mud pump pressure formula factory

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.

NOTE: Max RPM in the above equation varies according to type of pump, size of stroke, and other variables. Duplex pumps often run about 100 RPM Max. while triplex pumps will run somewhere between 100 RPM Max and 400 RPM Max.

I have a reciprocating pump and I know what my max rated rod load is (in foot pounds). I also know what size plunger size my pump has. What PSI will my pump produce?

Specific Gravity is used when sizing a centrifugal pump. Liquids with a specific gravity greater than 1.0 are heavier than water and conversely, liquids with a specific gravity lower than 1.0 are lighter weight than water and will generally float on water.

The pressure provided by the rig pump is the sum of all of the individual pressures in the circulating systems. All the pressure produced by the pump is expended in this process, overcoming friction losses between the mud and whatever it is in contact with:

Pressure losses in the annulus acts as a "back pressure" on the exposed formations, consequently the total pressure at the bottom of the annulus is higher with the pump on than with the pump off.

Assuming a circulating pump pressure is 3000 psi when pumping at 100 spm. The pump speed is increased to 120 spm. To approximate the new circulating pump pressure:

Assuming a circulating pump pressure in 3000 psi with a 10 ppg mud weight pumping at 100 spm. If the mud weight in the system was changed to 12 ppg. To approximate the new circulating pump pressure:

In this article provided pump related formulas like fluid flow rate and velocity, power calculation, Specific Speed of Pump (Nq), Total Head, Pump Torque and temperature rise, Net Positive Suction Head, Affinity laws for pump, Pump Efficiency & Overall Efficiency of the Pump

Specific Speed of pump (Nq) is identifies the geometrical similarity of pumps. It is useful to comparing different pump designs irrespective of pump size

In vertical pipe any liquid coloumn of water exerts a certain pressure (force per unit area) on a horizontal surface at the bottom area, this pressure is expressed in metres of liquid column or kg/cm2.

Pressure head calculated as per pumping system source tank is under some gauge pressure or vacuum open or open toatmosphericthan pressure head is calculated in metres of water column (MWC) of Feet of water column of liquid.

The amount of NPSH the pump requires to avoid cavitation is called Net Positive Suction Head Required (NPSHr). This value of the pump is determined based on actual pump test by the vendor.

The purpose of this article is to present some guidelines and simplified techniques to size pumps and piping typically used in mud systems. If unusual circumstances exist such as unusually long or complicated pipe runs or if very heavy or viscous drilling muds are used, a qualified engineer should analyze the system in detail and calculate an exact solution.

To write about pumps, one must use words that are known and well understood. For example, the label on the lefthand side of any centrifugal pump curve is Total Head Feet. What does this mean?

Total Head remains constant for a particular pump operated at a constant speed regardless of the fluid being pumped. However, a pump’s pressure will increase as the fluid density (mud weight) increases according to the following relationship:

Note that the pump pressure almost doubled. It follows that the required pump horsepower has increased by the same percentage. If the pump required 50 HP for water service, it will require the following horsepower for 16 lb/gal mud:

To summarize, a pump’s Total Head remains constant for any fluid pumped, only the pump pressure and pump horsepower will change. Therefore, a pump motor must be sized according to the heaviest weight mud to be pumped.

In our example problem, the required desilter pressure head is 75 ft. for any mud weight. However, the pressure would be 30.3 PSIG for water or 43.6 PSIG for 12 lb mud or 58.1 PSIG for 16 lb mud. A good rule of thumb is that the required pressure (PSIG) equals 4 times the mud weight (12 LB/GAL x 4 = 48 PSIG).

Determine the required pressure head and flow rate. If the pump is to supply a device such as a mud mixing hopper or a desilter, consult the manufacturer’s information or sales representative to determine the optimum flow rate and pressure head required at the device. (On devices like desilters the pressure head losses downstream of the device are considered negligible and are usually disregarded.)

Select the basic pump to pump the desired flow rate. Its best to refer to a manufacturer’s pump curve for your particular pump. (See example – Figure 3).

The pump’s impeller may be machined to a smaller diameter to reduce its pressure for a given application. Refer to the manufacturer’s pump curves or manufacturer’s representative to determine the proper impeller diameter. Excessive pressure and flow should be avoided for the following reasons:

The pump must produce more than 75 FT-HD at the pump if 75 FT-HD is to be available at the desilter inlet and the pump’s capacity must be at least 800 GPM. Therefore, we should consider using one of the following pumps from the above list: 4″ x 5″ Pump 1750 RPM – 1000 GPM at 160 FT-HD; or 5″ x 6″ Pump 1750 RPM – 1200 GPM at 160 FT-HD.

