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Flow rate

Water is running from a tap into a bath. How do you say how fast it is coming, and use that to know when the bath will be full?

You need
Liquid volume: liters and millilitersGrade 3Units of timeGrade 4
Where you are 2 in Rates from measurements
Measurement12
Rates from measurements2
Flow rate
Density

Why this exists

Turn on a tap over an empty bath and you can watch the water rise, but watching does not tell you when to turn it off. To plan, you need one number that says how fast the water is coming. “Fast” here is not how quickly the water moves through the pipe. It is how much water arrives: how many liters land in the bath in each minute.

That number is the flow rate, and it answers a whole family of practical questions. How long will a hose take to fill a pool? How much water does a garden’s drip line deliver overnight? How long must a pump run to clear a flooded basement? Each has two measured amounts, a volume of liquid and a length of time, and the flow rate is what ties them together.

It also works in reverse. You rarely see a liter of water go past; you see a full bucket, a meter reading or a stopwatch. Dividing the volume you collected by the time it took turns those into the flow rate, and the flow rate then predicts every other volume and every other time for that same flow.

Definition

The flow rate of a liquid is the volume of liquid that passes a point in one unit of time. You find it by dividing a volume by the time it took to pass:

\text{flow rate} = \frac{\text{volume}}{\text{time}}

With letters, using r for the rate, V for the volume and t for the time,

r = \frac{V}{t}

Its unit is a volume unit per time unit, where “per” means “in each”. A tap that delivers 12 liters in each minute has a flow rate of 12 liters per minute, written 12\ \text{L/min}. A drip that releases 2 milliliters each second has a flow rate of 2\ \text{mL/s}. Liters per hour (\text{L/h}) and liters per second (\text{L/s}) are common too.

A flow rate is a unit rate: it is the volume that goes with exactly one unit of time. In this lesson the flow is steady, meaning the same volume passes in every equal stretch of time. The Boundaries section below says what changes when it is not.

Intuition

Lay minutes and liters side by side on a double number line, so that each time sits directly above the volume that has poured out by then.

0
0
1
12
2
24
3
36
4
48
5
60
\text{minutes}
\text{liters}
Figure 1A tap running at 12 liters per minute. Every minute adds the same 12 liters.

Because the flow is steady, every minute brings the same 12 liters. Two minutes bring twice as much as one, five minutes five times as much. The volume poured is proportional to the time, and the constant that multiplies the time is the flow rate itself:

V = 12t

A faster tap has a bigger constant: more liters to each minute, so the numbers under the minute marks grow faster. A slow drip has a tiny one.

Notice that the flow rate belongs to the flow, not to whatever it is filling. The same tap fills a bucket and a pool at the same 12 liters each minute. The pool only takes longer because it holds more.

Results

One relationship, three forms. For a steady flow, the flow rate r, the volume V and the time t are tied together by

V = rt \qquad r = \frac{V}{t} \qquad t = \frac{V}{r}

Use the first to find how much flows in a given time, the second to find the rate from a measurement, and the third to find how long a given volume takes. All three need the units to match: a rate in liters per minute goes with a volume in liters and a time in minutes.

Changing the units. There are 60 seconds in a minute, 60 minutes in an hour and 1{,}000 milliliters in a liter, so

\begin{aligned} 1\ \text{L/s} &= 60\ \text{L/min} \\ 1\ \text{L/min} &= 60\ \text{L/h} \\ 1\ \text{L/min} &= 1{,}000\ \text{mL/min} \end{aligned}

A volume in cubic meters turns into liters the same way: a cube 10 cm along each edge holds exactly one liter, and a cube 1 m along each edge holds 10 \times 10 \times 10 of those, so

1\ \text{m}^3 = 1{,}000\ \text{L}

Changing the units of a rate treats these changes in general.

Flows at the same time combine. When two flows run into the same container at the same time, their rates add. When one flow runs in and another runs out, the container gains the difference:

\text{net rate} = \text{rate in} - \text{rate out}

Times do not add. Two hoses that each take an hour do not take two hours together, or one; they take half an hour, because together they deliver twice as much each minute.

Why it is true

Why V = rt. Suppose a steady flow delivers r liters each minute. After 1 minute, r liters have passed; after 2 minutes, r + r; after t whole minutes, t lots of r, which is rt. For part of a minute the same holds: split a minute into equal parts and, because the flow is steady, each part carries an equal share of the r liters. Half a minute carries \displaystyle \frac{1}{2}r, a quarter of a minute carries \displaystyle \frac{1}{4}r, and so on. So V = rt for every time t, whole or not.

