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Fluid Mechanics and Dynamics

8% of exam

Fluid properties and dimensionless numbers (Reynolds), the mechanical energy balance with pipe, valve, fitting, and packed-bed losses, the Bernoulli equation and hydrostatics, laminar and turbulent flow, flow measurement (orifices and Venturi meters), pumps, compressors, and vacuum systems, and compressible and non-Newtonian flow.

5 concepts

A. Fluid properties

Fluid Properties and Dimensionless Numbers

Define density, viscosity and surface tension, separate Newtonian from non-Newtonian behavior, and use the Reynolds and Froude numbers to classify flow.

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C. Mechanical energy balance

The Mechanical Energy Balance and Bernoulli

Write extended Bernoulli as heads, add pipe, valve and fitting friction losses, account for elevation, and size pump head or pressure drop with confidence.

Thermodynamics

Material and Energy Balances

Heat Transfer

Mass Transfer and Separation

Solids Handling

Chemical Reaction Engineering

Engineering Economics

Process Design

Process Control

Safety, Health, and Environment

Ethics and Professional Practice

Almost every flow problem on the FE Chemical exam is one statement applied between two points: the energy carried by a unit weight of fluid at point 1, plus any energy a pump adds, equals the energy at point 2 plus whatever friction destroyed along the way. Get fluent with that single bookkeeping line and you can size a pump, predict a pressure drop, find a flow rate, or read a hydraulic grade line without memorizing a dozen special cases. The relations here follow the FE Reference Handbook — Fluid Mechanics (Energy Equation, Bernoulli, Darcy-Weisbach, and Minor Losses), and the points lost are almost always bookkeeping: mixing pressure with head, dropping the velocity term, or forgetting that friction is strictly positive and always subtracts.

Bernoulli is energy per unit weight

Start with the frictionless, no-machine case. Each term of the Bernoulli equation has units of length — it is energy per unit weight of fluid, called head. Pressure head

P/\gamma

, velocity head

v^2/2g

, and elevation head

z

trade off against one another so their sum stays constant along a streamline. Where the pipe narrows,

v

rises, velocity head climbs, and pressure head must fall — the same exchange that lets a Venturi measure flow. Keep

\gamma=\rho g

straight: divide a pressure by specific weight to get head, multiply a head by specific weight to get pressure.

\frac{P_1}{\gamma}+\frac{v_1^{2}}{2g}+z_1=\frac{P_2}{\gamma}+\frac{v_2^{2}}{2g}+z_2

The extended (mechanical energy) balance

Real pipes have friction and real systems have pumps, so add a pump head

h_p

on the upstream side (energy in) and a total head loss

h_f

on the downstream side (energy out). This is the workhorse — the extended Bernoulli or mechanical energy balance. Solve it for whichever term is unknown: for pump sizing isolate

h_p

; for pressure drop in a straight run set

h_p=0

,

z_1=z_2

,

v_1=v_2

and you recover

P_1-P_2=\gamma h_f

. The head loss term is never negative — friction always removes mechanical energy.

\frac{P_1}{\gamma}+\frac{v_1^{2}}{2g}+z_1+h_p=\frac{P_2}{\gamma}+\frac{v_2^{2}}{2g}+z_2+h_f

Friction in straight pipe: Darcy-Weisbach

The major (skin-friction) loss in a length of straight pipe is the Darcy-Weisbach equation. The friction factor

f

is read from the Moody chart as a function of Reynolds number and relative roughness

\varepsilon/D

. Watch the convention war: chemical engineers often use the Fanning friction factor

f_{F}

, which is exactly one-quarter of the Darcy factor, so the Fanning form carries a factor of four. A loss computed with the wrong factor is off by 4x — always confirm which

f

a chart or problem gives you.

h_f=f\,\frac{L}{D}\,\frac{v^{2}}{2g}\qquad f_{\text{Darcy}}=4\,f_{\text{Fanning}}

Minor losses from valves and fittings

Every elbow, tee, valve, contraction, expansion, entrance and exit dissipates energy through flow separation. Each is charged a loss proportional to the velocity head through a dimensionless loss coefficient

C

(often written

K

). Sum the coefficients for all fittings on a run and add this to the straight-pipe loss to get the total

h_f

. A sharp pipe entrance from a reservoir costs about

0.5

velocity heads and an exit into a tank costs a full

1.0

; an open gate valve is small, a globe valve is large. Despite the name, in a fitting-heavy system the minor losses can dominate.

