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  • 18 March 2014 www.pump-zone.com PUMPS & SYSTEMS

    PUMPING PRESCRIPTIONS By Lev Nelik, Ph.D., P.E.Pumping Machinery, LLC

    P&S Editorial Advisory Board

    Stable Versus Thermal Minimum Continuous Flow for Centrifugal Pumps

    Centrifugal pumps have many advantages compared with positive displacement pumps. ey have the ability to run all over the curve. Open the valve to increase the ow, and close it to decrease the ow. is seems convenient and simple. However, this convenience can cause problems for end users. ere are many indications that they should not use a valve for ow controlor at least not to extremes. Staying close to the best e ciency point (BEP) is ideal to save energy and to avoid having a problem pump.

    Pump manufacturers note a minimum continuous stable ow (MCSF) line on their pump performance curves. A centrifugal pump can operate below (to the le of ) the MCSF, but it is not recommended. Hydraulic instabilities can become dramatic and cause pressure pulsations, vibra-tions, axial shuttling of the rotor and failures. e ideal MCSF depends on many factorssuch as

    design type and energy levels. A good rule of thumb is to avoid operating a pump below 50 percent of the BEP ow. Some speci cationssuch as American Petroleum Institute 610would limit the MCSF even more.Operating a pump below the MCSF, hydrau-

    lically stable, can cause damage. Operating below the minimum continuous thermal ow (MCTF), thermally stable, can be disastrous. e MCTF is much lower than the MCSF. When a discharge valve is closed or nearly closed, an oper-ator can literally boil the uid within the pump. e energy that the impeller generates into the uid trapped within the casing by the closed valve transfers into heat (see Equation 1).

    Q = cmT Equation 1

    Where:Q = heat addedc = speci c heatm = massT = the change in temperature

    CASE STUDY

    If the trapped uid is air, the pump can become hotter much faster. As an example, lets take a single-stage, overhung impeller centrifugal pump, driven by a 150-horsepower motor at 1,800 rpm, with 12-inch suction and discharge nozzles (see Image 1of the partially disassembled pump to show the impeller and casing).A 12-inch pipe, about 10 feet long, is between the pump

    suction ange and the supply tank, and a discharge valve is closed while the pump is operating. e motor power during the shut-valve condition depends on the design and type of a pump. For this article, assume it is less than half the power at the BEP, about 60 horsepower (40 kilowatts), which is reasonable for many pumps. is power has nowhere to go and must dissipate into the

    uid inside the pump casing. Eventually, through thermal convection, it heats the water in the suction pipe and in the supply tank. If thermal convection is quick, it will take a long time to heat the water, as the mass of water in suc-tion pipe and the tank will keep the temperature rise to a minimum. However, the internal hydraulics of the impeller-to-volute interaction could mean that the main portion of uid a ected by the rapid heating is only a portion of uid

    Figure 1. Pump performance curves and operating points Source: www.pump-magazine.com/pump_magazine/pump_articles/article_17/article_17.htm

    Image 1. A single-stage, overhung impeller centrifugal pump, driven by a

    150-horsepower motor at 1,800 rpm, with 12-inch suction and discharge nozzles

  • PUMPS & SYSTEMS www.pump-zone.com March 2014 19

    trapped between the impeller outside diameter (OD) and the casing outside boundary away from the impeller. If this is the case, the volume of uid subjected to heating is much smaller, and it will heat quickly.If the uid is water, the water mainly a ected thermally

    is within the space outside the 14-inch OD impeller vanes, which was roughly 6 inches wide with about a 2-inch gap from the OD to the casing wall (see Equation 2).

    3.14 x (14/12) x (6/12) x (2/12) = 0.5 cubic feet ( 3) of trapped volume Equation 2

    If the above formula (for water the speci c heat (Cp) = 4 kilojoules (kJ)/kilogram (kg) C for the mass of the trapped water:

    0.5 3 x 62.4 lb/ 3 = 31 lb = 14 kg

    Each second, 40 kW x 1 sec or 40 kJ are transferred into the trapped volume of water. Substituting the values into Equation 1 results in:

    40 = 4 x 14 x TT = 0.7 C or about 1 F

    In one minute, the temperature will increase 1 x 60 = 60 C.If the pump is on a li , and it started unprimed, the uid

    inside the volute is air. It has the same volume and will become hot faster because of airs lower value of speci c heat (Cp = 1.0 kJ/kg C). e energy entering the air is signi cantly less than it was

    for the water (by a multiple of speci c gravity), but the mass is lower, too:

    0.5 3 x 0.075 lb/ 3 = 0.04 lb = 0.02 kg

    e speci c heat of air is about 1 kJ/(kg C), and the shut-o power in the case of air, assuming direct correction by gravity, is:

    40 kW x (0.075 lb/ 3 / 62.4 lb/ 3 ) = 0.05 kW

    is results in an increase of about 0.05 kJ in one second. Using Equation 1, the temperature change can be calculated:0.05 = 1 x 0.02 x T, and T = 2.5 C or about 5 F rise

    in one second

    In one minute, the temperature increase will be:5 x 60 = 300 F

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  • PUMPING PRESCRIPTIONS

    20 March 2014 www.pump-zone.com PUMPS & SYSTEMS

    e numbers are approximated for simplicity, but the main point is that the temperature of water changes more slowly than the temperature of air because water has a higher heat capacity. Heat capacity is a property of a mate-rial that describes how much heat energy is required to change the temperature of a substance by 1 C. About one unit of heat energy is needed to warm the air 1 C. Four times the heat energy is required to warm water 1 C.One important aspect is what occurs if the water temper-

    ature reaches and exceeds vaporization temperature. When that happens, the phase transformation of liquid water into vapor includes a tremendous volume expansion, bringing with it more complications and potential damage to the equipment.Another factor that was mentioned at the beginning of

    the column is how fast the heat is transferred by internal convection from the uid in the casing to the rest of the piping. A simpli ed computational uid dynamics (CFD) model of a pipe lled with water and rapidly heated at one end could tell an interesting story.

    A simple CFD model of a 12-foot long 12-inch pipe, at zero seconds, one second, 10 seconds, one minute, ve minutes and so on would be informative. When will it reach a uniform temperature, if ever? For pump and piping enthusiasts, a complimentary seat

    for the next Pump School is reserved for you if you can show us this example. To make it more interesting, I pres-ent two cases:

    A system with an insulated pipe A system with a pipe with the insulation removed

    Will a terminal temperature be reached when the rate of heat supplied to the uid is equal to the rate of heat escaping through the pipe wall by conduction and radia-tion to the surrounding air? Will a terminal temperature be reached before the pump is damaged? I look forward to your responses. P&S

    Dr. Nelik (aka Dr. Pump) is president of Pumping Machinery, LLC, an Atlanta-

    based rm specializing in pump consulting, training, equipment troubleshoot-

    ing and pump repairs. Dr. Nelik has 30 years of experience in pumps and pump-

    ing equipment. He can be contacted at www.pump-magazine.com.

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