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drek231 [11]
3 years ago
8

Consider a cubic block whose sides are 6 cm long and a cylindrical block whose height and diameter are also 6 cm. both blocks ar

e initially at 22°c and are made of granite (k = 2.5 w/m · °c and a = 1.15x 10-6 m^2/s). now both blocks are exposed to hot gases at 550°c in a furnace on all of their surfaces with a heat transfer coefficient of 45 w/m2 · °c. determine the center temperature of each geometry after 10, 20, and 60 minutes.

Engineering
1 answer:
bija089 [108]3 years ago
5 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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Can you prove that the two bleu areas are the same without numbers please?
Svet_ta [14]

Answer:

\small{\boxed{\tt{\colorbox{green}{✓Verified\:answer}}}}\:

Just draw a line from point D join to point E

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6 0
2 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
miskamm [114]

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.

7 0
3 years ago
The regulated voltage of an alternator is stated as 13.6 to 14.6 volts at 3000 rpm with the
lions [1.4K]

Answer:

  d)  1 volt​

Explanation:

The allowable range is 1 volt​. The allowed tolerance (deviation from nominal) depends on what the nominal voltage is.

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