The inlet for a high-bypass-ratio turbofan engine has an area A1 of 6.0 m 2 and is designed to have an inlet Mach number M~ of 0
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:
all are zero.
Explanation:
We know that if flow is possible then it will satisfy the below equation
Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.
And velocity potential function given as follows
Rotationality of fluid is given by
So now putting value in the above equations ,we will find
So =0
Like this all all are zero.
That is why velocity potential flow is irroational flow.
Answer:
864 KN
Explanation:
Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.
Please kindly check attachment for the step by step solution of the given problem.
