Bernoulli’s Equation — Derivation, Applications & Solved Problems | EngineeringHulk

Bernoulli’s Equation — Derivation, Applications & Solved Problems

The single most important equation in fluid mechanics — energy conservation along a streamline, with full derivation, assumptions, modifications, and GATE CE numerical examples

Last Updated: April 2026

Key Takeaways

Bernoulli’s equation: p/ρg + V²/2g + z = constant along a streamline — sum of pressure head, velocity head, and elevation head is conserved.
Derived from Euler’s equation of motion along a streamline; valid for steady, incompressible, inviscid (frictionless) flow along a single streamline.
The continuity equation A₁V₁ = A₂V₂ is always applied together with Bernoulli to solve two-unknowns problems.
Modified Bernoulli (with head loss): p₁/ρg + V₁²/2g + z₁ = p₂/ρg + V₂²/2g + z₂ + h_L — used for real pipe flow.
When a pump is present: add pump head h_p to the left side; when a turbine is present: add turbine head h_t to the right side.
Key applications: venturimeter, pitot tube, orifice meter, free jet trajectory, and siphon analysis.
GATE CE tests Bernoulli + continuity together in 2–3 mark problems virtually every year — this is the highest-yield topic in Fluid Mechanics.

1. Euler’s Equation of Motion Along a Streamline

Bernoulli’s equation is derived by integrating Euler’s equation — Newton’s second law applied to an inviscid fluid element moving along a streamline.

Consider a fluid element of cross-sectional area dA and length ds along a streamline, inclined at angle θ to the horizontal. Applying Newton’s second law in the s-direction (along the streamline):

Euler’s Equation (along streamline):

–(∂p/∂s) – ρg(∂z/∂s) = ρ(V ∂V/∂s)

Rearranging:

(1/ρ) dp + V dV + g dz = 0

This holds for steady, inviscid, incompressible flow along a streamline.

Three forces act on the element: pressure force (net), gravity (body force), and inertia. Viscous forces are assumed zero (inviscid assumption).

2. Derivation of Bernoulli’s Equation

Integrate Euler’s equation along the streamline from point 1 to point 2, with ρ = constant (incompressible fluid):

∫(dp/ρ) + ∫V dV + ∫g dz = 0

p/ρ + V²/2 + gz = constant

Dividing throughout by g:

p/(ρg) + V²/(2g) + z = H = constant

Between two points on a streamline:

p₁/(ρg) + V₁²/(2g) + z₁ = p₂/(ρg) + V₂²/(2g) + z₂

2.1 Physical Meaning of Each Term

Term	Name	Unit	Physical Meaning
p/(ρg)	Pressure head	m	Energy per unit weight due to fluid pressure; height a fluid column of that pressure would rise
V²/(2g)	Velocity head (kinetic head)	m	Energy per unit weight due to fluid motion; height the fluid would rise if kinetic energy were converted to potential energy
z	Elevation head (datum head)	m	Potential energy per unit weight; height above the chosen datum
H = p/(ρg) + V²/(2g) + z	Total head (piezometric + velocity)	m	Total mechanical energy per unit weight — constant along a streamline in ideal flow

2.2 Pressure Forms of Bernoulli’s Equation

Energy per unit volume (multiply by ρg):

p + ½ρV² + ρgz = constant [Pa or N/m²]

Energy per unit weight (divide by g):

p/(ρg) + V²/(2g) + z = H [m of fluid]

Energy per unit mass (divide by ρ):

p/ρ + V²/2 + gz = constant [J/kg or m²/s²]

3. Assumptions & Limitations

3.1 Assumptions (All Must Hold for Standard Bernoulli)

#	Assumption	What It Means in Practice
1	Steady flow	Flow properties at any point do not change with time (∂/∂t = 0). Not valid during pump start-up, water hammer, or tidal flows.
2	Incompressible fluid	ρ = constant. Valid for liquids at normal pressures and for gases at low Mach numbers (Ma < 0.3).
3	Inviscid (frictionless) flow	No viscous losses. Real fluids have friction — this is addressed by the modified Bernoulli equation with head loss h_L.
4	Flow along a single streamline	Bernoulli applies between two points on the same streamline. Between different streamlines, it applies only if the flow is also irrotational (potential flow).
5	No energy addition or extraction	No pump or turbine between the two points. Modified Bernoulli adds these terms when present.

