One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isothermal expansion occurs at - 23C with a heat transfer to the air of 3.4 kj. The isothermal compression occurs at 27C to a final volume of 0.01m. Using the results of prob. 5.80 adapted to the case, Determine (a) the pressure, in Kpa, at each of the four principal states (b) the work, in KJ for each of the four processes (c) the coefficient of performance

Answer:

a) [tex]E_{m}[/tex] = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.001^{2}[/tex]) = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Area of Copper wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.002^{2}[/tex]) - 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Area of Copper wire = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Young's Modulus of Composite mixture:

[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] +  [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex]     Equation 1

here,

[tex]F_{st}[/tex] = Stress in Steel

[tex]F_{Cu}[/tex] = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = [tex]\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]

Ratio for area of steel = [tex]\frac{0.8}{3.2 }[/tex] = 0.25

Similarly, for Copper,

Ratio for area of copper = [tex]\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]

Ratio for area of copper = [tex]\frac{2.4 }{3.2}[/tex] = 0.75

Put these values in equation 1:

[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] +  [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex]    

[tex]E_{m}[/tex] = (0.25) [tex]E_{st}[/tex] +  (0.75)[tex]E_{Cu}[/tex]

We are given that,

  [tex]E_{st}[/tex] = 205 Gpa

[tex]E_{Cu}[/tex]  = 110 Gpa

So,

[tex]E_{m}[/tex] = (0.25) (205 Gpa) +  (0.75) (110 GPa)

[tex]E_{m}[/tex] = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

[tex]E_{m}[/tex] = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Acu = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]) + (140 Mpa) (2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex])

F = 224 + 336

Fnet = 560 N    ( because Mpa = [tex]10^{6}[/tex] N/[tex]m^{2}[/tex])

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

([tex]\alpha .[/tex]ΔT  + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Est}[/tex]) = [tex]\alpha .[/tex]ΔT  + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Ecu}[/tex])

Where [tex]\alpha[/tex] = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }[/tex] ) = ( 17 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) + [tex]\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )

(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) - 2.69 x [tex]10^{-6}[/tex]

combined thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C

Answer:

[tex]1.18\ \text{kPa}[/tex]

Explanation:

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

h = Height of reading = 12 cm

[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]

Pressure due to height difference is given by

[tex]P=\rho gh\\\Rightarrow P=1000\times 9.81\times 12\times 10^{-2}\\\Rightarrow P=1177.2\ \text{Pa}=1.1772\ \text{kPa}\approx 1.18\ \text{kPa}[/tex]

The pressure differential is [tex]1.18\ \text{kPa}[/tex].

According to the question,

The formula will be:

→ [tex]P = \rho g h[/tex]

By putting the values, we get

      [tex]= 1000\times 9.8\times 12\times 10^{-2}[/tex]

      [tex]= 1177.2 \ Pa[/tex]

      [tex]= 1.18 \ kPa[/tex]

Thus the answer above is correct.

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Answer:

2.29mHg

Explanation:

We first do a conversion of gage pressure to mmHg

= 204(10³) / 13595(9.807)

= 204000/133326.165

= 1.53mHg

755 mmhg = 0.755

The pressure is therefore

P = p(atmosphere) + Of

= 1.53 + 0.755

= 2.29 mHg

This is the absolute pressure on this tire. So this answers this question.

The absolute pressure in the tire is equal to 2,285.13 mmHg.

Given the following data:

To calculate the absolute pressure in the tire:

First of all, we would convert the value of gauge pressure in kPa to mmHg.

Conversion:

1 kPa = 7.50062 mmHg

204 kPa = [tex]204 \times 7.50062 = 1530.13\;mmHg[/tex]

Mathematically, absolute pressure is given by this formula:

[tex]P_{abs} = P_{atm} + P_{ga}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]P_{abs} = 755 + 1530.13[/tex]

Absolute pressure = 2,285.13 mmHg.

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Answer:

Uh- dude if you think I'm gonna download that, think twice...

Answer: the ripple factor is 0.005

Explanation:

Given the data in the question;

we know that expression of ripple factor is;

r = Vr(pp) / Vdc

where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)

and Vdc is the dc value of the filter output voltage ( 20 V)

so we substitute our given values;

r = 0.1 / 20

r = 0.005

Therefore; the ripple factor is 0.005

The ripple factor of the given power supply filter is; γ = 0.005

We are given;

Ripple peak to peak voltage; V_r,pp = 100 mV = 0.1 V

DC value of the filter output voltage; V_dc = 20 V

Now, formula for the ripple factor is given as;

γ = V_rms/V_dc

Now, our ripple peak to peak voltage is also known as the rms value of ripple voltage at the output. Thus;

γ = 0.1/20

γ = 0.005

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The correct answer is A. Earning a bachelor's degree in Civil Engineering from a four-year university, completing an internship, and seeking a job at a private firm.

Explanation:

In the U.S. and many countries, the best to start a career is to enroll in a formal educational program at a university or college. This helps students learn concepts, theories, methods, etc. they need for their profession. Moreover, a degree such as a bachelor's degree is required by employers. In this context, the first step for Connor is to earn a bachelor's degree in Civil Engineering.

Besides this, an internship is recommended after earning a degree because this is the way students can gain real-life work experience, which is considered positive by employers. This means the next step should be an internship.

Finally, Connor can seek a job to design bridges and other buildings because after the degree and internship he will have the experience and knowledge required by employers and by the job.

Answer:

team building exercises

Explanation:

Answer:   its not team building excercize

Explanation:

No amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.

This means a from of cylindal metal container that is used for shipping or storage of liquids.

When its surface is cracked, then, no amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.

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#SPJ2

Answer:

-3 dB point

Explanation:

A filter is a circuit deigned to pass or amplify certain frequencies while also preventing other frequencies from passing. There are different types of filter depending on the frequency range such as the low pass filter, high pass filter, bandstop filter, bandpass filter.

A low pass filter is a filter that allows the passage of frequencies below the cut off frequency and blocks frequencies above the cut off frequency. The cutoff frequency of a low pass filter is the frequency at the -3dB point.

Answer:

A

Explanation:

It is also a toxic material

Answer:

The correct answer is A. Lead.

Explanation:

I got it right on the Plato test.

Answer:

a pneumatic or electric power

Explanation:

The primary energy source for the controller in a typical control system is either "a pneumatic or electric power."

This is because a typical control system has majorly four elements which include the following:

1. Sensor: this calculates the controlled variable

2. Controller: this receives and process inputs from the sensor to the controlled device as output

3. Controlled device: this tweak the controlled variable

4. Source of energy: this is the energy used to power the control system. It could be a pneumatic or electric power

Answer:

the saturated density should be

Explanation:

918.63517 is the answer.

Answer:

b

Explanation:

Trust me. I am not the smartest though...

Answer:Animation can occur before an action. Prepares the audience for the action. ... A pose or action should clearly communicate to the audience the attitude, mood, reaction or idea of the character. The effective use of long, medium or close up shots, as well as camera angles helps tell the story.

Explanation:May i have brainlist plz only if u wanna give me brainlist though have an nice day and stay safe.