From Flow Rates to Changes in Volume
A flow rate describes how quickly the amount of a substance changes. In a tank, reservoir, or pipeline system, material may enter and leave at the same time. To estimate the change in stored volume, compare the inflow rate with the outflow rate over each time interval.
The central quantity is the net flow rate: inflow minus outflow. A positive net rate adds volume to the system; a negative net rate removes volume. When rates vary with time, a rectangle estimate treats a sampled rate as representative of its interval, multiplies by the interval’s duration, and adds the signed contributions.
For an interval of duration \(\Delta t\), a sampled net rate \(g(t^*)\) gives the rectangle estimate \(g(t^*)\Delta t\) for the volume change on that interval. The product has units of volume because the time units cancel.
Here, the time period is divided into \(n\) intervals, \(\Delta t_i\) is the duration of interval \(i\), and \(t_i^*\) is the selected sample time in that interval. This is an estimate when the rates change within the intervals. If the rates are constant on each interval, each rectangle gives the exact volume change for that interval.
A Practical Setup
For each time interval, find the net rate at the chosen sample time, multiply it by that interval’s own duration, and record the signed volume change. Then add the interval contributions. If an initial volume is given, add the estimated net change to it to obtain an estimated final volume.
Inflow and outflow rates must use compatible volume-per-time units. Convert them first if they do not.
Use the sample time specified in the problem or select a representative time that matches the available data.
Subtract outflow from inflow, then multiply by the interval’s duration. Keep negative contributions negative.
Add all the contributions, then add the initial volume if the question asks for a final amount.
The intervals need not have equal durations. An interval lasting three hours contributes three times as much as an interval lasting one hour when both have the same net rate. Ignoring that difference changes the estimate and usually gives the wrong units.
Worked Examples
Worked Example: A Tank with Midpoint Rate Data
A tank initially contains \(120\) L. During the next \(6\) min, divide the time into three \(2\)-min intervals. At the midpoint of each interval, the measured inflow and outflow rates are shown below. Estimate the net volume change and the final volume.
| Interval (min) | Midpoint (min) | Inflow (L/min) | Outflow (L/min) |
|---|---|---|---|
| \([0,2]\) | 1 | 14 | 5 |
| \([2,4]\) | 3 | 11 | 6 |
| \([4,6]\) | 5 | 9 | 8 |
For the first interval, the sampled net rate is \(14-5=9\) L/min. Its estimated contribution is \(9\cdot2=18\) L.
For the second interval, the sampled net rate is \(11-6=5\) L/min. Its estimated contribution is \(5\cdot2=10\) L.
For the third interval, the sampled net rate is \(9-8=1\) L/min. Its estimated contribution is \(1\cdot2=2\) L.
Adding the estimated change to the initial volume gives \(120+30=150\) L. The estimate is consistent with the positive net rate on all three intervals: the tank gains volume throughout this period.
Check: Each contribution is a rate in L/min multiplied by \(2\) min, so each is in litres. The contributions sum to \(30\) L, and \(120\text{ L}+30\text{ L}=150\text{ L}\).
Worked Example: Unequal Time Intervals
A reservoir contains \(500\) m³ at time \(0\). Rate measurements are taken at the left endpoint of each interval: \([0,1]\), \([1,4]\), and \([4,6]\) hours. Estimate the net volume change over the six hours and the final volume.
| Interval (h) | Left sample (h) | Inflow (m³/h) | Outflow (m³/h) |
|---|---|---|---|
| \([0,1]\) | 0 | 20 | 7 |
| \([1,4]\) | 1 | 18 | 8 |
| \([4,6]\) | 4 | 12 | 10 |
The interval widths are \(1\), \(4-1=3\), and \(6-4=2\) hours. The sampled net rates are \(20-7=13\), \(18-8=10\), and \(12-10=2\) m³/h.
The estimated contributions are \(13\cdot1=13\) m³, \(10\cdot3=30\) m³, and \(2\cdot2=4\) m³. Therefore, the estimated net change is \(13+30+4=47\) m³.
Check: The three durations add to \(1+3+2=6\) hours, the stated time period. Each rate-duration product is in m³, and adding the net gain to the initial volume gives \(547\) m³.