The other two forms say the same thing. Multiplication and division undo each other, so V = rt can be read as r = V \div t or as t = V \div r, provided the number you divide by is not zero.

Why rates add. In any one minute, the first hose puts in r_1 liters and the second puts in r_2 liters, and the container receives both. So it gains r_1 + r_2 liters in each minute, which is what a single flow at rate r_1 + r_2 would do. If a flow of r_{\text{out}} liters each minute is leaving at the same time, the container keeps only r_{\text{in}} - r_{\text{out}} of what arrives each minute.

Why the units change by 60. An hour is 60 minutes, and each of those minutes brings r liters. So an hour brings 60r liters, and a rate of r liters per minute is the same flow as 60r liters per hour. The flow has not changed; only the length of the time unit has.

Worked examples

Example 1

The bucket test

A garden hose fills a 10-liter bucket in 40 seconds. Find its flow rate in liters per second and in liters per minute.

Solution

The flow rate is the volume divided by the time it took:

r = \frac{10\ \text{L}}{40\ \text{s}} = 0.25\ \text{L/s}

A minute is 60 seconds, and each of those seconds brings 0.25 liters, so a minute brings

0.25 \times 60 = 15\ \text{L/min}

Check. 40 seconds is \displaystyle \frac{2}{3} of a minute, and \displaystyle 15 \times \frac{2}{3} = 10, which is the bucket.

This is how a flow rate is usually measured in practice: a container of known volume and a stopwatch.

Example 2

Filling a bath

The tap in Figure 1 runs at 12 liters per minute into an empty bath.

(a) How much water is in the bath after 9 minutes?

(b) The bath is full enough at 168 liters. How long does that take?

Solution

(a) Each minute adds 12 liters, so 9 minutes add nine times as much. With V = rt,

V = 12 \times 9 = 108\ \text{L}

(b) Now the volume is known and the time is not. The time is how many lots of 12 liters fit into 168 liters, which is t = V \div r:

t = \frac{168}{12} = 14\ \text{minutes}

Check. 12 \times 14 = 168.

Example 3

Drip irrigation

An emitter on a drip irrigation line releases 350 mL of water in 15 minutes.

(a) Find the flow rate in liters per hour.

(b) A vegetable bed has 24 of these emitters, and the line runs for 45 minutes. How much water does the bed get in total?

Solution

(a) An hour is 4 lots of 15 minutes, and each lot brings 350 mL, so an hour brings

4 \times 350 = 1{,}400\ \text{mL}

There are 1{,}000 mL in a liter, so the flow rate is 1.4\ \text{L/h}.

(b) 45 minutes is \displaystyle \frac{3}{4} of an hour, so one emitter releases

1.4 \times \frac{3}{4} = 1.05\ \text{L}

The 24 emitters run at the same time and each releases the same amount:

24 \times 1.05 = 25.2\ \text{L}

Check. 45 minutes is also 3 lots of 15 minutes, so one emitter releases 3 \times 350 = 1{,}050 mL, which is 1.05 L, as before.

Example 4

Reading a water meter

A home’s water meter counts the total number of liters that have passed through it, the way a car counts the total distance it has driven. With only the shower running, the meter reads 47{,}318 liters at 7{:}42 in the morning and 47{,}453 liters at 7{:}57. What is the shower’s flow rate?

Solution

The meter’s reading is a running total, so the water that passed is the change in the reading:

47{,}453 - 47{,}318 = 135\ \text{L}

The time that passed is from 7{:}42 to 7{:}57, which is 15 minutes. So

r = \frac{135}{15} = 9\ \text{L/min}

Neither reading on its own says anything about the flow. A total tells you how much water has ever passed; only the change in it over a known time gives a rate.

Example 5

Two hoses into one pond

A new garden pond holds 960 liters. One hose runs at 15 liters per minute and another at 25 liters per minute. How long does it take with both hoses filling it at once? How long would each take alone?

Solution

Both hoses pour into the pond at the same time, so in each minute the pond gains the water from both. The rates add:

15 + 25 = 40\ \text{L/min}
t = \frac{960}{40} = 24\ \text{minutes}

Alone, the first hose takes 960 \div 15 = 64 minutes and the second takes 960 \div 25 = 38.4 minutes.

