Pumps

6. PU PUMP MPS S AND AND PUM PUMP PIN ING G SY SYST STEM EM Syllabus Pumps and Pumping System: Types, Performance evaluation, Efficient system operation, Flow control strategies and energy conservation opportunities

6.1

Pump Types

Pumps come in a variety of sizes for a wide range of applications. They can be classified according to their basic operating principle as dynamic or displacement pumps. Dynamic pumps can be sub-classified as centrifugal and special effect pumps. Displacement pumps can be sub-classified as rotary or reciprocating pumps. In principle, any liquid can be handled by any of the pump designs. Where different pump designs could be used, the centrifugal pump is generally the most economical followed by rotary and reciprocating pumps. Although, positive displacement pumps are generally more efficient than centrifugal pumps, the benefit of higher efficiency tends to be offset by increased maintenance costs. Since, worldwide, centrifugal pumps account for the majority of electricity used by pumps, the focus of this chapter is on centrifugal pump. Centrifugal Pumps A centrifugal pump is of a very simple design. The two main parts of the pump are the impeller impeller and the diffuser. Impeller, which is the only moving part, is attached to a shaft and driven by a motor. Impellers are generally made of bronze, polycarbonate, cast iron, stainless steel as well as other materials. The diffuser diffuser (also called as volute) houses the impeller and captures and directs the water off the impeller. Water enters the center (eye) of the impeller and exits the impeller with the help of centrifugal force. As water leaves the eye of the impeller a low-pressure area is created, causing more water to flow into the eye. Atmospheric pressure and centrifugal force cause this to happen. Velocity Velocity is developed as the water flows through the impeller spinning at high speed. The water velocity is collected by the diffuser and converted to pressure by specially designed passageways that direct the flow to the discharge of the pump, or to the next impeller should the pump have a multi-stage configuration. The pressure (head) that a pump will develop is in Figuree 6.1 Figur 6.1 Centr Centrifuga ifugall pump pump direct relationship to the impeller diameter, the number of impellers, the size of impeller eye, and shaft speed. Capacity is determined by the exit width of the impeller. The head and capacity are the main factors, which affect the horsepower size of the motor to be used. The more the quantity of water to be pumped, the more energy is required. Bureau of Energy Efficiency

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A centrifugal pump is not positive positive acting; it will not pump the same volume volume always. The greater the depth of the water, the lesser is the flow from the pump. Also, Also, when it pumps against increasing pressure, the less it will pump. For these reasons it is important to select a centrifugal pump that is designed to do a particular job. Since the pump is a dynamic device, it is convenient to consider the pressure in terms of head i.e. meters of liquid column. The pump generates the same head of liquid whatever the density of the liquid being pumped. The actual contours of the hydraulic passages of the impeller and the casing are extremely important, in order to attain the highest efficiency possible. The standard convention for centrifugal pump is to draw the pump performance curves showing Flow on the horizontal axis and Head generated on the vertical axis. Efficiency Efficiency,, Power & NPSH Required (described later), are conventionally shown on the vertical axis, plotted against Flow, Flow, as illustrated in Figure 6.2.

Figuree 6.2 Pump Perfo Figur Performanc rmancee Curve Curve

Given the significant amount of electricity attributed to pumping systems, even small improvements in pumping efficiency could yield very significant savings of electricity. The pump is among the most inefficient of the components that comprise a pumping system, including the motor, transmission drive, piping and valves. Hydraulic power, power, pump shaft power and electrical input power power Total head, h d - hs (m) x ρ (kg/m3) x g (m/s 2) / 1000 Hydrauli Hydr aulicc power power Ph = Q (m3 /s) x Total Where hd – dis disch char arge ge he head ad,, hs – suc sucti tion on he head ad,, ρ – den densi sity ty of th thee flu fluid id,, g – acc accel eler erat atio ion n due due to gr grav avit ityy

efficiency, ηPump Pump shaft powe powerr Ps = Hydraulic power, P h / pump efficiency, Electrical input power = Pump shaft power P s ηMotor

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6.2 6. 2

Sys ysttem Char Charac actteri rist stic icss

In a pumping system, the objective, in most cases, is either to transfer a liquid from a source to a required destination, e.g. filling a high level reservoir, or to circulate liquid around a system, e.g. as a means of heat transfer in heat exchanger. A pressure is needed to make the liquid liquid flow at the required rate and this must overcome head 'losses' in the system. Losses are of two types: static and friction head. Static head is simply the difference in height of the supply and destination reservoirs, as in Figure 6.3. In this illustration, flow velocity in the pipe is assumed to be very small. Another example of a system with only static head is pumping into a pressurised vessel with short pipe runs. Static head is independent of flow and graphically would be shown as in Figure 6.4.

Figure 6.3 Static Head

Figure 6.4 Static Head vs. Flow

Friction head (sometimes called dynamic head loss) is the friction loss, on the liquid being moved, in pipes, valves and equipment in the system. Friction tables are universally available for various pipe fittings and valves. These tables show friction loss per 100 feet (or metres) of a specific pipe size at various flow rates. In case of fittings, friction is stated as an equivalent length of pipe of the same size. The friction losses are proportional to the square of the flow rate. A closed loop circulating system without a surface open to atmospheric pressure, would exhibit only friction losses and would have a system friction head loss vs. flow curve as Figure 6.5.

Figure Figu re 6.5 6.5 Frict Friction ion Head Head vs. vs. Flow Flow Bureau of Energy Efficiency

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Most systems have a combination of static and friction head and the system curves for two cases are shown in Figures 6.6 and 6.7. The ratio of static to friction head over the operating range influences the benefits achievable from variable speed drives which shall be discussed later.

Figure 6.6 System with High Static Head

Figure 6.7 System with Low Static Head

Static head is a characteristic of the specific installation and reducing this head where this is possible, generally helps both the cost of the installation and the cost of pumping the liquid. Friction head losses must be minimised to reduce pumping cost, but after eliminating unnecessary pipe fittings and length, further reduction in friction head will require larger diameter pipe, which adds to installation cost.

6.3

Pump Curves

The performance of a pump can be expressed graphically as head against flow rate. The centrifugal pump has a curve where the head falls gradually with increasing flow. This is called the pump characteristic curve (Head - Flow curve) -see Figure 6.8.

Figure 6.8 Head- Flow Curve Bureau of Energy Efficiency

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Pump operating point When a pump is installed in a system the effect can be illustrated graphically by superimposing pump and system curves. The operating point will always be where the two curves intersect. Figure 6.9.

Figure 6.9 Pump Operating Point

If the actual system curve is different in reality to that calculated, the pump will operate at a flow and head different to that expected. For a centrifugal pump, an increasing system resistance will reduce the flow, eventually to zero, but the maximum head is limited as shown. Even so, this condition is only acceptable for a short period without causing problems. An error in the system curve calculation is also likely to lead to a centrifugal pump selection, which is less than optimal for the actual system head losses. Adding safety margins to the calculated system curve to ensure that a sufficiently large pump is selected will generally result in installing an oversized pump, which will operate at an excessive flow rate or in a throttled condition, which increases energy usage and reduces pump life.

6.4

Factors Affecting Pump Performance

Matching Pump and System Head-flow Characteristics Centrifugal pumps are characterized by the relationship between the flow rate (Q) they produce and the pressure (H) at which the flow is delivered. Pump efficiency varies with flow and pressure, and it is highest at one particular flow rate. Bureau of Energy Efficiency

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The Figure 6.10 below shows a typical vendor-supplied head-flow curve for a centrifugal pump. Pump head-flow curves are typically given for clear water. The choice of pump for a given application depends largely on how the pump head-flow characteristics match the requirement of the system downstream of the pump.

Figure 6.10 Typical Centrifugal Pump Performance Curve

Effect of over sizing the pump As mentioned earlier, pressure losses to be overcome by the pumps are function of flow – the system characteristics – are also quantified in the form of head-flow curves. The system curve is basically a plot of system resistance i.e. head to be overcome by the pump versus various flow rates. The system curves change with the physical configuration of the system; for example, the system curves depends upon height or elevation, diameter and length of piping, number and type of fittings and pressure drops across various equipment - say a heat exchanger. A pump is selected based on how well the pump curve and system head-flow curves match. The pump operating point is identified as the point, where the system curve crosses the pump curve when they are superimposed on each other.


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The Figure 6.11 shows the effect on system curve with throttling.

Figure 6.11 Effect on System Curve with Throttling

In the system under consideration, water has to be first lifted to a height – this represents the static head. Then, we make a system curve, considering the friction and pressure drops in the systemthis is shown as the green curve. Suppose, we have estimated our operating conditions as 500 m 3 /hr flow and 50 m head, we will chose a pump curve which intersects the system curve (Point A) at the pump's best efficiency point (BEP). But, in actual operation, we find that 300 m 3 /hr is sufficient. The reduction in flow rate has to be effected by a throttle valve. In other words, we are introducing an artificial resistance in the system. Due to this additional resistance, the frictional part of the system curve increases and thus the new system curve will shift to the left -this is shown as the red curve. So the pump has to overcome additional pressure in order to deliver the reduced flow. Now, the new system curve will intersect the pump curve at point B. The revised parameters are 300 m3 /hr at 70 m head. The red double arrow line shows the additional pressure drop due to throttling. You may note that the best efficiency point has shifted from 82% to 77% efficiency. So what we want is to actually operate at point C which is 300 m 3 /hr on the original system curve. The head required at this point is only 42 meters. What we now need is a new pump which will operate with its best efficiency point at C. But there are other simpler options rather than replacing the pump. The speed of the pump can be reduced or the existing impeller can be trimmed (or new lower size impeller). The blue pump curve represents either of these options.


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Energy loss in throttling Consider a case (see Figure 6.12) where we need to pump 68 m 3 /hr of water at 47 m head. The pump characteristic curves (A…E) for a range of pumps are given in the Figure 6.12.

Figure 6.12 Pump Characteristic Curves


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6.5

Efficient Pumping System Operation

To understand a pumping system, one must realize that all of its components are interdependent. When examining or designing a pump system, the process demands must first be established and most energy efficiency solution introduced. For example, does the flow rate have to be regulated continuously or in steps? Can on-off batch pumping be used? What are the flow rates needed and how are they distributed in time? The first step to achieve energy efficiency in pumping system is to target the end-use. A plant water balance would establish usage pattern and highlight areas where water consumption can be reduced or optimized. Good water conservation measures, alone, may eliminate the need for some pumps. Once flow requirements are optimized, then the pumping system can be analysed for energy conservation opportunities. Basically this means matching the pump to requirements by adopting proper flow control strategies. Common symptoms that indicate opportunities for energy efficiency in pumps are given in the Table 6.1.

