# Trick Question?

1. Dec 13, 2014

### Zondrina

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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 $M(t_3)$.

2. Dec 13, 2014

### SteamKing

Staff Emeritus
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.

3. Dec 13, 2014

### Zondrina

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.

4. Dec 14, 2014

### SteamKing

Staff Emeritus
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.