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:
total weight of aggregate = 5627528 lbs = 2814 tons
Explanation:
we get here volume of space to be filled with aggregate that is
volume = 2000 × 48 × 0.5
volume = 48000 ft³
now filling space with aggregate of the density that is
density = 0.95 × 119.7
density = 113.72 lb/ft³
and dry weight of this aggregate is
dry weight = 48000 × 113.72
dry weight = 5458320 lbs
we consider here percent moisture is by weigh
so weight of moisture in aggregate will be
weight of moisture = 0.031 × 5458320
weight of moisture = 169208 lbs
so here total weight of aggregate is
total weight of aggregate = 5458320 + 169208
total weight of aggregate = 5627528 lbs = 2814 tons
Answer:
Find the answer in the attached in the order
Option a, Option c and Option b
Explanation:
Answer:
Thermal efficiency of cycle = 0.314
Quality of exhaust steam = 0.959
Explanation:
Answer:
more than 10 joules jjsjdnjdfjnfjfnjfnnfndnndjnfjgngjntjfjmrmdkdmjd