Volume of water flowing out a cylindrical pipe

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SUMMARY

The volume of water flowing out of a cylindrical pipe can be calculated using the formula V = v * A, where A is the cross-sectional area (A = π * r²) and v is the flow rate. For a flow rate of 3 cm/sec, the volume is V = 3πr² cm³/sec, and for a variable flow rate k cm/sec, the volume is V = kπr² cm³/sec. Graphing these equations reveals that V as a function of r produces a parabolic curve, while V as a function of k results in a linear relationship. It is essential to note that both r and k must be greater than or equal to zero.

PREREQUISITES
  • Understanding of the formula for the volume of a cylinder (V = π * r² * h)
  • Basic knowledge of graphing functions (parabolas and linear functions)
  • Familiarity with units of measurement (cm/sec and cm³/sec)
  • Concept of flow rate in fluid dynamics
NEXT STEPS
  • Learn about fluid dynamics principles and equations
  • Explore graphing techniques for quadratic and linear functions
  • Study the implications of variable flow rates on volume calculations
  • Investigate real-world applications of cylindrical flow in engineering
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Students in physics or engineering, fluid dynamics enthusiasts, and anyone interested in mathematical modeling of fluid flow through cylindrical structures.

Jacobpm64
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Water is flowing down a cylindrical pipe of radius r.

(a) Write a formula for the volume, V, of water that emerges from the end of the pipe in one second if the water is flowing at a rate of

(i) 3 cm/sec (ii) k cm/sec

(b) Graph your answer to part (a)(ii) as a function of
(i) r, assuming k is constant
(ii) k, assuming r is constant



All right, I just don't know where to start at all... I'm thinking that I'll probably need the formula for the volume of a cylinder. Besides that, I don't know what I have to do with the formula. I'm completely clueless on how to implement the rates of water flow into the volume of a cylinder formula.. if that's even what I have to do... I'm sure that if I can figure out part (a), part (b) won't be much of a problem. (just solving for different variables and graphing)
 
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The cross area of a cylinder equals A = r^2*PI, so the volume that flows through the cylinder in one second equals V = v*A [m^3/sec], where v is the speed of water.
 
you said the volume that flows through is in m^3/sec... does that mean I have to convert to m/sec, or will the rates work if i keep them as cm/sec?

Should work as cm/sec, right?

After you respond to this.. I'm going to give it a shot, and you can let me know how I'm doing.

Thanks a lot.
 
Nope, you don't have to convert anything, of course it works perfectly well with cm^3/sec, too. I just forgot that the speed was given in cm/sec.
 
ok, let's see if i did it correctly

(a) (i) V = 3pi*r2 (where V is in cm^3/sec)
(ii) V = kpi*r2 (where V is in cm^3/sec)

(b) (i) V on the vertical axis, r on the horizontal axis. It's just a parabola .. it'll look like x2.
(ii) V is on the vertical axis, k is on the horizontal axis. It's just a linear function with a positive slope.

How's this? (and is there any way to be more specific with the graphs, like to plot out any points, even if the points are just defined as variables?)
 
Jacobpm64 said:
How's this? (and is there any way to be more specific with the graphs, like to plot out any points, even if the points are just defined as variables?)

Looks great.

The graphs are perfectly ok, and I don't see a way to be more specific than you already are, except maybe to mention that for V(r) it makes sense if r > 0, since the cylinder radius can't be negative. As for V(k), k should be greater or equal than zero, since a negative volume doesn't make any sense.
 
thanks a lot
 

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