7.2 Bernoulli Equation and Mechanical Energy Balance

7.2 Bernoulli Equation and Mechanical Energy Balance

Quick Answer: Ideal Bernoulli: P/ρ + v²/2 + g z = constant along a streamline (steady, incompressible, frictionless, no shaft work). Head form: P/(ρg) + v²/(2g) + z = constant. Real systems use a mechanical energy balance with pump head and friction head added.

Section 7.1 gave velocity and Re. This section converts those velocities into pressure and elevation trade-offs—the core of tank draining, orifice intuition, and pump-required-head thinking on Domain C of the UPDA/MMUP Chemical exam.

Bernoulli Equation Along a Streamline

For a fluid particle moving along a streamline, under the classical restrictions listed below, mechanical energy per unit mass is conserved:

P/ρ + v²/2 + g z = constant

Between two points (1) and (2):

P₁/ρ + v₁²/2 + g z₁ = P₂/ρ + v₂²/2 + g z₂

Head Form (Meters of Fluid)

Divide by g:

P/(ρ g) + v²/(2 g) + z = constant

Energy per unit mass form uses J/kg (= m²/s²): P/ρ, v²/2, g z. Head form uses meters. Do not mix without converting by g.

Classic Assumptions (Memorize the Limits)

Exam rule: If the stem mentions roughness, fittings, long pipe, efficiency, or NPSH, you have left pure Bernoulli and need losses and/or pump work.

Intuition: Trading Heads

Torricelli’s idealization: Liquid drains from a large open tank (v₁ ≈ 0, P₁ = P₂ = P_atm) through a small orifice at depth h below the free surface: v₂ ≈ √(2 g h). Real orifices need a discharge coefficient C_d < 1 (v_actual ≈ C_d √(2 g h)).

Mechanical Energy Balance with Pump and Friction

Engineering practice writes a steady mechanical energy balance between points 1 and 2 (incompressible, per unit mass):

P₁/ρ + v₁²/2 + g z₁ + w_s = P₂/ρ + v₂²/2 + g z₂ + losses

where w_s is shaft work done on the fluid per unit mass (pump positive; turbine negative) and losses ≥ 0 account for friction and irreversible fittings.

Head Form (Most Common for Piping)

P₁/(ρg) + v₁²/(2g) + z₁ + h_p = P₂/(ρg) + v₂²/(2g) + z₂ + h_f + h_m

Rearranged for required pump head:

h_p = (P₂ − P₁)/(ρ g) + (v₂² − v₁²)/(2 g) + (z₂ − z₁) + h_f + h_m

Reading the formula:

• Pump must overcome static lift (z₂ − z₁), pressure rise, acceleration (velocity head increase), and all losses.
• If discharging to the same pressure and diameter as suction (P₂ ≈ P₁, v₂ ≈ v₁), h_p ≈ Δz + losses.

Worked Example: Ideal Bernoulli Through a Reducer

Problem. Incompressible liquid, ρ = 900 kg/m³. Horizontal pipe (z₁ = z₂), frictionless. Point 1: D₁ = 0.10 m, P₁ = 250 kPa. Point 2: D₂ = 0.05 m. Mean velocity v₁ = 1.5 m/s. Find P₂ (ideal).

Step 1 — Continuity for v₂.

A₂/A₁ = (D₂/D₁)² = 0.25 → v₂ = v₁ / 0.25 = 6.0 m/s

Step 2 — Bernoulli (mass form).

P₁/ρ + v₁²/2 = P₂/ρ + v₂²/2

P₂ = P₁ + ρ (v₁² − v₂²)/2

= 250000 + 900 × (2.25 − 36)/2

= 250000 + 900 × (−33.75)/2

= 250000 + 900 × (−16.875)

= 250000 − 15187.5 ≈ 234.8 kPa

Pressure fell because velocity head rose. With real friction, P₂ would be lower still.

Worked Example: Pump Head with Elevation and Losses

Problem. Pump water (ρ = 1000 kg/m³) from an open tank surface (point 1: P₁ = 1 atm, v₁ ≈ 0, z₁ = 0) to a pressurized vessel nozzle (point 2: P₂ = 1 atm + 150 kPa gauge on the liquid surface is not the nozzle—simplify: discharge into a line where P₂ = 250 kPa abs, v₂ = 2.0 m/s, z₂ = 25 m). Total loss head h_f + h_m = 8 m. Atmospheric pressure at tank = 101 kPa. Find h_p.

Use P₁ = 101 kPa, P₂ = 250 kPa, v₁ = 0, v₂ = 2.0 m/s, z₁ = 0, z₂ = 25 m, losses = 8 m.

h_p = (P₂ − P₁)/(ρ g) + (v₂² − 0)/(2 g) + 25 + 8

(P₂ − P₁)/(ρ g) = (149000)/(1000 × 9.81) ≈ 15.2 m

v₂²/(2 g) = 4 / 19.62 ≈ 0.20 m

h_p ≈ 15.2 + 0.20 + 25 + 8 ≈ 48.4 m

Elevation and losses dominate; velocity head is often small for liquid lines at a few m/s—but do not drop it automatically if the stem gives high gas or nozzle velocities.

Relating Pump Head to Power (Preview)

Fluid power delivered:

P_fluid = ṁ g h_p = ρ Q g h_p

Shaft power required:

P_shaft = P_fluid / η_pump

(Details and NPSH in Section 7.3.) Bernoulli/mechanical energy gives h_p; efficiency converts head into motor kW.

When Not to Use Bare Bernoulli

Link to Thermodynamic Energy Balances

Chapter 4’s open-system first law tracks thermal enthalpy and heat. Bernoulli tracks mechanical terms. For liquids with constant ρ and negligible internal-energy change from friction heating on short exams, mechanical head balances are the right tool for “what pressure and pump head do I need?” Friction ultimately becomes a small temperature rise, usually ignored in hydraulic MCQs.

UPDA Chemical Exam Workflow

1. Sketch points 1 and 2; mark P, v, z knowns/unknowns.
2. Check assumptions: incompressible? steady? shaft work? losses mentioned?
3. If ideal → apply Bernoulli / head equality.
4. If pump or friction → write mechanical energy / head balance and solve for h_p or P₂.
5. Convert head to pressure with ΔP = ρ g h when needed.
6. Keep one datum for z and consistent absolute or consistent gauge pressures (do not mix carelessly).

Common Traps

• Using Bernoulli without losses on a “100 m of commercial steel pipe” stem.
• Forgetting ρ g when converting between Pa and meters of head.
• Setting v = 0 at a nozzle exit (exit velocity is often the unknown).
• Measuring z from inconsistent datums at the two points.
• Treating gauge and absolute pressures inconsistently in (P₂ − P₁).
• Applying incompressible Bernoulli to choked gas flow.

Section 7.3 quantifies h_f via friction factors, estimates ΔP, and covers pump power, efficiency, and NPSH at exam depth.