When to use
Use these cards for the everyday arithmetic of a hydraulic job: checking whether a line is within the usual velocity, picking a first pipe size for a flow, converting a datasheet in psi or gpm to bar and m³/h, estimating the shaft power and the motor of a pump from its operating point, predicting what a VFD speed change or an impeller trim does to flow, head and power, estimating how long a tank takes to fill or drain, comparing a valve quoted in Cv with one quoted in Kv, and getting the order of magnitude of a water hammer surge. The calculator is open on this page — no sign-up, no project, no report. It does not give a verdict: to size a pump against a system curve, a control valve or a full water hammer analysis, use the dedicated calculators.
What the quick hydraulic calculations page does
The Quick hydraulic calculations page gathers eight one-line hydraulic calculations into a single screen. The calculator is live on this page, free and without sign-up: type a value and every result updates as you type. A shared density field (t/m³) feeds the cards that depend on the fluid — liquid column, mass flow, pump power and water hammer. For water, leave it at 1.
LOGOS runs these cards entirely in the browser, because they are direct formulas with no iteration and no equipment data. They deliberately give no verdict and no calculation report: they answer “how much is it?”, not “does the design pass?”. For a design check, LOGOS has dedicated calculators — for example pump sizing (system curve, operating point, NPSH) and water hammer (Korteweg wave speed, critical time, slow closure, pump trip).
Card 1 — Pipe velocity and the inverse: diameter for a target velocity
The velocity card computes v = Q/A = 4·Q/(π·D²) from the flow in m³/h and the inside diameter in mm. The inside diameter can be typed or taken from the pipe table, choosing the material (carbon steel, stainless steel or HDPE), the nominal size (DN) and the schedule or SDR/PN. The table is the same one LOGOS uses in its hydraulic calculators.
The inverse mode takes a target velocity and returns the inside diameter that gives exactly that velocity, DI = 1000·sqrt(4·(Q/3600)/(π·v)), plus the smallest pipe of the same material and schedule whose inside diameter is at least the required one. At 60 m³/h and 2.5 m/s the required ID is 92.13 mm and the suggested pipe is 4” STD/40 (ID 102.26 mm, v = 2.03 m/s).
The card shows a practical reference — up to about 1.5 m/s on suction and 1 to 2.5 m/s on discharge. It is a usual starting range, not a standard requirement: NPSH, the fluid, erosion and the project criterion decide.
Card 2 — Pressure conversion (bar, m of liquid, kPa, MPa, psi, kgf/cm²)
The pressure card converts a gauge or differential value among bar, m of liquid column, kPa, MPa, psi and kgf/cm². It uses exact unit definitions (1 psi = 6894.757 Pa, 1 kgf/cm² = 98.0665 kPa) and does not add atmospheric pressure. Only the liquid column depends on the density: 1 m of liquid = ρ·g, with g = 9.80665 m/s². That is why 10 bar is 101.97 m of water but 119.97 m of a liquid with ρ = 0.85 t/m³.
Card 3 — Flow conversion (m³/h, L/s, L/min, m³/s, gpm, t/h, kg/h)
The flow card converts volumetric and mass flows. The US gallon is 3.785411784 L, so 60 m³/h = 16.67 L/s = 1000 L/min = 264.17 US gpm. Mass flows (t/h, kg/h) use the density entered: for water at 1 t/m³, 60 m³/h is 60 t/h.
Card 4 — Pump power: hydraulic, shaft and minimum motor
The pumping power card takes the operating flow and head and returns three powers:
- Hydraulic power P_h = ρ·g·Q·H — the energy actually delivered to the fluid.
- Shaft power P_shaft = P_h/η — what the pump takes, with η from the manufacturer’s sheet at the operating point.
- Minimum motor power P_motor = P_shaft·(1 + margin) — with the margin from the project criterion or the standard adopted.
The efficiency and the margin have no built-in default: LOGOS does not guess equipment data. Without η, only the hydraulic power appears. For 60 m³/h at 45 m of water, P_h = 7.35 kW (10.0 cv); with η = 72 %, the shaft takes 10.22 kW; with a 15 % margin, the motor must be at least 11.75 kW, and the commercial motor is the first catalogue size above that.