Together is faster than either hose alone, as it must be. The time together is not the sum of the two times, and not their average either. Only the rates combine by adding.

Example 6

Pumping out a basement

A storm floods a basement whose floor is a rectangle 8 m by 6 m, leaving water 0.25 m deep everywhere. A pump removes 150 liters per minute.

(a) How many liters of water are in the basement?

(b) How long does the pump take to clear it?

(c) Water is in fact still seeping in through the walls at 30 liters per minute while the pump runs. How long does clearing the basement take now?

Solution

(a) The water fills a box 8 m long, 6 m wide and 0.25 m deep, so its volume is length times width times height:

8 \times 6 \times 0.25 = 12\ \text{m}^3

A cubic meter is 1{,}000 liters, so that is 12{,}000 liters.

(b) t = V \div r:

t = \frac{12{,}000}{150} = 80\ \text{minutes}

(c) The pump takes out 150 liters each minute, but 30 liters come back in each minute, so the basement loses only the difference:

150 - 30 = 120\ \text{L/min}
t = \frac{12{,}000}{120} = 100\ \text{minutes}

The seepage costs 20 extra minutes. The pump never works more slowly; it is the net rate that falls.

Example 7

A rain barrel that is not empty

A rain barrel holds 200 liters and already has 40 liters in it when a steady rain begins. The roof’s downpipe adds 2.5 liters per minute.

(a) How long until the barrel is full?

(b) Is the volume of water in the barrel proportional to the time since the rain began?

0 16 32 48 64 0 40 80 120 160 200 time (minutes) liters water in the barrel water added
Figure 2Both lines climb 2.5 liters every minute. Only the lower one starts at zero.
Solution

(a) The barrel has room for 200 - 40 = 160 more liters. At 2.5 liters per minute,

t = \frac{160}{2.5} = 64\ \text{minutes}

(b) No. Make a short table of the time t in minutes and the water W in the barrel in liters.

t01632
W4080120

For a proportional relationship every ratio W : t would be the same, but \displaystyle \frac{80}{16} = 5 while \displaystyle \frac{120}{32} = 3.75. The deeper reason shows at t = 0: no time has passed, yet the barrel holds 40 liters, not 0. In Figure 2 its line starts partway up the liters axis instead of at the origin.

What is proportional to the time is the water added, 2.5t: the lower line in Figure 2, which starts at (0, 0). The water in the barrel is that amount plus the starting 40:

W = 40 + 2.5t

The two lines are parallel because both climb by the flow rate, 2.5 liters, every minute. An amount that changes at a steady rate from a starting value other than zero is a constant rate of change, which is where this example leads.

Boundaries

The flow must be steady. Everything above assumes the same volume passes in every equal stretch of time. Real flows change: a tap is turned, the rain eases, a bath draining through its plughole slows as the water gets shallower. For a flow like that, volume divided by time still gives a number, but it is the average flow rate over that time: it gives the right total, not what was happening at each moment.

The volume in a container is proportional to time only if it starts empty. The volume that has flowed is always rt. The volume in the container is rt only if it started at zero, as Example 7 shows. If water is also leaving, r is the net rate: a bath that starts empty and gains a net 10 liters each minute still holds 10t liters.

Rates add only for flows running at the same time. Two hoses filling one pond at once combine. A hose that runs in the morning and another in the afternoon do not; there you add the volumes each delivers. Times never add to give the time together.

Flow rate counts volume, not depth. A steady flow adds the same number of liters each minute, but the water rises by the same height each minute only when the container has upright sides and the same shape all the way up, such as a box. In a basin that widens toward the top, the same liters spread over a bigger area higher up, so the depth rises more slowly there.

The units must match. Liters and milliliters, or seconds and minutes, cannot be divided or combined until they are in the same unit.

The net rate must be positive. The time t = V \div r makes sense only for r greater than 0. A pump that removes water no faster than it seeps in never clears the basement, however long it runs.

Where this goes next

Flow rate is one of the rates built from two measurements. Its sibling density divides a mass by a volume instead of a volume by a time, and the same three forms and the same care with units carry straight across.

Because the volume poured is proportional to the time, every steady flow has a straight-line graph through the origin, and the flow rate is how steep it is: that is the subject of proportional relationships in a graph. A container that does not start empty, like the rain barrel, leads to constant rates of change with a starting value.

