TRAINING DEPARTMENT
General
The science of the motions and equilibrium of a material system partly or wholly fluid is known as hydrodynamics called hydrostatics when the system is in equilibrium, or at rest and hydrokinetics when it is not. In the fire service, however, we use the term hydraulics to denote the science of hydrodynamics in general and to describe the study and behaviour of water, whether in motion or at rest.
Water when pure is colourless, odourless, liquid with a molecular composition to two atoms of hydrogen combined with one atom of oxygen. A liter of water has a mass of 1 kilogram (1 kg), corresponding to a downward force of 9.81 newtons (N). A cubic meter of water exerts a downward force of 9810 N, or 9.81 kilonewtons (kN). This is more commonly reckoned as 10 kN. The mass of water varies with the degree of purity. Ordinary sea water ‘weighs’ approximately 10.0 newtons per liter (N/litres).
Pure water has a freezing point of 0ºC and a boiling point of 100ºC, both at normal atmospheric pressure (1 bar approx.). Between these temperatures at atmospheric pressure, therefore, water exists as a liquid and exhibits all the characteristic properties of a fluid. It is virtually incompressible, and an increase of 1 bar only causes a decrease in volume of 0.000 002 per cent.
As a fluid, water has volume but is incapable of resisting change of shape, i.e. when poured into a container it will adjust itself irrespective of the shaper of the latter, and will come to rest with a level surface. This is because there is very little friction or cohesion between the individual molecules of which water is composed.
2. Principal Characteristics of Pressure
There are six basic rules governing the principal characteristics of pressure in liquids. These are:
a. Pressure is perpendicular to any surface on which it acts
If a vessel having flat sides, contains water, and that water has attained a position of rest, then the pressure on all sides of the vessel due to the weight is perpendicular to those sides, as shown by the direction of the arrows.
b. Pressure at any point of a fluid at rest is of the same intensity in all directions
In a line of piping or hose, two pressure gauges are inserted. If the water is at rest because a valve or hand-controlled branch has been shut down, the pressure gauges will register identical readings showing that the pressure at any point of a fluid at rest is the same in all directions.
c. Pressure applied from outside to a fluid contained in a vessel is transmitted in al directions
A hollow sphere with pressure gauges around the circumference has been filled with water, and pressure is applied. All the gauges will show the same pressure reading, providing that when pressure is applied to a fluid in a confined space, that pressure is transmitted equally in all directions.
d. Downward pressure of a fluid in an open vessel is proportional to its depth
If there are three vertical containers, each of 1 meter square cross-sectional area. The depth of water is 1 m, 2 m and 3 m respectively.
The weight of water in each container is 9810, 19 620 and 29 430 newtons respectively and since the cross-section of each container is 1 m², the corresponding pressure acting on the base of each container is, therefore, 9810, 19 620 and 29 430 N/m².
e. The downward pressure of a fluid in an open vessel is proportional to the density of the fluid.
If there are two containers, one holding mercury and the other water, the depth of the mercury is 10 mm and the depth of the water is 136 mm. The pressure at the bottom of each container is the same, as mercury is approximately 13.6 times as heavy as water. If the mercury container were filled to the same height as the
water, then the pressure at the base of the mercury container would be 13.6 times that at the base of the water container.
f. The downward pressure of a fluid on the bottom of a vessel is independent of the shape of that vessel.
A number of containers of varying shapes, but having the same cross-sectional area of 1 m² at the bottom. The pressure at the bottom of each is exactly the same providing that the depth of the liquid (or head) is the same in each case.
MEASUREMENT OF AREAS AND VOLUMES
The calculation of areas and volumes, especially the area of a circle, is a fundamental part of the mathematics required by firemen when making hydraulic calculations.
1. Area Of Regular Figures
The area of rectangle, a parallelogram, a triangle or a trapezium (a quadrilateral with one pair of sides parallel) is easily calculated:
Area of rectangle or square = base (b) x vertical height (h)
Area of parallelogram = base (b) x vertical height (h)
Area of trapezium = vertical height (h) x sum of parallel sides (a + b)
2
Area of triangle = base (b) x vertical height (h)
2
The area of triangle can also be calculated when the length of each side is known but when it is impossible to measure the vertical height, as for example, in the case of a large triangular-shaped pond or lake. The is found from the formula
Area of triangle = √s (s – a) (s – b) (s – c)
Where s = one half of the sum of the sides, a, b and c.
