Rate of Flow: Filling and Emptying Tanks

Use volume = rate of flow × time to find how long a tank takes to fill or empty, how deep the water is after a given time, the net rate of a tap and a drain together, and the flow through a pipe.

Before you start

A rate is an amount of one quantity per unit of another. For a constant rate, amount = rate × time and time = amount ÷ rate: a tap at 12 L/min lets out $12 \times 5 = 60$ L in 5 minutes (Rate and Average Rate).
When two taps fill the same tank, their rates add; an open drain takes its rate away (Problems Involving Rate). A rate changes unit one quantity at a time: 12 L/min is $12 \times 60 = 720$ L/h (Converting Units of Speed).
Volume of a cuboid = length × breadth × height; volume of a cylinder $= \pi r^{2}h$ . Water in a rectangular tank or an upright cylinder has the tank's inside base as its base and the depth of the water as its height, so its volume = base area × depth (Volume of Prisms, Volume of Cylinders).
$1 \text{ cm}^{3} = 1$ mL, $1000 \text{ cm}^{3} = 1$ L, $1 \text{ m}^{3} = 1000$ L $= 1\,000\,000 \text{ cm}^{3}$ (Converting Units of Area and Volume).

Rate of Flow

The rate of flow of water from a tap, a hose or a pump is the volume of water that flows in each unit of time: litres per minute (L/min), millilitres or cubic centimetres per second (mL/s, $\text{cm}^{3}$ /s), or cubic metres per hour ( $\text{m}^{3}$ /h). A tap that pours 8 L into a pail every minute has a rate of flow of 8 L/min.

If the rate stays the same, the volume that flows in is the same in every minute. So the volume after $t$ minutes is the rate times $t$ , and the time to pour in a given volume is how many times the rate goes into it. To fill an empty tank, the volume needed is its capacity, found from its inside measurements.

Rate of flow

Rate of flow = volume ÷ time

Volume = rate of flow × time

Time = volume ÷ rate of flow

Before you divide or multiply, have the volume and the rate in the same unit of volume (L with L/min, $\text{cm}^{3}$ with $\text{cm}^{3}$ /s). The time then comes out in the rate's unit of time.

Times in hours and minutes

A time of 75 min is 1 h 15 min (60 min make an hour). A time of 1.4 h is 1 h and $0.4 \times 60 = 24$ min, so 1 h 24 min, not 1 h 40 min. A time of 6000 s is $6000 \div 60 = 100$ min, which is 1 h 40 min.

Example 1: The time to fill a tank

A rectangular water tank is 1.5 m long, 1.2 m wide and 1 m high inside.

Starting at 7.50 a.m., the empty tank is filled with water from a tap at 24 L/min. How long does it take to fill, and at what time is it full?

Solution

Capacity of the tank $= 1.5 \times 1.2 \times 1 = 1.8 \text{ m}^{3}$

The rate is in litres, so change the volume to litres: $1.8 \text{ m}^{3} = 1.8 \times 1000 = 1800$ L.

Time = volume ÷ rate of flow $= 1800 \div 24 = 75$ min

75 min $= 60$ min $+ 15$ min $= 1$ h 15 min.

7.50 a.m. + 1 h = 8.50 a.m.; 8.50 a.m. + 15 min = 9.05 a.m.

The tank takes 1 hour 15 minutes to fill, and it is full at 9.05 a.m.

Example 2: Finding the rate of flow

A cylindrical pail has an internal radius of 14 cm and an internal height of 30 cm.

A tap fills the empty pail to the brim in 1 min 24 s. Taking $\pi = \frac{22}{7}$ , find the rate of flow of the tap in $\text{cm}^{3}$ /s.

Solution

Capacity of the pail $= \pi r^{2}h = \frac{22}{7} \times 14^{2} \times 30 = \frac{22}{7} \times 196 \times 30 = 22 \times 28 \times 30 = 18\,480 \text{ cm}^{3}$

The rate is wanted per second, so change the time to seconds: 1 min 24 s $= 60 + 24 = 84$ s.

Rate of flow = volume ÷ time $= 18\,480 \div 84 = 220 \text{ cm}^{3}\text{/s}$

Check: $220 \times 84 = 18\,480 \text{ cm}^{3}$ .