For a fixed volume, doubling the flow rate halves the filling time, so the rate and the time are inversely proportional. Changing liters per minute into milliliters per second, or gallons per hour, is the general job of conversion factors.

Much later, calculus deals with flows that do not stay steady. Totaling the water from a flow rate that keeps changing is accumulation, and asking how fast the depth rises in a cone or a cylinder while it fills is a related rates question. Both begin with the idea in this lesson: liters in each unit of time.

Exercises

Finding a flow rate

  1. A tap fills a 7-liter watering can in 20 seconds. Find its flow rate in liters per second and in liters per minute.
  2. A drip emitter releases 240 mL of water in 2 hours. Find its flow rate in milliliters per hour and in milliliters per minute.
  3. A pump removes 4{,}500 liters of water from a flooded basement in 1.5 hours. Find its flow rate in liters per hour and in liters per minute.
  4. During 18 minutes of steady rain, a rain barrel collects 27 liters. Find the flow rate into the barrel in liters per minute.
  5. A water meter reads 52{,}806 liters at 6{:}10 in the evening and 52{,}974 liters at 6{:}31, while only a garden sprinkler is running. Find the sprinkler’s flow rate.

Volume and time from a flow rate

  1. A hose runs at 14 liters per minute for 25 minutes. How much water does it deliver?
  2. A bath is to be filled to 176 liters from a tap running at 16 liters per minute. How long does that take?
  3. A drip line delivers 1.5 liters per hour. How much does it deliver in 3 hours 20 minutes?
  4. An inflatable pool holds 1{,}080 liters, and a hose fills it at 0.3 liters per second. How many minutes does it take?

Fractions in the measurements

These use unit rates with fractions.

  1. A hose fills \displaystyle \frac{3}{4} of a 20-liter bucket in \displaystyle \frac{1}{2} of a minute. Find its flow rate in liters per minute.
  2. In \displaystyle \frac{2}{3} of an hour, a drip line delivers \displaystyle 5\frac{1}{2} liters. Find its flow rate in liters per hour.

Changing units

  1. Write 18 liters per minute in milliliters per second.
  2. Write 250 milliliters per second in liters per hour.
  3. A pool holds 36{,}000 liters, and a hose fills it at 24 liters per minute. How many hours does it take?
  4. Tap A fills a 9-liter bucket in 45 seconds. Tap B runs at 13 liters per minute. Which tap is faster, and by how many liters per minute?

Flows together

  1. Two hoses, running at 12 and 18 liters per minute, fill a 600-liter trough together. How long does it take?
  2. A flooded basement holds 9{,}000 liters of water. A pump removes 120 liters per minute, while water seeps back in at 20 liters per minute. How long does clearing it take?
  3. A tap fills a bath at 15 liters per minute, but the plug leaks 3 liters per minute. How long does the bath take to reach 150 liters?
  4. Hose A alone fills a barrel in 20 minutes, and hose B alone fills it in 30 minutes. How long do they take together? (Pick a size for the barrel, or ask what fraction of it each hose fills in one minute.)
  5. A rain barrel holds 210 liters and already has 35 liters in it when steady rain starts adding 2.5 liters per minute. How much water is in it after 30 minutes, and when is it full?

