Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain.

Exercises · 9.12

Q. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain.

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Concept understanding — Bernoulli Equation Applicability

Bernoulli Equation Applicability

Imagine you're holding a garden hose with your thumb partially covering the opening. Water shoots out much faster than it flows inside the hose. You've just seen Bernoulli's principle in action — the pressure in the narrow opening drops, and speed increases. But here's the catch: if you kink the hose and water stops flowing entirely, Bernoulli's equation no longer describes what's happening. Why? Because the flow is no longer steady.

That's the heart of applicability: Bernoulli's equation is not a universal law of fluids. It's a powerful but conditional tool.

The Intuition First

Bernoulli's equation is really a statement of energy conservation for a moving fluid. Think of a streamline — an imaginary line that follows the path of a fluid particle. As that particle moves along, three forms of energy can change:

Kinetic energy (from speed)
Gravitational potential energy (from height)
Pressure energy (from the fluid being squeezed)

The equation says: along a streamline, the sum of these three remains constant — provided nothing adds or removes energy from the fluid (no friction, no pumps, no turbines).

Note

The "pressure energy" term is not a separate form of energy in thermodynamics. It's the work done by pressure forces to move the fluid. But for problem-solving, treat it as an energy term.

The Precise Statement

For an ideal fluid (incompressible, inviscid) flowing steadily along a streamline, Bernoulli's equation is:

P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}

Where:

$P$ = static pressure (the actual pressure in the fluid)
$\rho$ = fluid density (constant for incompressible flow)
$v$ = flow speed
$g$ = acceleration due to gravity
$h$ = height above a reference level

P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2

This form is used when comparing two points (1 and 2) on the same streamline.

The Four Conditions for Applicability

You cannot use Bernoulli's equation unless all four conditions are met:

1. Steady Flow

The velocity at any fixed point does not change with time. If you turn a tap on and off, or if a pump pulses, the flow is unsteady — Bernoulli fails.

2. Incompressible Flow

The density $\rho$ is constant. This is excellent for liquids (water, oil) and for gases at low speeds (below about 30% of the speed of sound). For high-speed gas flow (e.g., air around an aircraft wing at Mach 0.8), compressibility matters, and Bernoulli's equation is inaccurate.

3. Inviscid Flow (No Friction)

The fluid has zero viscosity — no internal friction. Real fluids have viscosity, but in many practical situations (flow far from walls, short distances), viscous effects are negligible. Never use Bernoulli's equation for flow through long pipes, where friction dominates.

Watch out

A common mistake: applying Bernoulli's equation to flow through a pipe with significant friction. The pressure drop due to friction is not accounted for. Use the Darcy-Weisbach equation or the Hagen-Poiseuille equation instead.

4. Along a Streamline

The equation holds along a single streamline. If you pick two points on different streamlines, the constant may differ. However, if the flow is irrotational (no vorticity), the constant is the same everywhere — a special case often used in aerodynamics.

When Can You Relax the Conditions?

Irrotational flow: If the flow is irrotational (curl of velocity = 0), Bernoulli's constant is the same for all streamlines. This is true for many idealised flows (e.g., flow around a sphere at low speeds, flow through a venturi meter).
Gradual changes: If friction is small and the flow is nearly steady, Bernoulli gives a good approximation. Engineers often use it with a "loss coefficient" to account for minor friction.

A Quick Decision Tree

Ask these questions in order:

Is the flow steady? (No time-dependent changes at a point)
Is the fluid incompressible? (Liquid? Gas at low speed?)
Is viscosity negligible? (Short distances, no long pipes, no boundary layers) …

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Bernoulli’s equation is a statement of conservation of mechanical energy along a streamline. Since it involves differences in pressure between two points, using gauge pressure instead of absolute pressure gives the same result — provided you are consistent and the reference (atmospheric) pressure cancels out.

Why this question matters

Many students get tangled in pressure units when applying Bernoulli’s equation. The equation itself doesn’t care whether you measure pressure relative to vacuum (absolute) or relative to the atmosphere (gauge) — what matters is that you use the same type at both points you compare. Let’s see why.

The core idea

Bernoulli’s equation for steady, incompressible, inviscid flow along a streamline is:

P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}

Here $P$ is the absolute pressure at a point. But if you switch to gauge pressure $P_g$ , defined as:

P_g = P - P_{\text{atm}}

then the constant on the right-hand side changes by $P_{\text{atm}}$ . However, when you write Bernoulli between two points 1 and 2, you compare:

P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2

If you substitute $P_1 = P_{g1} + P_{\text{atm}}$ and $P_2 = P_{g2} + P_{\text{atm}}$ , the $P_{\text{atm}}$ terms cancel on both sides. You get:

P_{g1} + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_{g2} + \frac{1}{2} \rho v_2^2 + \rho g h_2

So the equation holds identically with gauge pressures.

Step-by-step reasoning

Bernoulli’s equation is a difference equation

In any application, you always compare two points along a streamline. The constant on the right is never needed explicitly — only the equality of the sum at two locations matters. This is why any uniform shift in pressure (like subtracting $P_{\text{atm}}$ from every term) cancels out.
Gauge pressure is just a shifted scale

Gauge pressure = absolute pressure − atmospheric pressure. Since atmospheric pressure is the same at both points (assuming the fluid is open to the same atmosphere or the height difference is small enough that $P_{\text{atm}}$ doesn’t vary), the shift is identical for both sides.
Check a typical exam scenario

Suppose water flows from a tank open to the atmosphere (point 1 at the free surface) to a pipe exit also open to the atmosphere (point 2).
- With absolute pressures: $P_1 = P_{\text{atm}}$ , $P_2 = P_{\text{atm}}$ .
- With gauge pressures: $P_{g1} = 0$ , $P_{g2} = 0$ . Both give the same velocity result from Bernoulli. The atmospheric terms simply drop out. …

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Q.Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain.

Why this question matters

The core idea

Step-by-step reasoning

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