mud pump displacement calculator supplier

Rig pump output, normally in volume per stroke, of mud pumps on the rig is  one of important figures that we really need to know because we will use pump out put figures to calculate many parameters such as bottom up strokes,  wash out depth, tracking drilling fluid, etc. In this post, you will learn how to calculate pump out put for triplex pump and duplex pump in bothOilfield and Metric Unit.

Disclaimer: These calculators are for estimation purposes only. They are not intended to provide complete answers to complex problems, but rather for quick reference and estimations.

Although Power Zone strives to keep these calculators up to date and accurate, Power Zone is not liable for mistakes made as a result of errors or misinterpretations of these calculators.

Pump Output per Stroke (PO): The calculator returns the pump output per stroke in barrels (bbl).  However this can be automatically converted to other volume units (e.g. gallons or liters) via the pull-down menu.

A triplex mud (or slush) pump has three horizontal plungers (cylinders) driven off of one crankshaft. Triplex mud pumps are often used for oil drilling.

Impeller sizes are determined by calculating the TOR (sometimes called the time of rollover) for each compartment. This is the time, in seconds, required to completely move the fluid in a compartment (Table 10.1) and can be calculated by knowing the tank volume and impeller displacement:

We provide hydraulic components & repair services for industrial applications like paper mills, saw mills, steel mills, recycling plants, oil & gas applications and mobile applications, including construction, utility, mining, agricultural and marine equipment. This includes hydraulic pumps, motors, valves, servo/prop valves, PTOs, cylinders & parts.

Pumps tend to be one of the biggest energy consumers in industrial operations. Pump motors, specifically, require a lot of energy. For instance, a 2500 HP triplex pump used for frac jobs can consume almost 2000 kW of power, meaning a full day of fracking can cost several thousand dollars in energy costs alone!

So, naturally, operators should want to maximize energy efficiency to get the most for their money. Even a 1% improvement in efficiency can decrease annual pumping costs by tens of thousands of dollars. The payoff is worth the effort. And if you want to remotely control your pumps, you want to keep efficiency in mind.

In this post, we’ll point you in the right direction and discuss all things related to pump efficiency. We’ll conclude with several tips for how you can maintain pumping efficiency and keep your energy costs down as much as possible.

In simple terms, pump efficiency refers to the ratio of power out to power in. It’s the mechanical power input at the pump shaft, measured in horsepower (HP), compared to the hydraulic power of the liquid output, also measured in HP. For instance, if a pump requires 1000 HP to operate and produces 800 HP of hydraulic power, it would have an efficiency of 80%.

Remember: pumps have to be driven by something, i.e., an electric or diesel motor. True pump system efficiency needs to factor in the efficiency of both the motor AND the pump.

Consequently, we need to think about how electrical power (when using electric motors) or heat power (when using combustion engines) converts into liquid power to really understand pump efficiency.

Good pump efficiency depends, of course, on pump type and size. High-quality pumps that are well-maintained can achieve efficiencies of 90% or higher, while smaller pumps tend to be less efficient. In general, if you take good care of your pumps, you should be able to achieve 70-90% pump efficiency.

Now that we have a better understanding of the pump efficiency metric, let’s talk about how to calculate it. The mechanical power of the pump, or the input power, is a property of the pump itself and will be documented during the pump setup. The output power, or hydraulic power, is calculated as the liquid flow rate multiplied by the "total head" of the system.

IMPORTANT: to calculate true head, you also need to factor in the work the pump does to move fluid from the source. For example, if the source water is below the pump, you need to account for the extra work the pump puts in to draw source water upwards.

*Note - this calculation assumes the pump inlet is not pressurized and that friction losses are minimal. If the pump experiences a non-zero suction pressure, or if there is significant friction caused by the distance or material of the pipe, these should be factored in as well.

You"ll notice that the elevation head is minimal compared to the discharge pressure, and has minimal effect on the efficiency of the pump. As the elevation change increases or the discharge pressure decreases, however, elevation change will have a greater impact on total head.

Obviously, that’s a fair amount of math to get at the pump efficiency, considering all of the units conversions that need to be done. To avoid doing these calculations manually, feel free to use our simple pump efficiency calculator.

Our calculations use static variables (pump-rated horsepower and water source elevation) and dynamic variables (discharge flow and pressure). To determine pump efficiency, we need to measure the static variables only once, unless they change.

