CENTRIFULGAL PUMP

HEAD, FLOW, AND PRESSURE

Centrifugal Pumps:

Fundamentals

of Operation

HEAD

Centrifugal pumps are dynamic machines, which means that they convert velocity into feet of head. To explain this concept of converting speed or velocity into feet of head, let’s look at Fig. 29.1. This is the water intake section of the Chicago Water Treatment Plant. Water flows from Lake Michigan into a large concrete sump. The top of this sump is well above the level of Lake Michigan. The line feeding the sump extends three miles (3 mi) out into the lake. 

The long line is needed to draw water into the plant, away from the pollution along the Chicago lakefront. The pipeline diameter is 12 ft. Water flows in this line at a velocity of 8 ft/s. Six pumps, stationed atop the concrete sump, pump water into the water treatment plant holding tanks. One day, the plant experiences a partial power failure. Three of the six pumps shown in Fig. 1 shut down. A few moments later, the manhole covers on top of the sump blow off. Geysers of water spurt out of the manholes. What has happened?

FIGURE 1 Converting momentum into feet of head.

HYDRAULIC HAMMER

This is an example of water, or hydraulic, hammer. But what causes water hammer? Let’s consider the change in the water velocity in the 3-mi pipeline when half the pumps failed. The velocity in the pipeline dropped from 8 down to 4 ft/s. Velocity is a form of energy, called kinetic energy.

Energy can take several different forms:

• Heat (Btus)
• Elevation or potential energy
• Kinetic energy (miles per hour)
• Pressure (psig)
• Electrical power (amperes)
• Work (horsepower)
• Acceleration
• Chemical (heat of reaction)

About 250 years ago, Daniel Bernoulli first noticed two important things about energy:

• Energy in one form can be converted to energy in another form.
• While energy can be changed from one form to another, it cannot be created or destroyed.

This idea of the conservation of energy is at the heart of any process plant. In the case of the Chicago Water Treatment Plant, the reduced velocity of the water was converted into feet of head. That is,

the elevation of the water in the sump suddenly increased and blew

the manhole covers off the top of the sump.

MOMENTUM

If the length of the pipeline had been a few hundred feet, this incident would not have happened. It is not only the sudden reduction of the velocity of the water that caused an increase of the water level in the sump. It is also the mass of water in the 3-mi pipeline that contributed to the increased height of water in the sump. The combined effect of mass times velocity is called momentum.

The mass of water in our pipeline weighed 160 × 106 lb. This much water, moving at 8 ft/s, represents a tremendous amount of energy (about 500 million Btu per hour). If the flow of water is cut in half, then the momentum of the water flowing in the pipeline is also cut in half. This energy cannot simply disappear. It has to go somewhere. The energy is converted to an increase of feet of head in the sump; that is, the water level in the sump jumps up and blows the manhole covers off the top of the sump. Incidentally, this is a true story. Can you imagine what would have happened if all six pumps failed simultaneously due to an electric power failure? The result would be a dramatic lesson in the meaning of water hammer.

MY WASHING MACHINE

Figure 2 is a picture of our washing machine as it was originally installed. Whenever the shutoff valve on the water supply line closed, water hammer, or hydraulic shock, would shake the water piping. The momentum of the water flowing in the piping would be suddenly converted to pressure. If the end of a piping system is open (as into a sump), then the momentum of the water is converted to feet of head. But if the end of the piping system is closed, then the momentum of the water is converted to pressure. 

To fix this problem, I installed the riser tube shown in Fig. 3. The top of the riser tube is left full of air. Now when the water flow in the supply pipe is shut, the momentum of the water is converted to compression energy. That is, the air in the riser tube is slightly compressed, as indicated by the pressure gauge I installed at the top of the riser tube.

FIGURE 2 ConHydrolic Hammer Hits Home

FIGURE .3 Riser tube stops hydraulic hammer.

