Bernoulli's equation and flow measurement

Why it matters

Bernoulli's equation is the energy balance for flowing fluids. It explains why pressure falls where a flow speeds up, how a siphon or a spray gun works, and it is the basis of every head-type flow meter on a shop floor: venturi meters, orifice plates and Pitot tubes used to monitor coolant, compressed air and process water.

Key ideas

Derivation in one line. Apply Newton's second law along a streamline to a small fluid element in steady flow with no friction (Euler's equation): dp/ρ + V·dV + g·dz = 0. Integrating for constant density gives Bernoulli's equation.

Meaning of each term. Per unit weight of fluid (units of metres, called heads): p/(ρg) is the pressure head, V²/(2g) the velocity head and z the elevation head. Their sum is the total head H, which is constant along a streamline under the assumptions below. Multiply by ρg to get the per-unit-volume (pressure) form.

Assumptions — and what breaks them.

1. Steady flow.
2. Incompressible fluid (gas flows below a Mach number of about 0.3 are acceptable).
3. Inviscid (frictionless) flow — no head loss.
4. Along a single streamline (or anywhere in an irrotational flow).
5. No shaft work (pumps, turbines) and no heat transfer between the two points. For real pipes we add a head-loss term h_L and pump or turbine heads; this extended form is the energy equation used in pipe-flow problems.

Hydraulic and energy grade lines. The energy grade line (EGL) sits at height H = p/ρg + V²/2g + z; the hydraulic grade line (HGL) sits one velocity head below it, at p/ρg + z. Where the pipe rises above the HGL the gauge pressure is negative — the danger zone for air release and cavitation, for example at the summit of a siphon.

Applications.

• Free jet from a tank (Torricelli): a jet issuing under head h has ideal velocity V = √(2gh).
• Pitot tube: brings the flow to rest at its tip (stagnation point). Stagnation pressure p₀ = p + ½ρV², so V = √(2(p₀ − p)/ρ). A Pitot-static tube measures both pressures at once.
• Venturi meter: a converging section, a throat and a long gentle diverging section. The pressure drop between inlet and throat, combined with continuity, gives the flow rate. Because the diverging cone recovers most of the pressure, the permanent loss is small; the discharge coefficient C_d is typically about 0.95–0.99.
• Orifice meter: a thin plate with a sharp-edged hole. Cheap and compact, but the jet contracts (vena contracta) and separates, so the permanent pressure loss is large and C_d is only about 0.6–0.65. Take exact values from a standard or data book.
• Orifices and mouthpieces in tanks: C_d = C_c × C_v, where C_c (coefficient of contraction) is area of the jet at the vena contracta divided by orifice area and C_v (coefficient of velocity) is actual jet velocity divided by √(2gh).

Reading a differential manometer. When a U-tube containing a heavier liquid (relative density S_m) is connected across a meter carrying a lighter liquid (S), the piezometric head difference is h = x·(S_m/S − 1), where x is the manometer deflection. For a mercury–water manometer, h = 12.6x. This result automatically accounts for any difference in elevation between the tappings, so the same formula serves a vertical or inclined venturi.

Formulas

p/(ρ·g) + V²/(2·g) + z = constant

• p: static pressure (Pa); ρ: density (kg/m³); V: velocity (m/s); g = 9.81 m/s²; z: elevation (m). Steady, incompressible, frictionless, along a streamline.

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

• Energy equation for real pipe systems; h_p pump head, h_t turbine head, h_L head loss (all m).

V = √(2·g·h)

• Torricelli: ideal jet velocity under head h (m).

V = √(2·(p₀ − p)/ρ)

• Pitot tube; p₀: stagnation pressure (Pa).

Q = C_d · A₁·A₂·√(2·g·h) / √(A₁² − A₂²)

• Venturi or orifice meter; A₁: inlet area, A₂: throat (or orifice) area (m²); h: piezometric head difference (m of flowing fluid); C_d: discharge coefficient.

h = x·(S_m/S − 1)

• Differential manometer; x: deflection (m); S_m, S: relative densities of manometer liquid and flowing liquid.

