Fluid Mechanics · Chapter 5
Fluid Mechanics · Chapter 5

Mass, Bernoulli & Energy Equations

Three workhorse equations carry most of applied fluid mechanics. Conservation of mass balances what flows in and out; the Bernoulli equation trades pressure, kinetic, and potential energy along a streamline; and the energy equation, with its head-loss and pump/turbine terms, governs real piping systems. This chapter develops each, and shows exactly when Bernoulli is — and isn't — allowed.

Learning objectives

  • Choose a control volume and close steady or transient mass balances.
  • Distinguish Bernoulli, mechanical-energy, and full energy models by their assumptions.
  • Use gauge or absolute pressure consistently with a declared reference.
  • Compute flow rate, stagnation pressure, pump head, and shaft power.
  • Interpret hydraulic and energy grade lines and machine sign conventions.
  • Verify mass and energy closure, units, regime limits, and practical margins.
Section 5–1

Introduction: Conservation Principles

Start by drawing a system or control volume, ports, elevations, and outward normals. Mass conservation is universal. Bernoulli is a restricted mechanical-energy statement; the general energy equation admits heat, shaft work, internal energy, and loss. Method selection must precede substitution.

Problem-solving strategy

  1. Identify the target and draw a boundary that crosses each useful inlet and outlet once.
  2. Declare \(z\) positive upward, flow directions, pressure reference, and property temperature.
  3. Close mass first, then choose Bernoulli only if its restrictions survive an explicit audit.
  4. Use the energy equation when a machine, loss, heat transfer, or thermodynamic state change matters.
  5. Solve symbolically, then check units, signs, mass closure, energy closure, and regime validity.
Section 5–2

Conservation of Mass

Mass Flow Rate

The amount of mass crossing a section per unit time. For a uniform density and a velocity normal to the area, it is the product of density, average velocity, and cross-sectional area.

Physical meaning

Mass cannot disappear inside a junction. If density is constant, volume flow also balances; for a compressible gas, equal mass rates may correspond to different volume rates and average velocities.

Mass and Volume Flow Rate
\[ \dot{m} = \rho V_{\text{avg}} A_c = \rho \dot{V}, \qquad \dot{V} = V_{\text{avg}} A_c \]
Conservation of Mass (Continuity)

For a control volume, the net rate of mass flow in equals the rate of accumulation inside. For steady flow, what goes in must come out.

\[ \sum_{\text{in}}\dot{m} - \sum_{\text{out}}\dot{m} = \frac{dm_{CV}}{dt}; \qquad \text{(steady)}\ \ \sum_{\text{in}}\dot{m} = \sum_{\text{out}}\dot{m} \]
Steady Incompressible Flow (single stream)
\[ V_1 A_1 = V_2 A_2 \]
EXAMPLE 1Flow Split in a Cooling Manifold
Water enters a steady manifold through a \(50.0\ \text{mm}\) pipe at \(2.80\ \text{m/s}\). One branch carries \(2.10\ \text{L/s}\); find the average velocity in a \(32.0\ \text{mm}\) second branch.

Recognition strategy

A single inlet and two outlets with no storage identifies a steady continuity problem; pressure and energy equations are unnecessary.

Given information

  • Diameters are inside diameters; \(+x\) follows each outlet, and all reported outlet velocities are positive.
  • Water density is treated constant at the common but unspecified temperature. Pressure is unused, so gauge versus absolute reference is immaterial.

Model and interpretation

A fixed control volume encloses the junction. Steady incompressible one-dimensional section averages make \(Q_{\rm in}=Q_1+Q_2\); this is approximate insofar as measured profiles are replaced by averages.

Step-by-step solution
1. Find inlet area. \(A_{\rm in}=\pi(0.0500)^2/4=1.963\times10^{-3}\ \text{m}^2\).
2. Compute inlet flow. \(Q_{\rm in}=V_{\rm in}A_{\rm in}=(2.80)(1.963\times10^{-3})=5.498\times10^{-3}\ \text{m}^3/\text{s}\).
3. Close the volume balance. Converting \(2.10\ \text{L/s}=2.10\times10^{-3}\ \text{m}^3/\text{s}\) gives \(Q_2=3.398\times10^{-3}\ \text{m}^3/\text{s}\).
4. Find branch velocity. With \(A_2=\pi(0.0320)^2/4=8.042\times10^{-4}\ \text{m}^2\), \(V_2=Q_2/A_2=4.23\ \text{m/s}\).

