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pychu [463]
3 years ago
8

Use superpositions find​

Engineering
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

no

Explanation:

no

You might be interested in
Distinguish between systems analysis and systems design?
Zielflug [23.3K]

Answer:

System analysis can be defined as a deep analysis of a part of the structure of a module that has been designed before. System design means to make any module or a part of the structure from scratch and build it completely without estimation.

Explanation:

3 0
3 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0
2 years ago
1. A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450 MPa. Is the specimen lik
mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}

=  2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}

                 = 15 GPa

fracture will occur as the value is less than E/10 = 22.5

7 0
3 years ago
A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken o
irina [24]

Answer:

the heat transfer from the pipe will decrease when the insulation is taken off for r₂< r_{cr}

where;

r₂ = outer radius

r_{cr} = critical radius

Explanation:

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .

r_{cr} =\frac{k}{h}

The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than  critical radius (r₂< r_{cr}) 0,  and reaches a maximum when r₂ = r_{cr}, and starts to decrease for r₂< r_{cr}. Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< r_{cr} .

7 0
3 years ago
For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or
Alborosie

Answer:

Glass: Low-Loss dielectric

  α = 8.42*10^-11 Np/m

  β = 468.3 rad/m

  λ = 1.34 cm

  up = 1.34*10^8 m/s

  ηc = 168.5 Ω

Tissue: Quasi-Conductor

  α = 9.75 Np/m

  β = 12.16 rad/m

  λ = 51.69 cm

  up = 0.52*10^8 m/s

  ηc = 39.54 + j 31.72 Ω        

Wood: Good conductor

  α = 6.3*10^-4 Np/m

  β = 6.3*10^-4 Np/m

  λ = 10 km

  up = 0.1*10^8 m/s

  ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if  the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then  calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

                                σ / w*εr*εo

Where, w is the angular speed of wave

            εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

    Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\  

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

    Glass: Low-Loss dielectric

    Tissue: Quasi-Conductor

    Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

     Glass: Low-Loss dielectric

          α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

          α = 8.42*10^-11 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

          β = 468.3 rad/m

          λ = 2*pi / β = 2*pi / 468.3

          λ = 1.34 cm

          up = λ*f = 0.0134*10^10

          up = 1.34*10^8 m/s

          ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

          ηc = 168.5 Ω

     Tissue: Quasi-Conductor

          α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

          α = 9.75 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

          β = 12.16 rad/m

          λ = 2*pi / β = 2*pi / 12.16

          λ = 51.69 cm

          up = λ*f = 0.5169*100*10^6

          up = 0.52*10^8 m/s

          ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

          ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

          ηc = 39.54 + j 31.72 Ω

     Wood: Good conductor

          α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

          β = α = 6.3*10^-4 Np/m

          λ = 2*pi / β = 2*pi / 6.3*10^-4

          λ = 10 km

          up = λ*f = 10,000*1*10^3

          up = 0.1*10^8 m/s

          ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

          ηc = 6.28*( 1 + j )

         

           

         

8 0
3 years ago
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