If you want to measure the true efficiency of your pump, taking energy consumption into account, you could add an electrical meter. Your meter should consist of a current transducer and voltage monitor (if using DC) for electrical motors or a fuel gauge for combustion. This would give you a true understanding of how pump efficiency affects energy consumption, and ultimately your bank account.

Up until this point, we’ve covered the ins and outs of how to determine pump efficiency. We’re now ready for the exciting stuff - how to improve pump efficiency!

One of the easiest ways to improve pump efficiency is to actually monitor pumps for signs of efficiency loss! If you monitor flow rate and discharge (output power) along with motor current or fuel consumption, you’ll notice efficiency losses as soon as they occur. Simply having pump efficiency information on hand empowers you to take action.

Another way to increase efficiency is to keep pumps well-maintained. Efficiency losses mostly come from mechanical defects in pumps, e.g., friction, leakages, and component failures. You can mitigate these issues through regular maintenance that keeps parts in working order and reveals impending failures. Of course, if you are continuously monitoring your pumps for efficiency drops, you’ll know exactly when maintenance is due.

You can also improve pump efficiency by keeping pumps lubricated at all times. Lubrication is the enemy of friction, which is the enemy of efficiency (“the enemy of my enemy is my friend…”).

A fourth way to enhance pump efficiency is to ensure your pumps and piping are sized properly for your infrastructure. Although we’re bringing this up last, it’s really the first step in any pumping operation. If your pumps and piping don’t match, no amount of lubricant or maintenance will help.

In this post, we’ve given you the full rundown when it comes to calculating and improving pump efficiency. You can now calculate, measure, and improve pump efficiency, potentially saving your business thousands of dollars annually on energy costs.

For those just getting started with pump optimization, we offer purpose-built, prepackaged solutions that will have you monitoring pump efficiency in minutes, even in hazardous environments.

The pump horsepower calculator is used to estimate the pump power, i.e., the power transmitted to the shaft. A pump is one of the most common hydraulic machinery and is used to move fluid by the means of mechanical action by its impeller. Some of its application includes maintaining water supply across the city, heating, ventilation, and cooling systems (HVAC), hydraulics and pneumatics, and electricity generation (see hydroelectric power calculator)

The pump power is a function of hydraulic power and efficiency. Given the importance of this component, it is imperative to understand the basic characteristics of a pump to ensure greater efficiency of the larger processes. You can find more information about pump efficiency and pump power calculations in subsequent paragraphs.

The pump shaft power is defined as the power applied to achieve the head and the volumetric flow rate. It is a function of volumetric flow rate Q, differential head H, the density of fluid ρ, efficiency η, and the gravitational constant g. Mathematically, that"s:

We know that the pumps in most cases do not operate at an efficiency of 100%. Actually, cavitation drastically reduces it. The parameter of specific speed is used to compare the performance of the pump to the ideal case, i.e., a geometrically similar pump delivering 1 cubic meter of fluid per second against 1 m head. The specific speed NsN_\mathrm{s}Ns​ is a dimensionless quantity that is given by the equation:

Note that, while NsN_\mathrm{s}Ns​ is dimensionless, its value changes depending on the units system used for its inputs. The above version of the equation is used in the calculator that gives dimensionless output for specific speed. However, a simpler version of the equation was introduced without the acceleration due to gravity g to use with English units. Mathematically,

In our important role as hydraulic pump manufacturers, we are aware of the large number of variables that need to be considered when choosing the right pump for the specific application. The purpose of this first article is to begin to shed light on the large number of technical indicators within the hydraulic pump universe, starting with the parameter “pump head”.

The head of a pump is a physical quantity that expresses the pump’s ability to lift a given volume of fluid, usually expressed in meters of water column, to a higher level from the point where the pump is positioned. In a nutshell, we can also define head as the maximum lifting height that the pump is able to transmit to the pumped fluid. The clearest example is that of a vertical pipe rising directly from the delivery outlet. Fluid will be pumped down the pipe 5 meters from the discharge outlet by a pump with a head of 5 meters. The head of a pump is inversely correlated with the flow rate. The higher the flow rate of the pump, the lower the head.

What is the head of a pump? As mentioned earlier, the head corresponds to the actual energy that the pump delivers to the fluid. The Bernoulli equation is applied between the pump’s inlet and outlet sections:

However, during the design stage, P1 and P2 are never known (as there is no physical element yet and therefore it is not possible to effectively measure the pump’s inlet and outlet pressure).