3.2 Where Bernoulli Should NOT Be Applied Directly

Across a pump or turbine (use modified Bernoulli with h_p or h_t)
In highly turbulent flow regions (energy losses are significant)
At pipe entrances, sharp bends, or sudden expansions (use minor loss coefficients)
In compressible flow at high velocities (Ma > 0.3)
Across a hydraulic jump (use momentum equation instead)

4. Continuity Equation

Before applying Bernoulli, almost every problem requires the continuity equation to relate velocities at two sections. For steady, incompressible flow through a control volume:

A₁V₁ = A₂V₂ = Q = constant

where:

A₁, A₂ = cross-sectional areas at sections 1 and 2 (m²)
V₁, V₂ = mean flow velocities at sections 1 and 2 (m/s)
Q = volumetric flow rate / discharge (m³/s)

For a circular pipe of diameter D: A = πD²/4

Flow rate–velocity relationship: Q = AV = (πD²/4) × V

4.1 Continuity in 3D (General Form)

∂ρ/∂t + ∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z = 0

For steady incompressible flow: ∂u/∂x + ∂v/∂y + ∂w/∂z = 0

(Divergence of velocity field = 0)

Solving strategy for most GATE problems: Use continuity to find the unknown velocity (V₂ from V₁ and areas), then substitute both velocities into Bernoulli to find the unknown pressure or head.

5. Modified Bernoulli — With Head Loss, Pump & Turbine

5.1 Bernoulli with Head Loss (Real Fluid Flow)

For real flows with friction and minor losses, total head decreases in the direction of flow:

p₁/(ρg) + V₁²/(2g) + z₁ = p₂/(ρg) + V₂²/(2g) + z₂ + h_L

where h_L = total head loss between points 1 and 2 (m)

h_L = h_f (friction losses) + h_m (minor losses)

Friction loss (Darcy-Weisbach): h_f = fLV²/(2gD)

5.2 Bernoulli with a Pump

p₁/(ρg) + V₁²/(2g) + z₁ + h_p = p₂/(ρg) + V₂²/(2g) + z₂ + h_L

h_p = pump head added to the fluid (m)

Power delivered by pump to fluid: P = ρgQh_p (W)

Power input to pump: P_input = ρgQh_p / η_pump

5.3 Bernoulli with a Turbine

p₁/(ρg) + V₁²/(2g) + z₁ = p₂/(ρg) + V₂²/(2g) + z₂ + h_t + h_L

h_t = head extracted by the turbine (m)

Power output from turbine: P = ρgQh_t × η_turbine (W)

5.4 Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)

Line	Definition	Formula
Energy Grade Line (EGL)	Plot of total head H along the pipe length	H = p/(ρg) + V²/(2g) + z
Hydraulic Grade Line (HGL)	Plot of piezometric head (pressure + elevation) along the pipe length	HGL = p/(ρg) + z
Difference (EGL – HGL)	Velocity head at that section	V²/(2g)

In ideal flow (no losses), EGL is a horizontal line. HGL is below EGL by V²/(2g) — it dips where velocity is high (narrow pipe) and rises where velocity is low (wide pipe). The HGL must always be above the pipe centreline for positive pressure; if HGL drops below the pipe, the pressure becomes negative (sub-atmospheric) and cavitation may occur.

6. Applications — Venturimeter, Pitot Tube, Orifice, Siphon

6.1 Venturimeter

A venturimeter is a flow-measuring device inserted in a pipe. It has a converging section, a throat (minimum diameter), and a diverging (diffuser) section. By measuring the pressure difference between inlet and throat, flow rate is calculated using Bernoulli + continuity.

