A foot-pound is described as the amount of twisting force applied to a shaft by a perpendicular
1 answer:
You might be interested in
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:
the work done during this cooling is −28.7 kJ
Explanation:
Given data
mass (m) = 1 kg
r = 287 J/kg-K
pressure ( p) = 50 kPa
temperature (T) = 100°C = ( 100 +273 ) = 373 K
to find out
the work done during this cooling
Solution
we know the first law of thermodynamics
pv = mRT ....................1
here put value of p, m R and T and get volume v(a) when it initial stage in equation 1
50 v(a) = 1 × 0.287 × 373
v(a) = 107.051 / 50
v(a) = 2.1410 m³ .......................2
now we find out volume when temperature is 0°C
so put put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1
50 v(b) = 1 × 0.287 × 273
v(a) = 78.351 / 50
v(a) = 1.5670 m³ .......................3
by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.
work done =
integrate it and we get
work done = p ( v(b) - v(a) ) ................4
put the value p and v(a) and v(b) in equation 4 and we get
work done = p ( v(b) - v(a) )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 (−0.574)
work done = −28.7 kJ
here we can see work done is negative so its mean work done opposite in direction of inside air
'