Fluid Mechanics & Hydraulics

Energy Equation (Bernoulli with Machines & Losses)

The energy equation p₁/γ + V₁²/2g + z₁ + hp = p₂/γ + V₂²/2g + z₂ + ht + hL between two sections of one stream, rearranged for a single unknown: downstream pressure, downstream velocity, required pump head, head loss, downstream elevation, or pump power P = γQh/η. Velocities may come from flowrate + diameter via continuity. All pressures are gage. Use for: reservoir-to-nozzle problems, pump sizing, 'find the pressure at section two', velocity out of a tank (Torricelli), and ESPECIALLY tapering-pipe problems that ask for velocity AND pressure together (continuity + Bernoulli combined — give each section its diameter and the known velocity or flowrate; do NOT route those to continuity-flow, which cannot give pressure). Not for: compressible flow or branching networks.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

section1State of the upstream section (section 1)

z (optional)Elevation of the upstream section above the datum (omit to place the datum there, z = 0) [dimension: length]pressure (optional)Gage pressure at the upstream section (omit for a free surface or free jet at atmospheric, p = 0 gage) [dimension: pressure]velocity (optional)Flow velocity at the upstream section (omit to derive it from flowrate + diameter, or for a large reservoir where V ≈ 0) [dimension: velocity]diameter (optional)Pipe inside diameter at the upstream section — used with flowrate to get the velocity from continuity [dimension: length]

section2State of the downstream section (section 2)

z (optional)Elevation of the downstream section above the datum (omit to place the datum there, z = 0) [dimension: length]pressure (optional)Gage pressure at the downstream section (omit for a free surface or free jet at atmospheric, p = 0 gage) [dimension: pressure]velocity (optional)Flow velocity at the downstream section (omit to derive it from flowrate + diameter, or for a large reservoir where V ≈ 0) [dimension: velocity]diameter (optional)Pipe inside diameter at the downstream section — used with flowrate to get the velocity from continuity [dimension: length]

flowrate (optional)Volumetric flowrate through the system (needed for continuity velocities and for pump power) [dimension: flowrate]pump head (optional)Head ADDED by a pump between the sections (omit if none or if solving for it) [dimension: length]turbine head (optional)Head EXTRACTED by a turbine between the sections (omit if none) [dimension: length]head loss (optional)Total head loss between the sections (friction + minor); omit for zero or when solving for it [dimension: length]pump efficiency (optional)Pump efficiency η as a decimal in (0, 1]; used only for pumpPower (omitted = 1, ideal pump)

fluidFluid definition — give at most ONE of specificGravity, unitWeight, or density; omit all three for water (γ = 9.81 kN/m³)

specific gravity (optional)Specific gravity SG relative to water (γw = 9.81 kN/m³): 13.6 mercury, ~0.85 oil, 1 water. Give at most one of specificGravity, unitWeight, density.unit weight (optional)Fluid unit weight γ [dimension: unitweight]density (optional)Fluid mass density ρ (converted internally via γ = ρ·g) [dimension: density]

solve forThe single unknown to solve the energy equation for: downstream pressure, downstream velocity, required pump head, head loss, downstream elevation, or pump power (γQh/η)

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI: downstream pressure up a 5 m rise with 3 m of losses (equal velocities)

Given

section1: z 0 mpressure 200 kPavelocity 2 m/s
section2: z 5 mvelocity 2 m/s
head loss: 3 m
solve for: pressure2

Assumptions

No fluid was specified — fresh water (γ = 9.81 kN/m³) was assumed.
Steady, incompressible, single-stream flow with kinetic-energy correction factor α = 1.
All pressures are GAGE pressures — both sections share the same atmospheric reference.

Solution steps

Write the energy equation between the sections
Energy per unit weight (head) is conserved apart from pump work added, turbine work extracted, and losses; every term is a length.
p₁/γ + V₁²/2g + z₁ + hp = p₂/γ + V₂²/2g + z₂ + ht + hL
Energy head at the upstream section
Sum the pressure head, velocity head, and elevation head.
E₁ = p₁/γ + V₁²/2g + z₁
p₁/γ = 20.4 m; V₁²/2g = 0.204 m; z₁ = 0 m → E₁ = 20.59 m
= 20.59 m
Energy head at the downstream section
Only the known terms are evaluated here (velocity head, elevation head) — the remaining term is the unknown.
E₂ = p₂/γ + V₂²/2g + z₂
velocity head = 0.204 m; elevation z₂ = 5 m → known part of E₂ = 5.204 m
Solve for the downstream pressure
Rearrange the energy equation for the downstream pressure head, then multiply by the unit weight.
p₂/γ = E₁ + hp − ht − hL − V₂²/2g − z₂
p₂/γ = 20.59 m + 0 m − 0 m − 3 m − 0.204 m − 5 m = 12.39 m → p₂ = 121.5 kPa
= 121.5 kPa

Results

Pressure at the downstream section (gage)

121.5kPa

Velocity at the upstream section

2m/s

Velocity at the downstream section

2m/s

Where this answer was checked

source: Energy equation between pipe sections (NCEES FE Reference Handbook) — p₂ = p₁ + γ(z₁ − z₂ − hL) when V₁ = V₂
verified by: hand-recomputed
derivation: Equal velocity heads cancel. p₂ = 200 000 − 9810·(5 + 3) = 200 000 − 78 480 = 121 520 Pa = 121.52 kPa. (p₁/γ = 20.387 m.)