TABLE 6.1 SYMPTOMS THAT INDICATE POTENTIAL OPPORTUNITY FOR ENERGY SAVINGS

Symptom

Likely Reason

Best Solutions

Throttle valve-controlled systems

Oversized pump

Trim impeller, smaller impeller, variable speed drive, two speed drive, lower rpm

Bypass line (partially or completely) open

Oversized pump

Trim impeller, smaller impeller, variable speed drive, two speed drive, lower rpm

Multiple parallel pump system with the same number of pumps always operating

Pump use not monitored or controlled

Install controls

Constant pump operation in a batch environment

Wrong system design

On-off controls

High maintenance cost (seals, bearings)

Pump operated far away from BEP

Match pump capacity with system requirement


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Effect of speed variation As stated above, a centrifugal pump is a dynamic device with the head generated from a rotating impeller. There is therefore a relationship between impeller peripheral velocity and generated head. Peripheral velocity is directly related to shaft rotational speed, for a fixed impeller diameter and so varying the rotational speed has a direct effect on the performance of the pump. All the parameters shown in fig 6.2 will change if the speed is varied and it is important to have an appreciation of how these parameters vary in order to safely control a pump at different speeds. The equations relating rotodynamic pump performance parameters of flow, head and power absorbed, to speed are known as the Affinity Laws:

Where: Q = Flow rate H = Head P = Power absorbed N = Rotating speed Efficiency is essentially independent of speed Flow: Flow is proportional to the speed Q1 / Q2 = N1 / N2 Example: 100 / Q2 = 1750/3500 Q2 = 200 m3 /hr Head: Head is proportional to the square of speed H1 /H2 = (N12) / (N22) Example: 100 /H 2 = 17502 / 3500 2 H2 = 400 m Power(kW): Power is proportional to the cube of speed kW1 / kW2 = (N13) / (N23) Example: 5/kW2 = 17503 / 35003 kW2 = 40 As can be seen from the above laws, doubling the speed of the centrifugal pump will increase the power consumption by 8 times. Conversely a small reduction in speed will result in drastic reduction in power consumption. This forms the basis for energy conservation in centrifugal pumps with varying flow requirements. The implication of this can be better understood as shown in an example of a centrifugal pump in Figure 6.13 below. Bureau of Energy Efficiency

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Figure 6.13 Example of Speed Variation Effecting Effecting Centrifugal Pump Performance

Points of equal efficiency on the curves for the 3 different speeds are joined to make the isoefficiency lines, showing that efficiency remains constant over small changes of speed providing the pump continues to operate at the same position related to its best efficiency point (BEP). The affinity laws laws give a good approximation approximation of how pump performance curves change change with speed but in order to obtain the actual performance of the pump in a system, the system curve also has to be taken into account. Effects Effec ts of impe impeller ller diam diameter eter chan change ge Changing the impeller diameter gives a proportional change in peripheral velocity, velocity, so it follows that there are equations, similar to the affinity laws, for the variation of performance with impeller diameter D:

Efficiency varies when the diameter is changed within a particular casing. Note the difference in iso-efficiency lines in Figure 6.14 compared with Figure 6.13. The relationships shown here apply to the case for changing only the diameter of an impeller within a fixed casing geometry, which is a common practice for making small permanent adjustments to the performance of a centrifugal pump. Diameter changes are generally limited to reducing the diameter to about 75% of the maximum, i.e. a head reduction to about 50%. Beyond this, efficiency and NPSH are badly affected. However speed change can be used over a wider range without seriously reducing efficiency. For example reducing the speed by 50% typically results in a reduction of efficiency by 1 or 2 percentage points. The reason for the small loss of efficiency with the lower speed is that Bureau of Energy Efficiency

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mechanical losses in seals and bearings, which generally represent <5% of total power, are proportional to speed, rather than speed cubed. It should be noted that if the change in diameter is more than about 5%, the accuracy of the squared and cubic relationships can fall off and for precise calculations, the pump manufacturer's performance curves should be referred to.

Figuree 6.14 Exam Figur Example: ple: Impeller Impeller Diame Diameter ter Reduc Reduction tion on Centrif Centrifugal ugal Pump Pump Performanc Performancee

The illustrated curves are typical of most centrifugal pump types. Certain high flow, low head pumps have performance curve shapes somewhat different and have a reduced operating region of flows. This requires additional care in matching the pump to the system, when changing speed and diameter. Pump suction performance (NPSH) Liquid entering the impeller eye turns and is split into separate streams by the leading edges of the impeller vanes, an action which locally drops the pressure below that in the inlet pipe to the pump. If the incoming liquid is at a pressure with insufficient margin above its vapour pressure, then vapour cavities or bubbles appear along the impeller vanes just behind the inlet edges. This phenomenon is known as cavitation and has three undesirable effects: 1) The collapsing cavitation bubbles can erode the vane surface, especially especially when pumping water-based liquids. 2) Noise and vibration are increased, with possible shortened seal and bearing life. 3) The cavity areas will initially initially partially partially choke the impeller passages and reduce the pump performance. In extreme cases, total loss of pump developed head occurs. The value, by which the pressure in the pump suction exceeds the liquid vapour pressure, is expressed as a head of liquid and referred to as Net Positive Suction Head Available – (NPSHA). This is a characteristic of the system design. The value of NPSH needed at the pump suction to prevent the pump from cavitating is known as NPSH Required – (NPSHR). This is a characteristic of the pump design. The three undesirable effects of cavitation described above begin at different values of NPSHA and generally there will be cavitation cavitation erosion before there is a noticeable noticeable loss of pump Bureau of Energy Efficiency

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head. However for a consistent approach, manufacturers and industry standards, usually define the onset of cavitation as the value of NPSHR when there is a head drop of 3% compared with the head with cavitation free performance. At this point cavitation is present and prolonged operation at this point will usually lead to damage. It is usual therefore to apply a margin bywhich bywhi ch NPSHA shoul should d exceed NPSHR. As would be expected, the NPSHR increases as the flow through the pump increases, see fig 6.2. In addition, as flow increases in the suction pipework, friction losses also increase, giving a lower NPSHA at the pump suction, both of which give a greater chance that cavitation will occur. NPSHR also varies approximately with the square of speed in the same way as pump head and conversion of NPSHR from one speed to another can be made using the following equations. Q ∝ N NPSHR ∝ N 2 It should be noted however that at very low speeds there is a minimum NPSHR plateau, NPSHR does not tend to zero at zero speed It is therefore essential to carefully consider NPSH in variable speed pumping.

6.6 6. 6

Flow Fl ow Co Cont ntrrol Str trat ateg egie iess

Pump control by varying speed To understand how speed variation changes the duty point, the pump and system curves are over-laid. Two Two systems are considered, one with only friction loss and another where static head is high in relation to friction head. It will be seen that the benefits are different. In Figure 6.15,

Figure 6.15 6.15 Example of the the Effect of Pump Pump Speed Change in a System System With Only Friction Loss Loss Bureau of Energy Efficiency

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reducing speed in the friction loss system moves the intersection point on the system curve along a line of constant efficiency. The operating point of the pump, relative to its best efficiency point, remains constant and the pump continues to operate in its ideal region. The affinity laws are obeyed which means that there is a substantial reduction in power absorbed accompanying the reduction in flow and head, making variable speed the ideal control method for systems with friction loss. In a system where static head is high, as illustrated in Figure 6.16, the operating point for the pump moves relative to the lines of constant pump efficiency when the speed is changed. The reduction in flow is no longer proportional to speed. A small turn down in speed could give a big reduction in flow rate and pump efficiency, which could result in the pump operating in a region where it could be damaged if it ran for an extended period of time even at the lower speed. At the lowest speed illustrated, (1184 rpm), the pump does not generate sufficient head to pump any liquid into the system, i.e. pump efficiency and flow rate are zero and with energy still being input to the liquid, the pump becomes a water heater and damaging temperatures can quickly be reached.

Figure 6.16 Example for the Effect of Pump Speed Change with a System with High Static Head.

The drop in pump efficiency during speed reduction in a system with static head, reduces the economic benefits of variable speed control. There may still be overall benefits but economics should be examined on a case-by-case basis. Usually it is advantageous to select the pump such that the system curve intersects the full speed pump curve to the right of best efficiency, in order that the efficiency will first increase as the speed is reduced and then decrease. This can extend the useful range of variable speed operation in a system with static head. The pump manufacturer should be consulted on the safe operating range of the pump. Bureau of Energy Efficiency

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It is relevant to note that flow control by speed regulation is always more efficient than by control valve. In addition to energy savings there could be other benefits of lower speed. The hydraulic forces on the impeller, created by the pressure profile inside the pump casing, reduce approximately with the square of speed. These forces are carried by the pump bearings and so reducing speed increases bearing life. It can be shown that for a centrifugal pump, bearing life is inversely proportional to the 7 th power of speed. In addition, vibration and noise are reduced and seal life is increased providing the duty point remains within the allowable operating range. The corollary to this is that small increases in the speed of a pump significantly increase power absorbed, shaft stress and bearing loads. It should be remembered that the pump and motor must be sized for the maximum speed at which the pump set will operate. At higher speed the noise and vibration from both pump and motor will increase, although for small increases the change will be small. If the liquid contains abrasive particles, increasing speed will give a corresponding increase in surface wear in the pump and pipework. The effect on the mechanical seal of the change in seal chamber pressure, should be reviewed with the pump or seal manufacturer, if the speed increase is large. Conventional mechanical seals operate satisfactorily at very low speeds and generally there is no requirement for a minimum speed to be specified, however due to their method of operation, gas seals require a minimum peripheral speed of 5 m/s. Pumps in parallel switched to meet demand Another energy efficient method of flow control, particularly for systems where static head is a high proportion of the total, is to install two or more pumps to operate in parallel. Variation of flow rate is achieved by switching on and off additional pumps to meet demand. The combined pump curve is obtained by adding the flow rates at a specific head. The head/flow rate curves for two and three pumps are shown in Figure 6.17.

Figure 6.17 Typical Head-Flow Curves for Pumps in Parallel

The system curve is usually not affected by the number of pumps that are running. For a system with a combination of static and friction head loss, it can be seen, in Figure 6.18, that Bureau of Energy Efficiency

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the operating point of the pumps on their performance curves moves to a higher head and hence lower flow rate per pump, as more pumps are started. It is also apparent that the flow rate with two pumps running is not double that of a single pump. If the system head were only static, then flow rate would be proportional to the number of pumps operating. It is possible to run pumps of different sizes in parallel provided their closed valve heads are similar. By arranging different combinations of pumps running together, a larger number of different flow rates can be provided into the system. Care must be taken when running pumps in parallel to ensure that the operating point of the pump is controlled within the region deemed as acceptable by the manufacturer. It can be seen from Figure 6.18 that if 1 or 2 pumps were stopped then the remaining pump(s) would operate well out along the curve where NPSH is higher and vibration level increased, giving an increased risk of operating problems.

Figure 6.18 Typical Head-Flow Curves for Pumps in Parallel, With System Curve Illustrated.

Stop/start control In this control method, the flow is controlled by switching pumps on or off. It is necessary to have a storage capacity in the system e.g. a wet well, an elevated tank or an accumulator type pressure vessel. The storage can provide a steady flow to the system with an intermittent operating pump. When the pump runs, it does so at the chosen (presumably optimum) duty point and when it is off, there is no energy consumption. If intermittent flow, stop/start operation and the storage facility are acceptable, this is an effective approach to minimise energy consumption. The stop/start operation causes additional loads on the power transmiss ion components and increased heating in the motor. The frequency of the stop/start cycle should be within the motor design criteria and checked with the pump manufacturer. It may also be used to benefit from "off peak" energy tariffs by arranging the run times during the low tariff periods. To minimise energy consumption with stop start control it is better to pump at as low flow rate as the process permits. This minimises friction losses in the pipe and an appropriately small pump can be installed. For example, pumping at half the flow rate for twice as long can reduce energy consumption to a quarter. Bureau of Energy Efficiency

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Flow control valve With this control method, the pump runs continuously and a valve in the pump discharge line is opened or closed to adjust the flow to the required value.

Figure 6.19 Control of Pump Flow by Changing System Resistance Using a Valve.