C_d = C_c · C_v

• Orifice coefficients (dimensionless).

Worked examples

Example 1 (standard): venturi meter with a mercury manometer. Given: a horizontal venturi meter, inlet diameter 200 mm, throat diameter 100 mm, carries water. A mercury differential manometer (S_m = 13.6) shows a deflection of 150 mm. C_d = 0.98. Find the flow rate.

1. Head difference: h = x·(S_m/S − 1) = 0.15 × (13.6/1 − 1) = 0.15 × 12.6 = 1.89 m of water.
2. A₁ = π × 0.2²/4 = 0.031 416 m², A₂ = π × 0.1²/4 = 0.007 854 m².
3. √(2gh) = √(2 × 9.81 × 1.89) = 6.0895 m/s
4. A₁A₂/√(A₁² − A₂²) = (0.031 416 × 0.007 854)/√(0.031 416² − 0.007 854²) = 0.008 112 m²
5. Q = 0.98 × 0.008 112 × 6.0895 = 0.048 41 m³/s Answer: Q ≈ 0.0484 m³/s (48.4 L/s)

Example 2 (GATE level): siphon from a tank. Given: a 100 mm siphon pipe draws water from a large open tank and discharges to air through a 50 mm nozzle at its end. The nozzle exit is 5 m below the tank's free surface. The summit C of the siphon is 1 m above the free surface. Neglect losses. Find the discharge and the gauge pressure at C.

1. Bernoulli from the free surface (p = 0 gauge, V ≈ 0) to the nozzle exit (p = 0 gauge): V_exit = √(2g × 5) = √98.1 = 9.905 m/s.
2. Q = (π × 0.05²/4) × 9.905 = 0.019 45 m³/s
3. Velocity in the 100 mm pipe: V = 9.905 × (50/100)² = 2.476 m/s; velocity head V²/2g = 0.3125 m.
4. Bernoulli from the free surface (z = 0) to C (z = +1 m): 0 = p_C/(ρg) + 0.3125 + 1, so p_C/(ρg) = −1.3125 m.
5. p_C = −1.3125 × 1000 × 9.81 = −12 876 Pa Answer: Q ≈ 0.0194 m³/s; p_C ≈ −12.9 kPa (gauge), i.e. a partial vacuum at the summit

The summit pressure must stay well above the vapour pressure of water or the siphon will cavitate and break; this limits how high a siphon can rise above the supply level.

Common mistakes

• Using the manometer deflection x directly as h; for mercury–water it must be multiplied by 12.6.
• Applying Bernoulli across a pump, turbine or a region with large friction losses without adding those terms.
• Mixing gauge and absolute pressures on the two sides of the equation.
• Forgetting the (A₂/A₁)² term, i.e. assuming the inlet velocity is negligible in a venturi.
• Thinking the venturi throat pressure drop is a loss; most of it is recovered in the diffuser.
• Using C_d = 1 for an orifice meter.

For GATE PI

Typical questions: flow rate through a venturi or orifice meter with a differential manometer, velocity from a Pitot tube (including air with a water manometer), jet velocity and discharge from a tank orifice using C_c, C_v and C_d, and pressure at a point in a siphon or a pipe with changing elevation. One-mark questions test the assumptions of Bernoulli's equation and the comparison of venturi and orifice meters. Practise the h = x(S_m/S − 1) step until it is automatic.

Quick check

1. List the assumptions of Bernoulli's equation.
2. A Pitot tube in water shows a stagnation-minus-static head of 0.2 m. What is the velocity?
3. Why does a venturi meter have a higher C_d than an orifice meter?
4. A mercury–water manometer reads 100 mm. What head of water does that represent?

Answers: 1. Steady, incompressible, frictionless, along a streamline, no shaft work or heat transfer 2. √(2 × 9.81 × 0.2) ≈ 1.98 m/s 3. Its gradual contraction and diffuser avoid flow separation and jet contraction, so losses are small 4. 1.26 m of water