Theoretical discharge:

Q_th = (A₁A₂ / √(A₁² – A₂²)) × √(2g Δh)

Actual discharge:

Q_act = C_d × Q_th = C_d × (A₁A₂ / √(A₁² – A₂²)) × √(2g Δh)

where:

C_d = coefficient of discharge ≈ 0.96–0.99 for venturimeter (high, due to gradual convergence)
Δh = differential head = (p₁ – p₂)/(ρg) + (z₁ – z₂) (m of flowing fluid)
A₁ = inlet area; A₂ = throat area (m²)

For horizontal venturimeter (z₁ = z₂): Δh = (p₁ – p₂)/(ρg)

Differential manometer reading: Δh = x[(ρ_m/ρ_f) – 1]

where x = manometer deflection; ρ_m = manometric fluid density; ρ_f = pipe fluid density

6.2 Pitot Tube

A pitot tube measures the stagnation (total) pressure at a point in the flow by bringing the fluid to rest isentropically at its tip. The static pressure is measured separately (at a tapping in the pipe wall or via a piezometer tube alongside).

Applying Bernoulli between the free-stream point (1) and the stagnation point (2):

p₁ + ½ρV₁² = p₂ + 0 (V₂ = 0 at stagnation)

V₁ = √(2(p₂ – p₁)/ρ) = √(2g Δh)

where Δh = stagnation head – static head = (p_stag – p_static)/(ρg)

Pitot-static tube (combined): V = C_v√(2g Δh)

C_v ≈ 0.98–1.0 for a well-designed pitot-static tube

6.3 Orifice in a Tank (Torricelli’s Theorem)

Applying Bernoulli between the free water surface (1) and the vena contracta of the jet (2) — the narrowest cross-section of the jet just outside the orifice:

p₁/(ρg) + V₁²/(2g) + z₁ = p₂/(ρg) + V₂²/(2g) + z₂

With p₁ = p₂ = atmospheric, V₁ ≈ 0 (large tank), z₁ – z₂ = H (head above orifice):

V_th = √(2gH) (Torricelli’s theorem)

Actual jet velocity: V_act = C_v × √(2gH)

Actual discharge: Q = C_d × A₀ × √(2gH)

where C_d = C_v × C_c

C_v = coefficient of velocity ≈ 0.97–0.99
C_c = coefficient of contraction = A_{vena contracta}/A_orifice ≈ 0.61–0.64
C_d = coefficient of discharge ≈ 0.61–0.65 for sharp-edged orifice

6.4 Siphon

A siphon is an inverted U-tube that carries fluid over a barrier at a crest higher than the supply level, driven by atmospheric pressure. Bernoulli is applied from the supply surface to the discharge end, and again to the crest to find the pressure there (which must remain above vapour pressure to prevent cavitation).

At the crest C (height z_c above datum; flow velocity V_c):

p_c/(ρg) = H_atm – z_c – V_c²/(2g) – h_L,1→c

Condition for siphon to work: p_c > p_vapour

Maximum crest height (ignoring losses, with V_c ≈ 0):

z_c,max = H_atm/(ρg) ≈ 10.33 m of water (for water at 20 °C)

In practice (with losses): z_c,max ≈ 7–8 m of water

7. Worked Examples (GATE CE Level)

Example 1 — Venturimeter Discharge (GATE CE 2021 type)

Problem: A horizontal venturimeter has inlet diameter 30 cm and throat diameter 15 cm. The pressure difference between inlet and throat is 25 kPa. C_d = 0.98. Find the discharge through the pipe. (ρ_water = 1000 kg/m³)

Given:
D₁ = 0.30 m → A₁ = π(0.30)²/4 = 0.07069 m²
D₂ = 0.15 m → A₂ = π(0.15)²/4 = 0.01767 m²
Δp = 25,000 Pa → Δh = Δp/(ρg) = 25,000/(1000×9.81) = 2.548 m
C_d = 0.98

Step 1 — Throat velocity from Bernoulli + continuity:

From continuity: V₁ = (A₂/A₁)V₂ = (0.01767/0.07069)V₂ = 0.25 V₂

Bernoulli (horizontal, no losses): p₁/(ρg) + V₁²/(2g) = p₂/(ρg) + V₂²/(2g)

Δh = (V₂² – V₁²)/(2g) = V₂²(1 – 0.25²)/(2g) = V₂²(0.9375)/(2×9.81)

2.548 = 0.9375 V₂² / 19.62

V₂² = 2.548 × 19.62 / 0.9375 = 53.27

V₂ = 7.299 m/s

Step 2 — Theoretical discharge:

Q_th = A₂ × V₂ = 0.01767 × 7.299 = 0.12897 m³/s

Step 3 — Actual discharge:

