Calculating Volume of Fluid Flowing in Pipe

  • Thread starter Thread starter STEMucator
  • Start date Start date
  • Tags Tags
    Fluid Pipe Volume
Click For Summary
To estimate the volume of fluid flowing through a pipe between 10 seconds and 14 seconds, it is necessary to measure the instantaneous flow rate at three specific times, ideally at 10 s, 12 s, and 14 s. The volume can be calculated using integration, specifically the expression V = ∫(10 to 14) Q(t) dt, where Q(t) represents the flow rate. Utilizing numerical integration methods like Simpson's Rule or the trapezoidal rule can enhance the accuracy of the volume estimation, with Simpson's Rule generally providing better results. Ultimately, measuring the flow at these intervals allows for a more precise calculation of the fluid volume passing through the pipe.
STEMucator
Homework Helper
Messages
2,076
Reaction score
140

Homework Statement



A fluid flows through a pipe. The flow varies with time. We want to estimate the volume of fluid (##V##, in ##L##) that passes through the pipe between time ##t = 10 s## and ##t = 14 s## (i.e. we want to integrate the flow between these times).

The available instrumentation allows us to measure the instantaneous flow rate (in ##L/s##) at any three times of our choosing. We might, for example, decide to measure the flow at ##t = 10 s##, ##t = 12 s##, and ##t = 14 s##. At what three times ##t_1, t_2, t_3## would you choose to measure the flow? Keep in mind that we want to produce the best possible estimate.

Now assume the results of the measurements are ##M(t_1)##, ##M(t_2)##, and ##M(t_3)## in ##L/s##. Give an expression for the volume of fluid that passes through the pipe in the period of interest.

Homework Equations

The Attempt at a Solution



So we want to estimate ##V = \int_{10}^{14} \text{flow}(t) \space dt##.

I think the times ##t_1, t_2, t_3## that have been mentioned in the problem are appropriate, but it obviously can't be that easy right? I can't really see how to break up the interval any better than that.

Then I would want to measure ##\frac{d \text{flow}}{dt}## at ##t_1, t_2, t_3## and call those ##M(t_1)##, ##M(t_2)##, and ##M(t_3)##.
 
Physics news on Phys.org
Zondrina said:

Homework Statement



A fluid flows through a pipe. The flow varies with time. We want to estimate the volume of fluid (##V##, in ##L##) that passes through the pipe between time ##t = 10 s## and ##t = 14 s## (i.e. we want to integrate the flow between these times).

The available instrumentation allows us to measure the instantaneous flow rate (in ##L/s##) at any three times of our choosing. We might, for example, decide to measure the flow at ##t = 10 s##, ##t = 12 s##, and ##t = 14 s##. At what three times ##t_1, t_2, t_3## would you choose to measure the flow? Keep in mind that we want to produce the best possible estimate.

Now assume the results of the measurements are ##M(t_1)##, ##M(t_2)##, and ##M(t_3)## in ##L/s##. Give an expression for the volume of fluid that passes through the pipe in the period of interest.

Homework Equations

The Attempt at a Solution



So we want to estimate ##V = \int_{10}^{14} \text{flow}(t) \space dt##.

I think the times ##t_1, t_2, t_3## that have been mentioned in the problem are appropriate, but it obviously can't be that easy right? I can't really see how to break up the interval any better than that.

Then I would want to measure ##\frac{d \text{flow}}{dt}## at ##t_1, t_2, t_3## and call those ##M(t_1)##, ##M(t_2)##, and ##M(t_3)##.

Unless you are restricted from measuring the flow at all three times, more measurements = better results. If you can provide measurement at equally spaced time intervals, the total flow over the time interval can be computed by applying Simpson's Rule or some other numerical integration method.
 
Only three points are allowed I believe.

After doing some research I found my symbolism to be a bit off. So we want ##V = \int_{10}^{14} Q(t) \space dt##, where ##Q(t) = \frac{dV}{dt}## is the flow rate.

So if we measure ##\frac{dV}{dt}## (aka ##Q(t)##) at the three times mentioned in the problem statement, we obtain ##Q(10) = M(10), Q(12) = M(12), Q(14) = M(14)##. These are the ##y## values in this case.

So an expression for the volume of fluid is given by:

$$V = \int_{10}^{14} Q(t) \space dt = \frac{2}{3} \left[ M(10) + 4M(12) + M(14) \right]$$

Using Simpson's 1/3 rule with ##h = 2##. Alternatively:

$$V = \int_{10}^{14} Q(t) \space dt = \left[ M(10) + 2M(12) + M(14) \right]$$

Using composite trapezoidal integration with ##h = 2##.

I am unsure this is the best estimate, but for now it seems like it might be.
 
The simpson's rule will probably give you a more accurate result than the trapezoidal rule. With only three points, it's rather easy to check the flow with both.
 

Similar threads

Replies
15
Views
3K
Replies
1
Views
4K
Replies
31
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K