To understand how the flow rate is controlled, see Figure 6.19. With the valve fully open, the pump operates at "Flow 1". When the valve is partially closed it introduces an additional friction loss in the system, which is proportional to flow squared. The new system curve cuts the pump curve at "Flow 2", which is the new operating point. The head difference between the two curves is the pressure drop across the valve. It is usual practice with valve control to have the valve 10% shut even at maximum flow. Energy is therefore wasted overcoming the resistance through the valve at all flow conditions. There is some reduction in pump power absorbed at the lower flow rate (see Figure 6.19), but the flow multiplied by the head drop across the valve, is wasted energy. It should also be noted that, while the pump will accommodate changes in its operating point as far as it is able within its performance range, it can be forced to operate high on the curve, where its efficiency is low, and its reliability is affected. Maintenance cost of control valves can be high, particularly on corrosive and solids-containing liquids. Therefore, the lifetime cost could be unnecessarily high. By-pass control With this control approach, the pump runs continuously at the maximum process demand duty, with a permanent by-pass line attached to the outlet. When a lower flow is required the surplus liquid is bypassed and returned to the supply source. An alternative configuration may have a tank supplying a varying process demand, which is kept full by a fixed duty pump running at the peak flow rate. Most of the time the tank overBureau of Energy Efficiency

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flows and recycles back to the pump suction. This is even less energy efficient than a control valve because there is no reduction in power consumption with reduced process demand. The small by-pass line sometimes installed to prevent a pump running at zero flow is not a means of flow control, but required for the safe operation of the pump. Fixed Flow reduction Impeller trimming

Impeller trimming refers to the process of machining the diameter of an impeller to reduce the energy added to the system fluid. Impeller trimming offers a useful correction to pumps that, through overly conservative design practices or changes in system loads are oversized for their application. Trimming an impeller provides a level of correction below buying a smaller impeller from the pump manufacturer. But in many cases, the next smaller size impeller is too small for the pump load. Also, smaller impellers may not be available for the pump size in question and impeller trimming is the only practical alternative short of replacing the entire pump/motor assembly. (see Figures 6.20 & 6.21 for before and after impeller trimming). Impeller trimming reduces tip speed, which in turn directly lowers the amount of energy imparted to the system fluid and lowers both the flow and pressure generated by the pump. The Affinity Laws, which describe centrifugal pump performance, provide a theoretical relationship between impeller size and pump output (assuming constant pump speed):

Figure 6.20 Before Impeller trimming

Figure 6.21 After Impeller Trimming

Where: Q = flow H = head BHP = brake horsepower of the pump motor Subscript 1 = original pump, Subscript 2 = pump after impeller trimming D = Diameter Bureau of Energy Efficiency

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Trimming an impeller changes its operating efficiency, and the non-linearities of the Affinity Laws with respect to impeller machining complicate the prediction of pump performance. Consequently, impeller diameters are rarely reduced below 70 percent of their original size. Meeting variable flow reduction Variable Speed Drives (VSDs)

In contrast, pump speed adjustments provide the most efficient means of controlling pump flow. By reducing pump speed, less energy is imparted to the fluid and less energy needs to be throttled or bypassed. There are two primary methods of reducing pump speed: multiple-speed pump motors and variable speed drives (VSDs). Although both directly control pump output, multiple-speed motors and VSDs serve entirely separate applications. Multiple-speed motors contain a different set of windings for each motor speed; consequently, they are more expensive and less efficient than single speed motors. Multiple speed motors also lack subtle speed changing capabilities within discrete speeds. VSDs allow pump speed adjustments over a continuous range, avoiding the need to jump from speed to speed as with multiple-speed pumps. VSDs control pump speeds using several different types of mechanical and electrical systems. Mechanical VSDs include hydraulic clutches, fluid couplings, and adjustable belts and pulleys. Electrical VSDs include eddy current clutches, woundrotor motor controllers, and variable frequency drives (VFDs). VFDs adjust the electrical frequency of the power supplied to a motor to change the motor's rotational speed. VFDs are by far the most popular type of VSD. However, pump speed adjustment is not appropriate for all systems. In applications with high static head, slowing a pump risks inducing vibrations and creating performance problems that are similar to those found when a pump operates against its shutoff head. For Figure 6.22 Effect of VFD systems in which the static head repreBureau of Energy Efficiency

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sents a large portion of the total head, caution should be used in deciding whether to use VFDs. Operators should review the performance of VFDs in similar applications and consult VFD manufacturers to avoid the damage that can result when a pump operates too slowly against high static head. For many systems, VFDs offer a means to improve pump operating efficiency despite changes in operating conditions. The effect of slowing pump speed on pump operation is illustrated by the three curves in Figure 6.22. When a VFD slows a pump, its head/flow and brake horsepower (BHP) curves drop down and to the left and its efficiency curve shifts to the left. This efficiency response provides an essential cost advantage; by keeping the operating efficiency as high as possible across variations in the system's flow demand, the energy and maintenance costs of the pump can be significantly reduced. VFDs may offer operating cost reductions by allowing higher pump operating efficiency, but the principal savings derive from the reduction in frictional or bypass flow losses. Using a system perspective to identify areas in which fluid energy is dissipated in non-useful work often reveals opportunities for operating cost reductions. For example, in many systems, increasing flow through bypass lines does not noticeably impact the backpressure on a pump. Consequently, in these applications pump efficiency does not necessarily decline during periods of low flow demand. By analyzing the entire system, however, the energy lost in pushing fluid through bypass lines and across throttle valves can be identified. Another system benefit of VFDs is a soft start capability. During startup, most motors experience in-rush currents that are 5 – 6 times higher than normal operating currents. This high current fades when the motor spins up to normal speed. VFDs allow the motor to be started with a lower startup current (usually only about 1.5 times the normal operating current). This reduces wear on the motor and its controller.

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Energy Conservation Opportunities in Pumping Systems Ensure adequate NPSH at site of installation Ensure availability of basic instruments at pumps like pressure gauges, flow meters. Operate pumps near best efficiency point. Modify pumping system and pumps losses to minimize throttling. Adapt to wide load variation with variable speed drives or sequenced control of multiple units. Stop running multiple pumps - add an auto-start for an on-line spare or add a booster pump in the problem area. Use booster pumps for small loads requiring higher pressures. Increase fluid temperature differentials to reduce pumping rates in case of heat exchangers. Repair seals and packing to minimize water loss by dripping. Balance the system to minimize flows and reduce pump power requirements. Avoid pumping head with a free-fall return (gravity); Use siphon effect to advantage: Conduct water balance to minimise water consumption Avoid cooling water re-circulation in DG sets, air compressors, refrigeration systems, cooling towers feed water pumps, condenser pumps and process pumps.


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In multiple pump operations, carefully combine the operation of pumps to avoid throttling Provide booster pump for few areas of higher head Replace old pumps by energy efficient pumps In the case of over designed pump, provide variable speed drive, or downsize / replace impeller or replace with correct sized pump for efficient operation. Optimise number of stages in multi-stage pump in case of head margins Reduce system resistance by pressure drop assessment and pipe size optimisation


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QUESTIONS 1.

What is NPSH of a pump and effects of inadequate NPSH?

2.

State the affinity laws as applicable to centrifugal pumps?

3.

Explain what do you understand by static head and friction head?

4.

What are the various methods of pump capacity control normally adopted?

5.

Briefly explain with a diagram the energy loss due to throttling in a centrifugal pump.

6.

Briefly explain with a sketch the concept of pump head flow characteristics and system resistance.

7.

What are the effects of over sizing a pump?

8.

If the speed of the pump is doubled, power goes up by a) 2 times b) 6 times c) 8 times d) 4 times

9.

How does the pump performance vary with impeller diameter?

10.

State the relationship between liquid kW, flow and pressure in a pumping application.

11.

Draw a pump curve for parallel operation of pumps (2 nos).

12.

Draw a pump curve for series operation of pumps (2 nos).

13.

List down few energy conservation opportunities in pumping system.

REFERENCES 1. 2. 3.

British Pump Manufacturers' Association BEE (EMC) Inputs PCRA Literature


134

Boehm, R. F. “Pumps and Fans” The Engineering Handbook. Ed. Richard C. Dorf Boca Raton: CRC Press LLC, 2000

© 1998 by CRC PRESS LLC

39 Pumps and Fans 39.1 Pumps Centrifugal and Other Velocity-Head Pumps

•

Positive-Displacement Pumps • Pump/Flow Considerations

39.2 Fans

Robert F. Boehm University of Nevada, Las Vegas

Pumps are devices that impart a pressure increase to a liquid. Fans are used to increase the velocity of a gas, but this is also accomplished through an increase in pressure. The pressure rise found in pumps can vary tremendously, which is a very important design parameter along with the liquid flow rate. This pressure rise can range from simply increasing the elevation of the liquid to increasing the pressure hundreds of atmospheres. Fan applications, on the other hand, deal with generally small pressure increases. In spite of this seemingly significant distinction between pumps and fans, there are many similarities in the fundamentals of certain types of these machines, as well as in their application and theory of operation. The appropriate use of pumps and fans depends on the proper choice of device and the proper design and installation for the application. A check of sources of commercial equipment shows that many varieties of pumps and fans exist. Each of these has special characteristics that must be appreciated for achieving proper function. Preliminary design criteria for choosing between different types is given by Boehm [1987]. As one might expect, the wise application of pumps and fans requires knowledge of fluid flow fundamentals. Unless the fluid mechanics of a particular application is understood, the design could be less than desirable. In this section, pump and fan types are briefly defined. In addition, typical application information is given. Also, some ideas from fluid mechanics that are especially relevant to pump and fan operation are reviewed. For more details on this latter topic, see the section of this book that discusses fluid mechanics fundamentals.

39.1 Pumps The raising of water from wells and cisterns was the earliest form of pumping [a very detailed history of early applications is given by Ewbank (1842)]. Modern applications are much broader, and these find a wide variety of machines in use. Modern pumps function on one of two principles. By far the majority of pump installations are of the velocity-head type. In these devices the pressure rise is achieved by giving the fluid a


movement. At the exit of the machine, this movement is translated into a pressure increase. The other major type of pump is called a positive-displacement pump. These devices are designed to increase the pressure on the liquid while essentially trying to compress the volume. A categorization of pump types has been given by Krutzsch [1986]; an adaptation of this categorization is shown below. I. Velocity head A. Centrifugal 1. Axial flow (single or multistage) 2. Radial flow (single or double suction) 3. Mixed flow (single or double suction) 4. Peripheral (single or multistage) B. Special effect 1. Gas lift 2. Jet 3. Hydraulic ram 4. Electromagnetic II. Positive displacement A. Reciprocating 1. Piston, plunger a. Direct acting (simplex or duplex) b. Power (single or double acting, simplex, duplex, triplex, multiplex) 2. Diaphragm (mechanically or fluid driven, simplex or multiplex) B. Rotary 1. Single rotor (vane, piston, screw, flexible member, peristaltic) 2. Multiple rotor (gear, lobe, screw, circumferential piston) In the next section, some of the more common pumps are described.

Centrifugal and Other Velocity-Head Pumps Centrifugal pumps are used in more industrial applications than any other kind of pump. This is primarily because these pumps offer low initial and upkeep costs. Traditionally these pumps have been limited to low-pressure-head applications, but modern pump designs have overcome this problem unless very high pressures are required. Some of the other good characteristics of these types of devices include smooth (nonpulsating) flow and the ability to tolerate nonflow conditions. The most important parts of the centrifugal pump are the impeller and volute. An impeller can take on many forms, ranging from essentially a spinning disk to designs with elaborate vanes. The latter is usual. Impeller design tends to be somewhat unique to each manufacturer, and a variety of designs are available for a variety of applications. An example of an impeller is shown in Fig. 39.1. This device imparts a radial velocity to the fluid that has entered the pump perpendicularly to the impeller. The volute (there may be one or more) performs the function of slowing the fluid and increasing the pressure. A good discussion of centrifugal pumps is given by Lobanoff and Ross [1992].


Figure 39.1 Schematic of a centrifugal pump. The liquid enters perpendicular to the figure, and a radial velocity is imparted by clockwise spin of the impeller.

There are other types of velocity-head pumps. Jet pumps increase pressure by imparting momentum from a high-velocity liquid stream to a low-velocity or stagnant body of liquid. The resulting flow then goes through a diffuser to achieve an overall pressure increase. Gas lifts accomplish a pumping action by a drag on gas bubbles that rise through a liquid.

Positive-Displacement Pumps Positive-displacement pumps demonstrate high discharge pressures and low flow rates. Usually this is accomplished by some type of pulsating device. A piston pump is a classical example of a positive-displacement machine. The rotary pump is one type of positive displacement device that does not impart pulsations to the exiting flow [a full description of this type of pumps is given by Turton (1994)]. Several techniques are available for dealing with pulsating flows, including use of double-acting pumps (usually of the reciprocating type) and installation of pulsation dampeners. Positive-displacement pumps usually require special seals to contain the fluid. Costs are higher both initially and for maintenance, compared to most pumps that operate on the velocity-head basis. Positive-displacement pumps demonstrate an efficiency that is nearly independent of flow Figure 39.2 Typical pump applications, either in circuits or once-through arrangements, can be represented as combined fluid resistances, as sho wn. Resistances are determined from fluid mechanics analyses.


rate, in contrast to the velocity-head type. Very high head pressures (often damaging to the pump) can be developed if the downstream flow is blocked. For this reason a pressure-relief-valve bypass must always be used with positive-displacement pumps.