Use the operating point — the intersection of the pump curve with the system curve — and not the desired flow. If the pump runs to the right of the design point, it delivers more flow and takes more power.
Card 5 — Affinity laws: speed change or impeller trim
The affinity card applies Q₂ = Q₁·r, H₂ = H₁·r², P₂ = P₁·r³, with r = n₂/n₁ for a speed change (VFD) or r = D₂/D₁ for an impeller trim. For speed the relations hold at homologous points of the same pump; for the impeller diameter they are an approximation, good for small trims. Reducing a pump from 1750 to 1450 rpm (r = 0.8286) takes 60 m³/h, 45 m and 10.22 kW to 49.71 m³/h, 30.89 m and 5.81 kW — the power drops with the cube, which is why speed control saves energy.
The scaled point lies on the new pump curve; the real operating point still depends on the system curve, especially when the static head is a large part of the total head.
Card 6 — Tank filling or emptying time
The tank card divides the volume by the net flow: t = V / |Q_in − Q_out|. The volume can be typed or computed for a vertical cylinder between two levels, V = (π·D²/4)·h. If the inlet exceeds the outlet, the tank fills; if the outlet is larger, it empties; if they are equal, the card says the level does not change. A cylinder 4 m in diameter and 5 m between levels (62.83 m³) fills in 1.05 h at 60 m³/h, and empties in 2.51 h with 20 m³/h in and 45 m³/h out.
Card 7 — Cv ↔ Kv
The flow coefficient card converts the US Cv (gpm of water at 1 psi) and the metric Kv (m³/h at 1 bar). LOGOS derives the factor from the unit definitions rather than from a rounded table value: Kv = 0.865·Cv and Cv = 1.156·Kv (exactly 0.86498 and 1.15610). Cv = 100 is Kv = 86.50.
Card 8 — Joukowsky surge
The water hammer card gives the maximum surge of an instantaneous closure, ΔH = a·Δv/g, and the same value as pressure, Δp = ρ·a·Δv. The wave speed a is an input because it depends on the pipe material, the wall thickness and the fluid. With a = 1200 m/s and Δv = 2 m/s, ΔH = 244.7 m and Δp = 24.0 bar. This is the upper bound for a closure within the critical time 2L/a; for wave speed, slow closures and pump trip, use the water hammer calculator.
Limits: what the quick calculations do not do
- No head loss and no system curve. The velocity card does not compute friction; head loss, the system curve and the operating point are in the pump sizing calculator.
- No verdict and no report. The cards return numbers; they do not check limits, issue warnings or produce a Word calculation report. That is the job of the dedicated calculators in the logged-in area.
- No equipment defaults. Efficiency, motor margin and wave speed must be supplied by the user.
- Simplified models. The affinity law for impeller trims is an approximation; Joukowsky is the instantaneous-closure bound.
- Engineering responsibility stays with the engineer. A result from LOGOS supports the work; it does not replace the review and the signature of the professional responsible for the design.
Formulas and fundamentals
v = (Q/3600) / (π·(DI/1000)²/4) Mean velocity [m/s] from the flow Q [m³/h] and the inside diameter DI [mm]. In the table mode the DI comes from the pipe table (carbon steel by schedule, stainless steel by schedule S, HDPE by SDR/PN); in the manual mode you type the DI.
DI = 1000·sqrt( 4·(Q/3600) / (π·v_target) ) Inside diameter [mm] that gives exactly the target velocity v_target [m/s]. The calculator then lists the smallest pipe of the same material and schedule/SDR whose DI is ≥ the required one, i.e. whose velocity is ≤ the target.
1 bar = 100 kPa = 0.1 MPa; 1 psi = 6894.757 Pa; 1 kgf/cm² = 98.0665 kPa; 1 m of liquid = ρ·g = ρ[t/m³]·1000·9.80665 Pa Converts gauge or differential pressure among bar, m of liquid column (mca), kPa, MPa, psi and kgf/cm². Atmospheric pressure is not added. Only the liquid-column unit depends on the density ρ.
