Calculation of Pump Efficiency: Formula & Equation
The power of the water the pump efficiency produces divided by the input of shaft power is the pump efficiency. The quantity and pressure of water and the pressure at which a pump operates define its output power. The pump’s shaft is powered by an internal combustion engine or an electric motor.
How successfully a pump can transform one form of energy into another is determined by the differential in horsepower going into and out of the pump. To make the pumping system 100% efficient, the horsepower entering the pump should ideally match the horsepower exiting the pump. Pumps for Sale from a variety of Suppliers and Companies, as well as manufacturers and distributors, are available on INFO4ALL.
The website of INFO4ALL provides a thorough range of pump services to meet your unique needs. INFO4ALL vendors can assist you with this. Please get in touch with Pump Experts on INFO4ALL to learn more about how to connect with a wide range of Service Providers who regularly provide high-quality products.
It’s critical to pick a pumping system that uses less fuel or energy and has lower yearly pumping costs. Costs might rise annually dramatically with improperly selected and ineffective pumping systems. Additionally, the pumping plant might experience excessive wear, which would result in water waste.
For more reading, this article also includes a PDF file regarding pumps, their varieties, and their efficacy.
What Is the Ideal Pump Efficiency?
Even the smallest centrifugal pumps have efficiencies of 50 to 70%, while many medium and bigger ones have efficiencies of 75 to 93 percent. While any motor with ten horsepower or more may be developed to reach the 90% threshold, large AC motors may achieve a 97 percent efficiency.
Your pump will use more energy the further away it is from its ideal energy point, which will reduce the dependability and efficiency of your system. This increases operating costs and the likelihood of maintenance issues.
Basics of Pump Efficiency
Efficiency is the ability of a machine to transform one form of energy into another. Let’s say one energy unit is supplied to a machine, and its output is half that amount. Its effectiveness in this situation is 50%. Additionally, if a pump operates at 100% efficiency, the mechanical power input and water power output are equal. Since no pump is 100% efficient, the mechanical energy input will always be greater than the mechanical energy output. Lower efficiency are a result of frictional losses, leaks brought on by pressure differences inside the pump casing, and losses brought on by more complicated designs.
Centrifugal pumps and AC induction motors are two low-cost, high-performance equipment that are often employed in the pump sector. AC motors convert electrical energy into mechanical energy, whereas centrifugal pumps transform mechanical energy into hydraulic energy (flow, pressure, and velocity).
Smaller pumps normally fall within the range of 50 to 70 percent efficiency, while the majority of medium and larger centrifugal pumps have efficiencies in the range of 75 to 93 percent. On the other hand, big AC motors typically have an efficiency of around 97 percent, while motors of 10 horsepower or more can be made to have an efficiency of more than 90 percent.
The ratio of the water power output to the shaft power input may be used to calculate the total pump efficiency using the equation below.
Where η is the pump efficiency. In addition, PW and PS represent water power and shaft power, respectively.
Pump efficiency can be calculated by considering two values, including pump flow rate and total head.
Flow Rate
The amount of water flowing through a site at a particular moment is indicated by the flow rate. You can determine the flow rate with a flow metre. Although there are many different kinds of flow metres, an ultrasonic flow metre is often the most user-friendly.
Although there are no restrictions on the measurement units this flow metre may use, gallons per minute is the most common measurement unit.
Total Head
The distance from the source water’s surface to the pump’s output and the pressure present there must both be measured in order to calculate the total head. The accompanying picture demonstrates how to measure the distance from the water level to the pump output in order to compute this number for centrifugal pump efficiency.
The pressure at the pump exit must also be measured. This conversion is carried out using the knowledge that one pound per square inch (psi) of pressure is equal to 2.306 feet of water, or 10,000 Pa in SI units, or one meter of salt water. Consider, for instance, that the pressure at the pump outlet is 60 psi and that there is an 8-foot gap between the water’s surface and the pump’s exit. Consequently, it leads to the following total pressure head:
And this value can be expressed in metrics, knowing that 1 psi is equal to 0.0689476 Pa:
Determining Pump Efficiency
As mentioned before, pump efficiency can be calculated by dividing the water (PW) power by the shaft power (PS).
Water Power Output
Pumps apply energy to develop the discharge pressure and deliver flow. Thus, the hydraulic horsepower of the pump is influenced by two factors multiplied by each other:
p1 and p2 are the pressure of the pump at the suction and discharge lines, measured in N/m2 (or psi), and Q is the flow delivered by the pump in m3/s (or gallons per minute).
