Mechanical Engineer's Handbook

Fluid Dynamics

Fluid dynamics studies fluids in motion. To simplify calculations, we examine an ideal fluid, which is non-viscous and incompressible (constant density ρ\rho).

Volume Flow Rate and Velocity

Volume flow rate (QQ) measures the volume of fluid passing a cross-sectional area per unit time.

Q=dVdt=ddt(Ax)=Adxdt=AvQ = \frac{dV}{dt} = \frac{d}{dt}(A x) = A \frac{dx}{dt} = A v
(1)

The SI unit for flow rate is m3/s\text{m}^3/\text{s}. A common conversion is 1 L/min=1.67105 m3/s1 \text{ L/min} = 1.67 \cdot 10^{-5} \text{ m}^3/\text{s}.

Continuity Equations

Mass conservation dictates that the mass of fluid entering a conduit must equal the mass exiting. Mass flow rate is given by dmdt=ρAv\frac{dm}{dt} = \rho A v. Equating the mass flow rate at two points yields the general continuity equation:

ρ1A1v1=ρ2A2v2\rho_1 A_1 v_1 = \rho_2 A_2 v_2
(2)

For an incompressible fluid, density remains constant (ρ1=ρ2\rho_1 = \rho_2), reducing the expression to the incompressible continuity equation:

A1v1=A2v2    Q1=Q2A_1 v_1 = A_2 v_2 \implies Q_1 = Q_2
(3)

Fluid speed varies inversely with cross-sectional area (v1Av \propto \frac{1}{A}). For a circular pipe of radius rr, speed varies inversely with the square of the radius (v1r2v \propto \frac{1}{r^2}).

Example 1

Water flows through a hose of radius r1=0.900 cmr_1 = 0.900 \text{ cm} at a rate of Q=0.500 L/sQ = 0.500 \text{ L/s} and emerges through a nozzle of diameter d2=0.500 cmd_2 = 0.500 \text{ cm}. Calculate the water speed in the hose (v1v_1) and in the nozzle (v2v_2).

Convert QQ to m3/s\text{m}^3/\text{s}. Use Equation 1 for the hose and Equation 3 for the nozzle.

Converting flow rate yields Q=5.00104 m3/sQ = 5.00 \cdot 10^{-4} \text{ m}^3/\text{s}.

For the hose (r1=0.00900 mr_1 = 0.00900 \text{ m}):

v1=Qπr12=5.00104 m3/sπ(0.00900 m)21.96 m/sv_1 = \frac{Q}{\pi r_1^2} = \frac{5.00 \cdot 10^{-4} \text{ m}^3/\text{s}}{\pi (0.00900 \text{ m})^2} \approx 1.96 \text{ m/s}

For the nozzle (r2=0.00250 mr_2 = 0.00250 \text{ m}):

v2=(r1r2)2v1=(0.900 cm0.250 cm)2(1.96 m/s)25.4 m/sv_2 = \left(\frac{r_1}{r_2}\right)^2 v_1 = \left(\frac{0.900 \text{ cm}}{0.250 \text{ cm}}\right)^2 (1.96 \text{ m/s}) \approx 25.4 \text{ m/s}