2. Area of Circles
The area of a circle is found by squaring the radius and multiplying by the constant pi (π = 3.1416 or 3 1/7 ) i.e. πr2. This can easily be proved by drawing a circle on a squared paper and drawing a square ABCD on the radius of CD. By counting the number of small squares inside the whole of the circle it will be found that it is approximately 3 1/7 times as many as the number of small squares in the square ABCD on the radius
In hydraulics calculations, the size of a nozzle is usually given according to its diameter, and it is therefore more convenient to use.
d instead of r when calculating the area of a circle.
2
Thus, instead of πr2,we can use π d2 or d2 0.7854.
4
When working out the area of a figure, two lengths are multiplied. In each calculating volumes, three lengths must be multiplied. Basically this means multiplying the surface are by the depth.
a. Rectangular tanks
The volume of a rectangular tank is calculated by multiplying the length (l) by the breadth (b) by the depth (h). It is important to remember that all dimensions must be in the same units, such as meters or millimeters.
If the dimensions of the tank are length 7.5 meters, breadth 2 meters, depth 1 meter, the volume (in cubic meters)n will be:
7.5 x 2 x 1 = 15 cubic meters (m2)
As there are 1000 liters in a cubic meter, the total capacity of the tank in liters is obtained by multiplying the volume by 1000, i.e.
15 x 1000 = 15 000 liters
Capacity of a rectangular tank (liter) = l x b x 1000
b. Rectangular tanks with a sloping base
The capacity of rectangular tanks with uniformly sloping bases such as swimming pools, can be obtained by proceeding by multiplying by the average depth, which is ascertained by adding together the values for the deep and shallow ends and dividing by 2.
Capacity = l x b x h1 + h2 (cubic meters)
2
= l x b x h1 + h2 x 1000 (liters)
2
The volume of a tank of a triangular section is obtained by multiplying the length by the breadth by one-half the depth.
Volume of a triangular tank = l x b x h
2
c. Circular Tanks
The volume of a circular tank is determined by calculating the surface area and multiplying by depth (h), all measurements being in the same units. This can be expressed as
π r2h, or π D2h, or 0.7854 x D2h.
4
The capacity of a circular tank 10 meters in diameter and 4 meters deep would be:
0.7854 x 10 x 10 x 4 = 314.16 cubic meters
Multiplying by 1000 gives the capacity in liters = 314 160 liters.
(1) Quick method i
The capacity of a circular tank has been shown to be 0.7854 x D2h cubic meters,
But 0.7854 ~ 0.8, therefore an approximation would be:
Quick formula I: Capacity of circular tank (m3) ~ 0.8 D2h
where D = diameter and h = depth, both in meters.
In the above example, this would give
0.8 x 10 x 10 x 4 = 320 cubic meters
(which is about 2 per cent too high.)
Alternatively, the capacity in liters can be obtained by multiplying by 1000
Capacity of circular tank (liter) = 800 C2h
2
Quick
formula II: capacity of a circular tank (cubic meter = 8C2h
To obtain, therefore, an approximation of the number of liters in 1 meter of hose, all that is necessary is to multiply the square of the diameter (in millimeters) by 8 and then divide the result by 10 000.
Capacity of 1 meter of hose (liter) = 8d2
10 000
90 x 90 x 8 = 6.48 liters
10 000
Comparing this with the true value of 6.36 liters per meter, this quick formula gives an over-estimation of about 2 per cent.
Similarly, for 70 mm hose, the capacity per meter would be approximately
70 x 70 x 8 = 3.92 liters
10 000
which, again, is slightly more than the true value of 3.85 liters per meter.
Using the same approximation as in (1) above, the capacity of a 25 meter length of hose becomes
8d2 x 25 = d2 liters
Capacity of a 25 meter length of hose (liter) = d2 (1a)
50
Thus for a 25 meter length of 70 mm hose, the capacity is:
70 x 70 = 4900 ~ 100 liters
50 50
and for 45 mm hose, the capacity per 25 meter length is:
45 x 45 = 2025 = 40 liters.
