The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources

Answer:

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Explanation:

Answer:

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Answer:

the surface temperature of the plates when they come out of the oven is approximately 445 °C

Explanation:

Given the data in the question;

thickness t = 3 cm = 0.03 m

so half of the thickness L = 0.015 m

thermal conductivity of brass k = 110 W/m°C

Density p = 8530 kg/m³

specific heat [tex]C_p[/tex] = 380 J/kg°C

thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s

Temperature of oven T₀₀ = 700°C

initial temperature T[tex]_i[/tex] = 25°C

time t = 10 min = 600 s

convection heat transfer coefficient h = 80 W/m².K

Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.

So, using analytical one-term approximation method, the Fourier number > 0.2.

now, we determine the Biot number for the process

we know that; Biot number Bi =  hL / k

so we substitute

Bi =  hL / k

Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109

Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )

The interpolation method used to find the

λ₁ = 0.1039 and A₁ = 1.0018

so

The Fourier number т = ∝t/L²

we substitute

Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²

т = 0.02034 / 0.000225

т = 90.4

As we can see; 90.4 > 0.2

So,  analytical one-term approximation can be used.

∴ Temperature at the surface will be;

θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation

θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )

so we substitute

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )

θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998

θ(L,t)[tex]_{wall[/tex] = 0.3775

so we substitute into equation 1

θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)

0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )

0.3775 = ( T(L,t) - 700 ) / ( - 675 )

0.3775 × ( - 675 ) = ( T(L,t) - 700 )

- 254.8125 = T(L,t) - 700

T(L,t) = 700 - 254.8125

T(L,t) = 445.1875 °C ≈ 445 °C

Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C

Answer: hello your question lacks some data below is the missing data

Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol

H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.

H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg

Answer :

a) 34.98 lit/min

b) 1432.53 m^3/min

Explanation:

a) Calculate how much water is produced

density of water = 1 kg/liter

First we will determine the mass of condensed water using the relation below

inlet mass - outlet vapor mass =  0.0339508 * n * 18/1000 ----- ( 1 )

where : n = 57241.57

hence equation 1 = 34.98 Kg/min

∴ volume of water produced =  mass of condensed water / density of water

                                                =  34.98 Kg/min / 1 kg/liter

                                                = 34.98 lit/min

b) calculate the Volumetric flow rate of air entering the unit

applying the relation below

Pv = nRT

101325 *V = 57241.57 * 8.314 * 305  

∴ V = 1432.53 m^3/min