h_{f,\text{minor}}=\sum C\,\frac{v^{2}}{2g},\qquad h_f=f\frac{L}{D}\frac{v^{2}}{2g}+\sum C\,\frac{v^{2}}{2g}

Hydrostatics and the grade lines

When velocity is zero the balance collapses to hydrostatics: pressure rises linearly with depth,

P_2-P_1=-\gamma(z_2-z_1)=\gamma h

, which is the basis of manometers and submerged-surface forces. Two visual aids help in flowing systems. The hydraulic grade line (HGL) plots

P/\gamma+z

, the height a piezometer would reach; the energy line (EL) plots the full

P/\gamma+v^2/2g+z

. The EL sits a velocity head above the HGL and slopes downward in the flow direction by exactly

h_f

. A pump injects a vertical jump

h_p

into both lines.

P=P_{\text{atm}}+\rho g h\qquad(\text{absolute}),\qquad P_{\text{gauge}}=\rho g h

From head to power

Once you have the required pump head

h_p

, the rate of useful energy delivered to the fluid — the hydraulic (fluid) power — is weight flow times head,

\dot{W}_{\text{fluid}}=\rho g Q h_p=\gamma Q h_p

. Divide by pump efficiency to get brake power at the shaft, and again by motor efficiency for the electrical draw. This is the bridge from the energy balance to a motor nameplate, and it is a near-certain exam computation. Keep

Q

in

\mathrm{m^3/s}

and

h_p

in metres and the answer falls out in watts.

\dot{W}_{\text{fluid}}=\rho g Q h_p,\qquad \dot{W}_{\text{brake}}=\frac{\dot{W}_{\text{fluid}}}{\eta_{\text{pump}}}

Exam strategy

Draw the system, label points 1 and 2 at surfaces or gauges where you know the most (an open reservoir surface has

P=0

gauge and

v\approx0

), and write the extended balance once. Decide your basis up front — work entirely in heads (metres or feet) or entirely in pressures, never half and half. Compute the velocity head once and reuse it for every loss term, since all of

h_f

shares the same

v^2/2g

. Confirm

f

is Darcy (Moody) versus Fanning before substituting, and finish by checking that head loss came out positive and that your pump adds energy rather than removing it.

Key equations

Bernoulli equation (no friction, no machine)

\frac{P_1}{\gamma}+\frac{v_1^{2}}{2g}+z_1=\frac{P_2}{\gamma}+\frac{v_2^{2}}{2g}+z_2

Conservation of mechanical energy per unit weight along a streamline; each term in metres (or feet). $\gamma=\rho g$ .

Extended (mechanical energy) balance

\frac{P_1}{\gamma}+\frac{v_1^{2}}{2g}+z_1+h_p=\frac{P_2}{\gamma}+\frac{v_2^{2}}{2g}+z_2+h_f

Continuity

\dot{m}=\rho Q=\rho A v,\qquad Q=A_1 v_1=A_2 v_2

Darcy-Weisbach major loss

h_f=f\,\frac{L}{D}\,\frac{v^{2}}{2g}

Fanning–Darcy relation

f_{\text{Darcy}}=4\,f_{\text{Fanning}}

Chemical-engineering charts often give Fanning $f_F$

Minor (fitting) losses

h_{f,\text{minor}}=\sum C\,\frac{v^{2}}{2g}

Hydrostatic pressure

P_2-P_1=-\gamma(z_2-z_1)=\rho g h

Pressure drop in a straight horizontal pipe

P_1-P_2=\gamma\,h_f=f\,\frac{L}{D}\,\frac{\rho v^{2}}{2}

Hydraulic (fluid) power

\dot{W}_{\text{fluid}}=\rho g Q h_p=\gamma Q h_p

Brake and electrical power

\dot{W}_{\text{brake}}=\frac{\dot{W}_{\text{fluid}}}{\eta_{\text{pump}}},\qquad \dot{W}_{\text{elec}}=\frac{\dot{W}_{\text{brake}}}{\eta_{\text{motor}}}

Torricelli (frictionless efflux)

v=\sqrt{2gh}

Free jet velocity from a head $h$ when friction is negligible; the friction case puts $1+f\tfrac{L}{D}+\sum C$

Worked examples

Sizing a transfer pump

Problem. Water ( $\rho=1000\,\text{kg/m}^3$ ) is pumped at $Q=0.0200\,\text{m}^3/\text{s}$ through a $D=0.100\,\text{m}$ , $L=150\,\text{m}$ steel pipe from one open tank to another $25.0\,\text{m}$ higher. The Darcy friction factor is $f=0.0200$ and the fittings sum to $\sum C=5.0$ . Find the required pump head and the hydraulic power.

