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nlexa [21]
4 years ago
12

Consider a 20-cm X 20-cm X 20-cm cubical body at 477°C suspended in the air. Assuming the body closely approximates a blackbody,

determine
(a) the rate at which the cube emits radiation energy, in W and
(b) the spectral blackbody emissive power at a wavelength of 4 µm.
Engineering
1 answer:
Olin [163]4 years ago
8 0

Answer:

a) The rate at which the cube emits radiation energy is 704.48 W

b) The spectral blackbody emissive power is 194.27 W/m²μm

Explanation:

Given data:

a = side of the cube = 0.2 m

T = temperature = 477°C

Wavelength = 4 µm

a) The surface area is:

A_{s} =6a^{2} =6(0.2)^{2} =0.24m^{2}

According Stefan-Boltzman law, the rate of emission is:

E=\sigma T^{4} A_{s} =5.67x10^{-8} *(477)^{4} *0.24=704.48W

b) Using Plank´s distribution law to get the spectral blackbody emissive power.

E=\frac{C_{1} }{\lambda ^{5}(exp(\frac{C_{2} }{\lambda T}) -1 )} =\frac{3.743x10^{8} }{4^{5}(exp(\frac{1.4387x10x^{4} }{4*477})-1)  } =194.27W/m^{2} \mu m

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Answer:

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Explanation:

8 0
3 years ago
A classroom which normally contains 26 people is to be air- conditioned. A person in the classroom typically dissipates heat at
lutik1710 [3]

Answer:

Explanation:

Heat dissipated by 26 people per second

= 26 x 360 x 1000 / (60 x 60)

= 2600 J

Heat dissipated by bulbs per second

= 15 x 40

= 600 J

Heat dissipated by fans per second

200 J

Heat dissipated by air coming in

= 200 x 10 x 1000 / (60 x 60)

=555.55 J

Heat gain by walls

= 5000 x 1000 / (60 x 60)

= 1388.88 J

Total heat gain per second

= 2600 + 600+200+555.55

= 3955.55 J

Capacity of air-conditioner required

= 3955.55 J/s

= 3.9 kJ/s

= 3.9kW

= 4 kW

7 0
3 years ago
The following figure shows the tensile stress-strain curve for a brass alloy.
Aleksandr-060686 [28]
(A)-The Theoretical Strength Is Higher Than The Experimental Strength.

7 0
3 years ago
A shaft 70 mm in diameter is being pushed at a speed of 400 mm/s through a bearing sleeve 70.2 mm in diameter and 250 mm long. T
Allushta [10]

Answer:

F=989.6 N

Explanation:

Given that

Diameter of shaft = 70 mm

Diameter of bearing sleeve =70.2 mm

So clearance h=0.1 mm

Speed V= 400 mm/s

Length of shaft = 250 mm

\nu =0.005\ \frac{m^2}{s}\ , \rho=900\ \frac{kg}{m^3}

We know that

\mu =\rho\times\nu

μ= 900 x 0.005 Pa-s

μ= 4.5 Pa-s

As we know that

From Newton's law of viscosity ,the shear stress given as follows

\tau =\mu \dfrac{dU}{dy}

We also know that

Force = shear stress x area

Now by putting the values

\tau =4.5\times \dfrac{400}{0.1}

\tau=18,000 Pa

So force

F= 18,000 x π x 0.07 x 0.25

F=988.6 N

6 0
3 years ago
The task of framing a building has been estimated to take anaverage of 25 days with astandard deviation of 4 days. What duration
Nimfa-mama [501]

Answer:

30.128 days

Explanation:

Given that:

Mean = 25

Standard deviation = 4

Confidence interval = 90% = 0.9

Since the confidence interval should not exceed 90%

Then using z test table

P(z) = 0.9

For 0.8997 , we get = 1.28

For 0.9015, we get = 1.29

∴

\dfrac{0.9 - 0.8897}{0.9015 - 0.8997 }=\dfrac{ z - 1.28}{1.29 -1.28}

By solving

Z = 1.282

Thus, the duration to be used so that it will not exceed 90% C.I is:

Z = (x - μ)/σ

1.282= ( x - 25)/4

1.282 * 4 = x - 25

(1.282*4)+25 = x

x = 30.128 days

5 0
3 years ago
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