Answer:
The size of equalization basin is 6105.6 m³
Explanation:
The average flow is:
flow = ∑flow/n = 9.788/24 = 0.408 m³/s
Where n is the number or observations.
The inflow volume is:
where t is the time interval
in the same way it is calculated the inflow volume for each observation
The outflow volume is:
The volume of flow is:
in the same way it is calculated the volume of flow for each observation. According to the file attach, the highest volume is 6105.6 m³
Answer:
The third digit of the student ID = 5;
Explanation:
The Array is initially empty at the starting point.
ex) if x = 2, it is enqueue dequeue enqueue dequeue
Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084