Q_act = C_d × Q_th = 0.98 × 0.12897 = 0.1264 m³/s = 126.4 litres/s

Alternative using the direct formula:

Q_act = C_d × (A₁A₂/√(A₁²–A₂²)) × √(2gΔh)

= 0.98 × (0.07069×0.01767/√(0.07069²–0.01767²)) × √(2×9.81×2.548)

= 0.98 × (0.001249/0.06845) × 7.073 = 0.98 × 0.01824 × 7.073 = 0.1264 m³/s ✓

Answer: Q = 0.1264 m³/s

Example 2 — Pitot Tube Velocity Measurement

Problem: A pitot-static tube is inserted in a water pipe. The stagnation pressure is 120 kPa and the static pressure is 100 kPa. C_v = 0.98. Find the velocity of flow.

Given:
p_stag = 120 kPa; p_static = 100 kPa
Δp = 20,000 Pa → Δh = 20,000/(1000×9.81) = 2.039 m

Velocity:

V = C_v × √(2gΔh) = 0.98 × √(2 × 9.81 × 2.039)

= 0.98 × √(40.005) = 0.98 × 6.325 = 6.199 m/s ≈ 6.2 m/s

Answer: V = 6.2 m/s

Example 3 — Orifice in a Tank (Torricelli)

Problem: A tank has a sharp-edged circular orifice of 5 cm diameter at its base. The water head above the orifice centre is 4 m. C_d = 0.62. Find (a) the theoretical velocity, (b) the actual discharge.

Given:
D = 0.05 m → A₀ = π(0.05)²/4 = 1.9635 × 10⁻³ m²
H = 4 m; C_d = 0.62

(a) Theoretical velocity (Torricelli):

V_th = √(2gH) = √(2 × 9.81 × 4) = √78.48 = 8.859 m/s

(b) Actual discharge:

Q = C_d × A₀ × √(2gH) = 0.62 × 1.9635×10⁻³ × 8.859

= 0.62 × 0.01739 = 0.01078 m³/s = 10.78 litres/s

Answer: V_th = 8.86 m/s; Q = 10.78 L/s

Example 4 — Bernoulli Applied to a Tapering Pipe (GATE CE 2020 type)

Problem: Water flows through a horizontal pipe that tapers from 200 mm diameter at section 1 to 100 mm at section 2. The pressure at section 1 is 300 kPa and the velocity at section 1 is 2 m/s. Assuming ideal flow (no losses), find the pressure at section 2.

Given:
D₁ = 0.2 m → A₁ = π(0.2)²/4 = 0.03142 m²
D₂ = 0.1 m → A₂ = π(0.1)²/4 = 0.007854 m²
V₁ = 2 m/s; p₁ = 300,000 Pa; z₁ = z₂ (horizontal)

Step 1 — Velocity at section 2 (continuity):

A₁V₁ = A₂V₂

V₂ = (A₁/A₂) × V₁ = (0.03142/0.007854) × 2 = 4 × 2 = 8 m/s

Step 2 — Pressure at section 2 (Bernoulli, horizontal):

p₁ + ½ρV₁² = p₂ + ½ρV₂²

p₂ = p₁ + ½ρ(V₁² – V₂²)

p₂ = 300,000 + ½ × 1000 × (4 – 64)

p₂ = 300,000 + 500 × (–60)

p₂ = 300,000 – 30,000 = 270,000 Pa = 270 kPa

Answer: p₂ = 270 kPa. Pressure drops as velocity increases — consistent with Bernoulli.

Example 5 — Modified Bernoulli with Pump (GATE CE level)

Problem: A pump delivers water from a sump (point 1) to an overhead tank (point 2) 20 m above. The pipe is 150 mm diameter throughout. Flow velocity is 3 m/s. Total head loss h_L = 5 m. Find the power delivered by the pump to the fluid. (η_pump = 0.75)

Given:
z₁ = 0 (datum); z₂ = 20 m
V₁ = V₂ = 3 m/s (same diameter pipe throughout → from continuity)
p₁ = p₂ = 0 (both surfaces open to atmosphere — gauge pressures = 0)
h_L = 5 m

Modified Bernoulli (sump surface to tank surface):

0 + 0 + 0 + h_p = 0 + 0 + 20 + 5

h_p = 25 m

Note: Since diameter is constant, V₁ = V₂, so velocity heads cancel.

Flow rate:

Q = A × V = (π × 0.15²/4) × 3 = 0.01767 × 3 = 0.05301 m³/s

Power delivered to fluid:

P_fluid = ρgQh_p = 1000 × 9.81 × 0.05301 × 25 = 13,001 W ≈ 13.0 kW

Shaft power input (motor power required):

P_input = P_fluid / η = 13,001 / 0.75 = 17,334 W ≈ 17.3 kW

Answer: Pump head h_p = 25 m; Power to fluid = 13.0 kW; Motor input = 17.3 kW

Example 6 — Pressure at Crest of a Siphon

Problem: A siphon pipe of 100 mm diameter carries water from a reservoir. The crest of the siphon is 3.5 m above the reservoir water surface. The pipe discharges freely at 2 m below the reservoir surface. Neglect all losses and find the pressure at the crest. (p_atm = 101.325 kPa)

Set datum at reservoir water surface (point 1).
Crest C: z_c = +3.5 m above datum
Outlet (point 2): z₂ = –2 m below datum
p₁ = p₂ = p_atm (both open surfaces); V₁ ≈ 0 (large reservoir)

Step 1 — Velocity at outlet using Bernoulli (1 → 2):

0 + 0 + 0 = p_atm/(ρg) + V₂²/(2g) + (–2) … both sides have p_atm which cancels

0 = V₂²/(2g) – 2

V₂ = √(2×9.81×2) = √39.24 = 6.264 m/s

Step 2 — Velocity at crest (same diameter pipe → V_c = V₂ = 6.264 m/s)

Step 3 — Pressure at crest using Bernoulli (1 → C):

p₁/(ρg) + 0 + 0 = p_c/(ρg) + V_c²/(2g) + 3.5

p_atm/(ρg) = p_c/(ρg) + (6.264²/(2×9.81)) + 3.5

10.33 = p_c/(ρg) + 2.0 + 3.5

p_c/(ρg) = 10.33 – 5.5 = 4.83 m of water

p_c = 4.83 × 1000 × 9.81 = 47,378 Pa = 47.4 kPa absolute

Gauge pressure at crest = 47.4 – 101.3 = –53.9 kPa (sub-atmospheric)

Answer: p_c = 47.4 kPa absolute (–53.9 kPa gauge). Well above vapour pressure of water — siphon works safely.

8. Common Mistakes

Mistake 1 — Applying Bernoulli Across a Pump or Turbine Without Adding h_p or h_t

Error: Writing the standard Bernoulli equation between suction and delivery points of a pump without including the pump head h_p.
Root Cause: Forgetting that Bernoulli is an energy equation — a pump adds energy to the fluid, a turbine extracts it. Omitting these terms violates energy conservation.
Fix: Always identify every energy source (pump: +h_p on the inlet side) and sink (turbine: +h_t on the outlet side, head loss: +h_L on the outlet side) before writing the modified Bernoulli equation.

Mistake 2 — Forgetting to Square the Velocity in the Velocity Head Term

Error: Writing velocity head as V/(2g) instead of V²/(2g).
Root Cause: The velocity head comes from kinetic energy per unit weight = ½mV²/(mg) = V²/(2g). Omitting the square is a dimensional error — V/(2g) has units m/(m/s²) = s²/m, not metres.
Fix: Always dimension-check: velocity head [m] = V [m/s]² / (2 × g [m/s²]) = m²/s² ÷ m/s² = m ✓

Mistake 3 — Using Different Pressure Units on Each Side of Bernoulli

Error: Substituting p₁ in kPa on the left and p₂ in Pa on the right (or mixing absolute and gauge pressures).
Root Cause: Unit inconsistency. Bernoulli is written in metres (head) or Pascals — all terms must use the same unit system.
Fix: Convert everything to Pascals before substituting, or convert to metres of head (divide pressure in Pa by ρg). Gauge pressures can be used as long as both sides use gauge consistently — but be careful at free surfaces where p_gauge = 0.

