Answer: The answer should be space. I didn't know Chemical Engineers convert space. Hope this helps!
Explanation:
Term specifically describes small chunks of rocks and debris in space that burn up in Earth’s atmosphere is :
Meteors
Explanation:
- A meteor is a meteoroid or a particle broken off an asteroid or comet orbiting the Sun – that burns up as it enters the Earth's atmosphere, creating the effect of a "shooting star".
- Meteoroids that reach the Earth's surface without disintegrating are called meteorites.
- Due to Earth's escape velocity, the minimum impact velocity is 11 km/s with asteroid impacts averaging around 17 km/s on the Earth. The most probable impact angle is 45 degrees.
- Meteoroids have a pretty big size range. They include any space debris bigger than a molecule and smaller than about 330 feet space debris bigger than this is considered an asteroid.
- But most of the debris the Earth comes in contact with is "dust" shed by comets traveling through the solar system.
- The surface of a meteorite is generally very smooth and featureless, but often has shallow depressions and deep cavities resembling clearly visible thumbprints
- Most iron meteorites, like the example at right, have well-developed regmaglypts all over their surface.
Answer:
P=3.31 hp (2.47 kW).
Explanation:
Solution
Curve A in Fig1. applies under the conditions of this problem.
S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt
The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.
32.2
Fig. 32.2 Dimension of turbine agitator
The Reynolds number is calculated. The quantities for substitution are, in consistent units,
D a =2⋅ft
n= 90/ 60 =1.5 r/s
μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s
ρ = 93.5 lb/ft3 g= 32.17 ft/s2
NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600
From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c
The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).