Water can be moved either by carrying it or pumping it. Water is heavy and unstable substance and this poses great difficulty in carrying it. Hence, water is preferably moved by pumping it through pipes. Though water may be moved by gravity in exceptional cases where relief may permit that i.e where the source is on higher altitude than the destination. In that case, the water moves by gravity; otherwise, the water is moved by pumping using pumps.
To select the right type of pump for each pumping operation and to establish the power required to pump, the water engineer need to find out the optimum pipe diameter based on information about the amount of water to be pumped, the length of pipe, the static head and pipe frictional losses involved. In the choice of the pipe, balance is created between the cost of the pipe for larger diameter pipes and cost due to frictional losses/delay due to smaller diameter pipes and power requirements. This implies that there should be balance between capital costs and recurrent costs.
How to determine optimum pipe diameter to pump water
1. From the water friction loss chart, and using an assumed likely diameter, determine the coefficient of friction per 100 m (attach chart).
2. Calculate the friction head; friction head = (coefficient of friction x pipe length)/ 100.
3. Find the sum of friction head and static head.
4. Determine the percentage of friction in the pipe = (friction head/static head) x 100
Note: the value of the percentage of friction should preferably by below 15%.
5. Determine the power required to pump the water as follows: (rate of pumping x total head)/ 100 (kW) for fully efficient pump or (rate of pumping x total head)/ 50.5 for 50% efficient pump or using other more realizeable figures.
6. Use these figures to work out the corresponding pipe costs and pumping costs and make a choice often on the basis of 15% maximum friction over static head.
Types of pumps
This is double-acting pump (see Figure 1) that comprises essentially a cylinder barrel containing a plunger, suction and delivery valves, and a cross-head (a clever little mechanical device that converts motion from rotary to linear), to facilitate the pump being driven by a motor or engine. These pumps run at relatively slow speeds, and are useful where high heads and fairly low outputs are required, such as for animal or domestic use. They are made usually for outputs between 1 and 20 m3/h and heads up to 230 m. The hydraulic efficiency of piston is usually taken as 50% for power calculation.
Output from piston pumps, Q = [πD2l x 2N]/ [4 x 100]
D = diameter of the plunger (cm)
L = the stroke of the plunger in cm
N = revolution of the pulley in rpm
Q = output in l/m
For power requirement
hp = [QH/ 76] x [100/ E], where Q = flowrate or output (l/s) and H = head in m, E = efficiency in %
These types of pumps (see Figure 2) rely on the centrifugal or ‘outward flying’ effect created by the revolving water at high speed by means of a disc having blades (impeller) contained in a volute casing. The resulting ‘high pressure’ water is allowed to leave through a delivery branch.
Centrifugal pumps are produced in many different forms such as single stage, end suction pump, twin stage, horizontally split case pump, multistage turbine pump, self-priming mixed flow pump, and submersible electric or vertical spindle pump.
These pumps are compact and are suitable for higher output than reciprocating pumps. They are useful for irrigation, fire-fighting, rural and urban water supplies and can have hydraulic efficiencies as high as 85%.
Since the revolutions of the impeller in the pump cannot be altered, the periphery speed of the impeller is what determines its capable head. Thus, by reducing the diameter of the impeller, its periphery speed can be reduced. For practical purposes, the head, pump speed and impeller diameter can be taken to be in linear proportion to each other. Thus, v = πDN, where v = periphery velocity, πD = full diameter circumference and N = revolutions per minute.
To reduce the diameter based on the speed; (Required speed (n))/ (Committed Speed (N)) = (Reduced Impeller Diameter (d))/ (Full Diameter (D)) ≈ n/N = d/D ≈ d =(n/N) D.
Unlike the piston pumps, the hydraulic efficiency of centrifugal pumps varies according to the duty point. So, it is not easy to calculate output or power requirement. Each centrifugal pump has different characteristics and it is usual to read off the data required from the specific performance graph that has been operating from actual pump tests.
To start pumping;
a. Prime the system fully with water
b. Close the delivery valve
c. Startup power source
d. Gradually open up delivery valve
To close down;
a. Close delivery valve
b. Switch off power source
This type of pump (see Figure 3), also known as wing pumps are similar in principle to the piston pump having suction and delivery valves, but incorporating a semi-rotary vane instead of a plunger. The pumping efficiency depend on the small clearance between the vane and the casing which when worn out, it will cease to be able to lift water. They are also regarded as positive displacement pumps.
Hydraulic ram pumps
This pump (see Figure 4) makes use of the principle that water is incompressible. This, coupled with inertia describes how the pump works. This type of pump is not very efficient as large quantities of water are usually lost. However, it has no power requirement to drive it.
Mathematically, Efficiency = [wH x 100]/ Wh where W = weight of water flowing per second along the drive pipe; h = the head of water; w = weigh of water per second being delivered and H = the height of delivery of the water.
This pump (see Figure 5) works on the principle of Bernoulli which mathematically states thus:
Q = A x V1 = a x V2 where A = area of tube of large section of diameter, D and a = area of tube of smaller section of diameter, d; V1 and V2 are the corresponding velocities at these points; Q = flow rate of water.
Or (P1/w) + (V22/2g) = (P2/w) + (V22/2g), where w = density and g = acceleration due to gravity.
Jet pumps have poor hydraulic efficiency but are fairly cheap. They are cheap, in that no moving parts exist down the borehole, making its maintenance simple. They are often limited to pump 6 litres/second from about 40 m heights.
They are often called sludge pumps (see Figure 6) because of their ability to handle very dirty water. They belong to the class of positive displacement type of pump. The diaphragm can be made of rubber or special flexible plastic materials, depending on the fluid being handled.
Table 1: Information for choice of pump
In summary, choice of pump should consider purpose, hydraulic efficiency, running costs, capital costs, availability of spare parts, technical set up and maintenance.
How to determine power requirement of a pump
Power requirement to pump water depends on the weight of the water, the distance moved and the time taken to do it. These are usually quantified with measures of useful work done such as Horsepower (hp) and KiloWatts (kW).
Horsepower is defined as the power an engine produces.
One horsepower = 4500 kg-m/minute or 76.04 kg-m/s = 745.67 Watts = 0.746 kW.
It is the time taken to move 4500 kg of weight through 1 metre distance taking a minute to carry out the work.
When horse power and kW is related, it is called water horsepower (whp). whp is a measure of work done on water and can be applied to work done by water.
whp or water kilowatt = (litres per second x metres head)/ 101.97. The above value is usually theoretical because pumps are not 100% efficient. When the ratio of water power to actual power is expressed as a percentage, it is called efficiency.
Thus, efficiency = (water power/ actual power) x 100.
Actual power required = (water power x 100)/ efficiency, %.
Break horse power (Bhp) = (whp x 100)/ E, % or kW = (wkW x 100)/ E, %.
Willcock, D.B. Water Engineering. Evans Basic Technology