Answer:
v = 40 m/s
Explanation:
Difference in the height of the two columns:
Air density,
Water density,
The velocity of air flow is given by the equation:
The pressure forces on a submersed object will be (A)- Tangential to the objects body (B)- Parallel (C)- Normal (D)- None of the
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:
Given that
LHS of above given equation have dimension .
Now find the dimension of RHS
Dimension of P = .
Dimension of d= .
Dimension of μ = .
Dimension of L= .
So
It means that both sides have same dimensions.