Worked Example: Including an Interval of Net Outflow
A system initially contains \(50\) m³. On the interval \([0,2]\) min, its midpoint inflow and outflow rates are \(10\) and \(4\) m³/min. On \([2,4]\) min, the midpoint rates are \(5\) and \(7\) m³/min. Estimate the final volume.
On the first interval, the net rate is \(10-4=6\) m³/min. Over \(2\) min, the estimated change is \(6\cdot2=12\) m³.
On the second interval, the net rate is \(5-7=-2\) m³/min. The estimated change is \((-2)\cdot2=-4\) m³. The negative sign indicates a net loss on this interval.
The estimated final volume is \(50+8=58\) m³. The second interval removes \(4\) m³, but it does not erase the \(12\) m³ gained on the first interval.
Check: The signed contributions total \(8\) m³, and \(50\text{ m}^3+8\text{ m}^3=58\text{ m}^3\). The estimate remains positive, as expected for the stated initial volume and changes.
Why the Rectangle Calculation Works
First consider a time interval on which both flow rates are constant. If the inflow is \(I\) volume-units per unit time and the outflow is \(O\) volume-units per unit time, then over a duration \(\Delta t\), the amount entering is \(I\Delta t\) and the amount leaving is \(O\Delta t\). The change in stored volume is the amount entering minus the amount leaving:
This is exactly the rectangle calculation using the constant net rate \(I-O\). If the rates vary during the interval, a sampled value stands in for the changing rate. The product is then an estimate rather than an exact change, unless the chosen sample has a special property that makes the rectangle exact.
Across several intervals, the total change is the sum of the changes on the separate intervals. This follows because the end volume of one interval is the starting volume of the next: intermediate volumes cancel when the successive changes are combined. Thus, adding the signed rectangle contributions estimates the overall change. Adding that change to the initial volume estimates the final volume.
Choosing Samples and Reading the Estimate
Use the sample times the problem provides. If you must choose them, use data that represent the whole interval rather than selecting a rate merely because it makes the arithmetic convenient. Midpoint samples are often useful when rates are measured at interval centers or when a representative central value is appropriate. Left- and right-endpoint samples can be used when the data are given at those endpoints.
When the net rate is known to increase throughout an interval, a left-endpoint rectangle uses rates smaller than later rates, while a right-endpoint rectangle uses rates larger than earlier rates. For a decreasing net rate, these directions reverse. It is the behavior of the net rate that matters for the net-volume estimate: an increasing inflow alone does not determine whether the net rate increases if the outflow also changes.
For real data, an estimate should be reported with its units and with the sampling method identified when that information matters. If a problem gives only a few rate measurements, the rectangle calculation does not reveal every fluctuation between them. The result is an estimate based on the stated sampling choices, not a claim that the rates stayed constant.
A Common Mistake: Forgetting Interval Widths
In the unequal-interval example, one incorrect approach is to add the sampled net rates directly: \(13+10+2=25\). This is not a volume estimate. The rates have units of m³/h, so their sum still has units of m³/h; no elapsed time has been included.
The correction is to multiply each net rate by its own interval width: \(13\cdot1=13\) m³, \(10\cdot3=30\) m³, and \(2\cdot2=4\) m³. Then add the contributions to obtain \(13+30+4=47\) m³. Using one common width for all three intervals would also be incorrect because their durations differ.
Check Your Understanding
For each question, keep track of the sign of the net rate and the duration of every interval.
- A pipe brings water in at \(12\) L/min and a drain removes water at \(5\) L/min for \(3\) min. What is the net volume change if both rates are constant?
- A tank has midpoint rates \(8\) L/min inflow and \(11\) L/min outflow on a \(2\)-min interval. What signed volume contribution does the rectangle give?
- Why must unequal time intervals use their own widths in the calculation?
- Rate measurements at the left endpoints of intervals \([0,2]\) and \([2,5]\) are \(6\) and \(4\) m³/h for the net rate. What is the estimated net change?
- If the estimated net change is negative, what does that say about the final volume relative to the initial volume?
