Water Flow, Mass Flow through 5 Gallon Jug

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SUMMARY

The discussion focuses on calculating the minimum mass flow rate for a water system using a 5-gallon jug, aiming for a flow rate between 1 GPM and 4 GPM. The user calculated the velocity of water at atmospheric pressure to be 122.95 inches/sec, translating to 614.7 ft/min. Using the equation Q=VA, the required cross-sectional area was determined to be approximately 0.03024 in², leading to a minimum inner diameter of 0.19622 inches for the outlet. The user also noted that variations in water height would affect mass flow, suggesting the need for a differential equation (DE) to model the system accurately.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically flow rate calculations.
  • Familiarity with the equation of continuity (Q=VA) in fluid mechanics.
  • Basic knowledge of unit conversions between imperial and SI units.
  • Ability to apply differential equations to varying flow scenarios.
NEXT STEPS
  • Research "Fluid Dynamics and Flow Rate Calculations" for deeper insights into flow behavior.
  • Learn about "Differential Equations in Fluid Mechanics" to model varying mass flow rates.
  • Explore "Unit Conversion Techniques" for seamless transitions between measurement systems.
  • Investigate "Bernoulli's Principle" to understand pressure and velocity relationships in fluid flow.
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Engineers, fluid dynamics students, and anyone involved in designing water flow systems or optimizing fluid transport in practical applications.

bhaarat316
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Here is the run down first, I'm trying to find a min mass flow rate for my project which needs to be 1 GPM, and max 4 GPM. Now water is going out of a 5 gallon jug, that has the top cut off, and the water is flowing into a PVC pipe. The area is already set at 1.5 inches, which I know is to big. I need to find the velocity of water at atm pressure, and then the diameter of the circle which the water will funnel through. I did some work I just need help making sure I did it right, basically a check over. think of it as a Deer Park 5 gallon bottle, like the ones for water dispenser, but the top cut off, and it flowing out the nozzle.

Man how do I use the Latex thing?

So, I know V=\sqrt{2*19.5 inches * 387.6 inches/s^{2}}
V=122.95 inches/sec, 10.245 ft/s = 614.7 ft/min

Now the simple equation of Q=VA

1 GPM->.13ft^{3}/min = 614.7ft/min * A
.03024 in^2 = A(min)
sqrt(.03024/∏)= r(min) = .09811 inches

We would have to reduce our inner diameter to .19622 inches.
Now is this right? I can't believe that? I was initial thinking it would be a DE since our mass flow would be varying, depending the height of the water, gravity would be pushing it through the main system.
 
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A sketch would help to understand your calculations.
In addition, SI units would make unit conversions much easier.

I was initial thinking it would be a DE since our mass flow would be varying, depending the height of the water, gravity would be pushing it through the main system.
Based on your description of your system, I think this is true. Probably your 19.5 inches would vary.
 
mass flow.jpg


gravity will force the water through the tank and I need to figure out what the diameter will be, to get 1 gpm.
 

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