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AguaClara / ram_pump

Optimizing the hydraulic ram pump for more efficient use in providing plant plumbing.
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Finding Terminal Velocity #66

Closed cheertsang closed 5 years ago

cheertsang commented 5 years ago

7.35s 15.1416471 L = 2.06008804 L/s 5.85s 10.2L = 1.74358974 L/s 5.99 s 9.9 L = 1.8 L/s ^^ disregard these first three trials (was measured in quarts and also timing method was not precise)

6.09 s 9L = 1.47783251 L/s 5.72 s 9L = 1.57342657 L/s 5.63 s 8.6L =1.52753108 L/s 6.11 s 9.1L = 1.4893617 L/s 5.91 s 9L = 1.52284264 L/s

Average of the last five trials = 1.518199 L/s Diameter of drive pipe = 2.372 cm

Flow rate = velocity * area

cheertsang commented 5 years ago 
flow_rate = 1.518199*u.L/u.s
area = pc.area_circle(2.372*u.cm)
velocity = (flow_rate/area).to(u.m/u.s)

3.4356542382769817 meter/second

cheertsang commented 5 years ago

@monroews We were trying to measure the terminal velocity of the water in the drive pipe experimentally by measuring the flow rate and dividing by area of the drive pipe, but we were doing this without the ram pump and plate attached to the drive pipe. @mailemccann brought up the point that we should be measuring it with the ram pump and plate attached, to account for the headloss caused by the plate. Does it make more sense to measure terminal velocity with the ram pump and the plate attached?

monroews commented 5 years ago

You need to know the velocity of the water in the drive pipe when you slam the waste valve shut. Therefore you need to measure the velocity of the water with the ram pump attached and the valve in the same exact position as you will be using when you operate the ram pump.

cheertsang commented 5 years ago

250 mL/ 0.37 s = 0.675675676 L/s 630 mL/0.68 s = 0.926470588 L/s 620 mL/ 0.68 s = 0.911764706 L/s 480 mL/ 0.52 s = 0.923076923 L/s 560 mL/ 0.66 s = 0.848484848 L/s 780 mL/ 0.81 s = 0.962962963 L/s

Average of last 5 trials: 0.914552006 L/s

flow_rate = 0.914552006*u.L/u.s
area = pc.area_circle(.02372*u.m)
velocity = (flow_rate/area).to(u.m/u.s)

Terminal velocity = 2.0696130583267505 meter/second

monroews commented 5 years ago

The biggest error here is the time. Can you measuring larger volumes so you can have a sample time of a couple of seconds at least?

cheertsang commented 5 years ago

4.055 L/ 3.98 s 4.670 L/ 4.68 s 3.880 L/ 3.93 s 3.565 L/ 3.60 s 3.790 L/ 3.83 s

Q1=4.055*u.L/(3.98*u.s)
Q2=4.670*u.L/(4.68*u.s)
Q3=0.880*u.L/(3.93*u.s)
Q4=3.565*u.L/(3.60*u.s)
Q5=3.790*u.L/(3.83*u.s)

avg_Q = (Q1+Q2+Q3+Q4+Q5)/5

velocity2 = (avg_Q/area).to(u.m/u.s)

terminal velocity = 2.2556565988848085 meter/second