The components of an electronic system dissipating 90 W are located in a 1-m-long circular horizontal duct of 15-cm diameter. Th

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data:

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

To determine final temperature in an Isentropic process

----------- ( 1 )

T2 = 303 = 576.9K

The work performed in a piston-cylinder device is calculated using the subsequent formula

   ------- ( 2 )

where: cv = 3.1156 kJ/kg.k for helium gas

             T2 = 576.9K,    T1 = 303 K

substituting values into equation 2

= 853.4 kJ/kg

the work done in a steady flow compressor is determined using this

where: cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K

             T2 = 576.9 k, T1 = 303 K

plugging values back into equation 3

= 1422.3 kJ / kg

Answer:

total expense for the new boiler = $229706.825

total expense for new boiler = $127512

Explanation:

provided information

initial power p1 = 80 kW

price C = $160000

cost index CI 1 = 187

cost index CI 2= 194

capacity factor f = 0.6

subsequent power p2 = 120 kW

present cost = $18000

to determine

total expense and cost for 40 kW

solution

we evaluate CN cost for the new boiler and CO cost for the existing boiler

where x represents the capacity of the new boiler

first we compute the current cost of the old boiler which is

current cost CO = C × .............1

substituting the value here

current cost = 160000 ×

updated current cost = $165989.304

and

employing power sizing strategy for 124 kW

CN/CO = ...............2

insert value and calculate CN

CN/CO =  

CN / 165989.304 =  

CN = 211706.825

therefore the new expense = $211706.825

hence

total expense for the new boiler amounts to

total expense = new expense + current cost

total exp = 211706.825 + 18000

boiler expense total = $229706.825

and

concerning a 40 kW unit, the calculated new cost will be

applying equation 2

CN/CO =

CN / 165989.304 =  

CN = 109512

thus the new cost is $109512

therefore

total expense for the new boiler is

total exp = new cost + current cost

total expense = 109512 + 18000

total cost for the new boiler = $127512

Answer:

The air temperature should be similar in both regions.

Explanation:

Climate conditions differ globally. Additionally, factors such as distance from the ocean, landscape, and elevation influence these conditions.

If both areas are equidistant from the equator and close to the ocean, there is a high likelihood that they will have similar climate conditions, thus resulting in comparable temperature ranges.

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 classify the medium as follows:

                                σ / w*εr*εo

Where, w is the wave's angular frequency

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

- Now we categorize according to the following guidelines:

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

    Glass: Low-Loss dielectric

    Tissue: Quasi-Conductor

    Wood: Good conductor

- Next, we will apply the categorized material equations specified in 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 )