Storm Water Detention Tank with Two Orifices for Outflow

Classical Mechanics Level pending

A storm water detention tank collects storm water from a paved area at a maximum, constant rate of 5 m^3/s at t=0 seconds. The tank is 10 meters high and has an inside base area of 5 m by 5 m. The 1st orifice has a diameter of 50 cm and is 1 m above the tank base. The 2nd orifice has a diameter of 0.75 m and is 3.5 m above the tank base. What is the range from the tank the water from the 2nd orifice meets with the ground at t=30 seconds? Assume steady flow occurs within the tank and neglect any weir flow from orifices before orifice flow fully develops. The ground surrounding the tank is the same elevation as the inside of the tank bottom. Take g=9.81 m/s^2, and Cd=0.62 (entrance coefficient). Please provide your answer to the nearest 1/10th of a meter.


The answer is 2.7.

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2 solutions

Steven Chase
Mar 6, 2021

This was a fun problem. I ran a time simulation using Toricelli's Law and some conditional statements. Results are printed at the end. Water flows through an orifice if the water level is higher than the orifice. And then the flow speed and flow rate depend on the water level relative to the orifice center.

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import math

# constants

dt = 10.0**(-5.0)

Qin = 5.0

H = 10.0
Abase = 25.0

D1 = 0.5
y1 = 1.0

D2 = 0.75
y2 = 3.5

tf = 30.0

g = 9.81
Cd = 0.62

##################################

# derived constants

R1 = D1/2.0
R2 = D2/2.0

A1 = math.pi*(R1**2.0)
A2 = math.pi*(R2**2.0)

##################################

t = 0.0
y = 0.0             # water level

while t <= tf:

    if y > y1:
        v1 = Cd*math.sqrt(2.0*g*(y - y1))
        Q1 = A1*v1
    else:
        Q1 = 0.0

    if y > y2:
        v2 = Cd*math.sqrt(2.0*g*(y - y2))
        Q2 = A2*v2
    else:
        Q2 = 0.0

    Qout = Q1 + Q2

    Qnet = Qin - Qout

    # Qnet*dt = Abase*dy

    dy = Qnet*dt/Abase

    y = y + dy

    t = t + dt

####################################

print y
print ""

# y2 = 0.5*g*(t2**2.0)

t2 = math.sqrt(2.0*y2/g)

xf = v2*t2

print xf

#>>> 
#4.83150056172     # final water depth

#2.67686345291     # range from orifice 2
#>>> 

@Joe Byrne

A few questions about the new spring problem.

1) Are the springs un-stretched just prior to falling?
2) For the final distance from the right support, is this the total distance (accounting for both horizontal and vertical components)?

Steven Chase - 3 months ago

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Hey Steven, yes, the springs are un-stretched (I'll add note to clarify) and the distance is the resultant from x and y components.

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OK, thanks. I made those assumptions and it is not liking my answer. I will double-check my work

Steven Chase - 3 months ago

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@Steven Chase I probably made a mistake. I'll take down and check. Let you know what happened. Sorry.

@Steven Chase did you get 6.5628?

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@A Former Brilliant Member Assuming the left support is at (0,0) and the right support is at (6,0), I ended up with the mass at (2.2623, -6.1605), for a distance of about 7.2 from the right support.

Steven Chase - 3 months ago

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@Steven Chase Yes, that's right. I had the answer for the left support and not the right. Sorry, geez. I'll edit for the right support

@Steven Chase Corrected it. Sorry for the mishap. Hate it when that happens. It's still the right support and the distance is 7.2. I'm curious to see your program to do this.

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@A Former Brilliant Member No worries. I will post a solution in an hour or two. Interestingly, there are two equilibrium points. Thanks

Steven Chase - 3 months ago

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@Steven Chase No kidding, that is interesting.

Thanks to @Steven Chase and @Karan Chatrath for their insight on this problem.

Thank you for reposting the problem. This was fun

Karan Chatrath - 3 months, 1 week ago

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Glad you had fun! If you can find a solution for the DEQ in the first and second intervals(1.0 <h<3.5 and 3.5 < h < 4.8), will you please let me know. Thanks!

A Former Brilliant Member - 3 months, 1 week ago

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