ACCELERATION

Let us imagine that the six pumps in Fig. 1 have not run for a few days. The water level in the sump and the level of Lake Michigan will be the same. I now start all six pumps at the same time. An hour later, the water level in the sump is 12 ft below the water level in the lake. This 12 ft is called “feet of head loss.” If the pipeline is 3 mi long, we say we have “lost 4 ft of head per mile of pipeline.” If we look down into the sump, what would we see happening to the water level during this hour? Figure 4 is a graph of what we would observe. The water level in the sump would drop gradually to 15 ft below the level in the lake. Then the water level in the sump would come partly back up to its equilibrium level of 12 ft below the lake level. Why?

FIGURE 4 Effect of acceleration.

The loss of 12 ft of head as the water flows through the pipeline is due to friction; that is, 12 ft of head are converted to heat. But why do we have a temporary loss of an extra 3 ft? The answer lies in the concept of acceleration. Let’s say you are driving your car onto the expressway. To increase the speed of your car from 30 to 65 miles per hour, you press down on the accelerator pedal. Having reached a velocity of 65 miles per hour, you ease off the accelerator pedal to maintain a constant speed. 

Why? Well, according to Newton’s second law of motion, it takes more energy to accelerate your car than to keep it in motion. Referring again to Fig. 1, the water in the 3-mi pipeline is initially stagnant. Its velocity is zero. An hour later, the water has accelerated to 8 ft/s. When you accelerate your car, the extra energy required comes from the engine. But when we accelerate the water in the pipeline, where does the extra energy come from? Does this extra energy come from the pumps? Absolutely not! The pumps are downstream of the pipeline and the sump. 

They cannot contribute any energy to an upstream pipeline. No, dear reader, the energy to accelerate the water in the pipeline must come from Lake Michigan. But what is the only source of energy that the lake possesses? Answer—elevation or potential energy. To accelerate the water in the pipeline to 8 ft/s requires more energy than to keep the water flowing at that same velocity. And this extra energy comes from the 3 ft of elevation difference between the water level in the sump and the water level in the lake. Once the water has reached its steady-state velocity of 8 ft/s, the need for this extra conversion of feet of head to acceleration disappears, and the water level in the sump rises to within 12 ft of the lake’s level

STARTING NPSH REQUIREMENT

The need to accelerate the fluid in the suction of a pump is called the starting net positive suction head (NPSH) requirement. To calculate this starting NPSH requirement, let’s assume:

• Suction line = 100 ft
• Eight-in line
• Fluid = water @ 62 lbs/ft3
• Initial velocity is zero
• Final velocity is 10 ft/s Then proceed as follows:
• Mass in line is 2170 lb

PRESSURE

Loss of Suction Pressure The need to accelerate liquid in the suction line of a pump leads to a difficult operating problem, which occurs on start-up. Just before the pump shown in Fig. 5 is put on line, the velocity in the suction line is zero. The energy to increase the velocity (i.e., accelerate) the liquid in the suction line must come from the pressure of the liquid at the pump’s suction. As the pump’s discharge valve is opened, the velocity

FIGURE 29.5 Loss of suction pressure causes cavitation

In the suction line increases, reducing the pressure at the suction of the pump. The faster the discharge valve is opened, the greater the acceleration in the suction line, and the greater the loss in the pump’s suction pressure. If the pressure at the suction of the pump falls to its bubble or boiling point, the liquid will start to vaporize. This is called cavitation. A cavitating pump will have an erratically low discharge pressure and an erratically low flow. As shown in Fig. 5, the bubble-point pressure of the liquid is the pressure in the vessel.

We usually assume that the liquid in a drum is in equilibrium with the vapor. The vapor is then said to be at its dew point, while the liquid is said to be at its bubble point. To avoid pump cavitation on start-up, the experienced operator opens the pump discharge valve slowly. Slowly opening the discharge valve results in reduced acceleration of the liquid in the suction line and a slower rate of the conversion of suction pressure to velocity. Try this test. To illustrate what I have just explained, try this

experiment:

1. Open the suction valve to a pump completely.
2. Crack open the case vent to fill the case with liquid. This is called “priming the pump.”
3. Open the pump discharge valve completely.
4. Push the motor START button.
5. Observe the pump suction pressure.