Verification

Units: Area times velocity gives \(\text{m}^3/\text{s}\), and volume rate divided by area returns \(\text{m/s}\).

Signs and direction: The remaining branch flow is positive because inlet flow exceeds the specified first-branch outflow; it points outward as assumed.

Plausibility: \(2.10+3.398=5.498\ \text{L/s}\) closes volume and mass exactly under constant density, while the smaller branch area reasonably produces a higher speed.

Final result: Second-branch average velocity is \(V_2=4.23\ \text{m/s}\) outward.

Engineering interpretation

The continuity result fixes flow allocation but not branch pressures. A design review would still assess losses, Reynolds regime, balancing valves, and uncertainty in actual internal diameters and flow measurements.

Section 5–3

Mechanical Energy & Efficiency

Mechanical Energy

The portion of energy that can be converted to mechanical work completely and directly by an ideal device. Per unit mass it is the sum of flow (pressure) energy, kinetic energy, and potential energy.

Mechanical Energy (per unit mass)
\[ e_{\text{mech}} = \frac{P}{\rho} + \frac{V^2}{2} + gz \]

Pump & Turbine Efficiency

A pump adds mechanical energy to the fluid; a turbine extracts it. Real devices fall short of perfect conversion because of friction, so efficiency is defined as the ratio of useful mechanical energy transfer to the shaft work involved.

Pump and Turbine Efficiency
\[ \eta_{\text{pump}} = \frac{\dot{E}_{\text{mech,fluid}}}{\dot{W}_{\text{shaft,in}}}, \qquad \eta_{\text{turbine}} = \frac{\dot{W}_{\text{shaft,out}}}{|\Delta \dot{E}_{\text{mech,fluid}}|} \]

Overall efficiency multiplies device efficiency by motor or generator efficiency. The mechanical energy that is "lost" reappears as a small rise in the fluid's internal energy (temperature).

Physical meaning

A pump head is positive energy delivered to the fluid; turbine head is positive energy extracted from it. Shaft input exceeds useful pump power, while turbine shaft output is less than the mechanical-energy decrease of the fluid.

Section 5–4

The Bernoulli Equation

The Bernoulli Equation

Along a streamline, in regions where viscous effects are negligible, the sum of pressure, kinetic, and potential energy per unit mass is constant.

\[ \frac{P}{\rho} + \frac{V^2}{2} + gz = \text{constant} \]

Restrictions (when Bernoulli applies)

The equation holds only for flow that is steady, effectively frictionless (negligible viscous effects), incompressible (or low Mach number), with no shaft work (no pump, turbine, or fan in the section) and no significant heat transfer, applied along a streamline. When the flow is also irrotational, the constant is the same on every streamline.

EXAMPLE 2Air Speed from a Low-Speed Pitot Probe
A duct Pitot probe reads stagnation pressure \(156\ \text{Pa}\) above the local static pressure. Using supplied air density \(1.184\ \text{kg/m}^3\) at \(25^\circ\text{C}\), estimate speed and audit incompressibility with supplied sound speed \(346\ \text{m/s}\).

Recognition strategy

Static-to-stagnation pressure difference with no elevation change is a Bernoulli conversion from kinetic energy to pressure, followed by a Mach-number check.

Given information

  • \(\Delta P=156\ \text{Pa}\) is a gauge difference; density and sound speed are explicitly supplied at \(25^\circ\text{C}\).
  • Positive speed follows the duct flow; static and stagnation taps are at effectively equal elevation.

Model and interpretation

Apply steady inviscid incompressible Bernoulli along the probe streamline, with stagnation speed zero. The result is approximate because probe and viscous corrections are neglected; the Mach check tests compressibility.

