Most equipment is cooled by bringing cold air in the front and ducting the heat out of the back. What is the term for where the
1 answer:
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Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
Answer:
a.)
US Sieve no. % finer (C₅ )
4 100
10 95.61
20 82.98
40 61.50
60 42.08
100 20.19
200 6.3
Pan 0
b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4
c.) Cu = 3.33
d.) Cc = 1
Explanation:
As given ,
US Sieve no. Mass of soil retained (C₂ )
4 0
10 18.5
20 53.2
40 90.5
60 81.8
100 92.2
200 58.5
Pan 26.5
Now,
Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g
⇒ w = 421.2 g
As we know that ,
% Retained = C₃ = C₂×
∴ we get
US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)
4 0 0
10 4.39 4.39
20 12.63 17.02
40 21.48 38.50
60 19.42 57.92
100 21.89 79.81
200 13.89 93.70
Pan 6.30 100
Now,
% finer = C₅ = 100 - C₄
∴ we get
US Sieve no. Cummulative % retained (C₄) % finer (C₅ )
4 0 100
10 4.39 95.61
20 17.02 82.98
40 38.50 61.50
60 57.92 42.08
100 79.81 20.19
200 93.70 6.3
Pan 100 0
The grain-size distribution is :
b.)
From the diagram , we can see that
D10 = 0.12
D30 = 0.22
D60 = 0.12
c.)
Uniformity Coefficient = Cu =
⇒ Cu =
d.)
Coefficient of Graduation = Cc =
⇒ Cc = = 1
