point is approximately 27,778 pieces per year for both manual and automated methods.

To determine the break-even point (BEP) for the two production methods, we need to equate the total annual costs of the manual method with the total annual costs of the automated method. Let's denote:

- [tex]\( Q \)[/tex] as the annual production quantity (in pieces)

- [tex]\( C_{\text{manual}} \)[/tex] as the total annual cost of the manual method

- [tex]\( C_{\text{automated}} \)[/tex] as the total annual cost of the automated method

For the manual method:

[tex]\[ C_{\text{manual}} = \text{Fixed cost} + (\text{Hourly wage} \times \text{Hours per year}) \][/tex]

[tex]\[ C_{\text{manual}} = 5000 + (15 \times Q/10 \times 24 \times 365) \][/tex]

For the automated method:

[tex]\[ C_{\text{automated}} = \text{Fixed cost} + (\text{Variable cost per hour} \times \text{Hours per year}) \][/tex]

[tex]\[ C_{\text{automated}} = 55000 + (4.50 \times Q/25 \times 24 \times 365) \][/tex]

To find the break-even point, we set [tex]\( C_{\text{manual}} = C_{\text{automated}} \) and solve for \( Q \):[/tex]

[tex]\[ 5000 + (15 \times Q/10 \times 24 \times 365) = 55000 + (4.50 \times Q/25 \times 24 \times 365) \][/tex]

Let's solve this equation for [tex]\( Q \)[/tex]:

[tex]\[ 5000 + 54 \times Q = 55000 + 52.20 \times Q \][/tex]

[tex]\[ 54 \times Q - 52.20 \times Q = 55000 - 5000 \][/tex]

[tex]\[ 1.80 \times Q = 50000 \][/tex]

[tex]\[ Q = \frac{50000}{1.80} \][/tex]

[tex]\[ Q \approx 27777.78 \][/tex]

So, the break-even point for the two methods is approximately 27778 pieces per year.

Answer:

Find the attachment for the answer

Answer:

See attached image for diagrams and solution

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = [tex] \frac{1}{6}*334.8 = 55.8 ft^3 [/tex]

Volume of sand = [tex] \frac{2}{6}*334.8 = 111.6 ft^3 [/tex]

Volume of gravel = [tex] \frac{3}{6}*334.8 = 167.4 ft^3 [/tex]

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

Answer:

See attachment

Explanation:

Gain= Vo/Vin

      If we set Vout=9.62V corresponding to Vin=2.6V, then gain will be 3.7

Using above value of gain, let's design non-inverting op-amp configuration

Gain= 1+Rf/Rin

3.7= 1= Rf/Rin

2.7= Rf/Rin

If Rin=100Ω then Rf= 270Ω

Answer:

Answer : C ( Both A&B)

Explanation:

The level of illumination in the instrument panel lights will be determined by the panel dimming control and photocell. This panel dimming control consist of a potentiometer. The diameter positions existing in the diameter control acts like variable resistor. Based on its voltage drop BCM (Body Control Module) selects the intensity level by comparing the signal captured from photocell.

Another type is body control module receives the signal from head light rheostat which will be sent to instrument cluster. The instrument cluster controls the lamp intensity. Therefore, the statements said by both the technicians are correct.

Then, the correct option is C

The time of submersion of the steel plate in years is; 10 years

We are given;

The time of submersion of the steel plate is given by the formula;

t = KW/(ρA*CPR)

Now K is a constant and is equal to 534 provided CPR is in mpy and

A is in square inches.

Thus;

t = (534 * 2.6 × 10⁶)/(7.9 * 10 * 200)

t = 8.8 × 10⁴ hours

Now, 24 hours makes one day and there are 365 days in a year. Thus;

number of hours in a year = 24 * 365

Thus;

t in years = (8.8 × 10⁴)/(24 * 365)

t ≈ 10 years

Read more about corrosion at; https://brainly.com/question/5168322

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to get the step by step explanation to the question above.

The heat transfer while the reference information pertains to thermal expansion and physics-related work. Thermal expansion affects the volume, cross-sectional area, and height of objects, and these changes can be calculated using the coefficient of thermal expansion, initial dimensions, and temperature change.

The view factor calculations in heat transfer, specifically related to grooves machined from a solid block of material. While the initial question seems to relate to this topic, the provided reference information does not align with the question and seems to cover thermal expansion and work done by forces, which are different aspects of Physics.

However, to answer the student's question regarding thermal expansion, we can consider that when temperature changes, all dimensions of an object change. The volume change ΔV can be calculated using the formula ΔV = α·V·ΔT, where α is the coefficient of thermal expansion, V is the original volume, and ΔT is the temperature change. For block A with volume L·2L·L and block B with volume 2L·2L·2L, the change in volume will be proportional to each block's respective original volume.

The change in cross-sectional area, typically lw for block A and 2Lw for block B, and change in height h or 2L for each block, will be affected in a similar manner and can be calculated using α, the original dimensions, and the temperature change ΔT.

Complete Question:

Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.

Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.

Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!

Answer & Explanation:

[Find the attachments]

Step 1 :

Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.

Emitter and base, collector and base are reverse biased then BJT in cut off region.

Three sketches one below the other is shown in Figure 1.

