Answer:
Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally
updated information on model changes that do not affect the technician. Who is correct? the answer is c
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:
Shut off cylinder valve before turning off forklift.
Explanation:
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:
The formula should be 234.1 x (1 + 2.9%)^t
Explanation:
As the questions indicates, the CPI indicates what the value for money is in that given year compared to the base year. So a higher CPI amount is given as a result of inflation over the years. So the rate of inflation was such that the 100 dollars in 1983 (taken as the base year) is equivalent to 234.1 in 2015 as a result of inflation. So, essentially, the time value of money concept is being applied here. We can use the formula for calculating the future value of money over here as well.
The equation is: FV = PV × (1 + x%)^t
where, FV is future value, PV is the present value today, x% is the rate of change, and t is the time period in years.
With the values at hand, the formula to compute what the CPI should be t years after 2015,
CPI (t years after 2015) = 234.1 x (1.029)^t