Solution. Velocity:

v=Q/A=0.0200/(\tfrac{\pi}{4}(0.100)^2)=0.0200/0.007854=2.546\,\text{m/s}

h_p=\Delta z+\left(f\frac{L}{D}+\sum C\right)\frac{v^{2}}{2g}=25.0+(30+5.0)(0.3305)=36.6\,\text{m}

Pressure drop in a horizontal run

Problem. Water at $\rho=998\,\text{kg/m}^3$ flows at $v=2.50\,\text{m/s}$ through $L=80.0\,\text{m}$

Gravity discharge from a reservoir with friction

Problem. Water drains from a large open reservoir through a pipe whose outlet is $H=12.0\,\text{m}$ below the surface, discharging freely to atmosphere. The pipe is $L=50.0\,\text{m}$ , $D=0.0500\,\text{m}$ , $f=0.0200$

Common pitfalls

•Mixing head and pressure in the same equation. Every term must be in metres (or feet) OR every term in pascals — convert with $\gamma=\rho g$ before adding, never midway.
•Confusing Fanning and Darcy friction factors. $f_{\text{Darcy}}=4f_{\text{Fanning}}$ ; using a Fanning value in the Darcy formula underestimates $h_f$ by a factor of four.
•Dropping the velocity-head term when areas differ. In a contraction or at a free jet, $v_1\neq v_2$ and $\tfrac{v^2}{2g}$
•Putting head loss on the wrong side or giving it a sign. $h_f$ is always positive and always belongs with the downstream (point-2) terms; friction removes mechanical energy, it cannot add it.
•Confusing absolute and gauge pressure in hydrostatics and NPSH-type problems. Open surfaces are zero gauge; switch to absolute only when a vapor-pressure or vacuum reference demands it.
•Forgetting entrance and exit losses. A reservoir entrance costs $\sim 0.5\,\tfrac{v^2}{2g}$ and a submerged exit a full $1.0$ ; omitting them undersizes the pump.
•Using gauge pressure where the formula wants $\gamma=\rho g$ in inconsistent units (e.g., mixing $\rho$ in $\text{kg/m}^3$ with $h$ in feet) — keep one unit system end to end.

References

NCEES FE Reference Handbook — Fluid Mechanics
NCEES FE Reference Handbook — Fluid Mechanics (Energy Equation, Bernoulli, Darcy-Weisbach, Minor Losses)
Felder & Rousseau, Elementary Principles of Chemical Processes — fluid transport and mechanical energy balance

E. Laminar and turbulent flow

Laminar/Turbulent Flow, Friction Factor and Packed Beds

Tell laminar from turbulent flow by Reynolds number, get the friction factor from f=64/Re or the Moody chart, apply Hagen-Poiseuille, and size pressure drop through a packed bed with Ergun.

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G. Pumps, turbines, compressors, and vacuum systems

Pumps, NPSH, Compressors and Flow Measurement

Turn pump head into hydraulic and brake power, guard against cavitation with an NPSH margin, find the operating point on a system curve, and read flow from an orifice or Venturi.

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H. Compressible flow and non-Newtonian fluids

Compressible Flow and Non-Newtonian Fluids

Use the Mach number and isentropic relations for compressible gas flow, recognize choked and adiabatic vs isothermal pipe flow, and apply power-law and Bingham rheology.

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Adds pump head $h_p$ (energy in) and total head loss $h_f$ (energy out, always positive). The form used for pump and pressure-drop sizing.

Mass and (for constant density) volumetric flow conserved; $A=\tfrac{\pi}{4}D^2$ for round pipe. Units $\mathrm{m^3/s}$ , $\mathrm{m/s}$ .

Straight-pipe friction head; $f$ from Moody chart vs $\mathrm{Re}$ and $\varepsilon/D$ . $L$ , $D$ in m, $v$ in m/s.

; the Darcy form needs the factor of four (or use

h_f=4f_F\tfrac{L}{D}\tfrac{v^2}{2g}

).

$C$ (or $K$ ) is the dimensionless loss coefficient per fitting/valve/entrance/exit; sum over the run.

Static-fluid pressure variation with depth $h$ ; basis of manometers. Use absolute or gauge consistently.

Special case with $z_1=z_2$ , $v_1=v_2$ , no pump; converts head loss directly to a pressure drop in Pa.

Useful power delivered to fluid by a pump of head $h_p$ ; in watts when $Q$ in $\mathrm{m^3/s}$ , $h_p$ in m.

