All categories

4 years ago

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.

You might be interested in

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

3 years ago

How is a Doctor Who is a generalist different a specialist?

bekas [8.4K]

Answer:

specialist focus in one very specific thing, generalist focus and many things, they are like a jack of all trades

7 0

3 years ago

Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring

umka2103 [35]

Answer:

the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

Explanation:

Given the data in the question ;

flowrate Q = 400 cm³/s

cross section of the sprue is round

Diameter of sprue at the top d_top = 3.4 cm

Height of sprue = 20 cm = 0.2 m³

the proper diameter at its base so as to maintain the same volume flow rate = ?

first we determine the velocity at the sprue base

V_base = √2gh = √( 2×9.81×0.2) = √3.924 = 1.980908 m = 198.0908 cm

so, diameter of the sprue at the bottom will be

Q = AV = [ (( πd²_bottom)/4) × V_bottom ]

d_bottom = √(4Q/πV_bottom)

we substitute

d_bottom = √((4×400)/(π×198.0908 ))

d_bottom = √( 1600/622.3206)

d_bottom = √2.571022

d_bottom = 1.6034 cm

Therefore, the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

8 0

3 years ago

Some designers suggest that speech recognition should be used in a telephone menu system. This would allow users to interact wit

Ghella [55]

The designers suggest that speech recognition allows for spoken interaction or conversation and spoken prompts or commands.

Explanation:

The speech recognition is used when the user has physical impairments and hands are busy, eyes are occupied and the user cannot read.

We can save time with the usage of speech recognition system. The users are knowledgeable based on the actions available.

The problems faced in speech recognition system are in the noisy environment it has bad microphones and the commands should be learned and remembered.

The other obstacle is error correction is time consuming. the speech production has slow space to speech output provides privacy in public spaces. The disadvantage is it contains large amount of information.

5 0

4 years ago

Air enters the compressor of a regenerative gas-turbine engine at 300 K and 100 kPa, where it is compressed to 800 kPa and 580 K

Bad White [126]

Answer:

a) The amount of heat transfer in the regenerator, q = 114.12 kJ/kg

b) Thermal efficiency = 35.9%

Explanation:

The calculations are neatly handwritten and attached as files to this solution for easiness of expression and clarity. The cycle is also drawn on the T - S diagram and included in the attached files. Check the files below for the complete calculation.

I hope this helps!

3 0

3 years ago