1. Concept Of Pressure And Head
The fourth principle of the characteristics of pressure in fluids is that the downward pressure of a liquid in an open vessel is proportional to its depth, or the height of the surface above the point at which the pressure is measured.
2. Characteristics Of Suction Lift
a. Basic Facts
A pump is said to lift water when the water is taken from an open source below the inlet of the pump. Water has not tensile strength and cannot therefore be pulled upward, and it is the atmospheric pressure only which raises the water.
The work performed by a priming device when the water is being lifted on the suction side is to create a partial vacuum within the pump chamber and suction hose. As the impeller revolves in a centrifugal pump, a partial vacuum is maintained at the impeller inlet and in the suction hose. The atmosphere exerts pressure on the open surface of the water and so forces the water up through the suction hose and into the pump. The mechanical condition of the pump and its ability to create a partial vacuum also has a bearing on the total height of the suction lift.
b. Qualifying Factors
It has been shown that water cannot rise to a vertical height greater than approximately 10 m in a completely evacuated tube, and that water rises because it is forced up by the atmospheric pressure outside. When a fire pump is got to work from open water the factors to contend with are:
i. lifting the water from its existing level to the level of the inlet of the pump;
ii. overcoming frictional resistance to the water both on entering and on passing through strainers and suction hose;
iii. turbulence as the water enters the pump impeller (that is known as entry loss);
iv. creating flow. A certain proportion of the available atmospheric pressure is used in crating flow in the water. This varies according to the velocity in the suction hose, but in all cases represents a relatively small proportion of the available pressure.
Because of the factors (ii) and (iii), it is obvious that a static suction lift of 10m cannot be obtained in practice, and, whilst suction lifts of 8.5 are sometimes obtained under very good conditions, 8 m can be considered the approximate practical maximum.
c. Pump Entry Loss
Entry loss, i.e. loss due to the shock of the water entering the impeller, varies with the design of the pump.
b. For the same velocity, friction loss decreases directly with the increase in diameter
This is extremely important, for if the diameter of a pipe is doubled, the surface area is doubled, but the cross-sectional area is quadrupled (the cross-sectional area being proportional to the square of the diameter). Therefore, for any given velocity, if the diameter of the hose is doubled, the quantity of water is increased to four times, and the loss of pressure due to friction is consequently halved. This emphasizes the importance of always consequently halved. This emphasizes the importance of always using the largest diameter hose or pipeline which is practicable.
c. Friction loss increases directly with the square of the velocity
This is another very important factor, as, if the velocity of the water is halved, the loss of pressure due to friction is reduced to (½)² or one –quarter.
This has its most important application when pumping over long distances. If 2000 liters of water per minute are being delivered through one line of hose with a loss of pressure due to friction of 6 bar, the act of twinning the hose and passing 1000 liters/min through each line will result in the velocity being halved, but the loss of pressure in each line being reduced to 1.5 bar.
d. Friction loss increases with the roughness of the interior of the pipe
That friction increases with roughness is apparent when attempting to rub one’s finger over the sandpaper compared with a polished surface. Similarly, it is more difficult to force water through a pipe with a rough interior, than one with a smooth inside surface.
The degree of internal roughness of a particular type of hose or pipe is referred to as its friction factor (f), or co-efficient of friction.
e. Friction loss, or for all practical purposes, is independent of pressure
Experiments show that the loss of pressure due to friction is independent of the pressure or head at which the system is operating. Thus, if the friction loss for a given flow rate in a certain line of hose or pipe is such that a pump pressure of 7 bar is required to maintain a nozzle pressure of 3 bar, then it will be found that in order to maintain a nozzle pressure of 6 bar at the same flowrate (i.e. using a smaller nozzle), a pump pressure of 10 bar is required. The friction loss (4 bar) is the same pressure of 10 bar is the same whatever the overall pressure in the system.
With hose, increase in pressure may result in certain indirect effects, such as a slight stretching giving an increase in cross-sectional area and thus a slight reduction in friction loss. On the other hand, a rise in pressure may result in some increase in length and, therefore, a somewhat greater frictional loss. These effects are, however, negligible in practice.