Shaft power and electrical draw; efficiencies are fractions $\le 1$ .

in the denominator inside the root (the leading

1

is the exit velocity head).

. Velocity head:

v^2/2g=(2.546)^2/(2\cdot 9.81)=0.3305\,\text{m}

. Major loss:

h_{f,\text{major}}=f\tfrac{L}{D}\tfrac{v^2}{2g}=0.0200\cdot\tfrac{150}{0.100}\cdot 0.3305=30\cdot 0.3305=9.92\,\text{m}

. Minor loss:

\sum C\,\tfrac{v^2}{2g}=5.0\cdot 0.3305=1.65\,\text{m}

. Both tanks are open (

P_1=P_2=0

gauge) and the surfaces are nearly still (

v_1=v_2\approx 0

), so the balance gives

h_p=\Delta z+h_f=25.0+9.92+1.65=36.6\,\text{m}

. Hydraulic power:

\dot{W}=\rho g Q h_p=1000\cdot 9.81\cdot 0.0200\cdot 36.57=7.18\times 10^3\,\text{W}=\mathbf{7.18\,\text{kW}}

, with pump head

\mathbf{36.6\,\text{m}}

. Sanity check: of the

36.6\,\text{m}

, the static lift

25\,\text{m}

dominates and friction adds

\sim 11.6\,\text{m}

— reasonable for a long, fitting-rich line. Units:

(\text{kg/m}^3)(\text{m/s}^2)(\text{m}^3/\text{s})(\text{m})=\text{kg·m}^2/\text{s}^3=\text{W}

. Good.

of

D=0.0500\,\text{m}

pipe with

f=0.0250

(Darcy). The pipe is horizontal and of constant diameter. Find the pressure drop.

Solution. With

z_1=z_2

and

v_1=v_2

, the balance reduces to

P_1-P_2=\gamma h_f=f\tfrac{L}{D}\tfrac{\rho v^2}{2}

.

\dfrac{\rho v^2}{2}=\dfrac{998\,(2.50)^2}{2}=3119\,\text{Pa}

.

f\tfrac{L}{D}=0.0250\cdot\tfrac{80.0}{0.0500}=0.0250\cdot 1600=40.0

.

\Delta P=40.0\cdot 3119=1.25\times 10^5\,\text{Pa}=\mathbf{125\,\text{kPa}}

(equivalently

h_f=\Delta P/\gamma=12.7\,\text{m}

). Sanity check:

40

velocity-pressure units across

1600

diameters of pipe;

125\,\text{kPa}\approx 1.2\,\text{atm}

over

80\,\text{m}

is a stiff but realistic gradient for

2.5\,\text{m/s}

in a narrow line. Units:

\text{Pa}

throughout.

\Delta P=f\frac{L}{D}\frac{\rho v^{2}}{2}=40.0\,(3119\,\text{Pa})=125\,\text{kPa}

(Darcy), with a sharp entrance (

C=0.50

). Find the discharge velocity and volumetric flow.

Solution. Take point 1 at the still reservoir surface (

v_1\approx 0

,

P_1=0

gauge) and point 2 at the free jet (

P_2=0

gauge). The available head

H

supplies the exit velocity head plus all friction:

H=\dfrac{v^2}{2g}\left(1+f\tfrac{L}{D}+\sum C\right)

. Denominator:

1+0.0200\cdot\tfrac{50.0}{0.0500}+0.50=1+20.0+0.50=21.5

.

v=\sqrt{\dfrac{2gH}{21.5}}=\sqrt{\dfrac{2\cdot 9.81\cdot 12.0}{21.5}}=\sqrt{10.95}=3.31\,\text{m/s}

.

Q=vA=3.31\cdot\tfrac{\pi}{4}(0.0500)^2=3.31\cdot 1.963\times10^{-3}=6.50\times10^{-3}\,\text{m}^3/\text{s}=\mathbf{6.50\,\text{L/s}}

, exit velocity

\mathbf{3.31\,\text{m/s}}

. Sanity check: frictionless Torricelli would give

v=\sqrt{2\cdot 9.81\cdot 12}=15.3\,\text{m/s}

; friction over

1000

diameters slashes it to

3.31\,\text{m/s}

, as expected for a long small-bore line.

v=\sqrt{\frac{2gH}{1+f\frac{L}{D}+\sum C}}=\sqrt{\frac{2(9.81)(12.0)}{21.5}}=3.31\,\text{m/s}

matters — only cancel it when the diameter truly is constant.