The pump suction and discharge piping is generally the same diameter as the pump flange diameters. The resulting fluid velocities will then be within the recommended ranges of 4 to 10 FT/SEC for suction lines and 4 to 12 FT/

SEC for discharge lines. Circumstances may dictate that other pipe diameters be used, but remember to try to stay within the above velocity guidelines. Smaller pump discharge piping will create larger pressure drops in the piping

and the pump may not be able to pump the required amount of fluid. (For example, don’t use a 4″ discharge pipe on a 6″ x 8″ pump and expect the pump’s full fluid flow.)

6″ pipe may be used for the suction pipe since it is relatively short and straight and the pump suction is always flooded. 6″ pipe is fully acceptable for the discharge pipe and is a good choice since the desired header is probably 6″ pipe.

8″ pipe may be used for the suction pipe (V = 5.13 FT/SEC) since V is still greater than 4 FT/SEC. 8″ pipe would be preferred if the suction is long or the suction pit fluid level is low with respect to the pump.

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.

Every foot of water creates an additional 0.434 PSI of pressure, so we"ll find the elevation head by converting the change in elevation in feet to the suction pressure created by the water.

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.

Pipes have physical limits to how much fluid they can move at a particular pressure. If pipes aren’t sized properly, you’ll lose efficiency because your motor will have to work harder. It’s like air conditioning - if your ductwork isn’t sized appropriately for your home, you’ll end up paying more on your energy bill.

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.

When it comes to pumping terminology, one crucial term to know is GPM — a measurement that will help you determine if you’re choosing the right pump. So what is GPM, and how do you calculate it?

GPM stands for gallons per minute and is a measurement of how many gallons a pump can move per minute. It is also referred to as flow rate. GPM is variable based on another measurement known as the Head, which refers to the height the water must reach to get pumped through the system. It is also referred to as flow rate. GPM is variable based on another measurement known as the Head, which refers to the height the water must reach to get pumped through the system.

Pumps are typically measured by their GPM at a certain Head measurement. For example, a pump specification may read 150 GPM at 50 Feet of Head, which means the pump will work at 150 gallons per minute when pumping water at a height of 50 feet.

The GPM formula is 60 divided by the number of seconds it takes to fill a one gallon container. So if you took 10 seconds to fill a gallon container, your GPM measurement would be 6 GPM (60/10 seconds = 6 GPM). To most accurately calculate GPM, you use the pressure tank method and formula. For this calculation, you need to know the specifications of your pressure tank, including how many gallons it holds, the gallon drawdown and the PSI. The manufacturer specifies the gallon drawdown. Once you have that information, as well as a stopwatch to keep time, follow these steps:

For example, if it took four minutes for the pressure switch to turn off, and your gallon drawdown was 20 gallons, this would mean a GPM rate of five.

If you don’t have a pressure tank, you can also use a bucket or any other container, time how long it takes to fill up and then divide that by the volume the container holds.

GPM identifies the unique capabilities of a pump so you can select the right one for your specific needs. If you need a pump for a larger public area such as a golf course, marina or lake, you will need a pump with a much higher GPM than one used for your home’s well. Plus, choosing the correct pump is essential for reducing your costs and increasing your pump’s lifespan.

At GeoForm International, we are a leading manufacturer of high-quality submersible pumps, dredges, digester packages and aerators, all of which are made in the U.S. With our pump expertise, we know just how essential GPM is in the pumping and dredging industry from how much equipment costs to how long jobs will take.

Mud pump liner selection in today"s drilling operations seldom (at best) considers electrical implications. Perhaps, with more available useful information about the relationships between mud pump liner size and operational effects on the electrical system, certain potential problems can be avoided. The intent of this paper is to develop those relationships and show how they affect an electrical system on example SCR rigs.Introduction

There, seems to be little consideration for the relationships between liner size and demand on a rig"s engine/generator set(s). Yet, consideration for this relationship can prove to be very helpful to drillers and operators in efficiency of a rig"s electrical system. In order to develop the relationships and help drillers and operators understand the importance of each, relationships between liner size, pump speed, pump pressure, and electrical power will be developed. Only basic physical laws will be used to develop the relationships; and, once developed, the relationships are readily applied to realistic examples utilizing a mud pump manufacturer"s pump data. Finally, conclusions will be drawn from the examples.DEVELOPMENT OF RELATIONSHIPS BASIC RELATIONSHIPS

where HHP= Hydraulic horsepower, GPM = Mud pump volumetric flow rate in gallons per minute, and PST Mud pump output pressure in pounds peer square inch.

Hydraulic horsepower is reflected to the mud pump motor via a multiplier for mechanical efficiency. it follows that motor horsepower is then represented by

The discharge pressure of a pump is commonly referred to as head or pressure, but head and pressure are not the same across all liquids. It often causes confusion leading to a miscalculation of the correct pump specification to use.

