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kiruha [24]
4 years ago
12

What is a Planck Distribution and how is it used to solve for black body radiation problems?

Engineering
1 answer:
zvonat [6]4 years ago
4 0

Answer:

Planks law:

   Planks law gives the theoretical distribution for emissive power of black body.The emissive power for is given as follows

Planks distribution as a function of wavelength for different temperature.

E_{\lambda,b}=\dfrac{C_1}{\lambda^5 [exp{\frac{C_2}{\lambda T}-1}]}

Where C_1 andC_2 is the constant.

Important points for Planks distribution

1.At the given wavelength when temperature increases then emissive power will increase.

2.When temperature is increases then then distribution shift towards the left side.

3. We can assume that sun as a black body at  5800 K.

 

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A thick steel sheet of area 150 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
Bingel [31]

Answer:

Rate of corrosion = 24.95 mpy

Rate of corrosion = 0.63 mm/yr

Explanation:

given data

steel sheet area = 150 in²

weight loss = 485 g

density of steel = 7.9 g/cm³

time taken = 1 year

to find out

rate of corrosion in (a) mpy and (b) mm/yr

solution

we get here the rate of corrosion that is express as

rate of corrosion = (k × W) ÷ (D × A × T)     ..................1

here k is constant  and w is total weight lost and t is time taken for loss and A is surface area and D is density of steel

so put her value in equation 1 we get

Rate of corrosion = \frac{534*485*1000}{7.9*150*24hr*365day/yr}

Rate of corrosion = 24.95 mpy

and

Rate of corrosion = \frac{87.6*485*1000}{7.9*150*2.54^2*24hr*365days/yr}

Rate of corrosion = 0.63 mm/yr

5 0
3 years ago
Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo
noname [10]

Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

3 0
3 years ago
A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main
andreev551 [17]

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C  = 273 K

\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5

R = 287 J/kg K

C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s

C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s

V_2 = \dfrac{V_1}{C_1}\timesC_2

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 ×  147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁  )

= 165.49 - 147.5

= 17.5 m/s

6 0
4 years ago
Scheduling can best be defined as the process used to determine:​
shutvik [7]

Answer:

Overall project duration

Explanation:

Scheduling can best be defined as the process used to determine a overall project duration.

8 0
3 years ago
A disk with radius of 0.4 m is rotating about a centrally located axis with an angular acceleration of 0.3 times the angular pos
FrozenT [24]

Answer:

a₁= 1.98 m/s²   : magnitud of the normal acceleration

a₂=0.75  m/s²  : magnitud of the tangential acceleration

Explanation:

Formulas for uniformly accelerated circular motion

a₁=ω²*r : normal acceleration     Formula (1)

a₂=α*r:    normal acceleration     Formula (2)

ωf²=ω₀²+2*α*θ                             Formula (3)

ω : angular velocity

α : angular acceleration

r  : radius

ωf= final angular velocity

ω₀ : initial angular velocity

θ :   angular position theta

r  : radius

Data

r =0.4 m

ω₀= 1 rad/s

α=0.3 *θ , θ= 2π

α=0.3 *2π= 0,6π rad/s²

Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.

We calculate ωf with formula 3:

ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687

ωf=\sqrt{24.687} =4.97 rad/s

a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²    

a₂=α*r = 0,6π * 0.4 = 0.75  m/s²  

7 0
4 years ago
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