Pump/Flow Considerations Performance characteristics of the pump must be considered in system design. Simple diagrams of pump applications are shown in Fig. 39.2. First consider the left-hand figure. This represents a flow circuit, and the pressure drops related to the piping, fittings, valves, and any other flow devices found in the circuit, estimated using laws of fluid mechanics. Usually these resistances (pressure drops) are found to vary approximately with the square of the liquid flow rate. Typical characteristics are shown in Fig. 39.3. Most pumps demonstrate a flow-versus-pressure rise variation that is a positive value at zero flow and decreases to zero at some larger flow. Positive-displacement pumps, as shown on the right-hand side of Fig. 39.3, are an exception to this rule in that these devices usually cannot tolerate a zero flow. An important aspect to note is that a closed system can presumably be pressurized. A contrasting situation and its implications are discussed as follows. Figure 39.3 Overlay of the pump flow versus head curve with the circuit piping characteristics gives the operating state of the circuit. A typical velocity-head pump characteristic is shown on the left, while a positive-displacement pump curve is shown on the right.

The piping diagram shown on the right-hand side of Fig. 39.2 is a once-through system, another frequently encountered installation. However, the leg of piping represented in "pressure drop 1" can have some very important implications related to net positive suction head (NPSH). In simple terms, NPSH indicates the difference between the local pressure and the thermodynamic saturation pressure at the fluid temperature. If NPSH = 0 , the liquid will vaporize, which can result in a variety of outcomes from noisy pump operation to outright failure of components. This condition is also called cavitation. If it occurs, cavitation will first take place at the lowest pressure point within the piping arrangement. Often this point is located at, or inside, the inlet to the pump. Most manufacturers specify how much NPSH is required for satisfactory operation of their pumps. Hence, the actual NPSH (NPSHA) experienced by the pump must be larger than the manufacturer's required NPSH (NPSHR). If a design indicates insufficient NPSH, changes should be made in the system, possibly including alternative piping layout such as changes in pipe elevation and/or size, or use of a pump with smaller NPSH requirements.


Figure 39.4 A full range of performance information should be available from the pump manufacturer, and this may include the parameters shown.

The manufacturer should be consulted for a map of operational information for a given pump. A typical form is shown in Fig. 39.4. This information will allow the designer to select a pump that satisfies the circuit operational requirements while meeting the necessary NPSH and most efficient operation criteria. Several options are available to the designer for combining pumps in systems. Consider a comparison of the net effect between operating pumps in series or operating the same two pumps in parallel. For pumps with characteristics like centrifugal units, examples of this distinction are shown in Fig. 39.5. It is clear that one way to achieve high pumping pressures with centrifugal pumps is to place a number of units in series. This is an effect related to that found in multistage designs.


Figure 39.5 Series and parallel operation of centrifugal pumps. The resultant characteristics for two identical pumps are shown.

39.2 Fans As noted earlier, fans are devices that cause air to move. This definition is broad and can include, for instance, a flapping palm branch; the discussion here deals only with devices that impart air movement due to rotation of an impeller inside a fixed casing. In spite of this limiting definition, a large variety of commercial designs are included. Fans find application in many engineering systems. Along with heating and cooling equipment, they are the heart of heating, ventilating, and air conditioning (HVAC) systems. When the physical dimensions of a unit are not a significant limitation (usually the case), centrifugal fans are favored over axial flow units for HVAC applications. Many types of fans are found in power plants. Very large fans are used to furnish air to the boiler, as well as to draw or force air through cooling towers and pollution control equipment. Electronic cooling finds applications for small units. Because of the great engineering importance of fans, several organizations publish rating and testing criteria [see, for example, ASME (1990)]. Generally fans are classified according to how the air flows through the impeller. These flows may be axial (essentially a propeller in a duct), radial (conceptually much like the centrifugal pumps discussed earlier), mixed , and cross. Although there are many other fan designations, all industrial units are one of these classifications. Mixed-flow fans are so named because both axial and radial flow occurs on the vanes. Casings for these devices are essentially like those for axial-flow machines, but the inlet has a radial-flow component. On cross-flow impellers, the gas traverses the blading twice. A variety of vane types are found on fans, and each particular type is also used for fan classification. Axial fans usually have vanes of airfoil shape or vanes of uniform thickness. Some vane types that might be found on a centrifugal (radial-flow) fan are shown in Fig. 39.6.


Figure 39.6 Variety of vane types that might be used on a centrifugal fan.

Each type of fan has some specific qualities for certain applications. In general terms, most installations use centrifugal (radial-flow) fans. A primary exception is for very-high-flow, low-pressure-rise situations in which axial (propeller) fans are used. Similarities exist between fans and pumps because the fluid density essentially does not vary through either type of machine. Of course in pumps this is because a liquid can be assumed to be incompressible. In fans, a gas (typically air) is moved with little pressure change. As a result, the gas density can be taken to be constant. Since most fans operate near atmospheric pressure, the ideal gas equation can be used in determining gas properties. Flow control in fan applications, where needed, is a very important design concern. Methods for accomplishing this involve use of dampers (either on the inlet or on the outlet of the fan), variable pitch vanes, or variable speed control. Dampers are the least expensive to install but also the most inefficient in terms of energy use. Modern solid state controls for providing a variable frequency power to the drive motor is becoming the preferred control method when a combination of initial and operating costs is considered.

Defining Terms Actual net positive suction head (NPSHA): The NPSH at the given state of operation of a pump. Cavitation: A state in which local liquid conditions allow vapor voids to form (boiling). Net positive suction head (NPSH): The difference between the local absolute pressure of a liquid and the liquid's thermodynamic saturation pressure based on the liquid's temperature. Applies


to the inlet of a pump. Required net positive suction head (NPSHR): The amount of NPSH required by a specific pump for a given application.

References ASME. 1990. ASME Performance Test Codes, Code on Fans. ASME PTC 11-1984 (reaffirmed 1990). American Society of Mechanical Engineers, New York. Boehm, R. F. 1987. Design Analysis of Thermal Systems. John Wiley & Sons, New York, pp. 17 −26. Ewbank, T. 1842. A Description and Historical Account of Hydraulic and Other Machines for Raising Water, 2nd ed. Greeley and McElrath, New York. Krutzsch, W. C. 1986. Introduction: Classification and selection of pumps. In Pump Handbook, 2nd ed., ed. I. Karassik et al. McGraw-Hill, New York. Lobanoff, V. and Ross, R. 1992. Centrifugal Pumps: Design & Application, 2nd ed. Gulf, Houston, TX. Turton, R. K. 1994. Rotodynamic Pump Design. Cambridge University Press, England.

Further Information ASHRAE. 1992. Fans. Chapter 18 of 1992 ASHRAE Handbook, HVAC Systems and Equipment. American Society of Heating, Refrigerating, and Air Conditioning Engineers, Atlanta, GA. Dickson, C. 1988. Pumping Manual, 8th ed. Trade & Technical Press, Morden, England. Dufour, J. and Nelson, W. 1993. Centrifugal Pump Sourcebook, McGraw-Hill, New York. Garay, P. N. 1990. Pump Application Book. Fairmont Press, Liburn, GA. Krivchencko, G. I. 1994. Hydraulic Machines, Turbines and Pumps, 2nd ed. Lewis, Boca Raton, FL. Stepanoff, A. J. 1993. Centrifugal and Axial Flow Pumps: Theory, Design, and Application. Krieger, Malabar, FL.


TUTORIAL CENTRIFUGAL PUMP SYSTEMS by Jacques Chaurette p. eng.

copyright 2005

Fluide Design Inc., 5764 Monkland avenue, Suite 311, Montreal, Quebec, Canada H4A 1E9 Tel: 514.484.PUMP (7867) E-mail: [email protected] Web site: www.fluidedesign.com

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Table of contents

1. Different types of pump systems 2. Three important characteristics of a pump system: pressure, friction and flow 3. What is friction in a pump system 4. Energy and head in pump systems 5. Static head 6. Flow rate depends on elevation difference or static head 7. Flow rate depends on friction 8. How does a centrifugal pump produce pressure 9. What is total head 10 What is the relationship between head and total head 11. How to determine friction head 12. The performance or characteristic curve of the pump 13. How to select a centrifugal pump Examples of total head calculations - sizing a pump for a home owner application 14. Examples of common residential water systems 15. Calculate the pump discharge pressure from the pump total head

 Copyright . 2005----

www.lightmypump.com---------- Revised October 9, 2007

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Appendix A Flow rate and friction loss for different pipe sizes based at different velocities Appendix B Formulas and an example of how to do pipe friction calculations Appendix C Formulas and an example of how to do pipe fittings friction calculations Appendix D Formula and an example of how to do velocity calculation for fluid flow in a pipe Appendix E The relationship between pressure head and pressure



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Foreword

This tutorial is intended for anyone that has an interest in centrifugal pumps. There is no math, just simple explanations of how pump systems work and how to select a centrifugal pump. For those who want to do detail calculations, some examples have been included in the appendices. This tutorial answers the following questions: -

What are the important characteristics of a pump system?

-

What is head and how is it used in a pump system to make calculations easier?

-

What is static head and friction head and how do they affect the flow rate in a pump system?

-

How does a centrifugal pump produce pressure?

-

Why is total head and flow the two most important characteristics of a centrifugal pump?

-

What is meant by the pump rating? And what is the optimal operating point of a centrifugal pump?

-

How to do details calculations that will allow you to size and select a centrifugal pump?

-

How to verify that your centrifugal pump is providing the rated pressure or head?

-

What is density and specific gravity and how do they relate to pressure and head?



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1. Different types of pump systems

There are many types of centrifugal pump systems. Figure 1 shows a typical industrial pump system. There are many variations on this including all kinds of equipment that can be hooked up to these systems that are not shown. A pump after all is only a single component of a process although an important and vital one. The pumps’ role is to provide sufficient pressure to move the fluid through the system at the desired flow rate.

Figure 1 Typical industrial pump system.

Back in the old days domestic water supply was simpler...aaah the good old days. Goodnight John boy..

Figure 1a The old days.



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Domestic water systems take their water from various sources at different levels depending on the water table and terrain contours.

Figure 1b Water supply sources (source: The Ground Water Atlas of Colorado).

The system in Figure 2 is a typical domestic water supply system that takes it's water from a shallow well (25 feet down max.) using an end suction centrifugal pump. A jet pump works well in this application (see http://www.watertanks.com/category/43/) .

Figure 2 Typical residential pump system. The system in Figure 3 is another typical domestic water supply system that takes it's water from a deep well (200-300 feet) and uses a multi-stage submersible pump often called a turbine pump (http://www.webtrol.com/domestic_pumps/8in_turbine.htm ).



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Figure 2a Typical jet pump.

igure 3 Typical residential deep well pump system.



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Figure 3a Typical residential deep well pump system (source: The Ground Water Atlas of Colorado).

Figure 3b Another representation of a typical residential deep well pump system.



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Figure 3c Typical deep well residential submersible pump.

2. Three important characteristics of pump systems: pressure, friction and flow

Figure 4 Three important characteristics of pump systems.

Pressure, friction and flow are three important characteristics of a pump system. Pressure is the driving force responsible for the movement of the fluid. Friction is the force that slows down fluid particles. Flow rate is the amount of volume that is displaced per unit time. The unit of flow in North America, at least in the pump industry, is the US gallon per minute, USgpm. From now on I will just use gallons per minute or gpm. In the metric system, flow is in liters per second (L/s) or meters cube per hour (m 3/h).



