What does snow fall from?
1 answer:
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Answer:
a) 35%
b) yes it can be improved by moving the tray near the top
Tray should be located ( 1 to 2 meters below surface )
max removal efficiency ≈ 70%
c) The maximum removal will drop as the particle settling velocity = 0.5 m/h
Explanation:
Given data:
flow rate = 8000 m^3/d
Detention time = 1h
depth = 3m
Full length movable horizontal tray : 1m below surface
<u>a) Determine percent removal of particles having a settling velocity of 1m/h</u>
velocity of critical sized particle to be removed = Depth / Detention time
= 3 / 1 = 3m/h
The percent removal of particles having a settling velocity of 1m/h ≈ 35%
<u>b) Determine if the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray and the maximum removal efficiency</u>
The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the
Total Maximum removal efficiency
= percent removal + percent removal
= ( d,v
) .
+ ( d
,v
) .
= 100
hence max removal efficiency ≈ 70%
<u>c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?</u>
The maximum removal will drop as the particle settling velocity = 0.5 m/h
Answer:
a) y = -0.05*x^2
b) Range shorter, Radius of circular/spiral motion shorter
Explanation:
Given:
- The charge q = 1 C
- Magnetic Field = 1 k Tesla
- Mass of the charge m = 1 kg
- Acceleration due to Lorentz Force a = q ( v x B ) / m
- Acceleration due to Drag Force a = - c*v
- The velocity of the charge @ x = y = 0, v_i = 10 i
Find:
- Equation of path of the charge in the given magnetic field
- Describe the path with addition of Drag Force.
Solution:
- First we will determine the magnitude and direction of the acceleration due to Lorentz Force as follows:
a = q ( v x B ) / m
- Input values:
a = 1 ( 10 i x 1 k ) / 1 = -10 j
- We see that the acceleration acts in the negative y direction.
- To determine the path of the charge, we will use kinematic equation of motion for the particle as follows:
x = v_i*t
y = 0.5*a*t^2
- Determine the parametric equations for displacements in x and y directions:
x = 10*t
y = -0.5*10*t^2
Combine the parametric equations to determine the equation of path followed by the charge:
t = x / 10
y = -0.5*x^2 / 100
y = -0.05*x^2
- For the second case we have the acceleration due to drag force relation:
a = - 0.1*v_i
a = -0.1*10 i
a = - i
- There is a deceleration component of drag force acting in x direction, The parametric equation would be as follows:
x = 10*t - 0.5*1*t^2
x = 10*t - 0.5*t^2
- From this we can see that the x coordinate increases at a decreasing rate. Hence, the range of the projectile like motion will decrease. The total amount of distance traveled by the charge in x direction will decrease.
-Hence, after sufficient amount of time the charge moves in a circular motion. The radius of the circular motion will be shorter as compared to when we neglect the acceleration due to drag force.