The tap's rate of flow is $220 \text{ cm}^{3}$ /s (220 mL/s).

Changing the Units of a Rate of Flow

A rate of flow has two units, one of volume and one of time, and each is changed on its own, as a speed is. Change the volume: $1 \text{ L} = 1000 \text{ cm}^{3}$ , $1 \text{ m}^{3} = 1000$ L $= 1\,000\,000 \text{ cm}^{3}$ . Change the time: 1 min = 60 s, 1 h = 60 min = 3600 s.

Ask which way the number must go. In a minute, 60 times as much water flows as in a second, so a rate per second is multiplied by 60 to give the rate per minute; a rate per minute is divided by 60 to give the rate per second.

Example 3: Changing units of rate

Express (a) $250 \text{ cm}^{3}$ /s in L/min, (b) 15 L/min in $\text{m}^{3}$ /h, (c) $4.5 \text{ m}^{3}$ /h in L/min and in $\text{cm}^{3}$ /s.

Solution

(a) In 1 min (60 s): $250 \times 60 = 15\,000 \text{ cm}^{3}$ . In litres: $15\,000 \div 1000 = 15$ L. So $250 \text{ cm}^{3}$ /s $= 15$ L/min.

(b) In 1 h (60 min): $15 \times 60 = 900$ L. In $\text{m}^{3}$ : $900 \div 1000 = 0.9 \text{ m}^{3}$ . So 15 L/min $= 0.9 \text{ m}^{3}$ /h.

(c) $4.5 \text{ m}^{3} = 4.5 \times 1000 = 4500$ L in 1 h. In 1 min: $4500 \div 60 = 75$ L. So $4.5 \text{ m}^{3}$ /h $= 75$ L/min.

And $75 \text{ L} = 75\,000 \text{ cm}^{3}$ in 1 min, so in 1 s: $75\,000 \div 60 = 1250 \text{ cm}^{3}$ . So $4.5 \text{ m}^{3}$ /h $= 1250 \text{ cm}^{3}$ /s.

Check (c) in one step: $4.5 \text{ m}^{3} = 4\,500\,000 \text{ cm}^{3}$ in 3600 s, and $4\,500\,000 \div 3600 = 1250$ .

(a) 15 L/min (b) $0.9 \text{ m}^{3}$ /h (c) 75 L/min, which is $1250 \text{ cm}^{3}$ /s.

The Depth of Water After a Given Time

In a tank with a uniform cross-section standing upright (a rectangular tank, or a cylinder standing on its end), every centimetre of depth holds the same volume: the base area × 1 cm. So when water flows in at a steady rate, the level rises by the same amount every minute; when it flows out, the level falls by the same amount every minute.

To find the depth after a given time, find the volume that has flowed in (rate × time), then divide by the inside base area. Or find the rise in each minute (rate ÷ base area) first. Keep the volume and the area in matching units: $\text{cm}^{3}$ with $\text{cm}^{2}$ , $\text{m}^{3}$ with $\text{m}^{2}$ .

Depth of water in a tank of uniform cross-section

Volume of water = base area × depth, so the depth is found from base area × depth = volume.

Rise in level each minute = rate of flow ÷ base area (the rate in $\text{cm}^{3}$ /min with the area in $\text{cm}^{2}$ ).

This holds only for a tank whose cross-section is the same all the way up. Use the inside base.

Example 4: The depth after 15 minutes

An empty rectangular tank has an inside base 80 cm long and 50 cm wide, and is 60 cm high. Water flows into it from a tap at 12 L/min.

(a) By how much does the water level rise each minute? (b) How deep is the water after 15 minutes? (c) How many more minutes does the tap take to fill the tank?

Solution

Base area $= 80 \times 50 = 4000 \text{ cm}^{2}$ . Rate of flow: 12 L/min $= 12 \times 1000 = 12\,000 \text{ cm}^{3}$ /min.

(a) Each minute $12\,000 \text{ cm}^{3}$ spreads over $4000 \text{ cm}^{2}$ : rise $= 12\,000 \div 4000 = 3$ cm.

(b) Volume after 15 min $= 12\,000 \times 15 = 180\,000 \text{ cm}^{3}$ .