Reasoning

  1. In Exercise 20, is the volume of water in the barrel proportional to the time since the rain started? Explain, and name a quantity in that situation that is proportional to the time.
  2. Explain why two hoses filling a container together always finish sooner than the faster hose would alone.
  3. A basin that is wider at the top than at the bottom is filled from a tap at a steady 12 liters per minute. Does the water’s depth rise by the same amount every minute? Explain.
  4. Explain why the number in a flow rate written in liters per hour is 60 times the number in the same flow rate written in liters per minute, even though the flow is the same.
  5. A bath is always filled to the same volume V. Show that doubling the tap’s flow rate halves the filling time. What kind of relationship do the flow rate and the filling time have?
  6. A pump removes 25 liters per minute from a flooded basement while water seeps in at 25 liters per minute. What happens to the water in the basement? Explain using the net rate.
Answers
  1. 0.35 L/s and 21 L/min. 7 \div 20 = 0.35 liters each second, and a minute is 60 seconds: 0.35 \times 60 = 21.
  2. 120 mL/h and 2 mL/min. 240 \div 2 = 120 milliliters each hour, and an hour is 60 minutes: 120 \div 60 = 2.
  3. 3{,}000 L/h and 50 L/min. 4{,}500 \div 1.5 = 3{,}000 liters each hour, and 3{,}000 \div 60 = 50.
  4. 1.5 L/min. 27 \div 18 = 1.5.
  5. 8 L/min. The water that passed is 52{,}974 - 52{,}806 = 168 liters, in 21 minutes, and 168 \div 21 = 8.
  6. 350 liters. V = rt = 14 \times 25 = 350.
  7. 11 minutes. t = V \div r = 176 \div 16 = 11.
  8. 5 liters. 3 hours 20 minutes is \displaystyle 3\frac{1}{3} = \frac{10}{3} hours, and \displaystyle 1.5 \times \frac{10}{3} = 5.
  9. 60 minutes. 1{,}080 \div 0.3 = 3{,}600 seconds, and 3{,}600 \div 60 = 60 minutes.
  10. 30 L/min. The hose delivers \displaystyle \frac{3}{4} \times 20 = 15 liters in half a minute, and \displaystyle 15 \div \frac{1}{2} = 30.
  11. \displaystyle 8\frac{1}{4} L/h. \displaystyle 5\frac{1}{2} \div \frac{2}{3} = \frac{11}{2} \times \frac{3}{2} = \frac{33}{4}.
  12. 300 mL/s. 18 liters is 18{,}000 mL, spread over 60 seconds: 18{,}000 \div 60 = 300.
  13. 900 L/h. An hour is 3{,}600 seconds, so 250 \times 3{,}600 = 900{,}000 mL each hour, which is 900 liters.
  14. 25 hours. 36{,}000 \div 24 = 1{,}500 minutes, and 1{,}500 \div 60 = 25.
  15. Tap B, by 1 L/min. Tap A runs at 9 \div 45 = 0.2 liters per second, which is 0.2 \times 60 = 12 liters per minute, and 13 - 12 = 1.
  16. 20 minutes. Together the hoses deliver 12 + 18 = 30 liters per minute, and 600 \div 30 = 20.
  17. 90 minutes. The net rate is 120 - 20 = 100 liters per minute out, and 9{,}000 \div 100 = 90.
  18. 12.5 minutes, which is 12 minutes 30 seconds. The net rate is 15 - 3 = 12 liters per minute in, and 150 \div 12 = 12.5.
  19. 12 minutes. In one minute hose A fills \displaystyle \frac{1}{20} of the barrel and hose B fills \displaystyle \frac{1}{30} of it, so together they fill \displaystyle \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12} of it each minute, and the whole barrel takes 12 minutes. With a 60-liter barrel, say, the rates are 3 and 2 L/min, together 5 L/min, and 60 \div 5 = 12.
  20. 110 liters after 30 minutes, and full after 70 minutes. After 30 minutes the barrel holds 35 + 2.5 \times 30 = 35 + 75 = 110 liters. It has room for 210 - 35 = 175 more, and 175 \div 2.5 = 70.
  21. No. At t = 0 the barrel already holds 35 liters, not 0, and the ratios differ: after 10 minutes it holds 60 liters and after 30 minutes 110, and \displaystyle \frac{60}{10} = 6 while \displaystyle \frac{110}{30} = 3\frac{2}{3}. The water added since the rain started, 2.5t liters, is proportional to the time, with constant 2.5.
  22. Their rates add. If the faster hose runs at r_1 and the other at r_2, together they deliver r_1 + r_2 liters each minute, which is more than r_1 because r_2 is greater than 0. The same volume divided by a bigger rate gives a smaller time.
  23. No. Each minute adds the same 12 liters, but higher up the basin is wider, so those 12 liters spread over a bigger area and make a thinner layer. The depth rises quickly at first and more slowly as the basin fills.
  24. An hour is 60 minutes, and each minute brings the same volume, so one hour brings 60 times the volume of one minute. The flow is the same; the unit of time is 60 times as long, so the number of liters in it is 60 times as large.
  25. At rate r the time is \displaystyle \frac{V}{r}. At rate 2r it is \displaystyle \frac{V}{2r} = \frac{1}{2} \times \frac{V}{r}, half as long. The flow rate times the filling time is always V, a constant, so the two are inversely proportional.
  26. The amount of water stays the same. The net rate is 25 - 25 = 0 liters per minute, so the basement gains and loses nothing overall, and the pump never clears it however long it runs.
kg1128 · updated 2026-10-07 · CC-BY-SA-4.0