Much emphasis is laid on the necessity for slowly closing hydrants or shut-off branches, in order to avoid water hammer which might burst hose and damage water mains or pumps, and the reasons for its occurrence must, therefore, be understood. Water moving through a pipe presses both mass and velocity. Mathematically, it can be shown that the kinetic energy of a body varies with the square speed (kinetic energy = ½Mv², where M = mass (kg) and v = velocity (m/s)). Thus the effort required suddenly to arrest the movement of a body increases rapidly as its velocity rises.
This applies equally in the case of water moving through a pipe. If the flow is cut off suddenly, the kinetic energy possessed by the moving water is instantaneously converted into pressure energy which must be absorbed by the pipe and its fittings and the sudden shock may be so great that failure results, especially if the pipe is of brittle material.
1. Practical Considerations Of High Nozzle
The advantages of using a high nozzle pressure are:
i. great striking power;
ii. long reach;
iii. large volume of water.
Full advantage should be taken of branch holders, ground monitors, etc.
The pressure adopted will depend on:
iv. type and size of nozzle selected;
v. type of jet required;
vi. weather conditions, intensity of wind, etc.
2. Jet Reaction
When water is projected from a nozzle, a reaction equal and opposite to the force of the jets take place at the nozzle, and the nozzle tends to recoil in the opposite direction to the flow. Thus the man or men holding the branch must exert sufficient effort towards the jet to overcome this reaction.
1. Distribution Of Water Supplies
a. General
Water undertakers obtain their water from 3 main sources:
i. River intakes.
ii. Impounding reservoirs. These contain water collected from high ground, streams and general rainfall.
iii. Underground sources, e.g. wells, boreholes and springs.
About one-third of the total supply is drawn from each source but in each case the water is fed into main storage reservoirs, purified and then passed into the distribution system.
The distribution system conveys water to the consumer and, in general, consists of mains and pipes laid under public roads. There is no standard pattern for distribution system but it will usually consist of:
i. Trunk mains
ii. Secondary mains
iii. Service mains
iv. Service reservoirs
v. Booster pumps (where necessary).
b. Water Mains And Associated Features
i. Types Of Mains
Water mains fall into 3 groups: trunk, secondary and service. Due to the variations in supply systems, the difference between the groups is blurred. What is regarded as a trunk main by one undertaker may be deemed a secondary main by another.
The term trunk main is normally applied to a main which carries water from e.g. a pumping station to a reservoir or from a reservoir to a district.
Service mains are used to supply premises in the roads under which they are laid. Their carrying capacity usually is no larger than necessary to meet local demands.
ii. Service Reservoirs
3. Pressure And Flow In Mains
1. General Description
a. Principle of Operation
A centrifugal pump differs from positive displacement pumps in that it is unable to pump gases and does not work by displacement. It consists of two main parts, the impeller and the casing, each of which has a separate function. Basically, the impeller imparts a high velocity to the water, and then transforms most of the velocity energy into pressure energy.
b. The Impeller
The impeller is the spinning part of centrifugal pump. It is a circular metal casing with radial vanes, which rotates on a central shaft. Water received at its center is thrown outwards as it rotates and discharged at the periphery. As the water is thrown outwards, the pressure at the inlet decreases, creating a partial vacuum. This causes more water to be forced into the impeller from the supply source.
c. The Casing
The other main part of a centrifugal pump is the casing. The action of the impeller in thrusting water outwards naturally creates considerable turbulence and friction. As these factors represent wasted energy, it is necessary to reduce them to a minimum. The casing is designed to do this, by reducing the velocity of the water and imparting to it as steady and smooth a flow as possible. The casing may take the form of a volute and/or guide vanes.
(1) The Volute
Shaped like the shell of a snail, the volute is the simplest type of centrifugal pump casing. Its cross-sectional area gradually increases in a circular direction towards the outlet. The water leaving the impeller decreases in velocity and becomes less turbulent as it passes through the volute.
(2) Guide Vanes
Guide Vanes attached to the inner walls of the casing can also reduce turbulence. They are sometimes collectively known as a guide ring or diffuser.