You will see that the pump suction pressure will drop and then come partly back up. If the pump suction pressure were 15 psig to start with, it might drop to 12 psig and then come up to 14 psig. The permanent difference in the suction pressure between 15 and 14 psig is due to a 1-psig piping friction pressure loss. The temporary difference in the suction pressure between 14 and 12 psig is due to a 2-psig conversion of pressure to velocity or kinetic energy. 

To review—all the energy needed to accelerate the liquid to the suction of a pump comes from the pump’s suction pressure. None of this energy comes from the pump itself. Or, as one clever operator at the Unocal Refinery in San Francisco explained to me, “Pumps push, but they do not suck.”

Pump Discharge Pressure

Figure 6 illustrates the internal components of an “overhung, single-stage” centrifugal pump. The term “overhung” refers to the feature that the pump has only an inboard, but no outboard, bearing. The inboard side of a pump means the end closest to the driver. The term “single-stage” means that there is only one impeller. Multistage pumps can have five or six impellers

FIGURE 6 A centrifugal pump.

The main components of the pump shown in Fig. 6 are

• Shaft—used to spin the impeller
• Coupling—attaches the shaft to the turbine or motor driver
• Bearings—support the shaft
• Seal—prevents the liquid inside the pump from leaking out around the shaft
• Impeller wear ring—minimizes internal liquid leakage, from
• the pump discharge, back to the pump suction
• Impeller—accelerates the liquid
• Volute—converts the velocity imparted to the liquid by the impeller to feet of head The impeller is the working part of a centrifugal pump. 

The function of the impeller is to increase the velocity or kinetic energy of the liquid. The liquid flows into the impeller and leaves the impeller at the same pressure. The location of the oval dot

shown at the top of the impeller in Fig. 6 is called the “vane tip.” The pressure at the vane tip is the same as the pump’s suction pressure. However, as the high-velocity liquid escapes from the impeller and flows into the volute, its velocity decreases. The volute (which is also called the diffuser) is shaped like a cone. It widens out in the manner illustrated in Fig. 7. As the liquid flows into the wider section of the volute, its velocity is reduced, and the lost velocity is converted—well, not into pressure, but into feet of head.

FIGURE 7 A volute or diffuser converts velocity into feet of head.

FEET OF HEAD

A centrifugal pump develops the same feet of head regardless of the density of the liquid pumped, as long as the flow is constant. This statement is valid as long as the viscosity of the liquid is below 50 to 100 cP or 200 to 400 SSU (Saybolt Seconds Universal). But, as process operators or engineers, we are not interested in feet of head. We are interested only in pressure. Differential pressure is related to

differential feet of head as follows:

The increase in pressure ΔP is also called head pressure. For example, if I have 231 ft of water in the glass cylinder shown in Fig. 29.8, it would exert a head pressure of 100 psig on the pressure gauge at the bottom of the cylinder. The head pressure would be

FIGURE 8 Head pressure compared to feet of head.

proportionally reduced as the liquid becomes less dense. For example, 231 ft of kerosene (which has 0.75 s.g.) would exert a head pressure of only 75 psig, because it is lighter than water.

EFFECT OF SPECIFIC GRAVITY

Figure 9 shows a centrifugal pump that may pump water through valve B or naphtha through valve A. Question: When water is pumped, the pump’s discharge pressure is 110.4 psig. The pump’s suction pressure is 10.4 psig. When naphtha is pumped, what is the pump’s discharge pressure?

Answer: The head pressure developed by the pump is

and the pump discharge pressure = 10.4 + 60 = 70.4 psig Next question: The pump is driven by an ordinary alternatingcurrent electric motor. It is running at 3600 rpm. When we switch from pumping water to naphtha, what happens to the speed of the pump?

Answer: Nothing. An AC motor is a fixed-speed machine. It will continue to spin at 3600 rpm. Next question: The pump is pulling 100 amps of electrical power. The flow rate is 100 GPM. If we now close valve B and open valve A, what happens to the amp load on the motor driver?

FIGURE 9 Effect of specifi c gravity on pump performance.