Also, in the U.S. measurement system, PW can be expressed as:
Using this system, H (the head of water) and Q (the flow rate) are both measured in feet and in gallons per minute.
With a constant 3960, the product of flow and head (in GPM-feet) can be converted to BHP.
Shaft Power Input
The pump power is supplied by an electric motor. The formula for calculating the output power of an electric driver is:
V is the measured motor voltage in Volts, I is the measured motor current in Amperes, and PF is the power factor. The motor efficiency can be obtained by calculating it through testing or considering it as the design efficiency. A supplier manual provides information about coupling efficiency.
In the U.S. customary units, PS is the power given to the pump shaft in brake horsepower (BHP).
For example, using these equations, we can predict that a pump that generates 90 GPM at 35 feet of head and needs 1 BHP has an overall efficiency at that flow point as:
Heat balance across the pump can also be used to determine the actual power needed for the pump. The difference between the heat flow inlet and the heat flow outlet measures the pump power. This means that the fall in the efficiency of centrifugal pumps increases the liquid temperature at the pump discharge.
Different Categories Of Pump Efficiency
Mechanical (ηm), volumetric (ηv), and hydraulic efficiency (ηh) each add to the general efficiency of a pump. The bearing frame, stuffing box, and mechanical seals are the wellsprings of mechanical misfortunes. Volumetric efficiency is impacted by leakage through wear rings, adjusting openings, as well as clearances of the vanes in semi-open impellers. Liquid grating and volute and impeller misfortunes are considered as a feature of hydraulic efficiency.
However, despite mechanical and volumetric losses being important, hydraulic efficiency is more influential.
Volumetric Efficiency
The volumetric efficiency of a pump is determined by the proportion of the real flow conveyed by it to its hypothetical flow at a given strain. Pump displacement per transformation and driven speed are multiplied to determine the hypothetical flow. For example, in the event that the pump has a displacement of 200 cc/fire up and is turning at 500 RPM, the flow rate is 100L/min.
Flow meters must be utilized to measure the real flow. In the above example, assuming that the pump really flowed 90 liters each minute in the testing, we can say the pump is 90% effective.
Volumetric efficiency is the most common metric used to determine hydraulic pump condition — in light of an expansion in internal leakage brought about by wear or damage. On the off chance that the flow meter didn’t consider hypothetical flow, the genuine flow would be meaningless.
During the plan of a system, designing creators work out the flow capacity of a pump at a specific tension in view of its volumetric efficiency esteem. It is important to understand that leakage ways inside a pump normally remain steady while computing volumetric efficiency in view of flow tests. As a result, on the off chance that a pump is tried at not exactly its full displacement (or maximum RPM), this will inaccurately compute efficiency — except if leakage is treated as a steady and changed likewise.
Consider the instance of a variable displacement pump whose maximum flow rate is 100 liters/minute. On account of flow rate measurement of 90 liters/minute, the determined volumetric efficiency is 90%. For a pump that is flow tried at half displacement (50 L/min) at the same strain and flow temperature, leakage misfortunes will in any case remain 10 L/min, and so the determined volumetric efficiency is 80%.
The subsequent calculation isn’t inaccurate, yet it needs explanation: this pump is 80% productive at around 50% of its displacement. Because of the generally consistent leakage misfortunes of 10 L/min, a pump tried under the same circumstances will be 90% proficient at 100% displacement (100 L/min) — yet 0% productive at 10% displacement (10 L/min).
In the event that you picture the different leakage ways as fixed holes, you can make sense of why pump leakage is consistent at some random strain and temperature. The liquid flow rate through an opening is a component of its diameter (and shape), liquid strain drop, and consistency. Likewise, in the event that these variables don’t change, leakage will remain steady, paying little heed to displacement or shaft speed.
Mechanical/Hydraulic Efficiency
To determine the mechanical/hydraulic efficiency of a pump, the actual drive force required and its theoretical force are compared. Mechanical/hydraulic efficiency of 100% means the hydraulic pump needs no force to give flow at zero strain. In reality, mechanical and liquid grindings keep this from happening.
p is the discharge pressure of the pump, QT is the theoretical pump flow rate, TA is the actual torque delivered to the pump, and N is the rotational speed of the pump shaft.
The mechanical efficiency of a pump can also be calculated by considering torque:
This is sometimes called force efficiency, which is a component of speed and liquid consistency. As gooey grating is thick reliant, higher pump speeds bring about lower efficiency. Regardless of diminishing gooey misfortunes, lower consistency negatively impacts volumetric efficiency.
