How would I get this python code to run correctly? it's not working.
1 answer:
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The asymptotes of the open loop transfer are:
- Horizontal: y = 0
- Vertical: x = -10 and x = -100
The open loop transfer function is given as:
f(s) = 100(s + 1)/((s + 10)(s + 100))
Set the numerator of the function to 0.
So, we have:
f(s) = 0/((s + 10)(s + 100))
Evaluate
f(s) = 0
This means that, the vertical asymptote is y = 0
Set the denominator of the function to 0.
(s + 10)(s + 100) 0
Split
s + 10 = 0 and s + 100 = 0
Solve for s
s = -10 and s = -100
This means that, the horizontal asymptotes are s = -10 and s = -100
See attachment for the graph of the asymptotes
Read more about asymptotes at:
#SPJ1
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:
q = 0.1086 micro Coulombs
Explanation:
By Coulombs law, we have;
Where;
F = The electric force = 120 mgm
q₁ and q₂ = Charge
r = The separating distance = 30 cm = 0.3 m
k = 8.9876×10⁹ kg·m³/(s²·C²)
Where, q₁ and q₂, we have;
Whereby the force is the force of 120 milligram mass, we have;
0.00012 × 9.81 = 000011772 N
Substituting the values, we have;
q = 0.1086 micro Coulombs.