An air standard cycle with constant specific heats is executed in a closed pistoncylinder system and is composed of the following four processes: 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion with a volume ratio, r1=V4/V3 4-1 Constant pressure heat rejection with a volume ratio, r2=V4/V1 (a) Sketch the P-v and T-s diagrams for this cycle. (b)Find out T2/T1 as a function of k, r1, r2 only. (c) Find out T4/T1 as a function of k, r1, r2 only. (d)Find out T3/T4 as a function of k, r1, r2 only. (e) Find out T3/T2 as a function of k, r1, r2 only. (f) Obtain an expression for the back work ratio for a fixed minimum-tomaximum temperature ratio T1/T3. The expression should be of a function of T1/T3, k, r1, r2 only. (g)Obtain an expression for the cycle thermal efficiency as a function of k, r1, r2 only.

Answer:

P=8.44 kw

Explanation:

[Find the given attachment for solution]

Answer:

See attached image for diagrams and solution

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:

(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:

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

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:

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:

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:

Check the explanation

Explanation:

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

Answer:

See attached images

Answer:

NL = 207

Explanation:

Solution

Now,

The mean temperature is measured as:

Tm = (T₁ - T₀)/2

= (15 +  65)/2

= 40°C

So,

we find all the thermo-physical  properties of water from the table, that is properties of saturated water at T =40°C

Thermo conductivity, k = 0.631 W/m . K

Specific heat Cp = 4179 J/kg . K

Density р = 992. 1 kg/m³

The dynamic viscosity, μ = 0.653 * 10 ^⁻3 kg/m *s

Prandtl number, Pr = 4.32

At T = 15°C

рi = 992.1 kg/m³

At T = 90°C

Prandtl number, Prs = 1.96

Thus,

The maximum flow of velocity is known from the equation stated as:

Vmax = ST/ST - D *V

Here,

ST is refereed to as the transverse pitch for inline arrangements of the rods

so,

Vmax = 3/3-1 * 0.8

= 1.2 m/s

Now

The Reynolds number is determined from the equation given below

ReD =ρVmax D/μ

= 922.1 * 1.2 *(1 *10^⁻²)/ 0.653 * 10^⁻³

= 18231.55

From the table, The Nusselt number correlations fro cross flow over the tube banks for inline arrangement over the range of ReD  is shown as

1000 - 2 * 10⁵

Now, the Nusselt number is determined by

NuD = 0.27ReD ^0.63 Pr^ 0.36 (Pr/Prs)^0.25

= 0.27 * (18231.55)^0.63 (4.32)^0.36 * (4.32/1.96)^0.25

=269.32

Then,

The convective transfer of heat water coefficient  is determined  from the equation shown  by Diametral Nusselt Number

NuD =hD/k

So,

we re-write and solve for h

h = NuD * k/D

=269.32 * 0.631/(1 * 10 ^⁻2)

=16993.9 W/m² .K

Now,

The heat transfer surface area for a tube in a row is NT = 1

As = NT NLπDL

= 1*NL* π * (1 * 10^⁻2) * 4

= 0.1257NL

The logarithmic mean temperature of water is represented as

ΔTlm = Te - Ti/ln (Ts - Ti/Ts- Te)

= 65- 15/ln (90 -15/ 90 -65) = 45.51°C

Thus,

The rate of the heat transfer is determined  from the equation shown below,

Q =hAsΔTlm

=16993.9 *0.1257 * NL* 45.51.......equation (1)

The mas flow rate of water is determined by the equation below

m =ρiAcV

= ρi * (STL) * V

= 999.1 8 ( 3* 10^⁻2 * 4) * 0.8

= 95.91 kg/s

The rate of heat transfer of water is determined by the equation below

Q = mcp (Te- Ti)

= 95.91 * 4179 * (65-15)

=20041146.72 W..........(Equation 2)

Now,

The number of tube rows in the direction flow is determined  by measuring both equations 1 and 2 as

97219.61 NL = 20041146.72

NL =206.14

NL = 207

Therefore, the number of tube rows NL in the flow direction needed to achieve the indicated temperature rise is NL = 207

=

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:

https://brainly.com/question/29475799

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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:

See explaination

Explanation:

thermal conductivity is A measure of the ability of a material to transfer heat. Given two surfaces on either side of the material with a temperature difference between them, the thermal conductivity is the heat energy transferred per unit time and per unit surface area, divided by the temperature difference.

Please kindly check attachment for the step by step solution of the given problem.

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:

nmuda mudaf A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

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:

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:

See explaination

Explanation:

To compare the hollow and solid cylinder we need ro use torsional formula.

And since same material and length are given.

For same torque and angle of twist there will be same polar moment of area of the section for both the cylinder.

Please kindly check attachment for further solution

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:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

[tex] h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2} [/tex]

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

[tex] n = \frac{h_1 - h_2*}{h_1 - h_2} [/tex]

[tex] n = \frac{1278.8 - h_2*}{1278.8 - 1193.5} [/tex]

solving for h2, we have

[tex] h_2 = 1197.77 Btu/Ibm [/tex]

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

[tex] (1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}[/tex]

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

Answer:

x2 = 0.5056

Qin = 22.62Kj

Explanation:

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.

Answer:

Find the attachment for the answer

Answer:

a.  The work done by the compressor is 447.81 Kj/kg

b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg

c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg

d. The coefficient of performance is 2.746

e. The refrigerating efficiency is 71.14%

Explanation:

According to the given data we would need first the conversion of temperaturte from C to K as follows:

Temperature at evaporator inlet= Te=-16+273=257 K

Temperatue at condenser exit=Te=48+273=321 K

Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg

b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:

w=i4-i3

W/M=i4-i3

i4=W/M + i3

i4=2.5/0.05 + 397.81

i4=447.81 Kj/kg

a. Enthalpy at the compressor exit=447.81 Kj/kg

Therefore, the heat rejected from the condenser in kJ/kg=i4-i1

the heat rejected from the condenser in kJ/kg=447.81-260.51

the heat rejected from the condenser in kJ/kg=187.3 kJ/kg

c. Temperature at evaporator inlet= Te=-16+273=257 K

The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

d. To calculate the coefficient of performance we use the following formula:

coefficient of performance=Refrigerating effect/Energy input

coefficient of performance=137.3/50

coefficient of performance=2.746

the coefficient of performance is 2.746

e. The refrigerating efficiency = COP/COPc

COPc=Te/(Tc-Te)

COPc=255/(321-255)

COPc=3.86

refrigerating efficiency=2.746/3.86

refrigerating efficiency=0.7114=71.14%

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.

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]

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:

c. a delay or advance in time as compared to a pure cosine wave.

Explanation:

Electrical phase is measured in degrees, with 360° corresponding to a complete cycle. A sinusoidal voltage is proportional to the cosine or sine of the phase. Phase difference , also called phase angle , in degrees is conventionally defined as a number greater than -180, and less than or equal to +180.

The phase angle corresponds to delay or advance in time as compared to a pure cosine wave.

Answer:

Check the explanation

Explanation:

Kindly check the attached images for the step by step explanation to the question