1 m³/h = 1/3600 m³/s; 1 US gpm = 3.785411784 L/min; t/h = m³/h × ρ[t/m³] Converts among m³/h, L/s, L/min, m³/s, US gpm, t/h and kg/h. Mass flows use the density entered.
P_h = ρ·g·Q·H; P_shaft = P_h / η; P_motor,min = P_shaft·(1 + margin) P_h [kW] with ρ in kg/m³, g = 9.80665 m/s², Q in m³/s and H in m (÷1000). η is the pump efficiency at the operating point and the margin is the motor criterion of the project — both come from the user; without them only P_h is shown. The result is also given in cv (1 cv = 0.73549875 kW).
r = n₂/n₁ (or D₂/D₁); Q₂ = Q₁·r; H₂ = H₁·r²; P₂ = P₁·r³ Effect of a speed change (exact for the same impeller, at homologous points) or of an impeller trim (approximation, good for small trims) on flow, head and power.
t = V / |Q_in − Q_out|; V_cylinder = (π·D²/4)·h Time [h] with V in m³ and flows in m³/h. If Q_in > Q_out the tank fills; if Q_out > Q_in it empties; if they are equal the level does not change. The volume can be typed or computed for a vertical cylinder between two levels.
Kv = 0.865·Cv; Cv = 1.156·Kv Cv is the US flow coefficient (US gpm of water at 1 psi drop); Kv the metric one (m³/h at 1 bar). The factor comes from the definitions themselves: K = (0.227125 m³/h per gpm)·sqrt(100000/6894.757) = 0.86498, and 1/K = 1.15610.
ΔH = a·Δv / g; Δp = ρ·g·ΔH = ρ·a·Δv Maximum surge [m] for a closure within the critical time 2L/a. a = pressure wave speed [m/s] (entered by the user), Δv = velocity change [m/s]. Δp is shown in bar.
Standards & methods
- Exact unit definitions: g = 9.80665 m/s² (standard gravity), 1 psi = 6894.757293168 Pa, 1 kgf/cm² = 98.0665 kPa, 1 US gallon = 3.785411784 L
- IEC 60534-1 / ANSI/ISA-75.01.01 — definitions of the flow coefficients Kv and Cv
- Pump affinity (similarity) laws — Q ∝ n, H ∝ n², P ∝ n³
- Joukowsky (1900) — pressure surge of an instantaneous velocity change
- Pipe inside diameters: ASME B36.10M (carbon steel), ASME B36.19M (stainless steel), SDR/PN series (ISO 4427) for HDPE
Typical reference values
| Quantity | Typical range | Note |
|---|---|---|
| Pipe velocity — suction (usual practice) | up to ~1.5 m/s | Practical reference shown in the calculator, not a standard requirement. The project criterion (NPSH margin, fluid, standard adopted) governs. |
| Pipe velocity — discharge (usual practice) | 1 to 2.5 m/s | Same reference. Higher velocities raise head loss and the Joukowsky surge (ΔH ∝ Δv). |
| Kv / Cv factor | Kv = 0.865·Cv (Cv = 1.156·Kv) | Derived from the unit definitions; not an empirical value. |
| 1 bar of water column (ρ = 1 t/m³) | 10.197 m | With ρ = 0.85 t/m³ the same 1 bar is 11.997 m of liquid — the column depends on density. |
| Pressure wave speed a (Joukowsky card) | user input | Depends on pipe material, wall thickness and fluid. The water hammer calculator computes it with Korteweg's equation. |
| Pump efficiency η and motor margin | user input | Take η from the manufacturer's sheet at the operating point and the margin from the project criterion; the calculator has no built-in default. |
Worked example
One duty point, six cards: 60 m³/h of water, H = 45 m, carbon steel pipe
Inputs
- Density ρ (all cards)
- 1.0 t/m³
- Velocity: flow / pipe
- 60 / 4" Sch STD/40 (ID 102.26 mm) m³/h / -
- Velocity, inverse: target velocity
- 2.5 m/s
- Pump power: Q / H / η / motor margin
- 60 / 45 / 72 / 15 m³/h / m / % / %
- Affinity: speed 1750 → 1450 rpm, with Q₁ / H₁ / P₁
- 60 / 45 / 10.22 m³/h / m / kW