Answer: The power demand will go down to 60 amp. Amp(ere)s are a form of electrical work. The units of work are foot-pounds. The feet of head developed by the pump is not affected by the specific gravity of the liquid. But the weight of liquid pumped is proportional to the specific gravity. If the specific gravity drops by 40 percent, and the liquid volume (GPM) stays constant, then the pounds lifted by the pump drops by 40 percent and so does the electrical work.

Last question: Water is being pumped to a tank on a hill at 100 GPM. If we now switch to pumping a lighter fluid at the same rate, can we pump the lighter fluid to a higher elevation, only to a lower elevation, or the same elevation?

Answer: The same elevation. Centrifugal pumps develop the same feet of head at a given volumetric flow rate regardless of the specific gravity of the liquid pumped. This means the ability of the pump to push liquid uphill is the same even if the density of the liquid changes

PUMP CURVE

The feet of head developed by a pump is affected by the volume of liquid pumped. Figure 10 is a typical pump curve. As the flow of liquid from a centrifugal pump increases, the feet of head developed by the pump goes down, as does the pump discharge pressure. A pump curve has two general areas. These are the flat portion of the curve and the steep portion of the curve. We normally design and operate a pump to run toward the end of the flat portion of its curve. A centrifugal pump operating on the flat portion of its curve loses only a small portion of its discharge pressure when flow is increased. 

This is a desirable operating characteristic. A centrifugal pump operating on the steep portion of its curve loses a large portion of its discharge pressure when flow is increased. Pumps operating quite far out on their curve will have an erratic discharge pressure, as the flow through the pump varies. For one such pump, I have seen the discharge pressure drop from 210 to 90 psig as the flow increased by 40 percent. The flow through this pump was regulated by a downstream level-control valve. The discharge pressure varied so erratically that the operators thought the pump was cavitating.

DEVIATION FROM PUMP

Pumps always perform on their performance curve. It is just that they do not always perform on the manufacturer’s performance curve

FIGURE 29.10 Centrifugal pump curve.

The dotted line in Fig. 7 compares an actual performance curve to the manufacturer’s performance curve. Why the deviation?

• The manufacturer’s curve was generated using water. We are pumping some other fluid, perhaps with a much higher viscosity.
• The clearance between the impeller and the impeller wear ring (Fig. 6) may have increased as a result of wear.
• The impeller itself may be worn, or the vane tips at the edge of the impeller might be improperly machined.
• If the pump is driven by a steam turbine, the pump speed could be lower than in the design.
• Available NPSH may be marginal.

DRIVER HORSEPOWER

Normally, increasing the flow from a centrifugal pump increases the amperage load on the motor driver, as shown in Fig. 11. Driver horsepower is proportional to GPM times feet of head. As shown in Fig. 10, as the flow increases the feet of head developed by the pump decreases. On the flat part of the pump curve, the flow increases rapidly, while the head slips down slowly. Hence the product of GPM times feet of head increase. On the steep part of the pump curve, the flow increases slowly, while the head drops off rapidly. Hence, the product of GPM times feet of head remains the same, or goes down.

FIGURE 11 Horsepower requirement for a centrifugal pump.

The response of the amp load on the motor driver is an indication of pump performance. A pump operating on the proper, flat portion of its curve will pull more motor amps as flow increases. A pump operating on the poor, steep portion of its curve will not pull more amps. If the manufacturer’s horsepower/flow curve indicates that more horsepower is required as flow is increased, but the motor amp load is not increasing, then something is wrong with the pump. The amp meter for a pump is best located right on the start-stop box next to the motor. Unfortunately, more often, the only amp meter is located on the electrical breaker box, which is at a remote location.

PUMP IMPELLER

Impeller Diameter

The pump impeller resembles a hollow wheel with radial vanes. The diameter of this wheel can be trimmed down on a lathe. If an impeller with a 10-in diameter is trimmed down to 9 in, it will have a new performance curve below its former performance curve. Both the flow and head of the pump will be reduced. Also, the amp load on the motor driver will decrease. As a rough rule of thumb, reducing a pump’s impeller diameter by 10 percent will reduce the amp load by 25 percent. Not only will this save electricity, but the motor will run much cooler. 