Example 2

SI: pump head and power, reservoir to reservoir 20 m up, hL = 2.5 m, η = 0.75

Given

section1: z 0 m
section2: z 20 m
flowrate: 0.05 m^3/s
head loss: 2.5 m
pump efficiency: 0.75
solve for: pumpPower

Assumptions

No fluid was specified — fresh water (γ = 9.81 kN/m³) was assumed.
Steady, incompressible, single-stream flow with kinetic-energy correction factor α = 1.
All pressures are GAGE pressures — both sections share the same atmospheric reference.
No velocity information at the upstream section — it was treated as a large reservoir / free surface with negligible velocity (V ≈ 0).
No velocity information at the downstream section — it was treated as a large reservoir / free surface with negligible velocity (V ≈ 0).
No pressure given at the upstream section — it was taken as atmospheric (p = 0 gage: free surface or free jet).
No pressure given at the downstream section — it was taken as atmospheric (p = 0 gage: free surface or free jet).

Solution steps

Write the energy equation between the sections
Energy per unit weight (head) is conserved apart from pump work added, turbine work extracted, and losses; every term is a length.
p₁/γ + V₁²/2g + z₁ + hp = p₂/γ + V₂²/2g + z₂ + ht + hL
Energy head at the upstream section
Sum the pressure head, velocity head, and elevation head.
E₁ = p₁/γ + V₁²/2g + z₁
p₁/γ = 0 m; V₁²/2g = 0 m; z₁ = 0 m → E₁ = 0 m
= 0 m
Energy head at the downstream section
Sum the pressure head, velocity head, and elevation head.
E₂ = p₂/γ + V₂²/2g + z₂
pressure head p₂/γ = 0 m; velocity head = 0 m; elevation z₂ = 20 m → known part of E₂ = 20 m
Solve for the required pump head
The pump must supply the downstream energy plus turbine extraction and losses, minus what the upstream section already has.
hp = E₂ + ht + hL − E₁
h_p = 20 m + 0 m + 2.5 m − 0 m = 22.5 m
= 22.5 m
Pump power
Brake power is the water power γ·Q·hp divided by the pump efficiency.
P = γ·Q·hp/η
P = 9.81 kN/m^3 × 0.05 m^3/s × 22.5 m / 0.75 = 14.71 kW
= 14.71 kW

Results

Required pump head hp

22.5m

Required pump power (brake)

14.71kW

Velocity at the upstream section

0m/s

Velocity at the downstream section

0m/s

Where this answer was checked

source: Pump sizing by the energy equation, P = γQh/η (NCEES FE Reference Handbook) — two free surfaces at rest: hp = Δz + hL; brake power γQhp/η
verified by: hand-recomputed
derivation: hp = 20 + 2.5 = 22.5 m. P = 9810·0.05·22.5/0.75 = 11 036.25/0.75 = 14 715 W = 14.715 kW.

Example 3

US: Torricelli outflow velocity from a tank 15 ft above the outlet

Given

section1: z 15 ft
section2
solve for: velocity2

Assumptions

No fluid was specified — fresh water (γ = 9.81 kN/m³) was assumed.
Steady, incompressible, single-stream flow with kinetic-energy correction factor α = 1.
All pressures are GAGE pressures — both sections share the same atmospheric reference.
No velocity information at the upstream section — it was treated as a large reservoir / free surface with negligible velocity (V ≈ 0).
No pressure given at the upstream section — it was taken as atmospheric (p = 0 gage: free surface or free jet).
No pressure given at the downstream section — it was taken as atmospheric (p = 0 gage: free surface or free jet).
No elevation given at the downstream section — the datum was placed there (z = 0).

Solution steps

Write the energy equation between the sections
Energy per unit weight (head) is conserved apart from pump work added, turbine work extracted, and losses; every term is a length.
p₁/γ + V₁²/2g + z₁ + hp = p₂/γ + V₂²/2g + z₂ + ht + hL
Energy head at the upstream section
Sum the pressure head, velocity head, and elevation head.
E₁ = p₁/γ + V₁²/2g + z₁
p₁/γ = 0 ft; V₁²/2g = 0 ft; z₁ = 15 ft → E₁ = 15 ft
= 15 ft
Energy head at the downstream section
Only the known terms are evaluated here (pressure head, elevation head) — the remaining term is the unknown.
E₂ = p₂/γ + V₂²/2g + z₂
pressure head p₂/γ = 0 ft; elevation z₂ = 0 ft → known part of E₂ = 0 ft
Solve for the downstream velocity
Rearrange for the downstream velocity head, then take the square root.
V₂²/2g = E₁ + hp − ht − hL − p₂/γ − z₂; V₂ = √(2g·(that head))
V₂²/2g = 15 ft → V₂ = 31.07 ft/s
= 31.07 ft/s

Results

Velocity at the downstream section

31.07ft/s

Velocity at the upstream section

0ft/s

Where this answer was checked

source: Torricelli's theorem V = √(2gh) as the loss-free energy equation — free surface (V ≈ 0, p = 0) to free jet (p = 0), h = 15 ft
verified by: hand-recomputed
derivation: h = 15 ft = 4.572 m. V₂ = √(2·9.80665·4.572) = √89.672 = 9.4695 m/s = 31.068 ft/s (US check: √(2·32.174·15) = √965.2 = 31.07 ft/s).