- Tank: vertical cylinder D × h, inlet / outlet flow
- 4 × 5 ; 60 / 0 m ; m³/h
- Cv ↔ Kv: Cv
- 100 -
- Joukowsky: wave speed a / Δv
- 1200 / 2 m/s
Results
- Velocity in the 4" Sch 40
- 2.03 m/s
- Required ID for 2.5 m/s → smallest pipe
- 92.13 mm → 4" STD/40 -
- Hydraulic / shaft / minimum motor power
- 7.35 / 10.22 / 11.75 kW
- Same powers in cv
- 10.0 / 13.9 / 16.0 cv
- Affinity (r = 0.8286): Q₂ / H₂ / P₂
- 49.71 / 30.89 / 5.81 m³/h / m / kW
- Tank volume / filling time
- 62.83 / 1.05 (≈ 1 h 03 min) m³ / h
- Kv
- 86.50 -
- Joukowsky surge ΔH / Δp
- 244.7 / 24.0 m / bar
All numbers come from the calculator's own engine. At 60 m³/h, the 4" Sch 40 line runs at 2.03 m/s — inside the usual 1 to 2.5 m/s discharge range shown by the calculator, and the inverse card confirms that 4" is the smallest Sch 40 pipe that keeps the velocity at or below 2.5 m/s (it needs 92.13 mm; the 3" Sch 40, with 77.92 mm, would run at 3.50 m/s). At 45 m of head the fluid receives 7.35 kW; with 72 % pump efficiency the shaft takes 10.22 kW and, with a 15 % margin, the motor must be at least 11.75 kW — the next catalogue size above that. Slowing the pump from 1750 to 1450 rpm on a VFD (r = 0.8286) moves the homologous point to 49.71 m³/h and 30.89 m, and the power falls with the cube, to 5.81 kW. Note that the new point is a homologous point on the pump curve, not the new operating point: where the pump actually runs depends on the system curve. A 4 m × 5 m tank (62.83 m³) fills in 1.05 h at 60 m³/h. A valve rated Cv = 100 is Kv = 86.50. And if that same line were stopped instantly, with a = 1200 m/s and Δv = 2 m/s, the surge would reach 244.7 m (24.0 bar) over the operating pressure — the cue to open the full water hammer calculator.
Common mistakes
- Using the nominal size instead of the inside diameter. A 4" pipe has an ID of 102.26 mm in Sch 40 and 97.18 mm in Sch 80 (carbon steel table): at 60 m³/h that is 2.03 m/s versus 2.25 m/s.
- Forgetting to divide m³/h by 3600 in v = Q/A. The mistake gives a velocity 3600 times too large and is the most common slip in hand calculations.
- Sizing the motor on the hydraulic power. P_h = ρ·g·Q·H is the power delivered to the fluid; the shaft takes P_h/η, and the motor must cover the shaft power plus the project margin.
- Using the desired flow instead of the operating point in the power card. The pump runs where its curve meets the system curve; if that flow is higher, the power is higher too.
- Applying the impeller-trim affinity law to large trims. For speed it is exact at homologous points; for diameter it is an approximation that loses accuracy as the trim grows.
- Converting m of liquid to bar with the water density when the fluid is not water. 10 bar is 101.97 m of water but 119.97 m of a liquid with ρ = 0.85 t/m³.
- Treating the Joukowsky value as the surge of any closure. ΔH = a·Δv/g is the upper bound for a closure within 2L/a; slower closures produce less, and a pump trip needs the full water hammer analysis.
Frequently asked questions
How do I calculate the velocity of water in a pipe?
Divide the flow by the internal cross-section: v = Q/A = 4·Q/(π·D²), with Q in m³/s and D as the inside diameter in meters. For Q in m³/h, divide by 3600 first. Example: 60 m³/h = 0.01667 m³/s; in a pipe with an inside diameter of 102.26 mm (4" Sch 40), A = 0.008213 m² and v ≈ 2.03 m/s. The calculator on this page does it as you type and can take the inside diameter straight from the pipe table.