Let the depth be $d$ cm. Base area × depth = volume: $4000d = 180\,000$ , so $d = 180\,000 \div 4000 = 45$ .

Check with (a): 3 cm each minute for 15 minutes is $3 \times 15 = 45$ cm.

(a) The level rises 3 cm each minute. (b) After 15 minutes the water is 45 cm deep. (c) The tap takes 5 more minutes.

Example 5: Emptying a cylindrical tank

A cylindrical tank has an internal diameter of 1.4 m and a height of 1.5 m. It holds water 1.2 m deep.

A pump drains the water at 22 L/min. Taking $\pi = \frac{22}{7}$ , find (a) how long the pump takes to empty the tank, in hours and minutes, (b) the depth of the water after the pump has run for 35 minutes.

Solution

The internal radius is $1.4 \div 2 = 0.7$ m. Base area $= \pi r^{2} = \frac{22}{7} \times 0.7^{2} = \frac{22}{7} \times 0.49 = 1.54 \text{ m}^{2}$ .

(a) Volume of water $= 1.54 \times 1.2 = 1.848 \text{ m}^{3} = 1.848 \times 1000 = 1848$ L.

Time $= 1848 \div 22 = 84$ min $= 1$ h 24 min.

(b) In 35 min the pump takes out $22 \times 35 = 770$ L $= 0.77 \text{ m}^{3}$ .

Let the fall in level be $f$ m: $1.54f = 0.77$ , so $f = 0.77 \div 1.54 = 0.5$ .

Depth after 35 min $= 1.2 - 0.5 = 0.7$ m.

Check: the water left is $1848 - 770 = 1078$ L $= 1.078 \text{ m}^{3}$ , and $1.54 \times 0.7 = 1.078$ .

(a) The pump empties the tank in 1 hour 24 minutes. (b) After 35 minutes the water is 0.7 m deep.

Example 6: From one depth to another

A cylindrical tank of internal radius 35 cm and height 1 m holds water 20 cm deep. A tap pours more water in at 7 L/min.

Taking $\pi = \frac{22}{7}$ , find how long it takes (a) for the water to be 80 cm deep, (b) to fill the tank.

Solution

Work in centimetres: height 1 m $= 100$ cm. Base area $= \frac{22}{7} \times 35^{2} = \frac{22}{7} \times 1225 = 3850 \text{ cm}^{2}$ .

Only the water still to come matters, not the water already there.

(a) The level must rise $80 - 20 = 60$ cm. Volume needed $= 3850 \times 60 = 231\,000 \text{ cm}^{3} = 231$ L.

Time $= 231 \div 7 = 33$ min.

(b) To fill, the level must rise $100 - 20 = 80$ cm. Volume $= 3850 \times 80 = 308\,000 \text{ cm}^{3} = 308$ L.

Time $= 308 \div 7 = 44$ min.

(a) The water is 80 cm deep after 33 minutes. (b) The tank is full after 44 minutes.

Filling and Draining Together

Two taps filling the same tank pour in the sum of their rates each minute, as you have seen for two taps. If a drain is open at the same time, it takes its rate away. What is left is the net rate: how much the volume of water in the tank goes up (or down) each minute.

If the rate in is greater than the rate out, the tank fills, more slowly than with the drain shut. If the rate out is greater, the water level falls, even with the tap running.

Net rate of flow

Net rate = total rate in − total rate out

Time = volume ÷ net rate (add the rates, never the times)

Example 7: A tap with the drain left open

An empty rectangular tank is 1.2 m long, 0.8 m wide and 1 m high inside.

A tap fills it at 30 L/min, but the drain at the bottom has been left open and lets out 6 L/min. (a) How long does the tank take to fill? (b) How deep is the water after 25 minutes?

Solution

Net rate $= 30 - 6 = 24$ L/min: the tank gains 24 L each minute.

(a) Capacity $= 1.2 \times 0.8 \times 1 = 0.96 \text{ m}^{3} = 960$ L. Time $= 960 \div 24 = 40$ min.

(With the drain shut it would take $960 \div 30 = 32$ min.)

(b) Volume after 25 min $= 24 \times 25 = 600$ L $= 0.6 \text{ m}^{3}$ . Base area $= 1.2 \times 0.8 = 0.96 \text{ m}^{2}$ .