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Pressure is often expressed in pounds per square inch (psi) in the Imperial system and kiloPascals (kPa) in the metric system. In the Imperial system of measurement, the unit psig or pounds per square inch gauge is used, it means that the pressure measurement is relative to the local atmospheric pressure, so that 5 psig is 5 psi above the local atmospheric pressure. The kPa unit scale is intended to be a scale of absolute pressure measurement and there is no kPag, but many people use the kPa as a relative measurement to the local atmosphere and don't bother to specify this. This is not a fault of the metric system but the way people use it. The term pressure loss or pressure drop is often used, this refers to the decrease in pressure in the system due to friction. In a pipe or tube that is at the same level, your garden hose for example, the pressure is high at the tap and zero at the hose outlet, this decrease in pressure is due to friction and is the pressure loss. As an example of the use of pressure and flow units, the pressure available to domestic water systems varies greatly depending on your location with respect to the water treatment plant. It can vary between 30 and 70 psi or more. The following table gives the expected flow rate that you would obtain for different pipe sizes assuming the pipe or tube is kept at the same level as the connection to the main water pressure supply and has a 100 feet of length (see Figure 4a).

Table 1 Expected flow rates for 100 feet of pipe of various diameters based on available pressure.



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Figure 4a A typical garden hose connection, see Table 1 for flow rate vs. pressure.



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The unit of friction is....Sorry, I think I need to wait 'til we get closer to the end to explain the reasoning behind this unit.

Figure 5 A typical pump system.

The pump provides the energy necessary to drive the fluid through the system and overcome friction and any elevation difference. Pressure is increased when fluid particles are forced closer together. For example, in a fire extinguisher work or energy has been spent to pressurize the liquid chemical within, that energy can be stored and used later. Is it possible to pressurize a liquid within a container that is open? Yes. A good example is a syringe, as you push down on the plunger the pressure increases, and the harder you have to push. There is enough friction as the fluid moves through the needle to produce a great deal of pressure in the body of the syringe.

Figure 6 Pressure produced by fluid friction in a syringe.

If we apply this idea to the pump system of Figure 5, even though the discharge pipe end is open, it is possible to have pressure at the pump discharge because there is sufficient friction in the system and elevation difference.



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3. What is friction in a pump system

Friction is always present, even in fluids, it is the force that resists the movement of objects.

Figure 7 Friction, the force that resist movement.

When you move a solid on a hard surface, there is friction between the object and the surface. If you put wheels on it, there will be less friction. In the case of moving fluids such as water, there is even less friction but it can become significant for long pipes. Friction can be also be high for short pipes which have a high flow rate and small diameter as in the syringe example. In fluids, friction occurs between fluid layers that are traveling at different velocities within the pipe (see Figure 8). There is a natural tendency for the fluid velocity to be higher in the center of the pipe than near the wall of the pipe. Friction will also be high for viscous fluids and fluids with suspended particles.

Figure 8 Friction between layers of fluid and the



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pipe wall. Another cause of friction is the interaction of the fluid with the pipe wall, the rougher the pipe, the higher the friction.

Friction depends on: -

average velocity of the fluid within the pipe viscosity pipe surface roughness

An increase in any one of these parameters will increase friction. The amount of energy required to overcome the total friction loss within the system has to be supplied by the pump if you want to achieve the required flow rate. In industrial systems, friction is not normally a large part of a pump's energy output. For typical systems, it is around 25% of the total. If it becomes much higher then you should examine the system to see if the pipes are too small. However all pump systems are different, in some systems the friction energy may represent 100% of the pump's energy, This is what makes pump systems interesting, there is a million and one applications for them. In household systems, friction can be a greater proportion of the pump energy output, maybe up to 50% of the total, this is because small pipes produce higher friction than larger pipes for the same average fluid velocity in the pipe (see the friction chart later in this tutorial). Another cause of friction are the fittings (elbows, tees, y's, etc) required to get the fluid from point A to B. Each one has a particular effect on the fluid streamlines. For example in the case of the elbow (see Figure 9), the fluid streamlines that are closest to the tight inner radius of the elbow lift off from the pipe surface forming small vortexes that consume energy. This energy loss is small for one elbow but if you have several elbows and other fittings it can become significant. Generally speaking they rarely represent more than 30% of the total friction due to the overall pipe length.

Figure 9 Streamline flow patterns for typical fittings such an elbow and a tee. 4. Energy and head in pump systems

Energy and head are two terms that are often used in pump systems. We use energy to describe the movement of liquids in pump systems because it is easier than any other



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method. There are four forms of energy in pump systems: pressure, elevation, friction and velocity. Pressure is produced at the bottom of the reservoir because the liquid fills up the container completely and its weight produces a force that is distributed over a surface which is pressure. This type of pressure is called static pressure. Pressure energy is the energy that builds up when liquid or gas particles are moved slightly closer to each other. A good example is a fire extinguisher, work was done to get the liquid into the container and then to pressurize it. Once the container is closed the pressure energy is available for later use. Any time you have liquid in a container, even one that is not pressurized, you will have pressure at the bottom due to the liquid’s weight, this is known as static pressure. Elevation energy is the energy that is available to a liquid when it is at a certain height. If you let it discharge it can drive something useful like a turbine producing electricity. Friction energy is the energy that is lost to the environment due to the movement of the liquid through pipes and fittings in the system. Velocity energy is the energy that moving objects have. When a pitcher throws a baseball he gives it velocity energy. When water comes out of a garden hose, it has velocity energy.

Figure 10 The relationship between height, pressure and velocity.

In the figure above we see a tank full of water, a tube full of water and a cyclist at the top of a hill. The tank produces pressure at the bottom and so does the tube. The cyclist has elevation energy that he will be using as soon as he moves.



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As we open the valve at the tank bottom the fluid leaves the tank with a certain velocity, in this case pressure energy is converted to velocity energy. The same thing happens with the tube. In the case of the cyclist, the elevation energy is gradually converted to velocity energy. The three forms of energy: elevation, pressure and velocity interact with each other in liquids. For solid objects there is no pressure energy because they don’t extend outwards like liquids filling up all the available space and therefore they are not subject to the same kind of pressure changes. The energy that the pump must supply is the friction energy plus the difference in height that the liquid must be raised to which is the elevation energy. PUMP ENERGY = FRICTION ENERGY + ELEVATION ENERGY

Figure 11 Pump energy equals elevation energy plus friction energy.

You are probably thinking where is the velocity energy in all this. Well if the liquid comes out of the system at high velocity then we would have to consider it but this is not a typical situation and we can neglect this for the systems discussed in this article. The last word on this topic, it is actually the velocity energy difference that we would need to consider. In Figure 11 the velocities at point 1 and point 2 are the result of the position of the fluid particles at points 1 and 2 and the action of the pump. The difference between these two velocity energies is an energy deficiency that the pump must supply but as you can see the velocities of these two points will be quite small. Now what about head? Head is actually a way to simplify the use of energy. To use energy we need to know the weight of the object displaced. Elevation energy E.E. is the weight of the object W times the distance d: EE = W x d



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Friction energy FE is the force of friction F times the distance the liquid is displaced or the pipe length l: FE = F x l Head is defined as energy divided by weight or the amount of energy used to displace a object divided by its weight. For elevation energy, the elevation head EH is: EH = W x d / W = d For friction energy, the friction head FH is the friction energy divided by the weight of liquid displaced: FH = FE/W = F x l

/W

The friction force F is in pounds and W the weight is also in pounds so that the unit of friction head is feet. This represents the amount of energy that the pump has to provide to overcome friction. I know you are thinking: “…this doesn’t make sense” , how can feet represent energy? If I attach a tube to the discharge side of a pump, the liquid will rise in the tube to a height that exactly balances the pressure at the pump discharge. Part of the height of liquid in the tube is due to the elevation height required (elevation head) and the other is the friction head and as you can see both can be expressed in feet and this is how you can measure them.

Figure 12 Measuring elevation head and friction head.



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5. Static head

Webster’s dictionary definition of head is: “a body of water kept in reserve at a height”.

Figure 13 The definition of head.

It is expressed in terms of feet in the Imperial system and meters in the metric system. Because of its height and weight the fluid produces pressure at the low point and the higher the reservoir the higher the pressure (see Figure 13).

Figure 14 Pressure depends on the height of the liquid surface.

The amount of pressure at the bottom of a reservoir is independent of its shape, for the same liquid level, the pressure at the bottom will be the same. This is important since in complex piping systems it will always be possible to know the pressure at the bottom if we know the height (see Figure 15). To find out how to calculate pressure from height go to section 14.



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Figure 15 The pressure level at the bottom of a tank depends on the liquid surface height.

When a pump is used to displace a liquid to a higher level it is usually located at the low point or close to it. The head of the reservoir, which is called static head, will produce pressure on the pump that will have to be overcome once the pump is started. To distinguish between the pressure energy produced by the discharge tank and suction tank, the head on the discharge side is called the discharge static head and on the suction side the suction static head (see Figure 16).

Figure 16 The static head on the pump when the Usually the liquid is displaced suction tank is full. from a suction tank to a discharge tank. The suction tank fluid provides pressure energy to the pump that helps the pump. We want to know how much pressure energy the pump itself must supply so therefore we subtract the pressure energy provided by the suction tank. The static head is then the difference in height of the discharge tank fluid surface minus the suction tank fluid surface. Static head is sometimes called total static head to indicate that the pressure energy available on both sides of the pump has been considered (see Figure 16).



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Since there is a difference in height between the suction and discharge flanges or connections of a pump by convention it was decided that the static head would be measured with respect to the suction flange elevation (see Figure 17).

Figure 17 The static head on the pump is measured with respect to the pump suction.

end is open to atmosphere then the static head is measured with respect to the pipe end (see Figure 18).

Figure 18 The static head on the pump with the discharge pipe end open to atmosphere.



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Sometimes the discharge pipe end is submerged such as in Figure 19, then the static head will be the difference in elevation between the discharge tank fluid surface and suction tank fluid surface. Since the fluid in the system is a continuous medium and all fluid particles are connected via pressure, the fluid particles that are located at the surface of the discharge tank will contribute to the pressure built up at the pump discharge. Therefore the discharge surface elevation is the height that must be considered for static head. Avoid the mistake of using the discharge pipe end as the elevation for calculating static head if the pipe end is submerged (see Figure 20). Note: if the discharge pipe end is submerged then a check valve on the pump discharge is required to avoid backflow when the pump is stopped.

Figure 19 The height of the discharge pipe end is in this case the correct height for static head.

Figure 20 Incorrect difference in elevation for static head.

You can change the static head by raising the surface of the discharge tank (assuming the pipe end is submerged) or suction tank or both. All of these changes will influence the flow rate.



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To correctly determine the static head follow the liquid particles from start to finish, the start is almost always at the liquid surface of the suction tank, this is called the inlet elevation. The end will occur where you encounter an environment with a fixed pressure such as the open atmosphere, this point is the discharge elevation end or outlet elevation. The difference between the two elevations is the static head. The static head can be negative because the outlet elevation can be lower than the inlet elevation.

6. Flow rate depends on elevation difference or static head

For identical systems, the flow rate will vary with the static head. If the pipe end elevation is high, the flow rate will be low (see Figure 21). Compare this to a cyclist on a hill with a slight upward slope, his velocity will be moderate and correspond to the amount of energy he can supply to overcome the friction of the wheels on the road and the change in elevation.

Figure 21 The effect of pipe end elevation on flow.

If the liquid surface of the suction tank is at the same elevation as the discharge end of the pipe then the static head will be zero and the flow rate will be limited by the friction in the system. This is equivalent of a cyclist on a flat road, his velocity depends on the amount of friction between the wheels and the road and the air resistance (see Figure 22).



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Figure 22 Flow rate is limited by friction in the system when the static head is zero.

In Figure 23, the discharge pipe end is raised vertically until the flow stops, the pump cannot raise the fluid higher than this point and the discharge pressure is at its maximum. Similarly the cyclist applies maximum force to the pedals without getting anywhere.

