Which type of engineering is more of an exploratory process that does not need to create a product that fits limitations and specifications?


reverse engineering

materials engineering

documentation process

engineering design process

Answer:

2nd one

Explanation:

on edge 21

The option that identifies the engineer should choose the reverse engineering process in the following scenario is there will be changes to the existing product. The correct option is b.

The distinction between the reverse engineering process and the engineering design process is quite slight. The engineering design process is a method that requires months of planning and design in order to provide a blueprint for a specific project.

The fundamental skill of learning to make something by working backward from an existing inspiration for your project is called reverse engineering.

Three contrasts are that the engineering design process is based more on mental and written learning than it is on physical learning (unlike the reverse engineering process).

Therefore, the correct option is b, There will be changes to the existing product.

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Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

[tex]A_1 \ and \ A_2[/tex]

Thus, using the summation rule, the view factor [tex]F_{11[/tex] and [tex]F_{12[/tex] is as follows:

[tex]F_{11}+F_{12}=1[/tex]

Let assume the surface (1) is flat, the [tex]F_{11} = 0[/tex]

Now:

[tex]0+F_{12}=1[/tex]

[tex]F_{12}=1[/tex]

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder [tex]A_2[/tex] to the flat base surface [tex]A_1[/tex]; we have:

[tex]A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}[/tex]

Suppose, we replace DL for [tex]A_1[/tex] and

[tex]A_2[/tex] =  [tex]\dfrac{\pi D}{2}[/tex]

Then:

[tex]F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64[/tex]

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

[tex]Q_{21} = Q_{evaporation}[/tex]

But, before that;  let's find the radian heat exchange occurring among the dome and the flat base surface:

[tex]Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)[/tex]

where;

[tex]\sigma = Stefan \ Boltzmann's \ constant[/tex]

[tex]T_1 = base \ temperature[/tex]

[tex]T_2 = temperature \ of \ the \ dome[/tex]

∴

[tex]Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m[/tex]

Recall the energy balance formula;

[tex]Q_{21} = Q_{evaporation}[/tex]

where;

[tex]Q_{evaporation} = mh_{fg}[/tex]

here;

[tex]h_{fg}[/tex] = enthalpy of vaporization

m = the water mass flow rate

∴

[tex]83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}[/tex]

The drying rate per unit length is 0.037 kg/S.m

Given data;

From the attached diagram, the surface 1 is flat, it is a view factor, f[tex]_1_1[/tex] = 0.

Applying summation rule and solving the view factor from the base surface A[tex]_1[/tex] to the cylindrical dome A[tex]_2[/tex].

[tex]f_1_1+f_1_2=1[/tex]

Put F[tex]_1_2=0[/tex]

[tex]0+f_1_2=1[/tex]

This makes [tex]f_1_2=1[/tex]

Applying reciprocal rule and solving the view factor from the cylindrical dome A[tex]_2[/tex] to the base surface A[tex]_1[/tex].

[tex]A_2F_2_1=A_1F_1_2\\F_2_1=(\frac{A_1}{A_2})F_1_2[/tex]

Where A is the area of the surface.

Substitute DL for A[tex]_1[/tex] and [tex]\frac{\pi D}{2}[/tex] for A[tex]_2[/tex]

[tex]F_2_1 = \frac{DL}{(\frac{\pi D}{2}})L *1 = \frac{2 }{\pi } =2/3.14 = 0.64[/tex]

Using the energy balance equation to the dryer,

[tex]Q_2_1=Q_e_v_a_p[/tex]

Let's calculate the radiation heat exchange between the dome and the base surface per unit length by using the equation below

[tex]Q_2_1=F_2_1A_2[/tex]δ[tex](T_2^4-T_1^4)[/tex]

[tex]Q_2_1= F_2_1 * \frac{\pi D}{2}[/tex]δ[tex](T_2^4-T_1^4)[/tex]

substitute the respective values into the equation

[tex]Q_2_1=0.64*(\frac{\pi }{2}*1.5)*5.67*10^-^8*(1000^4-370^4)\\Q_2_1=8.3899.15W/m[/tex]

Let's calculate the mass flow rate of water using the amount of heat required for drying up.

[tex]Q_2_1=Q_e_v_a_p\\Q_e_v_a_p=mh_f_g\\[/tex]

where [tex]h_f_g= 2257*10^3J/kg[/tex]

and this is the enthalpy of vaporization and mass flow rate of water.

[tex]83899.15=m*2257*10^3\\m=0.037kg/S.m[/tex]

The drying rate per unit length is 0.037kg/S.m

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Answer:

Explanation:

Brake systems are most likely to fail at the ends.  One of the most common problems is a trailer lighting splitter is left on. With time and usage, the fuse can blow interrupting the signal path from the switch to the bulb.

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Explanation:

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I am not sure it's confusing

Answer: i believe it’s acoustics

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This is not a clear indication of the hot zone as the information of the radioactivity of the background is not provided clearly.

Explanation:

According to IAEA as well as NRCP, the hot area is defined on the basis of the radioactivity reading it shows instead of contrast or comparative reading from the background. The value of radiation activity which will be required to declare an area as hot zone is if it is greater than 0.1 mSv/h or [tex]1.5091\times 10^{29} kg^{-1} s^{-1}[/tex].