At this point we can easily calculate the head losses of the system, and therefore choose the correct size of the pump to achieve the desired flow rate at the resulting equivalent head.

The pump head indicator is present and can be found in the data sheets of all our main products. To obtain more information on the technical data of our pumps, please contact the technical and sales team.

When choosing a size and type of mud pump for your drilling project, there are several factors to consider. These would include not only cost and size of pump that best fits your drilling rig, but also the diameter, depth and hole conditions you are drilling through. I know that this sounds like a lot to consider, but if you are set up the right way before the job starts, you will thank me later.

Recommended practice is to maintain a minimum of 100 to 150 feet per minute of uphole velocity for drill cuttings. Larger diameter wells for irrigation, agriculture or municipalities may violate this rule, because it may not be economically feasible to pump this much mud for the job. Uphole velocity is determined by the flow rate of the mud system, diameter of the borehole and the diameter of the drill pipe. There are many tools, including handbooks, rule of thumb, slide rule calculators and now apps on your handheld device, to calculate velocity. It is always good to remember the time it takes to get the cuttings off the bottom of the well. If you are drilling at 200 feet, then a 100-foot-per-minute velocity means that it would take two minutes to get the cuttings out of the hole. This is always a good reminder of what you are drilling through and how long ago it was that you drilled it. Ground conditions and rock formations are ever changing as you go deeper. Wouldn’t it be nice if they all remained the same?

Centrifugal-style mud pumps are very popular in our industry due to their size and weight, as well as flow rate capacity for an affordable price. There are many models and brands out there, and most of them are very good value. How does a centrifugal mud pump work? The rotation of the impeller accelerates the fluid into the volute or diffuser chamber. The added energy from the acceleration increases the velocity and pressure of the fluid. These pumps are known to be very inefficient. This means that it takes more energy to increase the flow and pressure of the fluid when compared to a piston-style pump. However, you have a significant advantage in flow rates from a centrifugal pump versus a piston pump. If you are drilling deeper wells with heavier cuttings, you will be forced at some point to use a piston-style mud pump. They have much higher efficiencies in transferring the input energy into flow and pressure, therefore resulting in much higher pressure capabilities.

Piston-style mud pumps utilize a piston or plunger that travels back and forth in a chamber known as a cylinder. These pumps are also called “positive displacement” pumps because they literally push the fluid forward. This fluid builds up pressure and forces a spring-loaded valve to open and allow the fluid to escape into the discharge piping of the pump and then down the borehole. Since the expansion process is much smaller (almost insignificant) compared to a centrifugal pump, there is much lower energy loss. Plunger-style pumps can develop upwards of 15,000 psi for well treatments and hydraulic fracturing. Centrifugal pumps, in comparison, usually operate below 300 psi. If you are comparing most drilling pumps, centrifugal pumps operate from 60 to 125 psi and piston pumps operate around 150 to 300 psi. There are many exceptions and special applications for drilling, but these numbers should cover 80 percent of all equipment operating out there.

The restriction of putting a piston-style mud pump onto drilling rigs has always been the physical size and weight to provide adequate flow and pressure to your drilling fluid. Because of this, the industry needed a new solution to this age-old issue.

As the senior design engineer for Ingersoll-Rand’s Deephole Drilling Business Unit, I had the distinct pleasure of working with him and incorporating his Centerline Mud Pump into our drilling rig platforms.

In the late ’90s — and perhaps even earlier —  Ingersoll-Rand had tried several times to develop a hydraulic-driven mud pump that would last an acceptable life- and duty-cycle for a well drilling contractor. With all of our resources and design wisdom, we were unable to solve this problem. Not only did Miller provide a solution, thus saving the size and weight of a typical gear-driven mud pump, he also provided a new offering — a mono-cylinder mud pump. This double-acting piston pump provided as much mud flow and pressure as a standard 5 X 6 duplex pump with incredible size and weight savings.

The true innovation was providing the well driller a solution for their mud pump requirements that was the right size and weight to integrate into both existing and new drilling rigs. Regardless of drill rig manufacturer and hydraulic system design, Centerline has provided a mud pump integration on hundreds of customer’s drilling rigs. Both mono-cylinder and duplex-cylinder pumps can fit nicely on the deck, across the frame or even be configured for under-deck mounting. This would not be possible with conventional mud pump designs.

The second generation design for the Centerline Mud Pump is expected later this year, and I believe it will be a true game changer for this industry. It also will open up the application to many other industries that require a heavier-duty cycle for a piston pump application.