Step-by-step solution
1. Reduce Bernoulli. Equal elevation and zero stagnation-point speed give \(P_0-P=\rho V^2/2\).
2. Solve symbolically. The positive speed magnitude is \(V=\sqrt{2\Delta P/\rho}\); the negative root would contradict the declared flow direction.
3. Substitute supplied data. \(V=\sqrt{2(156\ \text{Pa})/(1.184\ \text{kg/m}^3)}=16.23\ \text{m/s}\).
4. Check the regime. \(M=V/a=16.23/346=0.0469\), well below the usual low-Mach threshold of \(0.3\).

Verification

Units: \(\text{Pa}/(\text{kg/m}^3)=\text{m}^2/\text{s}^2\); the square root therefore yields \(\text{m/s}\).

Signs and direction: Stagnation pressure exceeds static pressure, so \(\Delta P>0\) and the positive root agrees with the declared downstream direction.

Plausibility: Substitution back gives \(\rho V^2/2=156\ \text{Pa}\); \(M=0.0469\) supports negligible density change, though Reynolds number cannot be checked without length and viscosity.

Final result: \(V\approx16.2\ \text{m/s}\) and \(M\approx0.0469\), using the supplied \(25^\circ\text{C}\) properties.

Engineering interpretation

The pressure signal is consistent with low-speed incompressible operation. Calibration, probe alignment, turbulence, blockage, and sensor uncertainty can dominate the ideal-model error and should be included in reported airspeed uncertainty.

Section 5–5

Hydraulic & Energy Grade Lines (HGL & EGL)

Heads

Dividing the Bernoulli equation by \(g\) expresses each term as a height, or "head": the pressure head \(P/\rho g\), the velocity head \(V^2/2g\), and the elevation head \(z\). Their sum is the total head \(H\).

Physical meaning

The EGL is mechanical energy per unit weight; the HGL omits velocity head. Loss lowers the EGL in the flow direction, a pump raises it, and a turbine lowers it abruptly. Local HGL below atmospheric datum means negative gauge pressure, not necessarily cavitation.

Bernoulli in Head Form
\[ \frac{P}{\rho g} + \frac{V^2}{2g} + z = H = \text{constant} \]

HGL and EGL

The hydraulic grade line plots \(P/\rho g + z\) (the level liquid reaches in a piezometer). The energy grade line plots the total head \(P/\rho g + V^2/2g + z\) (the level in a Pitot tube). The vertical gap between them is the velocity head \(V^2/2g\). For the ideal Bernoulli flow the EGL is horizontal; friction makes it slope downward.

The full balance is exact when all relevant energy modes and boundary transfers are represented. Head-loss forms are reduced mechanical-energy models; coefficients, friction factors, roughness, and machine curves must come from stated correlations, measurements, or vendor data rather than invention.

EGL (total head) HGL V²/2g flow → (friction tilts both lines down)
Figure 5–1. The EGL traces the total head and the HGL traces pressure-plus-elevation head; the gap between them is the velocity head, and friction lowers both along the pipe. (Original schematic.)
Section 5–6

The General Energy Equation

Conservation of Energy

The rate of energy added to a control volume by heat and work equals the rate of accumulation inside plus the net rate at which energy leaves with the flowing mass. It follows from the RTT with \(B = E\).

\[ \dot{Q}_{\text{net,in}} + \dot{W}_{\text{shaft,net,in}} = \frac{d}{dt}\!\int_{CV}\! e\rho\,dV + \int_{CS}\!\left(\frac{P}{\rho}+e\right)\!\rho(\vec{V}\cdot\vec{n})\,dA \]

Head Loss

Real flows convert some mechanical energy irreversibly into internal energy through friction. Expressed as a head, this is the head loss \(h_L\) — distinct from the inefficiencies of pumps and turbines, which are accounted for separately.

Section 5–7

Energy Analysis of Steady Flows

For a single-inlet, single-outlet steady flow, the energy equation takes its most useful engineering form — a head balance with explicit pump, turbine, and loss terms.