Let the depth be $d$ m: $0.96d = 0.6$ , so $d = 0.6 \div 0.96 = 0.625$ .

(a) The tank takes 40 minutes to fill. (b) After 25 minutes the water is 0.625 m (62.5 cm) deep.

Example 8: When more flows out than in

A rectangular tank with an inside base 70 cm by 50 cm holds water 48 cm deep. A tap pours water in at 9 L/min while the drain lets water out at 15 L/min.

Does the water level rise or fall? How long is it until the tank is empty?

Solution

Rate out − rate in $= 15 - 9 = 6$ L/min: more flows out than in, so the level falls, and the tank loses 6 L each minute.

Volume of water $= 70 \times 50 \times 48 = 168\,000 \text{ cm}^{3} = 168$ L.

Time to empty $= 168 \div 6 = 28$ min.

The water level falls, and the tank is empty after 28 minutes.

Water Flowing Through a Pipe

Often the rate of flow is not given; instead you are told the speed of the water along a pipe, say $v$ cm/s, and the pipe's internal radius $r$ cm. In 1 second the water moves $v$ cm along the pipe, so the water that comes out of the end in that second is the water that was in the last $v$ cm of the pipe: a cylinder of radius $r$ and length $v$ .

Flow through a pipe

Volume of water each second = area of cross-section of the pipe × speed of the water

$= \pi r^{2} \times v$ , where $r$ is the pipe's internal radius and $v$ the speed of the water.

With $r$ in cm and $v$ in cm/s, the volume is in $\text{cm}^{3}$ /s. Change a speed in m/s to cm/s (1 m/s = 100 cm/s), or the radius to metres, so that both lengths are in the same unit.

Example 9: Filling a tank through a pipe

Water flows at 0.5 m/s through a pipe of internal diameter 2.8 cm into an empty rectangular tank 1.2 m long, 1.1 m wide and 1.4 m high inside.

Taking $\pi = \frac{22}{7}$ , find (a) the rate of flow in $\text{cm}^{3}$ /s and in L/min, (b) the time taken to fill the tank, in hours and minutes.

Solution

(a) Radius $= 2.8 \div 2 = 1.4$ cm. Speed: 0.5 m/s $= 0.5 \times 100 = 50$ cm/s.

Volume each second $= \pi r^{2}v = \frac{22}{7} \times 1.4^{2} \times 50 = \frac{22}{7} \times 1.96 \times 50 = 308 \text{ cm}^{3}$

In L/min: $308 \times 60 = 18\,480 \text{ cm}^{3}$ in a minute, and $18\,480 \div 1000 = 18.48$ L, so 18.48 L/min.

(b) Capacity $= 1.2 \times 1.1 \times 1.4 = 1.848 \text{ m}^{3} = 1\,848\,000 \text{ cm}^{3}$ .

Time $= 1\,848\,000 \div 308 = 6000$ s. In minutes: $6000 \div 60 = 100$ min $= 1$ h 40 min.

Check with the rate in L/min: $1848 \div 18.48 = 100$ min.

(a) The rate of flow is $308 \text{ cm}^{3}$ /s, which is 18.48 L/min. (b) The tank takes 1 hour 40 minutes to fill.

Example 10: The speed of the water

A garden hose has an internal diameter of 2 cm. It delivers 18 L of water a minute.

Find the speed of the water in the hose, in cm/s, correct to 3 significant figures. Use the $\pi$ button on your calculator.

Solution

Radius $= 2 \div 2 = 1$ cm. Rate of flow in $\text{cm}^{3}$ /s: 18 L $= 18\,000 \text{ cm}^{3}$ in 60 s, so $18\,000 \div 60 = 300 \text{ cm}^{3}$ /s.

Let the speed be $v$ cm/s. Volume each second $= \pi r^{2}v$ :

$\pi \times 1^{2} \times v = 300$

$\pi v = 300$

$v = 300 \div \pi = 95.492\ldots \approx 95.5$

Check: $\pi \times 1^{2} \times 95.492\ldots = 300.0 \text{ cm}^{3}$ each second.

The water flows at about 95.5 cm/s (just under 1 m/s).