Oil and Gas drilling process - Pupm output for Triplex and Duplex pumpsTriplex Pump Formula 1 PO, bbl/stk = 0.000243 x ( in) E.xample: Determine the pump output, bbl/stk, at 100% efficiency for a 7" by 12". triplex pump: PO @ 100%,= 0.000243 x 7 x12 PO @ 100% = 0.142884bbl/stk Adjust the pump output for 95% efficiency: Decimal equivalent = 95 + 100 = 0.95 PO @ 95% = 0.142884bbl/stk x 0.95 PO @ 95% = 0.13574bbl/stk Formula 2 PO, gpm = [3(D x 0.7854)S]0.00411 x SPM where D = liner diameter, in. S = stroke length, in. SPM = strokes per minute Determine the pump output, gpm, for a 7" by 12". triplex pump at 80 strokes per minute: PO, gpm = [3(7 x 0.7854) 1210.00411 x 80 PO, gpm = 1385.4456 x 0.00411 x 80 PO = 455.5 gpm

Example:Duplex Pump Formula 1 0.000324 x (liner diameter, in) x ( stroke lengh, in) = ________ bbl/stk -0.000162 x (rod diameter, in) x ( stroke lengh, in) = ________ bbl/stk Pump out put @ 100% eff = ________bbl/stk Example: Determine the output, bbl/stk, of a 5 1/2" by 14" duplex pump at 100% efficiency. Rod diameter = 2.0": 0.000324 x 5.5 x 14 = 0.137214bbl/stk -0.000162 x 2.0 x 14 = 0.009072bbl/stk Pump output @ 100% eff. = 0.128142bbl/stk Adjust pump output for 85% efficiency: Decimal equivalent = 85 100 = 0.85 PO@85%)= 0.128142bbl/stk x 0.85 PO@ 85% = 0.10892bbl/stk Formula 2

PO. bbl/stk = 0.000162 x S[2(D) - d] where S = stroke length, in. D = liner diameter, in. d = rod diameter, in. Example: Determine the output, bbl/stk, of a 5 1/2". by 14". duplex pump @ 100% efficiency. Rod diameter = 2.0in.: PO@100%=0.000162 x 14 x [ 2 (5.5) - 2 ] PO @ 100%)= 0.000162 x 14 x 56.5 PO@ 100%)= 0.128142bbl/stk Adjust pump output for 85% efficiency: PO@85%,= 0.128142bb/stkx 0.85 PO@8.5%= 0.10892bbl/stk Metric calculation Pump output, liter/min = pump output. liter/stk x pump speed, spm. S.I. units calculation Pump output, m/min = pump output, liter/stk x pump speed, spm. Mud Pumps Mud pumps drive the mud around the drilling system. Depending on liner size availability they can be set up to provide high pressure and low flow rate, or low pressure and high flow rate. Analysis of the application and running the Drill Bits hydraulics program will indicate which liners to recommend. Finding the specification of the mud pumps allows flow rate to be calculated from pump stroke rate, SPM. Information requiredo Pump manufacturer o Number of pumps o Liner size and gallons per revolution Weight As a drill bit cutting structure wears more weight will be required to achieve the same RoP in a homogenous formation. PDC wear flats, worn inserts and worn milled tooth teeth will make the bit drill less efficiently. Increase weight in increments of 2,000lbs approx. In general, weight should be applied before excessive rotary speed so that the cutting structure maintains a significant depth of cut to stabilise the bit and prevent whirl. If downhole weight measurements are available they can be used in combination with surface measurements to gain a more accurate representation of what is happening in the well bore.

Honda Power Equipment offers pumps to meet a wide variety of applications. Use the chart and information below to select the right pump for your specific needs or

For general de-watering needs, Honda"s popular WB series pumps offer the best features at a value price. These models offer commercial grade components like silicon carbide seals, anti-vibration mounts, and a fixed-mount cast iron

The WH series are perfect for applications needing high pressure, such as sprinklers or nozzles. These pumps are lightweight and compact, making them highly portable. Possible applications include irrigation and fire suppression,

Honda Trash Pumps are the ultimate choice for contractors and rental applications. The WT series can handle solids up to 1 1/16” in diameter. These pumps are designed to move water, and lots of it – up to 433 gallons

Pumping water from excavations, crawl spaces, underground passageways, grain elevators, farm stock tanks, construction sites, cooling towers and parking lot sumps.