There are many similarities between the plan phases of the centrifugal pump and the enlistment motor. They have similar features in that each can be modified by the planner by changing just two major components. A motor comprises of a rotor and a stator. Centrifugal pumps, notwithstanding, utilize their impeller and volute (or diffuser) to move liquid. The impeller is the primary component to investigate while assessing centrifugal pump efficiency.
For any impeller, we realize that its head varies as a square of the change in speed. By multiplying the speed, the head will increase four times as much. The same applies to small changes in diameter assuming that you keep the speed constant. Similar standards apply to flow through an impeller, however here it is straightforwardly proportional to the change in diameter or speed — twofold either, and the flow will twofold. The change in rotational speed or diameter of an impeller actually alludes to the peripheral speed of a point at its outskirts. This speed determines how much head and flows an impeller can create.
For any impeller, we realize that its head varies as a square of the change in speed. By multiplying the speed, the head will increase four times as much. The same applies to small changes in diameter assuming that you keep the speed constant. The performance of an impeller is also affected by factors, for example, vane shape.
Today, the starting place of the plan of a pump is “explicit speed.” It is communicated based on the rotational speed, head, and flow rate, as follows:
Taking everything into account, overall efficiency is determined by multiplying efficiencies together. Based on
input flow and tension, the overall efficiency can be utilized to determine how much drive power is required for a hydraulic pump. The more proficient the pump, the less energy it will use for a given flow rate and tension, and the less effective, the more energy will be lost to heat.
η=ηvηm
eta =frac{pQ_A}{T_AN}times 100%η=TANpQA×100%
To learn more about pumps, you can read the following PDF.
Factors Affecting Pump Efficiency
The efficiency of a centrifugal pump is influenced by various factors:
- The minimum thermal flow is maintained in recirculation lines to prevent cavitation during low-flow pump operation. As an inevitable result, the pump becomes less efficient.
- The internal surface roughness is an important consideration. Smooth surfaces in pump internals contribute to high efficiency.
- The efficiency of centrifugal pumps decreases as wear rings clearance increases. The wear rings reduce the clearance between the pump casing and the impeller.
- A higher fluid viscosity reduces pump efficiency.
- The efficiency will decrease due to mechanical losses in couplings, bearings, packings, and so on.
- The trimming of impellers will decrease efficiency.
Heat Pump Efficiency
What is Heat Pump Efficiency
An Air Source Heat Pump (ASHP) typically creates about 3kW thermal energy for each 1kW of electrical power consumed, giving a successful “efficiency” of 300%.
It is thermodynamically impossible to have more than 100% efficiency, as this indicates that more energy is being generated than is being placed in. As a reason, the performance is displayed as a Coefficient of Performance (COP) instead of efficiency.
The case above would be addressed as having a COP of 3.
Apparently, more energy is being given than is consumed because as it were “valuable” input energy is power used to run the compressor and circulating pumps.
The remainder of the energy transferred from a heat source that would somehow not be used (like the ground, ambient air, or a waterway) isn’t viewed as an energy input.
While the title Coefficient of Performance is usually utilized, it is essential to understand the distinctions and to have the option to compare different types that might be utilizing different measures.
- Seasonal Coefficient of Performance _ Used principally with ASHPs to measure heat pump performance in a year, accounting for varying air temperatures. We have three climatic zones set across Europe, defined in BSEN14825.
- Seasonal Energy Efficiency Ratio (SEER)_ Used principally with ASHPs to measure heat pump cooling performance in a year, accounting for varying air temperatures. We have three climatic zones set across Europe, defined in BSEN14825.
- Energy Efficiency Ratio (EER)_ It is a measure of the ratio of given thermal power to total electrical energy. Electrical energy includes ancillaries such as all the pumps, fans, and controls.
- Seasonal Performance Factor (SPF)_ It gives a measure of the ratio of produced thermal energy over the year to total electrical consumption.
Equipment Efficiency
Among other forms of “renewable” energy, where the fuel source is virtually unlimited and free, it is the total cost of production rather than the efficiency that matters.
For comparison, other kinds of heat generation have the subsequent efficiencies:
- Conventional gas/oil boiler: 70–80% efficiency
- Condensing gas/oil boiler: 90–96% efficiency
- Direct electric heating: 35–45% efficiency (with losses in generation and distribution).
The compressor pressure must increase to drive it as the temperature difference between the output and input rises; as a result, the COP declines.
It is crucial to comprehend the temperature at which it is measured and the range across which it will operate as a result.