Unloading the motor by 25 percent may increase the life of the motor windings by 10 years. When the control valve downstream of a pump is operating in a mostly closed position, the pump is a good candidate to have its impeller trimmed. Sometimes, the pressure drop across a control valve is so huge (≥100 psi) that it makes a roaring sound. The energy represented by this wasteful ΔP is coming from the electricity supplied to the pump’s motor. 

Hence, trimming the impeller also reduces wear because of erosion in downstream control valves. We frequently increase the size of impellers to increase pump capacity and discharge pressure. However, many chemical plants break the law when they do so. State boiler codes require that vessels and heat exchangers downstream of pumps be rated for the

• Maximum pump discharge pressure
• At the maximum specific gravity that can reasonably be expected
• Unless the equipment is protected by a relief valve

Increasing the size of an impeller always increases its maximum discharge pressure. Hence, increasing the impeller diameter may be unlawful, depending on the maximum allowable working

pressure (MAWP) of downstream heat exchangers. The MAWP is shown on the exchanger name plate. Increasing the size of the impeller by 10 percent will increase the amperage load on the motor driver by 30 percent. For many process pumps, this would require a new motor and breaker to support the larger impeller.

Trimming the pump impeller on a lathe also requires the reshaping of the tips of the impeller vanes. Most often, this requirement is ignored, and the revamped pump operates below its expected curve.

Rather than worry about these fine points, it is best to order new impellers directly from the manufacturer. Installing a larger impeller may require an upgrade in the pump bearings and the mechanical seal. While it is good practice to purchase a pump with a seal and bearings rated for the

maximum impeller size, some pumps can have bearings and seals rated only for their purchased impeller size. 

For process people, the only practical method to check that a pump’s seal and bearings are suitable for the larger impeller is to check with the manufacturer. Drilling pressure balancing holes in the hub of the impeller helps to reduce the extra strain on the bearings of a larger impeller. Sometimes installing a full-sized impeller causes the pump to vibrate. This happens because of a disturbance to the liquid as the vanes spin past the volute or diffuser (i.e., the case outlet opening).

Exxon had, at one time, a standard of keeping impeller sizes at not more than 92 percent of the full impeller diameter. This seems a conservative practice to me. I have usually kept impeller sizes in my

revamp designs 0.25 to 0.5 in below maximum. I don’t recall any of my clients complaining that such pumps vibrated as a result of the enlarged impeller. On the other hand, for a new installation, now that I’ve learned of the Exxon standard, I plan to follow it in my process design work.

IMPELLER ROTATION

Motor-driven pumps often are found running backward. If the motor is wired up backward (this is called reversing the polarity of the leads), the pump will run with the impeller spinning backward. This will reduce the pump’s discharge pressure, sometimes by a little (10 percent) and sometimes by a lot (90 percent). It depends on the design of the impeller. You cannot see the direction of rotation of a pump, but if you touch a pencil to the spinning shaft, you can feel the direction of rotation. The correct direction of rotation is indicated by an arrow stamped on the top of the pump case. There are right- and left-handed pumps. So two otherwise identical pumps may have opposite directions of rotation.

Capacity EFFECT OF TEMPERATURE ON PUMP CAPACITY

We wish to increase the capacity of a centrifugal pump. Should we make the liquid hotter or colder? Let’s make a few assumptions:

• Cooling the liquid will not increase its viscosity above 30 or 40 cSt.
• We are pumping hydrocarbons. Cooling many hydrocarbons by 100°F increases their density by 5 percent.
• The pump’s motor is somewhat oversized.

On this basis, cooling the liquid by 100°F will raise the pump’s discharge pressure by 5 percent. If the pump is developing 1000 psig of differential pressure, cooling the liquid by 200°F will increase its discharge pressure to 1100 psig. If the discharge control valve is now opened, the pump’s discharge pressure will drop back to 1000 psig as the flow increases. The discharge pressure drops because the pump develops less feet of head at a higher flow rate.

Source : Digital Engineering Library @ McGraw-Hill