Answer:
Some general principles are given below in the explanation segment.
Explanation:
Sewage treatment seems to be a method to extract pollutants from untreated sewage, consisting primarily of domestic sewage including some solid wastes.
<u>The principles are given below:</u>
- Unless the components throughout the flow stream become greater than the ports or even the gaps throughout the filter layer, those holes would be filled as either a result of economic detection.
- The much more common element of filtration would be the use of gravity to extract a combination.
- Broadcast interception or interference.
- Inertial influence.
- Sieving seems to be an excellent method to distinguish particulates.
Answer and Explanation:
Flame Front Generator: It is a ignition system which is very useful in flaring system .In this system the air and gases are mixed together and make a combustible air gas mixture. There is a flame front region where the combustion reaction takes place , it is the region where gases as like hydrogen and air mixed with each other and form combustible gases.
Answer:
C. underground road
Explanation:
Generally compound curves are not filtered and recommended for use in an underground road. However, they are best used in the road, water way, and rail way.
Given:
diameter of sphere, d = 6 inches
radius of sphere, r = 
= 3 inches
density, 
= 493 lbm/ 
S.G = 1.0027
g = 9.8 m/ 
= 386.22 inch/ 
Solution:
Using the formula for terminal velocity,
= 
(1)
where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A = 
Volume of sphere, V = 
Using the above formulae in eqn (1):
= 
= 
= 
Therefore, terminal velcity is given by:
= 
inch/sec
Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q = 
<em>d</em>T
= (T₂ - T₁)
From the above equations, the underlying assumption is that remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let 
be heat constant of calorimeter
Q₂ = ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m ' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹
