The population MTBR relationship

The product M × T is the population’s cumulative exposure when every included machine is observed for the full reporting interval. Dividing that exposure by the repair-event count gives MTBR = M × T / R. The answer uses the same time unit selected for T.

Solve any fourth variable

• MTBR — enter machine count, repairs, and reporting time.
• Machine count (M) — enter MTBR, repairs, and reporting time.
• Repair count (R) — enter MTBR, machine count, and reporting time.
• Reporting time (T) — enter MTBR, machine count, and repairs.

Worked example

A population of 100 machines is tracked for 12 months and records 20 qualifying repairs. Fleet exposure is 1,200 machine-months, so MTBR = 100 × 12 / 20 = 60 months per repair. This is equivalent to one repair per machine every five reporting intervals on average.

Define the population and repair event first

MTBR is only comparable when the population, observation window, and definition of a repair remain consistent. Decide whether standby machines, planned interventions, inspections, seal replacements, upgrades, or repeated work orders count before collecting the data. A longer MTBR generally indicates fewer repairs per unit of machine exposure, but it is not automatically the same as MTBF.

Assumptions and limits

• Every machine in M contributes the full reporting interval T; otherwise use summed machine exposure outside this simplified model.
• All machines belong to a meaningfully comparable population and repair events are counted consistently.
• MTBR is a historical fleet average, not a prediction of the next repair date for one machine.
• Small repair counts and short reporting windows can produce volatile results; use consistent rolling periods for trending.
• Calculated machine or repair counts may be fractional for planning scenarios, while observed counts are whole events.

The post Mean Time Between Repairs Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The belt-drive relationship

With no slip, the belt has the same linear speed at both pulleys. Therefore NH × DH = NL × DL, and the speed ratio is R = NH/NL = DL/DH. A larger driven pulley turns more slowly; a smaller driven pulley turns faster.

Solve any fourth variable

• High-speed diameter (DH) — enter DL, NH, and NL.
• Low-speed diameter (DL) — enter DH, NH, and NL.
• High pulley speed (NH) — enter DH, DL, and NL.
• Low pulley speed (NL) — enter DH, DL, and NH.

Worked example

A 4-inch high-speed pulley turning at 1,800 rpm drives a 12-inch low-speed pulley. The diameter ratio is 12/4 = 3, so the low-speed shaft turns at 600 rpm. The ideal belt speed is approximately 1,884.96 ft/min or 9.57557 m/s.

Use pitch or effective diameter

Use the pulley diameter at the belt pitch line—the pitch diameter or effective diameter specified by the pulley manufacturer. Outside diameter is not always the same value, especially for V-belt sheaves, and substituting it can shift the predicted speed ratio. For synchronous belts, pulley tooth-count ratio is normally the preferred exact ratio.

Assumptions and limits

• The drive has two pulleys connected by one belt with no intermediate or idler ratio effect.
• The relation is ideal and assumes no belt slip, creep, or elastic speed loss.
• Diameters are compatible effective or pitch diameters measured in the same units.
• The calculator does not size belt section, pulley grooves, center distance, wrap angle, belt length, tension, service factor, power capacity, or shaft loading.
• Confirm allowable pulley and belt speeds with the component manufacturer.

The post Belt-Drive Pulleys Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The sinusoidal relationship

For sinusoidal displacement x(t) = Xpk sin(2πft), peak velocity is Vpk = 2πfXpk. Because the entered displacement is peak-to-peak, Xpk = Xpp/2, so Vpk = 2πfXpp/2. The inverse forms are Xpp = 2Vpk/(2πf) and f = Vpk/(πXpp). Frequency in hertz is CPM divided by 60.

Solve in any direction

• Peak velocity (Vips) — enter peak-to-peak displacement and frequency.
• Peak-to-peak displacement (Xpp) — enter peak velocity and frequency.
• Frequency (CPM or Hz) — enter peak velocity and peak-to-peak displacement.

Worked example

A filtered sinusoidal vibration of 1 mil peak-to-peak at 3,600 CPM (60 Hz) has a peak displacement amplitude of 0.5 mil and a peak velocity of 0.188496 in/sec, equivalent to 4.78779 mm/s. Solving backward with that velocity and frequency returns 1 mil peak-to-peak.

Amplitude conventions matter

This tool pairs peak velocity with peak-to-peak displacement. Do not enter RMS velocity without converting it to peak for a sinusoid (peak = RMS × √2). Likewise, do not enter zero-to-peak displacement in the peak-to-peak field without doubling it.

Discrete-frequency assumption

The conversion is valid for a single sinusoidal component at a known frequency, such as a filtered spectral line. It does not convert broadband, random, transient, shock, or unfiltered overall vibration. For a spectrum, convert each frequency component separately while keeping consistent amplitude conventions.

Assumptions and limits

• The vibration is sinusoidal and occurs at one discrete, filtered frequency.
• Velocity is peak amplitude and displacement is peak-to-peak excursion.
• CPM equals shaft rpm only for a 1× rotational vibration component.
• The mathematical result does not account for sensor bandwidth, integration noise, filtering, mounting, or calibration uncertainty.

The post Velocity in in/sec Peak to Displacement in Mils Peak-to-Peak Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The sinusoidal relationship

For a sinusoidal velocity waveform, peak acceleration is Apk = ωVpk = 2πfVpk. The inverse forms are Vpk = Apk/(2πf) and f = Apk/(2πVpk). Frequency in hertz is CPM divided by 60. The calculator converts velocity to metres per second and acceleration to metres per second squared before solving.

Solve in any direction

• Peak acceleration (A) — enter peak vibration velocity and frequency.
• Peak velocity (Vips) — enter peak acceleration and frequency.
• Frequency (CPM or Hz) — enter peak acceleration and peak velocity.

Worked example

A filtered sinusoidal vibration of 0.1 in/sec peak at 3,600 CPM (60 Hz) has a peak acceleration of 0.097644 g, equivalent to 0.95756 m/s². Solving backward with that acceleration and frequency returns 0.1 in/sec peak.

Amplitude conventions matter

This tool pairs peak velocity with peak acceleration. Do not mix peak and RMS values. For a pure sinusoid, peak = RMS × √2; apply that conversion to both quantities if the source instrument reports RMS.

Discrete-frequency assumption

The conversion is valid for a single sinusoidal component at a known frequency, such as a filtered spectral line. It does not convert broadband, random, transient, shock, or unfiltered overall vibration. For a spectrum, convert each frequency component separately using consistent amplitude conventions.

Assumptions and limits

• The vibration is sinusoidal and occurs at one discrete, filtered frequency.
• Acceleration and velocity are both peak amplitudes.
• CPM equals shaft rpm only for a 1× rotational vibration component.
• The mathematical result does not account for sensor bandwidth, integration noise, filtering, mounting, or calibration uncertainty.

The post Velocity in in/sec Peak to Acceleration in G’s Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The sinusoidal relationship

For displacement x(t) = Xpk sin(2πft), differentiating twice gives a peak acceleration magnitude of Apk = (2πf)²Xpk. Because the entered displacement is peak-to-peak, Xpk = Xpp/2, so Apk = (2πf)²Xpp/2. Frequency in hertz is CPM divided by 60.

Solve in any direction

• Acceleration (A) — enter peak-to-peak displacement and frequency.
• Displacement (Xpp) — enter peak acceleration and frequency.
• Frequency (CPM or Hz) — enter peak acceleration and peak-to-peak displacement.

Worked example

A filtered sinusoidal vibration of 1 mil peak-to-peak at 3,600 CPM (60 Hz) has a peak displacement amplitude of 0.5 mil and a peak acceleration of 0.18405 g, equivalent to 1.80495 m/s². Solving backward with 0.18405 g and 3,600 CPM returns 1 mil peak-to-peak.

Amplitude conventions matter

This tool pairs peak acceleration with peak-to-peak displacement. Do not enter RMS acceleration without converting it to peak for a sinusoid (peak = RMS × √2). Likewise, do not enter zero-to-peak displacement in the peak-to-peak field without doubling it.

Discrete-frequency assumption

The conversion is valid for a single sinusoidal component at a known frequency, such as a filtered spectral line. It does not convert broadband, random, transient, shock, or unfiltered overall vibration. For a spectrum, convert each frequency component separately while keeping a consistent amplitude convention.

Assumptions and limits

• The vibration is sinusoidal and occurs at one discrete, filtered frequency.
• Acceleration is peak amplitude and displacement is peak-to-peak excursion.
• CPM means cycles per minute. It equals shaft rpm only for a 1× rotational vibration component.
• The mathematical result does not account for sensor bandwidth, integration noise, filtering, mounting, or calibration uncertainty.

The post Acceleration in G’s to Displacement in Mils Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The formulas

Rotational mechanical power is P = Tω, where T is torque and ω is angular velocity. For speed N in revolutions per minute, ω = 2πN/60. Therefore T = P/ω, P = Tω, and N = 60P/(2πT). The calculator converts inputs to watts, newton-metres, and radians per second before solving.

Solve in any direction

• Torque (T) — enter mechanical shaft power and rotational speed.
• Shaft power (HP or P) — enter torque and rotational speed.
• Rotational speed (N) — enter mechanical shaft power and torque.

Worked example

A shaft transmitting 100 hp at 1,750 rpm has an angular velocity of 183.26 rad/s and transmits 300.12 lbf·ft of torque, equivalent to 406.91 N·m. Solving backward with that torque and speed returns 100 hp.

Use mechanical shaft power

HP or P must be the mechanical power at the shaft section where torque is required. Electrical motor input includes motor losses and will overstate shaft power unless efficiency is applied. Likewise, pump hydraulic power is lower than shaft power because the pump has hydraulic and mechanical losses. Use brake horsepower or measured shaft power when available.

Keep torque, power, and speed at the same point

A gearbox changes torque and speed while approximately conserving power minus losses. Do not combine motor-side rpm with pump-side torque, or power measured before a lossy transmission with values after it. Convert all three variables to the same shaft location and operating condition.

Assumptions and limits

• Power, torque, and speed are steady or consistently averaged values from the same operating point.
• The calculation represents transmitted mechanical power and does not add drivetrain efficiency or service factors.
• It does not calculate starting, accelerating, stall, jam, impact, or torsional-vibration peak torque.
• Mechanical horsepower is used in US mode, not metric horsepower (PS).
• The result is operating torque, not a shaft, coupling, gearbox, or motor torque rating.

The post Torque with Horsepower and Speed Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The formulas

For a solid circular shaft under pure torque, the polar moment of inertia is J = πD⁴/32. The torsion relation τ = Tr/J, evaluated at the outside radius r = D/2, gives the maximum surface stress Ss = τmax = 16T/(πD³). Shaft torque follows from power and angular speed: T = P/ω and ω = 2πN/60.

Solve in any direction

• Shaft shear stress (Ss) — enter power, rotational speed, and solid-shaft diameter.
• Shaft power (HP or P) — enter nominal shear stress, speed, and diameter.
• Rotational speed (N) — enter nominal shear stress, power, and diameter.
• Shaft diameter (D) — enter nominal shear stress, power, and speed.

Worked example

A solid 2 in (50.8 mm) shaft transmitting 100 hp at 1,750 rpm carries 406.9 N·m of torque. The nominal maximum surface shear stress is 2,293 psi, or 15.81 MPa. Solving backward with that stress, speed, and diameter returns 100 hp.

Assumptions

• The shaft is solid, circular, straight, and experiencing pure torque.
• There is no bending, axial load, transverse shear, shock, fatigue, or residual stress.
• There are no stress risers such as keyways, shoulders, grooves, splines, holes, or surface defects.
• Power and rotational speed describe the same operating condition and neglect drivetrain losses between the stated power point and shaft section.

Accounting for a stress riser

The calculator reports the nominal smooth-shaft stress. If an applicable torsional stress concentration factor Kt is known, estimate the local elastic peak as τlocal = Kt × Ss. Obtain Kt from a validated geometry-specific reference; do not assume one generic multiplier covers every keyway, shoulder, groove, or spline. Fatigue design may require a fatigue stress concentration factor rather than the theoretical elastic Kt.

Diameter sensitivity

Nominal shear stress varies with 1/D³. At the same transmitted torque, a 10% reduction in diameter raises nominal stress by about 37%. Use the actual minimum load-carrying diameter at the section being checked, not a nominal bearing or coupling size.

Limits

• This solid-shaft formula does not apply to hollow shafts; use their polar moment of inertia instead.
• The result is stress, not allowable capacity. Compare it with an appropriate material limit and design factor for static or fatigue service.
• Transient startup, trip, jam, and torsional-vibration torque can exceed the steady torque inferred from horsepower and rpm.
• Final shaft design should include combined loading, stress concentrations, fatigue, deflection, critical speed, and applicable code or manufacturer requirements.

The post Shaft Shear Stress Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The formulas

One point on the impeller rim travels one circumference, πD, during each revolution. Multiplying by rotational speed and converting minutes to seconds gives V = πDN/60 when D is in metres, N is in rpm, and V is in metres per second. The inverse forms are N = 60V/(πD) and D = 60V/(πN).

Solve in any direction

• Peripheral velocity (V) — enter the impeller outside diameter and rotational speed.
• Rotational speed (N) — enter the required peripheral velocity and impeller diameter.
• Impeller diameter (D) — enter the required peripheral velocity and rotational speed.

Worked example

A 12 in (0.3048 m) impeller rotating at 1,750 rpm has a peripheral velocity of π × 0.3048 × 1750 ÷ 60 = 27.93 m/s, or 91.63 ft/s. Reversing the calculation with that velocity and diameter returns 1,750 rpm.

Which diameter should you use?

Use the diameter of the circular path at the point whose velocity you need. For impeller outlet tip speed, this is normally the current outside diameter after any trim—not the casing diameter, shaft diameter, or the maximum untrimmed impeller size on a family curve.

Interpret the result carefully

Peripheral velocity is a kinematic result, not a permissible-speed rating. KSB notes that typical impeller-outlet circumferential speeds are about 20 to 60 m/s, with exceptional cases up to 140 m/s. The acceptable value for a real pump depends on impeller material, geometry, balance, liquid, cavitation performance, noise, stresses, and the manufacturer’s published limits.

Assumptions and limits

• N is actual impeller rotational speed, not merely motor nameplate synchronous speed.
• D and V refer to the same radius on the rotating impeller.
• The relationship describes solid-body rim motion; it does not calculate the liquid’s absolute or relative velocity.
• Confirm any speed or diameter change against the manufacturer curve and mechanical speed rating.

The post Peripheral Velocity of an Impeller Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The formulas

For a full circular pipe, cross-sectional area is A = πD²/4. Continuity gives Q = A · Vf, so Vf = Q/A and D = √[4Q/(πVf)]. The calculator converts every input to SI internally before solving, then displays the result in the selected units.

Use actual inside diameter

Enter the pipe’s actual internal diameter, not just nominal pipe size. Schedule, material, and wall thickness change the bore. Because velocity varies with 1/D², a modest diameter difference can materially change the answer. If the nominal size and schedule are known, use the Pipe Size & Velocity Calculator to look up the inside diameter.

Solve in any direction

• Fluid velocity (Vf) — enter volumetric flow and actual pipe inside diameter.
• Volumetric flow (Q) — enter average velocity and actual inside diameter.
• Pipe inside diameter (D) — enter volumetric flow and the target average velocity.

Worked example

A full 4 in inside-diameter pipe carrying 100 gpm has an area of 12.566 in². The average velocity is 2.553 ft/s (0.7782 m/s). Reversing the calculation with 2.553 ft/s and a 4 in diameter returns approximately 100 gpm.

Assumptions and limits

• The pipe is full and has a circular, constant inside diameter.
• Flow is steady, and Vf is the bulk average velocity across the section rather than the local centerline velocity.
• Q is actual volumetric flow at the pipe conditions. For gases, do not mix standard flow with actual pipe volume without correcting for pressure and temperature.
• The calculation does not determine pressure drop, friction loss, or whether the resulting velocity is suitable for the service.

The post Velocity to Flow Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.

The formulas

In US units, H (ft) = 2.31 · P (psi) / SG. In metric units, H (m) = 10.2 · P (bar) / SG. Rearranging gives P = H · SG / K and SG = K · P / H, where K is 2.31 ft/psi or 10.2 m/bar.

Why specific gravity matters

A pressure gauge measures force per area, while head expresses energy as the height of a fluid column. At the same pressure, a lighter fluid produces a taller column and a denser fluid produces a shorter one. For example, 100 psi corresponds to 231 ft of water at SG = 1.0, but about 289 ft of a light hydrocarbon at SG = 0.8.

Keep the pressure basis consistent

The relationship works with gauge pressure, absolute pressure, or a pressure difference, but the pressure and head must describe the same basis. Gauge pressure produces gauge head; absolute pressure produces absolute head. In pump calculations, pressure rise normally means discharge pressure minus suction pressure using readings on the same basis.

Solve in any direction

• Head (H) — enter pressure and fluid specific gravity.
• Pressure (P) — enter fluid head and specific gravity.
• Specific gravity (SG) — enter corresponding pressure and head values.

Worked example

A pump raises the pressure of a liquid with SG = 0.85 by 100 psi. The equivalent head is H = 2.31 × 100 ÷ 0.85 = 271.8 ft. Entering 271.8 ft and SG = 0.85 in the reverse calculation returns approximately 100 psi.

Head versus pressure

A centrifugal pump develops essentially the same head at a given operating point regardless of fluid density, but the pressure rise changes with specific gravity. That is why pump curves are normally plotted in head rather than pressure. Use the result with the Total Dynamic Head Calculator to build a complete system requirement.

The post Pressure to Head Calculator appeared first on PumpCalcs — Free Pump Calculators & Hydraulics Reference.