Figure 23 The pump produces zero flow at its maximum outlet pressure.

If the discharge pipe end is lower than the liquid surface of the suction tank then the static head will be negative and the flow rate high (see Figure 24). If the negative static



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head is large then it is possible that a pump is not required since the energy provided by the difference in elevation may be sufficient to move the fluid through the system without the use of a pump as in the case of a siphon (see the pump glossary). By analogy, as the cyclist comes down the hill he looses his stored elevation energy which is transformed progressively into velocity energy. The lower he is on the slope, the faster he goes.

Figure 24 Negative static head increases flow rate. Pumps are most often rated in terms of head and flow. In Figure 23, the discharge pipe end is raised to a height at which the flow stops, this is the head of the pump at zero flow. We measure this difference in height in feet (see Figure 25). Head varies depending on flow rate, but in this case since there is no flow and hence no friction, the head of the pump is THE MAXIMUM HEIGHT THAT THE FLUID CAN BE LIFTED TO WITH RESPECT TO THE SURFACE OF THE SUCTION TANK. Since there is no flow the head (also called total head) that the pump produces is equal to the static head.

Figure 25 Highest possible total head of a pump.



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In this situation the pump will deliver its maximum pressure. If the pipe end is lowered as in Figure 21, the pump flow will increase and the head (also known as total head) will decrease to a value that corresponds to the flow. Why? Let’s start from the point of zero flow with the pipe end at its maximum elevation, the pipe end is lowered so that flow begins. If there is flow there must be friction, the friction energy is subtracted (because it is lost) from the maximum total head and the total head is reduced. At the same time the static head is reduced which further reduces the total head.

Figure 26 Variation of total head vs. pipe end elevation or static head.

When you buy a pump you don’t specify the maximum total head that the pump can deliver since this occurs at zero flow. You instead specify the total head that occurs at your required flow rate. This head will depend on the maximum height you need to reach with respect to the suction tank fluid surface and the friction loss in your system. For example, if your pump is supplying a bathtub on the 2 nd floor, you will need enough head to reach that level, that will be your static head, plus an additional amount to overcome the friction loss through the pipes and fittings. Assuming that you want to fill the bath as quickly as possible, then the taps on the bath will be fully open and will offer very little resistance or friction loss. If you want to supply a shower head for this bathtub then you will need a pump with more head for the same flow rate because the shower head is higher and offers more resistance than the bathtub taps. Luckily, there are many sizes and models of centrifugal pumps and you cannot expect to purchase a pump that matches exactly the head you require at the desired flow. You will probably have to purchase a pump that provides slightly more head and flow than you require and you will adjust the flow with the use of appropriate valves. Note: you can get more head from a pump by increasing its speed or impeller diameter or both. In practice, homeowners cannot make these changes and to obtain a higher total head, a new pump must be purchased.



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7. Flow rate depends on friction

For identical systems, the flow rate will vary with the size and diameter of the discharge pipe. A system with a discharge pipe that is generously sized will have a high flow rate. This is what happens when you use a large drain pipe on a tank to be emptied, it drains very fast.

Figure 27 A large pipe produces low friction.

The smaller the pipe, the less the flow. How does the pump adjust itself to the diameter of the pipe, after all it does not know what size pipe will be installed? The pump you install is designed to produce a certain average flow for systems that have their pipes sized accordingly. The impeller size and its speed predispose the pump to supply the liquid at a certain flow rate. If you attempt to push that same flow through a small pipe the discharge pressure will increase and the flow will decrease. Similarly if you try to empty a tank with a small tube, it will take a long time to drain (see Figure 28).



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Figure 28 A small pipe produces high friction.

when the discharge pipe is long, the friction will be high and the flow rate low (see Figure 29) and if the pipe is short the friction will be low and the flow rate high (see Figure 30)



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Figure 29 A long pipe produced high friction.

Later on in the tutorial, a chart will be presented giving the size of pipes for various flow rates.

Figure 30 A short pipe produces low friction.



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8. How does a centrifugal pump produce pressure

Fluid particles enter the pump at the suction flange or connection. They then turn 90 degrees into the plane of the impeller and fill up the volume between each impeller vane. The next image shows what happens to the fluid particles from that point forward for an animated version go to http://www.fluidedesign.com/downloads.htm in the middle of the page.

Figure 31 Fluid particle paths in a centrifugal pump.

Figure 32 Different parts of a centrifugal pump.



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Figure 33 Cut-away view of a centrifugal pump.

A centrifugal pump is a device whose purpose is to produce pressure by accelerating fluid particles to a high velocity providing them with velocity energy. What is velocity energy? It's a way to express how the velocity of objects can affect other objects, you for example. Have you ever been tackled in a football match? The velocity at which the other player comes at you determines how hard you are hit. The mass of the player is also an important factor. The combination of mass and velocity produces velocity energy. Another example is catching a hard baseball pitch, ouch, there can be allot of velocity in a small fast moving baseball. Fluid particles that move at high speed have velocity energy, just put your hand on the open end of a garden hose. The fluid particles in the pump are expelled from the tips of the impeller vanes at high velocity, they then hit the inner casing of the pump and are decelerated lowering the velocity energy and raising the pressure energy. Unlike friction that wastes energy, the decrease in velocity energy serves to increase pressure energy; this is the principal of energy conservation in action. The same thing happens to a cyclist that starts at the top of a hill, his speed gradually increases as he looses elevation. The cyclist’s elevation energy is transformed into velocity energy; in the pump’s case the velocity energy is transformed into pressure energy. Try this experiment, find a plastic cup or other container that you can poke a small pinhole in the bottom. Fill it with water and attach a string to it, and now you guessed it, start spinning.



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The faster you spin, the more water comes out the small hole, the water is pressurized inside the cup using centrifugal force in a similar fashion to a centrifugal pump. In the case of a pump, the rotational motion of the impeller projects fluid particles at high speed into the volume between the casing wall and the impeller tips. Prior to leaving the pump, the fluid particles slow down to the velocity at the inlet of the discharge pipe (see Figures 31 and 32) which will be the same velocity throughout the system if the pipe diameter does not change. How does the flow rate change when the discharge pipe end elevation is changed or when there is an increase or decrease in pipe friction? These changes cause the pressure at the pump outlet to increase when the flow decreases, sounds backwards doesn’t it. Well Figure 34 An experiment with centrifugal it’s not and you will see why. How does the force. pump adjust to this change in pressure? Or in other words, if the pressure changes due to outside factors, how does the pump respond to this change.

Pressure is produced by the rotational speed of the impeller vanes. The speed is constant. The pump will produce a certain discharge pressure corresponding to the particular conditions of the system (for example, fluid viscosity, pipe size, elevation difference, etc.). If changing something in the system causes the flow to decrease (for example closing a discharge valve), there will be an increase in pressure at the pump discharge because there is no corresponding reduction in the impeller speed. The pump produces excess velocity energy because it operates at constant speed, the excess velocity energy is transformed into pressure energy and the pressure goes up.

All centrifugal pumps have a performance or characteristic curve that looks similar to the one shown in Figure 35 (assuming that the level in the suction tank remains constant), this shows how the discharge pressure varies with the flow rate through the pump.



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Figure 35 Typical curve of discharge pressure vs. flow rate for a pump.

At 200 gpm, this pump produces 20 psig discharge pressure, and as the flow drops the pressure increases, and will be 40 psig at zero flow. Note: this applies to centrifugal pumps, many home owners have positive displacement pumps, often piston pumps. These pumps produce constant flow no matter what changes are made to the system. This is why all such pump have an internal relief valve that opens to relieve pressure and protect the pump from excessive pressure cause by the closing of a water tap for example. Also typically, the pump has a pressure switch that shuts the motor down when the pressure gets too high saving energy in the process. 9. What is total head

Total head and flow are the main criteria that are used to compare one pump with another or to select a centrifugal pump for an application. Total head is related to the discharge pressure of the pump. Why can't we just use discharge pressure? Pressure is a familiar concept, we are familiar with it in our daily lives. For example, fire extinguishers are pressurized at 60 psig (410 kPa), we put 35 psig (240 kPa) air pressure in our bicycle and car tires. Pump manufacturers do not use discharge pressure as criteria for pump selection. One of the reasons is that they do not know how you will use the pump. They do not know what flow rate you require and the flow rate of a centrifugal pump is not fixed as it is in a positive displacement pump. The discharge pressure depends on the pressure available on the suction side of the pump. If the source of water for the pump is below or above the pump suction, for the same flow rate you will get a different discharge pressure. Therefore to eliminate this problem, it is preferable to use the difference in pressure between the inlet and outlet of the pump. Pump manufacturers have taken this a step further, the amount of pressure that a pump can produce will depend on the density of the fluid, for a salt water solution which is denser than pure water, the pressure will be higher for the same flow rate. Once again, the manufacturer doesn't know what type of fluid is in your system, so that a criteria that



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does not depend on density is very useful. There is such a criteria and it is called TOTAL HEAD, and this is defined as the difference in head between the inlet and outlet of the pump. You can measure the discharge head by attaching a tube to the discharge side of the pump and measuring the height of the liquid in the tube with respect to the suction of the pump. The tube will have to be quite high for a typical domestic pump. If the discharge pressure is 40 psi the tube would have to be 92 feet high. This is not a practical method but it helps explain how head relates to total head and how head relates to pressure. You do the same to measure the suction head. The difference between the two is the total head of the pump.

Figure 36 A method for measuring total head.

For these reasons the pump manufacturers have chosen total head as the main parameter that describes the pump’s available energy.

The fluid in the measuring tube of the discharge or suction side of the pump will rise to the same height for all fluids regardless of the density. This is a rather astonishing statement, here’s why. The pump doesn’t know anything about head, head is a concept we use to make our life easier. The pump produces pressure and the difference in pressure across the pump is the amount of pressure energy available to the system. If the fluid is dense, such as a salt solution for example, more pressure will be produced at the pump discharge than if the fluid were pure water. Compare two tanks with the same cylindrical shape, the same volume and liquid level, the tank with the denser fluid will have a higher pressure at the bottom. But the static head of the fluid surface with respect to the bottom is the same. Total head behaves the same way as static head, even if the fluid is denser the total head as compared to a less dense fluid such as pure water will be the same. This is a surprising fact, see this experiment on video on the web that shows this idea in action (http://www.fluidedesign.com/video1.htm#vid0.9).



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10. What is the relationship between head and total head

Total head is the height that the liquid is raised to at the discharge side of the pump less the height that the liquid is raised to at the suction side (see Figure 36). Why less the height at the suction side? Because we want the energy contribution of the pump only and not the energy that is supplied to it.

What is the unit of head? First let's deal with the unit of energy. Energy can be expressed in foot-pounds which is the amount of force required to lift an object up multiplied by the vertical distance. A good example is weight lifting. If you lift 100 pounds (445 Newtons) up 6 feet (1.83 m), the energy required is 6 x 100= 600 ft-lbf (814 N-m). Head is defined as energy divided by the weight of the object Figure 37 Energy is required to lift weights. displaced. For the weight lifter, the energy divided by the weight displaced is 6 x 100 / 100= 6 feet (1.83 m), so the amount of energy per pound of dumbbell that the weight lifter needs to provide is 6 feet. This is not terribly useful to know for a weight lifter but we will see how very useful it is for displacing fluids.

You may be interested to know that 324 foot-pounds of energy is equivalent to 1 calorie. This means that our weight lifter spends 600/324 = 1.8 calories each time he lifts that weight up 6 feet, not much is it. The following figure shows how much energy is required to displace vertically one gallon of water.

Figure 38 Energy required to lift 1 gallon of water up 10 feet.



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This next figure shows how much head is required to do the same job.

Figure 39 Head vs. energy.