What pipe diameter do I need for a given flow and velocity?
Invert the velocity equation: D = sqrt(4·Q/(π·v)). For 60 m³/h at 2.5 m/s the required inside diameter is 92.13 mm. The inverse card then picks the smallest pipe of the chosen material and schedule whose inside diameter is at least that — here, the 4" Sch 40 (102.26 mm), which runs at 2.03 m/s.
How do I convert Cv to Kv?
Kv = 0.865 × Cv, and Cv = 1.156 × Kv. The factor is not empirical: Cv is defined in US gpm of water at 1 psi of pressure drop and Kv in m³/h at 1 bar, so K = 0.227125 × sqrt(100000/6894.757) ≈ 0.86498. A valve with Cv = 100 has Kv ≈ 86.5; one with Kv = 100 has Cv ≈ 115.6.
What power does a pump need for 60 m³/h at 45 m of head?
The hydraulic power is P = ρ·g·Q·H = 1000 × 9.80665 × (60/3600) × 45 ≈ 7.35 kW (10.0 cv). The shaft power is that divided by the pump efficiency at the operating point: with η = 72 %, 10.22 kW. With a 15 % motor margin the motor must be at least 11.75 kW, and you choose the next catalogue size up. Efficiency and margin must come from the pump sheet and the project criterion.
What happens to flow, head and power if I reduce the pump speed?
By the affinity laws, with r = n₂/n₁: flow scales with r, head with r² and power with r³. Going from 1750 to 1450 rpm (r = 0.8286) turns 60 m³/h, 45 m and 10.22 kW into 49.71 m³/h, 30.89 m and 5.81 kW. These are homologous points on the pump curve; the real operating point at the new speed is where the scaled pump curve meets the system curve.
How long does it take to fill a tank?
Divide the volume by the net flow: t = V / (Q_in − Q_out). A vertical cylinder 4 m in diameter with 5 m between levels holds π × 4²/4 × 5 = 62.83 m³; at 60 m³/h with no outlet it fills in 1.05 h (about 1 h 03 min). If the outlet is larger than the inlet, the same formula gives the emptying time: at 20 m³/h in and 45 m³/h out, 2.51 h.
Glossary
- Inside diameter (ID, DI)
- The actual bore of the pipe, which depends on the nominal size and the wall thickness (schedule or SDR). It is the diameter that goes into v = Q/A.
- Schedule (Sch) / SDR
- Wall-thickness series. Steel pipes use schedules (40, 80, STD, XS...); HDPE uses SDR, the ratio of outside diameter to wall thickness, linked to the pressure class PN.
- Hydraulic power
- Power delivered to the fluid, P_h = ρ·g·Q·H. It does not include pump losses.
- Shaft power (brake power)
- Power the pump takes at the shaft, P_h/η, where η is the pump efficiency at the operating point.
- Affinity laws
- Similarity relations for centrifugal pumps: flow proportional to speed, head to its square and power to its cube.
- Flow coefficient Kv / Cv
- Flow of water that passes through a valve at a reference pressure drop: Kv in m³/h at 1 bar, Cv in US gpm at 1 psi.
- Wave speed (a)
- Speed at which a pressure wave travels in the fluid + pipe system. It is the input of the Joukowsky equation.
- Meters of liquid column (mca)
- Pressure expressed as the height of a liquid column, h = p/(ρ·g). It depends on the liquid's density.
Understand the concept
Free interactive lessons from LOGOS Learning: change the inputs and watch the physics behind this calculator.
- LOGOS Learning · LessonPump operating point: why flow drops as static head risesSingle pump: static head, head loss, total head and the operating point.
- LOGOS Learning · LessonTop-entry discharge: tank levels and predictable pump flowTank levels, free discharge and predictable flow.
- LOGOS Learning · LessonWater hammer: wave speed, Joukowsky and the critical closing timeWave speed, Joukowsky and the critical closing time.