Answer:
The answers and explanations are shown in the attachment
Explanation:
The step by step approach is as shown in he attachment.
Answer:
v₂ = 1.25 m/s
Explanation:
given,
inlet velocity, v₁ = 5 ft/s
inlet Area = A = 10 ft²
outlet velocity, v₂ = ?
outlet Area = 4 A
= 4 x 10 = 40 ft²
Using continuity equation
A₁ v₁ = A₂ v₂
10 v₁ = 40 v₂


v₂ = 1.25 m/s
Hence, the exit velocity of the water through the duct is equal to 1.25 m/s.
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Assessing risks. ...
Preventing or controlling risks. ...
Control measures. ...
Mitigation. ...
Preparing emergency plans and procedures. ...
Providing information, instruction and training for employees. ...
Places where explosive atmospheres may occur ('ATEX' requirements)
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Answer:
Hello there, please check step by step explanations to get answers.
Explanation:
Given that:
5-hp single-cylinder engine is used. At most, the belt transmits 60 percent of this power. The driving sheave has a diameter of 6.2 in. and the driven, 12.0 in. The belt selected should be as close to 92 in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt, and specify it using the standard designation.
See attached documents for clearity and step by step procedure to answer
Answer:
a) 47500 mg/L
b) 5250366.444 kg/year
Explanation:
Given data:
suspended solids removal efficiency = 20%
Flowrate in the primary clarifier ( Q ) = 20 MGD ( change to Liters/day
Q = 20* 10^6 * 3.785412 Liters /day
settled concentration ( St ) = 950mg/L * 0.2 = 190 mg/L
amount of settled solid = Q * St
= ( 20* 10^6 * 3.785412 ) * 190 = 14384.5656 kg/day
∴ Amount going into sludge with a flowrate of 0.08 MGD = 14384.5656 kg/day
<u>a) concentration of solid in sludge ( leaving the clarifier )</u>
= amount of settled solid / flow rate out of the clarifier in liters/day
= 14384.5656 / ( 0.08 * 10^6 * 3.785412 )
= 0.0475 kg/L
= 47500 mg/L
<u>b) Determine mass of solids that is removed annually </u>
= 14384.5656 kg/day * 365 days
= 5250366.444 kg/year