Common Mistakes

Mixing units. A rate in L/min with a volume in $\text{cm}^{3}$ or $\text{m}^{3}$ gives nonsense. Change the volume to the rate's unit first: $1 \text{ m}^{3} = 1000$ L (not 100 L), $1 \text{ L} = 1000 \text{ cm}^{3}$ .
Changing the time the wrong way. A rate per second is multiplied by 60 to give the rate per minute (more water flows in a minute than in a second). $250 \text{ cm}^{3}$ /s is $15\,000 \text{ cm}^{3}$ /min, not $4.17 \text{ cm}^{3}$ /min.
Reading a decimal of an hour as minutes. 1.4 h is 1 h 24 min, not 1 h 40 min; 100 min is 1 h 40 min, not 1.40 h.
Filling the whole tank when part is full. If the tank already holds water, only the volume still needed is divided by the rate.
Forgetting the drain, or adding the times. With a tap and an open drain, use the net rate (in − out). Never add or average the times of two taps.
Using the diameter in $\pi r^{2}v$ , or a speed in m/s with a radius in cm. Halve the diameter, and change the speed to cm/s or the radius to m, so that both are in the same unit of length.

Practice

Try each question before you open its solution. Give answers that are not exact correct to 3 significant figures unless the question says otherwise.

Question 1

Express (a) 6 L/min in $\text{cm}^{3}$ /s, (b) $540 \text{ cm}^{3}$ /s in $\text{m}^{3}$ /h, (c) $2.7 \text{ m}^{3}$ /h in L/min.

ShowHide solution

(a) 6 L $= 6000 \text{ cm}^{3}$ in 60 s, so $6000 \div 60 = 100 \text{ cm}^{3}$ /s.

(b) In 1 h (3600 s): $540 \times 3600 = 1\,944\,000 \text{ cm}^{3}$ , and $1\,944\,000 \div 1\,000\,000 = 1.944 \text{ m}^{3}$ . So $1.944 \text{ m}^{3}$ /h.

(a) $100 \text{ cm}^{3}$ /s (b) $1.944 \text{ m}^{3}$ /h (c) 45 L/min

Question 2

Mei Ling's fish tank is 60 cm long, 30 cm wide and 40 cm high inside. She fills the empty tank with water to 5 cm below the top, using a hose that delivers 4.5 L/min.

How long does this take?

ShowHide solution

Depth of water $= 40 - 5 = 35$ cm.

Volume of water $= 60 \times 30 \times 35 = 63\,000 \text{ cm}^{3} = 63$ L.

Time $= 63 \div 4.5 = 14$ min.

It takes 14 minutes.

Question 3

An empty cylindrical tank of internal radius 0.6 m is filled with water to a depth of 1.4 m in 45 minutes.

Find the rate of flow in L/min. Use the $\pi$ button on your calculator.

ShowHide solution

Volume of water $= \pi r^{2}h = \pi \times 0.6^{2} \times 1.4 = 0.504\pi \text{ m}^{3} = 1.583\,36\ldots \text{ m}^{3}$

In litres: $1.583\,36\ldots \times 1000 = 1583.36\ldots$ L.

Rate of flow $= 1583.36\ldots \div 45 = 35.185\ldots \approx 35.2$ L/min

The rate of flow is about 35.2 L/min.

Question 4

An empty rectangular tank has an inside base 50 cm by 40 cm and is 45 cm high. A tap fills it at 5 L/min.

(a) By how much does the water level rise each minute? (b) How deep is the water after 12 minutes? (c) How long does the tap take to fill the tank?

ShowHide solution

Base area $= 50 \times 40 = 2000 \text{ cm}^{2}$ . Rate: 5 L/min $= 5000 \text{ cm}^{3}$ /min.

(a) Rise each minute $= 5000 \div 2000 = 2.5$ cm.

(b) Depth after 12 min $= 2.5 \times 12 = 30$ cm. (Check: $5000 \times 12 = 60\,000 \text{ cm}^{3}$ , and $60\,000 \div 2000 = 30$ cm.)

(a) 2.5 cm each minute (b) 30 cm (c) 18 minutes

Question 5

A rainwater barrel at a community garden is a cylinder of internal radius 35 cm and height 90 cm. It holds water 60 cm deep. A tap at the bottom lets the water out at 7.7 L/min.

