6 triplex mud pump output calculator for sale

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.

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.

Bourgoyne, A.J.T., Chenevert , M.E. & Millheim, K.K., 1986. SPE Textbook Series, Volume 2: Applied Drilling Engineering, Society of Petroleum Engineers.

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.

Technical Notes: This pages uses an iFrame on Internet Explorer and all other browsers use an embed tag. Mobile devices with a screen width smaller than 600px will download the PDF directly instead of using an iFrame or Embed tag to preview.

This approach works well but relying on a printed reference is not without the risk since the wrong value can still be selected from the fine print of a reference table, or the reference document can be damaged or lost (e.g., dropped in the mud pit) altogether.

The intermediate casing can be sealed using the pressure grouting technique (Figure 3) to pump cement slurry down through the drill pipe and out to the annulus through a float shoe (a drillable check valve connected to the base of the casing). The inside of the intermediate casing is kept full of water during the cement placement to equilibrate hydraulic pressures inside and outside the casing. After the intermediate casing is sealed with the pressure grouted cement, the float shoe can be drilled out and the borehole advanced for installation of the screen and filter pack in the lower part of the well.

This is counterintuitive because it would not seem possible for a heavy steel casing filled with water (weighing tens of thousands of pounds) to be capable of floating. Nonetheless, because the water-filled casing is immersed in a much denser fluid (neat cement slurry generally weighs about 15.6 lb/gallon), it may float, just as with other heavy objects will float under the right conditions (e.g., icebergs and battleships).

For the example shown in Figure 4, we would imagine a 15.6 lb/gallon cylinder of neat cement that is within a borehole filled with 15.6 lb/gallon cement. Thus, the imaginary cylinder of cement is surrounded by the same material, so it will be in complete equilibrium and will neither float nor sink.

This imaginary situation enables us to characterize the two forces at play in this scenario: gravity (the downward force) and buoyancy (the upward force). The imaginary cylinder of 15.6 lb/gallon cement shown in Figure 4 is in equilibrium, so we know that the downward and upward forces are equal. We can calculate the downward force like this:

If we assume that we’ve got a 400-foot-long intermediate casing with a 16-inch OD and a 0.3125-inch wall thickness, the volume of the imaginary cement cylinder will be:

Therefore, 65,177 pounds is the buoyancy force since it is equal to the weight of the imaginary cement cylinder as calculated above. If the weight of the real-world steel water-filled casing is less than 65,177 pounds, then that casing will float.

If you apply the weight calculations for a 400-foot-long steel casing with a 16-inch diameter and a 5/16-inch wall thickness, which is filled with water, you’ll see that the downward force in this example is only 52,982 pounds. Thus, the casing in this example will float. The lesson from this counterintuitive scenario is that a casing can actually float. (I’ve seen it happen, and trust me, you don’t want to).

There are several calculations that are commonly applied by drilling fluid engineers (mud engineers) to determine the time period required for the fluid to move from one location in the borehole to another. Some of the more common equations are described below.

The uphole velocity calculation provides a determination of the speed at which the drilling mud will flow as it moves up the borehole. For direct air rotary or reverse circulation drilling methods, the uphole velocity is high, so this calculation is generally applicable only for the direct mud-rotary drilling method. The formula for uphole velocity is:

Thebottoms-up time calculation enables us to determine the time period for the drilling fluid (and the cuttings it is carrying) to travel from the drill bit up to the land surface. This is illustrated in Figure 6(A).

We can calculate the bottoms-up time by using the uphole velocity formula with the borehole depth and drilling mud flow rate plugged in, but that flow rate is being generated by the mud pump, and positive displacement mud pumps (duplex or triplex) are almost never equipped with a flow meter. To determine the flow coming from the mud pump, we can use the formulas:

Remember the strokes are counted in both the forward and backward directions on a duplex pump, but only in the forward direction on a triplex pump. Drillers often have reference charts that provide oilfield barrels per stroke (bbl/stroke), which can be converted to gpm by timing the strokes per minute and converting barrels to gallons (1 barrel = 42 gallons).

The round-trip time enables us to see the result of drilling fluid additives, as indicated by the return flow of fluids at the land surface, as is illustrated in Figure 6(B). The round-trip time calculation is the same as bottoms-up time, but with the travel time of fluid to displace the drill pipe added in.

A specified volume of drilling fluids (called a pill) can be circulated to a particular depth interval within the borehole (called spotting), so that the additives in the pill of drilling mud can address the borehole problem at a particular depth of the borehole. This is shown in Figure 6(C).

The calculation for time required to spot a pill of drillingfluid involves determining the pumping time (at the calculated flow rate) required to displace the fluid so that the drilling mud additives are located adjacent to the problematic interval. This approach is used by mud engineers to address problems such as lost circulation or stuck drill pipe.