What is pressure?Pressure is fluid dependent and relative to fluid density. A calculation of the pressure will change according to the fluid’s specific gravity.

Manometers on pumps indicate suction and delivery pressures, and the corresponding differential pressure, which does not take into account the specific gravity of the fluid being pumped meaning any readings provided need to be checked against the fluids specific gravity to calculate the differential pressure being produced.

The absorbed power is the power consumed by the pump during operation at the required duty point. Although a pump may be fitted with a larger motor than is required for the duty, the power absorbed is usually indicated on the pump curve with an intersecting line of where the pump is expected to perform.

The density and pressure directly affect the power absorbed by the motor during operation. The amount of power absorbed by a motor during operation is multiplied by the SG to calculate the power absorbed.

As a pump operates according to the system it is installed in, care will need to be taken to ensure the absorbed power at the highest point of the curve is larger than the density multiplied by the SG.

In the above example, the highest point is 7.5Kw so for water (7.5Kw x 1) max absorbed power will be 7.5Kw, and for a fluid, with a density of 1.3, the maximum absorbed power would be 7.5Kw x 1.3sg = 9.75Kw. A motor greater than 10Kw would be advisable. Read more about why a pump may be fitted with a larger motor.

Head loss refers to the pressure loss (friction) imparted on a fluid as it travels through components in a system. All parts from straight pieces of pipe, to bends, valves, heat exchangers and couplings create different levels of friction loss.

Different angled bends such as 45° bend compared to a 90° will create different levels of loss in a system. Each item has a pressure loss value known as a K factor detailing the losses caused by each part.

Geodetic HeadThis is the physical difference in height between the liquid level and the highest discharge point. This figure can often change when pumping from tanks or pits where the level can fluctuate. There is often a misconception that the Geodetic head is measured from the depth of the suction pipe, but this is not something considered when calculating this but is instead the difference in fluid levels.Pump DeratingAs pump curves are based on water, a correction factor must be applied to a curve to correct it against the liquid being pumped. This is applied when a fluid is more viscous than water, viscosity fluctuates with temperature or if solids are present.

There are several factors which must be considered when correcting pump curves from particle size, temperature, to the properties of a liquid, so it is always better to discuss curve correction with one of our engineers.

Whether onshore or offshore, well drilling sites rely on a multitude of systems to successfully perform the drilling operation. The mud pump is a key component tasked with circulating drilling fluid under high pressure downhole. The mud pump can be divided into two key sections: the power end or crosshead and the fluid end. Proper alignment of the pump’s crosshead to the fluid end liner is necessary to maximizing piston and liner life. Misalignment contributes to

accelerated wear on both the piston and the liner, and replacing these components requires downtime of the pump. Traditional methods of inspecting alignment range from using uncalibrated wooden rods, Faro Arms and micrometers to check the vertical and horizontal alignment of the piston rod OD to the piston liner ID. These are time consuming and cumbersome techniques that are ultimately not well suited to troubleshoot and solve alignment issues.

A “Mud Pump Laser Alignment Kit” enables you to measure where the piston will run through the liner at various positions along the pump’s stroke. It will also project a laser centerline from the fluid end back towards the rear power end of the pump that can be used to determine how much shimming is required to correct any alignment issues. The kit can include either a 2-Axis receiver or a 4-Axis which accepts the laser beam and documents where it falls on the active surface of the receiver. The 4-Axis receiver can decrease alignment time by as much as 50% as it will measure angularity as well as X and Y while the 2-Axis does not and will need multiple measurement locations to get the same information. In addition, the alignment system is a non-intrusive service requiring the removal of only the piston rod which allows for much quicker service and less down time on the pump. As the mud pumps in question are located globally both on and offshore, having a small, portable system is another great advantage. Our recommendation would be Pinpoint laser System’s “Mud Pump Alignment Kit”. They are being used by many of the leading repair service companies and have been their main alignment tool for over 15 years. Manufacturers are also utilizing these for new pump set-up.

A measure of the energy per unit of time that is being expended across the bit nozzles. It is commonly calculated with the equation HHP=P*Q/1714, where P stands for pressure in pounds per square in., Q stands for flow rate in gallons per minute, and 1714 is a conversion factor necessary to yield HHP in terms of horsepower. Bit manufacturers often recommend that fluid hydraulics energy across the bit nozzles be in a particular HHP range, for example 2.0 to 7.0 HHP, to ensure adequate bit tooth and bottom-of-hole cleaning (the minimum HHP) and to avoid premature erosion of the bit itself (the maximum HHP).