Taking $\pi = \frac{22}{7}$ , find (a) how long the barrel takes to empty, (b) the depth of the water after 12 minutes.

ShowHide solution

Base area $= \frac{22}{7} \times 35^{2} = 3850 \text{ cm}^{2}$ . Rate: 7.7 L/min $= 7700 \text{ cm}^{3}$ /min.

Fall in level each minute $= 7700 \div 3850 = 2$ cm.

(a) Time to empty $= 60 \div 2 = 30$ min. (Check: volume $= 3850 \times 60 = 231\,000 \text{ cm}^{3} = 231$ L, and $231 \div 7.7 = 30$ min.)

(b) In 12 min the level falls $2 \times 12 = 24$ cm, so the depth is $60 - 24 = 36$ cm.

(a) 30 minutes (b) 36 cm

Question 6

An empty rectangular tank is 1 m long, 0.8 m wide and 0.75 m high inside. Tap A pours in 18 L/min and tap B pours in 12 L/min, but the drain is open and lets out 5 L/min.

How long does the tank take to fill?

ShowHide solution

Capacity $= 1 \times 0.8 \times 0.75 = 0.6 \text{ m}^{3} = 600$ L.

Net rate $= 18 + 12 - 5 = 25$ L/min.

Time $= 600 \div 25 = 24$ min.

The tank takes 24 minutes to fill.

Question 7

The tank of a fountain in a park holds 900 L of water when full. Starting with the tank full, the fountain's pump lets water out at 27 L/min while a tap puts water back at 15 L/min. How long is it until the tank is empty? Give your answer in hours and minutes.

ShowHide solution

Rate out − rate in $= 27 - 15 = 12$ L/min, so the tank loses 12 L each minute.

Time $= 900 \div 12 = 75$ min $= 1$ h 15 min.

The tank is empty after 1 hour 15 minutes.

Question 8

Mr Tan starts filling an empty rectangular tank, 1.2 m long, 1 m wide and 1 m high inside, at 8.40 a.m., with a tap at 40 L/min. After 15 minutes the drain is opened by mistake, and from then on it lets out 10 L/min.

(a) How deep is the water when the drain is opened? (b) At what time is the tank full?

ShowHide solution

Capacity $= 1.2 \times 1 \times 1 = 1.2 \text{ m}^{3} = 1200$ L. Base area $= 1.2 \times 1 = 1.2 \text{ m}^{2}$ .

(a) In the first 15 min: $40 \times 15 = 600$ L $= 0.6 \text{ m}^{3}$ . Depth: $1.2d = 0.6$ , so $d = 0.5$ m.

(b) Still needed: $1200 - 600 = 600$ L. Net rate $= 40 - 10 = 30$ L/min, so $600 \div 30 = 20$ min more.

Total time $= 15 + 20 = 35$ min. 8.40 a.m. + 35 min = 9.15 a.m.

(a) The water is 0.5 m deep. (b) The tank is full at 9.15 a.m.

Question 9

Water flows at 0.8 m/s through a pipe of internal diameter 3 cm.

Find (a) the rate of flow in $\text{cm}^{3}$ /s, (b) the rate of flow in L/min, (c) the time the pipe takes to fill a 2000 L tank, to the nearest minute. Use the $\pi$ button on your calculator.

ShowHide solution

Radius $= 3 \div 2 = 1.5$ cm. Speed $= 0.8 \times 100 = 80$ cm/s.

(a) Volume each second $= \pi r^{2}v = \pi \times 1.5^{2} \times 80 = 180\pi = 565.486\ldots \approx 565 \text{ cm}^{3}$ /s

(b) In a minute: $180\pi \times 60 = 10\,800\pi \text{ cm}^{3}$ , which is $10.8\pi = 33.929\ldots \approx 33.9$ L/min.

(a) about $565 \text{ cm}^{3}$ /s (b) about 33.9 L/min (c) about 59 minutes

Question 10

Water flows along a pipe at 40 cm/s and comes out at 6 L/min. Find the internal radius of the pipe in cm. Use the $\pi$ button on your calculator.