Steady-Flow Energy Equation (head form)
\[ \frac{P_1}{\rho g} + \alpha_1\frac{V_1^2}{2g} + z_1 + h_{\text{pump},u} = \frac{P_2}{\rho g} + \alpha_2\frac{V_2^2}{2g} + z_2 + h_{\text{turbine},e} + h_L \]

Here \(h_{\text{pump},u}\) is the useful head delivered by a pump, \(h_{\text{turbine},e}\) the head extracted by a turbine, \(h_L\) the irreversible head loss in the piping, and \(\alpha\) the kinetic-energy correction factor. When there is no machine and no loss, this reduces exactly to the Bernoulli head equation.

EXAMPLE 3Transfer Pump Duty and Cavitation Screen
Water at \(20^\circ\text{C}\) is pumped at \(0.0180\ \text{m}^3/\text{s}\) between open reservoirs \(14.0\ \text{m}\) apart in elevation. Supplied total loss is \(6.20\ \text{m}\), pump efficiency is \(78.0\%\), suction absolute-pressure head is \(8.90\ \text{m}\), vapor-pressure head is \(0.238\ \text{m}\), and suction velocity head is \(0.55\ \text{m}\). Find shaft power and available NPSH.

Recognition strategy

A pump and supplied losses prohibit ideal Bernoulli. Use the steady mechanical-energy head equation, then separately compare absolute suction head with vapor-pressure head.

Given information

  • Both reservoir surfaces are at atmospheric pressure, so their gauge pressures are zero; suction pressure data are absolute.
  • Supplied water density is \(998\ \text{kg/m}^3\) at \(20^\circ\text{C}\), \(g=9.81\ \text{m/s}^2\), \(+z\) upward, and pump head is positive into the fluid.

Model and interpretation

Use steady incompressible flow, negligible reservoir velocities, no turbine, and the supplied aggregate loss. The calculation is approximate because losses and efficiency are operating-point data, not derived curves.

Step-by-step solution
1. Reduce the head balance. Surface pressures cancel and surface velocities are negligible, so \(h_p=\Delta z+h_L=14.0+6.20=20.20\ \text{m}\).
2. Find useful fluid power. \(\dot W_{\rm fluid}=\rho gQh_p=(998)(9.81)(0.0180)(20.20)=3.56\ \text{kW}\).
3. Apply pump efficiency. With efficiency defined useful fluid power over shaft input, \(\dot W_{\rm shaft,in}=3.56/0.780=4.57\ \text{kW}\).
4. Screen cavitation. Using supplied absolute heads at the suction datum, \(\mathrm{NPSH}_A=8.90+0.55-0.238=9.21\ \text{m}\).

Verification

Units: \((\text{kg/m}^3)(\text{m/s}^2)(\text{m}^3/\text{s})(\text{m})=\text{W}\); all NPSH terms are meters of liquid.

Signs and direction: Elevation and loss both demand positive pump head, and shaft input exceeds useful fluid power because efficiency is below unity.

Plausibility: Multiplying \(4.57\ \text{kW}\) by \(0.780\) recovers \(3.56\ \text{kW}\). Positive NPSH available is necessary but adequacy cannot be claimed without vendor \(\mathrm{NPSH}_R\) and margin criteria.

Final result: Required shaft input is approximately \(4.57\ \text{kW}\); \(\mathrm{NPSH}_A\approx9.21\ \text{m}\).

Engineering interpretation

This defines one duty point, not a pump selection. Engineers must intersect system and vendor pump curves, verify efficiency and motor margin, compare available and required NPSH, and propagate level, loss, and property uncertainty.

Chapter recap

  • Mass conservation closes every control volume before energy analysis begins.
  • Bernoulli is restricted to steady, incompressible, effectively inviscid flow without machine work along a streamline.
  • The mechanical-energy equation tracks pressure, velocity, elevation, machines, and irreversible loss.
  • Pressure references, property basis, correction factors, and machine signs must be explicit.

Before moving on: you should be able to select the proper balance, solve a multiport continuity problem, interpret HGL/EGL, calculate pump power, and verify mass, energy, Mach regime, and cavitation margin requirements.