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:
fluid power
Explanation:
fluids commonly used in fluid power are Oil, Water, Air, CO², and Nitrogen gas, fluid power is commonly confused with hydraulic power, which only uses liquids, fluid power uses either liquids or gases
Answer:
radius = 0.045 m
Explanation:
Given data:
density of oil = 780 kg/m^3
velocity = 20 m/s
height = 25 m
Total energy is = 57.5 kW
we have now
E = kinetic energy+ potential energy + flow work
![E = \dot m ( \frac{v^2}{2] + zg + p\nu)](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%20%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p%5Cnu%29)
![E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p_%7Batm%7D%20%5Cfrac%7B1%7D%7B%5Crho%7D%29)

solving for flow rate
![\dot m = 99.977we know that [tex]\dot m = \rho AV](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%2099.977%3C%2Fp%3E%3Cp%3Ewe%20know%20that%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cdot%20m%20%20%3D%20%5Crho%20AV)

solving for d

d = 0.090 m
so radius = 0.045 m