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SCORPION-xisa [38]
4 years ago
15

In order to impress your neighbors and improve your vision in traffic jams, you decide to mount a cylindrical periscope 2.0 m hi

gh and 0.1 m in diameter on top of your car. You consult the literature and find that the drag coefficient (based on frontal area) is 1.1 for Re<200,000 (Re based on diameter) and drops to 0.35 for Re>200,000. Using this information, answer the three questions given. (HINT: make a sketch)The periscope's drag at your maximum speed of 75 mph is about:(A) 45 N(B) 4 N(C) 450 N(D) 5 N
Engineering
1 answer:
kondaur [170]4 years ago
5 0
Follow @richard.gbe on Instagram for the answer
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What line separates two lanes traveling in the same direction
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White lane lines separate lanes of traffic moving in the same direction. (UK)

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Technician A says automotive gasoline engines run on the 2 stroke principal. Technician B says Diesel engine hace less compressi
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both are incorect

Explanation:

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3 years ago
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
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Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

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