ShowHide solution

Rate in $\text{cm}^{3}$ /s: 6 L $= 6000 \text{ cm}^{3}$ in 60 s, so $6000 \div 60 = 100 \text{ cm}^{3}$ /s.

Let the radius be $r$ cm. Volume each second $= \pi r^{2}v$ :

$\pi \times r^{2} \times 40 = 100$

$40\pi r^{2} = 100$

$r^{2} = \frac{100}{40\pi} = 0.795\,77\ldots$ (divide both sides by $40\pi$ )

$r = \sqrt{0.795\,77\ldots} = 0.892\,06\ldots \approx 0.892$ (a length, so the positive root)

Check: $\pi \times 0.892\,06\ldots^{2} \times 40 = 100.0 \text{ cm}^{3}$ /s.

The internal radius is about 0.892 cm.

Question 11

Water flows at 25 cm/s through a pipe of internal radius 1.4 cm into a cylindrical tank of internal radius 35 cm.

Taking $\pi = \frac{22}{7}$ , find how long the water level takes to rise by 48 cm.

ShowHide solution

Volume each second from the pipe $= \frac{22}{7} \times 1.4^{2} \times 25 = \frac{22}{7} \times 1.96 \times 25 = 154 \text{ cm}^{3}$ .

Base area of the tank $= \frac{22}{7} \times 35^{2} = 3850 \text{ cm}^{2}$ . Volume for a rise of 48 cm $= 3850 \times 48 = 184\,800 \text{ cm}^{3}$ .

Time $= 184\,800 \div 154 = 1200$ s $= 1200 \div 60 = 20$ min.

Check: each minute $154 \times 60 = 9240 \text{ cm}^{3}$ raises the level $9240 \div 3850 = 2.4$ cm, and $48 \div 2.4 = 20$ min.

The level rises 48 cm in 20 minutes.

Question 12

The pool at a community club is 25 m long and 10 m wide, with a flat floor. It is filled to a depth of 1.2 m through a pipe of internal diameter 10 cm, in which the water flows at 2 m/s.

(not drawn to scale)

(a) Find the rate of flow in $\text{m}^{3}$ /h. (b) How long does it take to fill the empty pool to 1.2 m? Give the time in hours and minutes, to the nearest minute. (c) How deep is the water after 2 hours? Use the $\pi$ button on your calculator.

ShowHide solution

Work in metres: radius $= 10 \div 2 = 5$ cm $= 0.05$ m; speed 2 m/s.

(a) Volume each second $= \pi \times 0.05^{2} \times 2 = 0.005\pi \text{ m}^{3}$ . In an hour (3600 s): $0.005\pi \times 3600 = 18\pi = 56.548\ldots \approx 56.5 \text{ m}^{3}$ /h.

(b) Volume of water $= 25 \times 10 \times 1.2 = 300 \text{ m}^{3}$ . Time $= \frac{300}{18\pi} = 5.305\,16\ldots$ h.

$0.305\,16\ldots \times 60 = 18.3\ldots$ min, so the time is about 5 h 18 min.

Depth: $250d = 36\pi$ , so $d = 36\pi \div 250 = 0.452\,38\ldots \approx 0.452$ m.

(a) about $56.5 \text{ m}^{3}$ /h (b) about 5 hours 18 minutes (c) about 0.452 m (45.2 cm)

Key points

Rate of flow = volume ÷ time; volume = rate of flow × time; time = volume ÷ rate of flow.
Put the volume and the rate in the same unit of volume first: $1 \text{ m}^{3} = 1000$ L, $1 \text{ L} = 1000 \text{ cm}^{3}$ .
Change a rate one unit at a time: per second × 60 gives per minute, per minute × 60 gives per hour.
Give times in hours and minutes: 75 min = 1 h 15 min; 1.4 h = 1 h 24 min.
In a tank of uniform cross-section, base area × depth = volume of water, and the level rises (or falls) by rate ÷ base area each minute.
A tank already partly full needs only the volume still to come.
Net rate = rates in − rates out; add the rates, never the times. If more flows out than in, the level falls.
Water through a pipe: volume each second $= \pi r^{2}v$ , with the internal radius and the speed in the same unit of length.
Use the value of $\pi$ the question gives, and round only at the end.