[find the figure in attachment]

Step 2:

Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.

Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.

[For  Steps 3 4 5 6 and 7 find attachments]

Answer:

ARPANET is the direct precedent for the Internet, a network that became operational in October 1969 after several years of planning.

Its promoter was DARPA (Defense Advanced Research Projects Agency), a US government agency, dependent on the Department of Defense of that country, which still exists.

Originally, it connected research centers and academic centers to facilitate the exchange of information between them in order to promote research. Yes, being an undertaking of the Department of Defense, it is understood that weapons research also entered into this exchange of information.

It is also explained, without being without foundation, that the design of ARPANET was carried out thinking that it could withstand a nuclear attack by the USSR and, hence, probably the great resistance that the network of networks has shown in the face of major disasters and attacks.

It was the first network in which a packet communication protocol was put into use that did not require central computers, but rather was - as the current Internet is - totally decentralized.

Explanation:

Below I present as a summary some of the most relevant aspects exposed on the requested website about the origin and authors of ARPANET:

1. Licklider from MIT in August 1962 thinking about the concept of a "Galactic Network". He envisioned a set of globally interconnected computers through which everyone could quickly access data and programs from anywhere. In spirit, the concept was very much like today's Internet. He became the first head of the computer research program at DARPA, and from October 1962. While at DARPA he convinced his successors at DARPA, Ivan Sutherland, Bob Taylor and MIT researcher Lawrence G. Roberts, of the importance of this network concept.

2.Leonard Kleinrock of MIT published the first article on packet-switching theory in July 1961 and the first book on the subject in 1964. Kleinrock convinced Roberts of the theoretical feasibility of communications using packets rather than circuits, That was an important step on the road to computer networking. The other key step was to get the computers to talk together. To explore this, in 1965, working with Thomas Merrill, Roberts connected the TX-2 computer in Mass. To the Q-32 in California with a low-speed phone line creating the first wide-area (albeit small) computer network built . The result of this experiment was the understanding that timeshare computers could work well together, running programs and retrieving data as needed on the remote machine, but that the circuitry switching system of the phone was totally unsuitable for the job. Kleinrock's conviction of the need to change packages was confirmed.

3.In late 1966 Roberts went to DARPA to develop the concept of a computer network and quickly developed his plan for "ARPANET", and published it in 1967. At the conference where he presented the document, there was also a document on a concept of UK packet network by Donald Davies and Roger Scantlebury of NPL. Scantlebury told Roberts about NPL's work, as well as that of Paul Baran and others at RAND. The RAND group had written a document on packet switched networks for secure voice in the military in 1964. It happened that work at MIT (1961-1967), in RAND (1962-1965) and in NPL (1964-1967) all they proceeded in parallel without any of the investigators knowing about the other work. The word "packet" was adopted from the work in NPL and the proposed line speed to be used in the ARPANET design was updated from 2.4 kbps to 50 kbps.

Answer:

a) [tex]\dot W = 0.417\,kW[/tex], b) [tex]COP_{R} = 2.198[/tex], c) Irreversible.

Explanation:

a) The power input required by the refrigerator is:

[tex]\dot W = \dot Q_{H} - \dot Q_{L}[/tex]

[tex]\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)[/tex]

[tex]\dot W = 0.417\,kW[/tex]

b) The Coefficient of Performance of the refrigerator is:

[tex]COP_{R} = \frac{\dot Q_{L}}{\dot W}[/tex]

[tex]COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]

[tex]COP_{R} = 2.198[/tex]

c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:

[tex]COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}[/tex]

[tex]COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}[/tex]

[tex]COP_{R,ideal} = 6.218[/tex]

The refrigeration cycle is irreversible, as [tex]COP_{R} < COP_{R,ideal}[/tex].

Answer:

3.305 * 10 ^ ⁻4

Explanation:

Solution

Given:

We calculate the value of k for which is the dependent variable in Avrami equation

y = 1 - exp (-kt^n)

exp (-kt^n) =1-y

-kt^n = ln (1-y)

so,

k =ln(1-y)/t^n

Now,

we substitute 1.1 for n, 0.50 for y, and 129 s for t

k = ln (1-0.50)/129^1.1

k= 3.305 * 10 ^ ⁻4

Answer:

We want to determine the location after 10mins

Explanation:

The release of diborane is continuous

It is release at a rate of 500lb per 20mins

Then, let find the rate in mg/s

1 lb = 453592.37 mg

So, the mass rate Q is

Q = 500lb / 20mins

Q = 500 × 453592.37 mg / 20 × 60sec

Q = 188,996.82 mg/s

Given that mass concentration of

m~ = 5mg/m³

Then,

Rate of volume is

V~ = Q / m~

V~ = 188,996.82 / 5

V~ = 37,799.364 m³/s

The wind speed is

V = 5mph

Let convert to m/s

1 mph = 0.447 m/s

Then, 5mph = 2.235 m/s

From Pasquill Gilford, the cloud atmosphere characteristic is class A

To know the location, we will divide the velocity by heat rate

X = V / Q

X = 2.235 / 188,996.82 mg/s

X = 2.235 / 0.188996kg/s

X = 11.83 m / kg

The location is 0.000001183 m per mg of diborane

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.