Home / Calculators / Energy Equation (Bernoulli with Machines & Losses) Fluid Mechanics & Hydraulics
Solve the energy equation between two flow sections for one unknown, including pressure, velocity, pump head, head loss, elevation, or pump power.
Direct solving is free and needs no account. Have a word problem instead? Submit it as text .
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 m pressure 200 kPa velocity 2 m/s
section2 z 5 m velocity 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).