Answer:
A selective surface with large absorption for solar radiation and high reflectance for thermal infrared radiation was produced by use of surface oxidation of stainless steel. The surfaces were studied for use with concentrated light in a solar power plant at temperatures of 400°C and higher.
In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.
Sample surfaces were subjected to long temperature ageing tests in order to gain some idea of the thermal stability of the surfaces. The results promise better-performing surface and the production of durable selective finishes at, possibly, lower cost than competing processes.
Explanation:
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
Answer:
//The program prompts user to input three integers and it displays them, adds and gets their average
//begin
public class Test
{
public static void Main()
{
//input intergers
int[] score = new int[3];
int avg,rem,sum = 0;
for(int i=0;i<3;i++)
{
Console.WriteLine("Enter an integer score ");
score[i] = Convert.ToInt32(Console.ReadLine());
sum = sum + score[i];
}
avg = sum/3;
rem = sum%3;
Console.WriteLine("The average of "+score[0]+","+score[1]+","+score[2]+" is "+avg +" with a remainder of "+rem);
}
}
Answer:
your answer is correct
Explanation:
You have the correct mapping from inputs to outputs. The only thing your teacher may disagree with is the ordering of your inputs. They might be written more conventionally as ...
A B Y
0 0 1
0 1 0
1 0 0
1 1 1
That is, your teacher may be looking for the pattern 1001 in the last column without paying attention to what you have written in column B.