Mistake 4 — Wrong Differential Head Formula for Manometer-Equipped Venturimeter

Error: Using Δh = x × (ρ_m/ρ_f) instead of Δh = x × (ρ_m/ρ_f – 1) for the differential manometer reading.
Root Cause: The manometer deflection x is in a heavier fluid (mercury), but the pipe fluid column also contributes — the net head is the difference, giving the –1 term.
Fix: Derive from first principles once: balance pressures at the lower meniscus, and the –1 always emerges. For mercury–water: Δh = x(13.6 – 1) = 12.6x metres of water.

Mistake 5 — Taking V₁ ≈ 0 When the Tank Area is NOT Much Larger Than the Orifice

Error: Assuming V₁ = 0 (large tank) even when the problem gives a tank diameter only slightly larger than the orifice.
Root Cause: The V₁ ≈ 0 approximation is valid only when A_tank ≫ A_orifice (typically A_tank/A_orifice > 10 is sufficient). If the ratio is small, V₁ from continuity (V₁ = (A₀/A_tank)V₂) is not negligible.
Fix: Always check the area ratio. If given tank and orifice diameters, compute both areas and use continuity to get V₁ before applying Bernoulli.

9. Frequently Asked Questions

Q1. Why does pressure decrease when velocity increases in a pipe — isn’t that counter-intuitive?

This is one of the most common conceptual questions in fluid mechanics. The key is to think in terms of energy, not force. Bernoulli’s equation states that total mechanical energy per unit weight (pressure head + velocity head + elevation head) is constant along a streamline. When a pipe narrows, continuity forces the velocity to increase. Since total energy is conserved and the elevation hasn’t changed, the pressure head must fall to compensate for the rise in velocity head. The fluid isn’t being “pushed” by the higher-velocity region — it’s being accelerated into the constriction by the higher pressure behind it (just like a river speeds up through a gorge because there’s a pressure gradient driving it). The high-velocity, low-pressure region at the throat of a venturimeter is a direct consequence of this trade-off.

Q2. What is the difference between stagnation pressure and static pressure?

Static pressure (p) is the pressure of the fluid as measured by an instrument moving with the fluid — it is the thermodynamic pressure, the one that acts on the walls of a pipe or channel. Stagnation pressure (p₀ = p + ½ρV²) is the pressure that would be reached if the fluid were brought to rest isentropically (without loss). It is the sum of static pressure and dynamic pressure (½ρV²). A pitot tube measures stagnation pressure at its stagnation point (tip). The difference between stagnation and static pressures equals the dynamic pressure: p₀ – p = ½ρV², which allows the velocity to be computed. In incompressible flow, stagnation pressure is the maximum pressure a flow can exert — it represents the total kinetic energy per unit volume that can be converted to pressure.

Q3. How is Bernoulli’s equation related to the energy equation for real pipe flow?

Bernoulli’s equation is a special case of the general energy equation, valid only for ideal (inviscid, incompressible, steady) flow along a streamline. The full energy equation for real pipe flow adds head loss h_L (representing energy dissipated by viscous friction and turbulence) and work terms for pumps and turbines. This gives the modified Bernoulli: p₁/(ρg) + V₁²/(2g) + z₁ + h_p = p₂/(ρg) + V₂²/(2g) + z₂ + h_t + h_L. For laminar pipe flow, h_L is given by the Hagen-Poiseuille equation; for turbulent flow, it is computed from the Darcy-Weisbach equation using the Moody friction factor. In most GATE CE pipe flow problems, you apply this extended Bernoulli rather than the frictionless version.

Q4. Can Bernoulli’s equation be applied to open channel flow?

Yes — Bernoulli is applied to open channel flow, but the pressure term at the free surface is always atmospheric (p = p_atm ≡ 0 gauge), so the pressure head at the surface is zero. The total energy per unit weight at any section of an open channel is E = y + V²/(2g) + z (where y is the water depth and z is the channel bed elevation above datum) — this is the basis of the specific energy concept used in hydraulic jump analysis. Bernoulli is implicit in the derivation of the discharge equations for broad-crested weirs, sharp-crested weirs, and channel transitions. The main limitation is that real channels have friction losses (Manning’s or Chezy’s equations quantify these) and the simple Bernoulli must be supplemented with energy slope corrections for gradually-varied flow profiles.