If we use energy to describe how much work the pump needs to do to displace a volume of liquid, we will need to know the weight. If we use head, we only need to know the vertical distance of movement. This is very useful for fluids because pumping is a continuous process, usually when you pump you leave the pump on for a long time, you don't worry about how many pounds of fluid have been displaced we are mainly interested in the flow rate. The other very useful aspect of using head is that the elevation difference or static head can be used directly as one part of the value of total head, the other part being friction head as shown in Figure 40.

Figure 40 Discharge static head plus friction head equals pump discharge head.  Copyright . 2005----


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How much static head is required to pump water up from the ground floor to the second floor, or 15 feet up? Remember that you must also take into consideration the level of the water in the suction tank. If the water level is 10 feet below the pump suction connection then the static head will be 10 + 15 = 25 feet. Therefore the total head will have to be at least 25 feet plus the friction head loss of the fluid moving through the pipes.

Figure 41 Static head. 11. How to determine friction head

Friction head is the amount of energy loss due to friction caused by fluid movement through pipes and fittings. It takes a force to move the fluid against friction, in the same way that a force is required to lift a weight. The force is exerted in the same direction as the moving liquid and energy is expended. In the same way that head was calculated to lift a certain weight, the friction head is calculated with the force required to overcome friction times the displacement (pipe length) divided by the weight of fluid displaced. These calculations have been done for us and you can find the values for friction head loss in Table 2 for different pipe sizes and flow rates.



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Table 2 Table 2 gives the flow rate and the friction head loss for water being moved through pipes of different diameter at a velocity of 10 ft /s. I have chosen 10 ft/s because it is a typical value for velocity in pipes, it is not too large which would create allot of friction and not too small which would slow things down, it’s just right. Velocity depends on the flow rate and the pipe diameter, you will find friction loss charts for flow rates of 5 ft/s and 15 ft/s in Appendix A, imperial and metric. If you wish to do you own calculation of velocity, you can find out how in Appendix D. If the velocity you are using is less than 10 ft/s then the friction loss will be less and if the velocity is higher the loss will be greater. A velocity of 10 ft/s is normal practice for sizing pipes on the discharge side of the pump. For the suction side of the pump, it is desirable to be more conservative and size pipes for a lower velocity, for example between 4 and 7 feet/second. This is why you normally see a bigger pipe size on the suction side of the pump than on the discharge. A rule of thumb is to make the suction pipe the same size or one size larger than the suction connection of the pump. Why bother with velocity, isn’t flow rate enough information to describe fluid movement through a system. It depends how complicated your system is, if the discharge pipe has a constant diameter then the velocity though out will be the same. Then if you know the flow rate, based on the friction loss tables, you can calculate the friction loss with the flow rate only. If the discharge pipe diameter changes then the velocity will change for the same flow rate and a higher or lower velocity means a higher or lower friction loss in that portion of the system. You will then have to use the velocity to calculate the friction head loss in this part of the pipe. You can find a velocity calculator at this web site http://www.fluidedesign.com/applets.htm#applets4.



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Those who would like to do pipe friction calculations will find the information in Appendix B and pipe fittings friction loss calculations are in Appendix C. A calculator for pipe friction loss is available here ( http://www.fluidedesign.com/applets.htm#applets13) and for fittings friction loss here( http://www.fluidedesign.com/applets.htm#applets15). 12. The performance or characteristic curve of the pump

The pump characteristic curve has a similar appearance to the previous curve of discharge pressure vs. flow (Figure 35). As I mentioned this is not a practical way of describing the performance because you would have to know the suction pressure used to generate the curve. Figure 42 shows a typical total head vs. flow rate characteristic curve. This is the type of curve that all pump manufacturers publish for each model pump for a given operating speed. Not all manufacturers will provide you with the pump characteristic curve. However, the curve does exist and if you insist you can probably get it. Generally speaking the more you pay, the more technical information you get.

Figure 42 Characteristic curve of a pump.

13. Elevation changes between the suction tank surface and the discharge point can be disregarded

Surprising statement! You can have the discharge pipe going up or down (within reason) as much as you like and this has no effect on the static head of the system. The fluid looses its elevation energy as it goes up but regains it without loss as it comes back down. If the discharge elevation of the system does not change then any changes prior to this point will cancel out as you reach the discharge point of the system. What does make a difference is the length of pipe between these two points, by going up and down you increase the length of the pipe which increases the friction head which will increase the total head but the static head will remain the same. There is a limit as to how high you can go in between the pump discharge and the pump system discharge point. This height will depend on the available shut-off head of the



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pump. Shut-off head is the maximum head that a pump can produce at zero flow. When you start the pump in a system with fluid that is going upwards, the velocity in the pipe will be low until the fluid completely fills the system. When the system is filled equilibrium is reached between the pump’s ability to push the fluid through the system and the resistance that the system offers, at this equilibrium the flow rate is established. If the shut-off head is insufficient then there will be insufficient energy to get the liquid past the high point. If the shut-of head is high enough to get the liquid past the high point then the system can be filled up which results in a lowering of the static head allowing the pump to operate at a lower total head. In the characteristic curve shown in Figure 42 the shut-off head is 125 feet. If your system has a high point higher than 125 feet between the suction surface and the high point then it will be impossible to fill this system up and operate it at the required flow rate.

13. How to select a centrifugal pump

It is unlikely that you can buy a centrifugal pump off the shelf, install it in an existing system and expect it to deliver exactly the flow rate you require. The flow rate that you obtain depends on the physical characteristics of your system such as friction which depends on the length and size of the pipes and elevation difference which depends on the building and location. The pump manufacturer has no means of knowing what these constraints will be. This is why buying a centrifugal pump is more complicated than buying a positive displacement pump which will provide its rated flow no matter what system you install it in. The main factors that affect the flow rate of a centrifugal pump are: -

friction, which depends on the length of pipe and the diameter static head, which depends on the difference of the pipe end discharge height vs. the suction tank fluid surface height fluid viscosity, if the fluid is different than water.

The steps to follow to select a centrifugal pump are: 1. Determine the flow rate To size and select a centrifugal pump, first determine the flow rate. If you are a home owner, find out which of your uses for water is the biggest consumer. In many cases, this will be the bathtub which requires approximately 10 gpm (0.6 L/s). In an industrial setting, the flow rate will often depend on the production level of the plant. Selecting the right flow rate may be as simple as determining that it takes 100 gpm (6.3 L/s) to fill a tank in a reasonable amount of time or the flow rate may depend on the interaction between processes.



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2. Determine the static head This a matter of taking measurements of the height between the suction tank fluid surface and the discharge pipe end height or the discharge tank fluid surface elevation. 3. Determine the friction head The friction head depends on the flow rate, the pipe size and the pipe length. This is calculated from the values in the tables presented here (see Table 2). For fluids different than water the viscosity will be an important factor and Table 1 is not applicable. 4. Calculate the total head The total head is the sum of the static head (remember that the static head can be positive or negative) and the friction head. 5. Select the pump You can select the pump based on the pump manufacturer’s catalogue information using the total head and flow required as well as suitability to the application. Example of total head calculation - Sizing a pump for a home owner application

Experience tells me that to fill a bath up in a reasonable amount of time, a flow rate of 10 gpm is required. According to Table 2, the copper tubing size should be somewhere between 1/2" and 3/4", I choose 3/4". I will design my system so that from the pump there is a 3/4" copper tube main distributor, there will be a 3/4" take-off from this distributor on the ground floor to the second floor level where the bath is located. On the suction, I will use a pipe diameter of 1”, the suction pipe is 30 ft long.

Figure 43 Typical domestic water system.



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Friction loss on the suction side of the pump

According to calculation or the use of tables which is not presented here the friction loss for a 1" tube is has a friction loss of 0.068 feet per feet of pipe. In this case, the distance is 30 feet. The friction loss in feet is then 30 x 0.068 = 2.4 feet. There is some friction loss in the fittings, let's assume that a conservative estimate is 30% of the pipe friction head loss, the fittings friction head loss is = 0.3 x 2.4 = 0.7 feet. If there is a check valve on the suction line the friction loss of the check valve will have to be added to the friction loss of the pipe. A typical value of friction loss for a check valve is 5 feet. A jet pump does not require a check valve therefore I will assume there is no check valve on the suction of this system. The total friction loss for the suction side is then 2.4 + 0.7 = 3.1 feet. You can find the friction loss for a 1” pipe at 10 gpm in the Cameron Hydraulic data book of which the next figure is an extract:

Figure 43a Friction loss data from the Cameron Hydraulic data book (available at http://www.cameronbook.com/Merchant2/Merchant.mv ).

Friction loss on the discharge side of the pump

According to calculation or the use of tables which is not presented here the friction loss for a 3/4" tube is has a friction loss of 0.23 feet per feet of pipe. In this case, the distances are 10 feet of run on the main distributor and another 20 feet off of the main distributor up to the bath, for a total length of 30 feet. The friction loss in feet is then 30 x 0.23 = 6.9 feet. There is some friction loss in the fittings, let's assume that a conservative estimate is 30% of the pipe friction head loss, the fittings friction head loss is = 0.3 x 6 .9 = 2.1 feet. The total friction loss for the discharge side is then 6.9 + 2.1 = 9 feet.



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You can find the friction loss for a 0.75” pipe at 10 gpm in the Cameron Hydraulic data book of which the next figure is an extract:

Figure 43b Friction loss data from the Cameron Hydraulic data book.

You can also do pipe friction calculation on the Fluide Design web site: http://www/fluidedesign.com/applets.htm.

The total friction loss for piping in the system is then 9 + 3.1 = 12.1 feet. The static head as per Figure 41 is 35 feet. Therefore the total head is 35 + 12.1 = 47 feet. We can now go to the store and purchase a pump with at least 47 feet of total head at 10 gpm. Sometimes total head is called Total Dynamic Head (T.D.H.), it has the same meaning. The pump’s rating should be as close as possible to these two figures without splitting hairs. As a guideline, allow a variation of plus or minus 15% on total head. On the flow, you can also allow a variation but you may wind up paying for more than what you need.



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What is the pump rating? The manufacturer will rate the pump at its optimum total head and flow, this point is also known as the best efficiency point or B.E.P. At that flow rate, the pump is at its most efficient and there will be minimal amount of vibration and noise. Of course, the pump can operate at other flow rates, higher or lower than the rating but the life of the pump will suffer if you operate too far away from its normal rating. As a guideline, aim for a variation of plus or minus 15% on total head.

Figure 44 Best efficient point of a pump. 14. Examples of common residential water systems

This next figure shows a typical small residential water system. The red tank is an accumulator (see http://www.fluidedesign.com/pump_glossary.htm).

Figure 44a Typical small residential plumbing system.



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The following figures show various common water systems and indicates what the static head, the friction head and the pump total head.

Figure 45 Fountain water supply with nozzle.

This web site (http://www.atlanticfountains.com/spray_patterns.htm ) will tell you the flow requirement of each nozzle and the nozzle head requirement.

Figure 46 Fountain water supply (no nozzle).



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Figure 47 Domestic water supply.

Figure 48 Domestic water supply (shallow well).



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Figure 49 Domestic water supply (deep well).


www.lightmypump.com---------www.lightmypump.com---------- Revised October 9, 2007

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15. Calculate the pump discharge pressure from the pump total head

To calculate the pressure at the bottom of a pool, you need to know the height of the water above you. It doesn’t matter if it’s a pool or a lake, the height is what determines how much fluid weight is above and therefore the pressure. Pressure can be applied to a small object or a big object. For example if we take tanks of different sizes from small to big, the pressure will be the same everywhere at the bottom no matter how big the surface as long as the surface is at the same level.

Figure 50 Pressure at the bottom of tanks of various sizes. Press ure is equal to a force divided by a surface. It is often expressed in pounds per square inch or psi. The force is the weight of water. The density of water is 62.3 pounds per cubic foot.

Figure 51 Pressure is the weight of the liquid divided by the cross-sectional area.

The weight of water in tank A is the density times it’s volume.

W A = dens.× volume



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The volume of the tank is the cross-sectional area A times the height H. The cross-sectional area is pi or π times the diameter squared divided by 4.

A =

π

× D 2 4

The cross-sectional area of tank A is:

A =

× 12

3.1416 4

= 0.78 ft 2

The volume V is A x H:

V = A × H = 0.78×10 = 7.8 ft 3 The weight of the water W A is:

W A = dens.×V = 62.3× 7.8 = 485.9 lb Therefore the pressure is:

p =

W A A

=

485.9 0.78

= 622.9

lb / ft 2

This is the pressure in pounds per square feet, one more step is required to get the pressure in pounds per square inch or psi. There is 12 inches to a foot therefore there is 12x12 = 144 inches to a square foot. The pressure p at the bottom of tank A in psi is:

p = p (lb / ft ) × 2

1 ft 2 144 in

2

=

622.9 144

= 4.3 lb / in 2 = 4.3 psi

If you do the calculation for tanks B and C you will find exactly the same result, the pressure at the bottom of all these tanks is 4.3 psi. The general relationship for pressure vs. tank height is:

p ( psi ) =

1 2.31

SG × H ( ft )

SG or specific gravity is another way of expressing density, it is the ratio of a fluid’s density to that of water, so that water will have an SG =1. Denser liquids will have a value greater than 1 and lighter liquids a value less than 1. The usefulness of specific



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gravity is that it has no units since it is a comparative measure of density or a ratio of densities therefore specific gravity will have the same value no matter what system of units we are using, Imperial or metric. For those of you who would like to see how this general relationship is found go to Appendix E at the end of this article. We can measure head at the discharge side of the pump by connecting a tube and measuring the height of liquid in the tube. Since the tube is really only a narrow tank we can use the pressure vs. tank height equation

p ( psi) =

1 2.31

SG × H ( ft )

to determine the discharge pressure. Alternatively, if we put a pressure gauge at the pump discharge, we can then calculate the discharge head.

Figure 52 Discharge static head plus friction head equals pump discharge head.

We can calculate the discharge pressure of the pump based on the total head which we get from the characteristic curve of the pump. This calculation is useful if you want to troubleshoot your pump or verify if it is producing the amount of pressure energy that the manufacturer says it will at your operating flow rate.

Figure 53 Typical pump system.  Copyright . 2005----


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For example if the characteristic curve of the pump is as shown in Figure 55 and the flow in the system is 20 gpm. The total head is then 100 feet. The installation is as shown in Figure 53, a domestic water system that takes its water from a shallow well 15 feet lower than the pump suction. The pump will have to generate lift to get the water up to its suction connection. This means that the pressure will be less than the atmosphere pressure at the pump suction. Why is this pressure less than atmospheric pressure or low? If you take a straw, fill it with water, cover one end with your fingertip and turn it upside down you will notice that the liquid does not come out of the straw, try it!. The liquid is pulled downward by gravity and creates a low pressure under your fingertip. The liquid is maintained in balance because the low pressure and the weight of the liquid is exactly balanced by the force of atmospheric pressure that is directed upwards. The same phenomenon occurs in the pump suction which is pulling up liquid from a low source. Like in the straw, the pressure close to the pump suction connection must be low for the liquid to be supported.

Figure 54 There is low pressure at the pump suction when the liquid surface is below the pump.

To calculate the discharge head, we determine the total head from the characteristic curve and subtract that value from the pressure head at the suction, this gives the pressure head at the discharge which we then convert to pressure. We know that the pump must generate 15 feet of lift at the pump suction, lift is negative static head. It should in fact be slightly more than 15 feet because a higher suction lift will be required due to friction. But let’s assume that the pipe is generously sized and that the friction loss is small.



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Figure 55 Characteristic pump curve.

TOTAL HEAD = 100 = HD - HS or HD = 100 + HS The total head is equal to the difference between the pressure head at the discharge H and the pressure head at the suction H S. HS is equal to –15 feet because it is a lift therefore: HD = 100 + (-15) = 85 feet The discharge pressure will be:

p

=

1 2.31

×1.0 × 85 = 37 psig

Now you can check your pump to see if the measured discharge pressure matches the prediction. If not, there may be something wrong with the pump.

Figure 56 Head vs. pressure measurement.  Copyright . 2005----


D

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Note: you must be careful where you locate the pressure gauge, if it is much higher than the pump suction, say higher than 2 feet, you will read less pressure than actually is there at the pump. Also the difference in velocity head of the pump discharge vs. the suction should be accounted for but this is typically small.



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APPENDIX A

Flow rate and friction loss for different pipe sizes based on 5 ft/s velocity Flow rate and friction loss for different pipe sizes based on 15 ft/s velocity Flow rate and friction loss for different pipe sizes based on 4.5 m/s velocity Flow rate and friction loss for different pipe sizes based on 1.5 m/s velocity



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FRICTION HEAD LOSS FOR VARIOUS FLOW RATES



page….55




page….56




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PIPE FRICTION HEAD LOSS FOR VARIOUS FLOW RATES



page….58




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APPENDIX B

Formulas and an example of how to do pipe friction calculations



page….60

PIPE FRICTION CALCULATION

The average velocity v in a pipe is calculated based on the formula [1] and the appropriate units are indicated in parentheses. (see the last page for a table of all the symbols)

q (USgal . / min)

v ( ft / s )

= 0 .4085

v ( m / s )

= 1273 .23

2

D (in )

[1]

2

Or in the metric system [1a]

q ( L / s ) 2

D ( mm )

2

The Reynolds Re number is calculated based on formula [2].

Re

. = 77458

Re

= 1000

v ( ft / s) D(in )

[2]

ν ( cSt )

Or in the metric system

v ( m / s) D( mm)

[2a]

ν ( cSt )

If the Reynolds number is below 2000 than the flow is said to be in a laminar regime. If the Reynolds number is above 4000 the regime is turbulent. The velocity is usually high enough in industrial processes and homeowner applications to make the flow regime turbulent. The viscosity of many fluids can be found in the Cameron Hydraulic data book. The viscosity of water at 60F is 1.13 cSt. If the flow is laminar then the friction parameter f is calculated with the laminar flow equation [3].

f =


64 Re


[3]

page….61

If the flow is turbulent then the friction parameter f is calculated based on the SwameeJain equation [4].

f

=

[4]

0 .2 5

  ε  lo g 10  +  3.7 D 

  R e 0 . 9   5.7 4

2

In the turbulent flow regime the friction factor f depends on the absolute roughness of the pipe inner wall. Table B1 provide some values for various materials. PIPE MATERIAL

Absolute roughness ε (ft) 0.00015 0.0004 0.0005

Steel or wrought iron Asphalt-dipped cast iron Galvanized iron Table B1

The friction factor ∆HFP/L is calculated with the Darcy-Weisback equation [5]

∆ H FP  ft fluid    = 1200 L  100 ft pipe 

2

f

(v( ft / s)) D (in)

×

[5]

2 g ( ft / s2)

g=32.17 ft/s2 or in the metric system

∆ H FP  m fluid    = 10 5 L  100 m pipe 

f

(v ( m / s )) D ( mm )

×

2

[5a]

2 g ( m / s 2)

g=9.8 m/s2 The pipe friction loss ∆HFP is calculated with equation [6]

∆ H FP  ft fluid  L( ft pipe)  × L  100 ft pipe  100

[6]

∆ H  fluid  L(m pipe) × ∆ H FP (m fluid ) = FP  L  100 m pipe  100

[6a]

∆ H FP ( ft fluid ) = or in the metric system



page….62

Example calculation

Calculate the pipe friction loss of a 2 1/12” schedule 40 (2.469” internal pipe diameter) new steel pipe with a flow rate of 149 gpm for water at 60F and a pipe length of 50 feet. The roughness is 0.00015 ft and the viscosity is 1.13 cSt.

The average velocity v in the pipe is:

v ( ft / s )

= 0.4085 ×

149 2 .469 2

[1’]

= 9 .98

The Reynolds Re number is:

Re

= 7745.8×

9.98× 2.469 1.13

[2’]

=1.69 ×10 5

The friction parameter f is:

f =

0 . 25

   0 . 00015 × 12 5 . 74  log 10    + 5 0 .9   3 . 7 × 2 . 469 ( 1 . 69 × 10 )    

2

= 0 . 02031

[4’]

The friction factor ∆HFP/L is calculated with the Darcy-Weisback equation [5]

∆ H FP  ft fluid    = 1200× 0.02031× L  100 ft pipe 

2

9.98 =15.34 2.469 × 2 × 32.17

[5’]

The pipe friction loss ∆HFP is:

∆ H FP ( ft fluid ) =15.34 ×


50 100

= 7.67


[6’]

page….63

Symbols

Variable nomenclature

Imperial system (FPS units)

D Re q ∆HFP

in (inch) non dimensional USgpm (gallons per minute) ft (feet) CSt (centistokes) Ft (feet) ft/s (feet/second) ft (feet) Non dimensional feet of fluid/100 ft of pipe ft/s2 (feet per second square)

pipe diameter Reynolds number flow rate friction head loss in pipes ν viscosity ε pipe roughness v velocity L pipe length f friction parameter ∆HFP/L friction factor g acceleration due to gravity (32.17 ft/s 2)



page….64

APPENDIX C

Formulas and an example of how to do pipe fittings friction calculations



page….65

PIPE FITTING FRICTION CALCULATION

The friction loss for fittings depends on a K factor which can be found in many sources such as the Cameron Hydraulic data book or the Hydraulic Institute Engineering data book, the charts which I reproduce here are shown Figures C1 and C2. The fittings friction ∆HFF can be calculated based on the following formula where K is a factor dependant on the type of fitting, v is the velocity in feet/second, g is the acceleration due to gravity (32.17 ft/s 2 or 9.8 m/s2).

2

2

2

2

( ft / s) ∆ H FF ( ft fluid ) = K v 2 g ( ft / s2) In the metric system

( m / s) ∆ H FF (m fluid ) = K v 2 g ( m / s 2)

For example a 2 ½” inch screwed elbow has a K factor of 0.85 according to Figure C1 and using a velocity of 10 ft/s (this is determined from the flow rate). The fittings friction loss will be:

∆ H FF ( ft fluid ) = 0.85


10 2 × 32 .17 2

= 1 .3


page….66

Figure C1 Pressure head loss K coefficients for fittings (source the Hydraulic Institute Standards book www.pumps.org).



page….67

Figure C2 Pressure head loss K coefficients for manual valves and other devices (source the Hydraulic Institute Standards book www.pumps.org).



page….68

APPENDIX D

Formula and an example of how to do velocity calculation for fluid flow in a pipe



page….69

PIPE AVERAGE VELOCITY CALCULATION

The average velocity in a pipe can be calculated based on the following formula where v is the velocity in feet/second, D the internal diameter in inches and q the flow rate in USgallons per minute.

v ( ft / s )

= 0 .4085

q (USgal . / min ) D 2 (in ) 2

In the metric system:

v ( m / s )

= 1273 .2

q ( L / s ) D 2 ( mm ) 2

For example a 2 ½” inch schedule 40 pipe has an internal diameter of 2.469 in, what is the average pipe velocity for a flow rate of 105 gpm.

v ( ft / s )


= 0.4085

105 2 .469 2

= 9.98


page….70

APPENDIX E

The relationship between pressure and pressure head



page….71

THE RELATIONSHIP BETWEEN PRESSURE HEAD AND PRESSURE

We would like to be able to calculate at any time the head that corresponds to a certain pressure anywhere in the system. One place that’s of particular interest is the pressure at the pump discharge. If we connect a tube to the pump discharge the pressure at the connection will make the liquid in the tube rise to a height that balances the force of pressure at the bottom. This will be the pressure head that corresponds to this pressure.

Figure E1

The measuring tube acts a s a small narrow tank so that this is the same process as calculating the pressure at the bottom of a tanks. The weight of a fluid column of height (z) is:

F = g V =γ V =γ z A since V = z A The pressure (p) is equal to the fluid weight (F) divided by the cross-sectional area (A) at the point where the pressure is calculated :

p = F = A


γ z A

A

= γ z


Pumps

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