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Fluid mechanics models

Connect conservation laws to hydrostatics, pipe flow, boundary forces and compressibility with twenty worked examples.

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Matter pathway: atom → solid → liquid → gas → plasma. Quantum mechanics and quantum field theory provide foundations across the pathway; they are not additional phases. This is a connected modeling route, not a universal heating curve. Actual phases depend on pressure, composition, and kinetics.

1. Mass and momentum

Definitions & inputs. ρ density, u velocity, p pressure, μ dynamic viscosity, g body acceleration.

  1. Local mass conservation balances storage and flux.

    ∂tρ+∇⋅(ρu)=0\partial_t\rho+\nabla\cdot(\rho\mathbf u)=0

  2. For incompressible flow, pressure, viscous and body forces set acceleration.

    ρDu/Dt=−∇p+μ∇2u+ρg\rho D\mathbf u/Dt=-\nabla p+\mu\nabla^2\mathbf u+\rho\mathbf g

  3. Constant density reduces continuity and gives the uniform-section volume flux.

    ∇⋅u=0,Q=Av\nabla\cdot\mathbf u=0,\quad Q=Av

Interpretation. Compressible flow retains density variation and generally needs an energy equation.

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2. Hydrostatics and energy

Definitions & inputs. z elevation, g gravity, V displaced volume, v speed.

  1. Static momentum balance relates pressure to elevation.

    dp/dz=−ρgdp/dz=-\rho g

  2. Integrating hydrostatic traction gives buoyancy.

    Fb=ρgVF_b=\rho gV

  3. Dot inviscid momentum with the streamline displacement and integrate.

    p+ρv2/2+ρgz=constantp+\rho v^2/2+\rho gz=\mathrm{constant}

Interpretation. Pumps, viscous losses and heat/compressibility effects require a generalized energy balance.

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3. Viscous pipe flow

Definitions & inputs. R radius,L pipe length,Δp pressure drop,D=2R,vmean average velocity.

  1. Axial momentum reduces to a radial differential equation.

    μr−1d(r du/dr)/dr=−Δp/L\mu r^{-1}d(r\,du/dr)/dr=-\Delta p/L

  2. Integrate using finite centreline gradient and zero wall velocity.

    u(r)=Δp(R2−r2)/(4μL)u(r)=\Delta p(R^2-r^2)/(4\mu L)

  3. Integrate velocity over the cross section and express the result as a Darcy friction factor.

    Q=πR4Δp/(8μL),fD=64/ReQ=\pi R^4\Delta p/(8\mu L),\quad f_D=64/Re

Interpretation. Do not use the laminar relation indiscriminately in transitional or turbulent flow.

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4. Similarity and force

Definitions & inputs. Re Reynolds number, Cd drag coefficient, A frontal area, a sound speed, γ specific heat ratio.

  1. Compare inertia with viscosity and flow speed with wave speed.

    Re=ρvD/μ,Ma=v/aRe=\rho vD/\mu,\quad Ma=v/a

  2. Define drag relative to dynamic pressure and reference area.

    FD=CDρv2A/2F_D=C_D\rho v^2A/2

  3. Linearize pressure-density response for an ideal gas under isentropic small disturbances.

    a=γRsTa=\sqrt{\gamma R_sT}

Interpretation. Similarity parameters guide model selection; drag coefficients are not universal constants.

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Graphical worked example

Steady fully developed laminar circular-pipe Poiseuille flow, no slip. X axis: Radius / pipe radius (dimensionless). Y axis: Axial velocity / mean velocity (dimensionless).
Steady fully developed laminar circular-pipe Poiseuille flow, no slip. Related worked calculation · Download SVG · Plot data

Twenty worked examples

Open a problem to see its defined inputs, assumptions, equation, numerical substitution, result, and interpretation. Values are illustrative analytical exercises.

Example 01. Hydrostatic gauge pressure

Definitions & inputs. Water ρ=1000 kg/m³,depth 2 m,g=9.81 m/s².

  1. Choose the governing model and isolate the requested quantity.

    p=ρghp=\rho gh

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    p=1000(9.81)(2)p=1000(9.81)(2)

  3. Evaluate the expression; the result uses the units shown.

    Result=19620 Pa\mathrm{Result}=19620\ {\rm Pa}

Interpretation. Absolute pressure additionally includes surface pressure.

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Example 02. Buoyant force

Definitions & inputs. Displaced water volume .003 m³,ρ=1000 kg/m³.

  1. Choose the governing model and isolate the requested quantity.

    Fb=ρgVF_b=\rho gV

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Fb=1000(9.81)(0.003)F_b=1000(9.81)(0.003)

  3. Evaluate the expression; the result uses the units shown.

    Result=29.43 N\mathrm{Result}=29.43\ {\rm N}

Interpretation. Compare this upward force with weight and other loads.

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Example 03. Mass flow rate

Definitions & inputs. ρ=1000 kg/m³,Q=.002 m³/s.

  1. Choose the governing model and isolate the requested quantity.

    m˙=ρQ\dot m=\rho Q

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    m˙=1000(0.002)\dot m=1000(0.002)

  3. Evaluate the expression; the result uses the units shown.

    Result=2 kg s−1\mathrm{Result}=2\ {\rm kg\,s}^{-1}

Interpretation. Density must match the section where Q is specified.

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Example 04. Pipe velocity

Definitions & inputs. Q=.001 m³/s,D=.02 m.

  1. Choose the governing model and isolate the requested quantity.

    v=4Q/(πD2)v=4Q/(\pi D^2)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    v=4(0.001)/(π0.022)v=4(0.001)/(\pi0.02^2)

  3. Evaluate the expression; the result uses the units shown.

    Result=3.183099 m s−1\mathrm{Result}=3.183099\ {\rm m\,s}^{-1}

Interpretation. This is area-averaged velocity.

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Example 05. Contraction velocity

Definitions & inputs. A1=.02 m²,v1=1 m/s,A2=.005 m².

  1. Choose the governing model and isolate the requested quantity.

    v2=A1v1/A2v_2=A_1v_1/A_2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    v2=0.02(1)/0.005v_2=0.02(1)/0.005

  3. Evaluate the expression; the result uses the units shown.

    Result=4 m s−1\mathrm{Result}=4\ {\rm m\,s}^{-1}

Interpretation. Constant density and no branches are assumed.

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Example 06. Dynamic pressure

Definitions & inputs. Air ρ=1.2 kg/m³,v=20 m/s.

  1. Choose the governing model and isolate the requested quantity.

    q=ρv2/2q=\rho v^2/2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    q=1.2(20)2/2q=1.2(20)^2/2

  3. Evaluate the expression; the result uses the units shown.

    Result=240 Pa\mathrm{Result}=240\ {\rm Pa}

Interpretation. This is not the static pressure.

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Example 07. Ideal efflux speed

Definitions & inputs. Liquid head h=2 m,large tank,negligible losses.

  1. Choose the governing model and isolate the requested quantity.

    v=2ghv=\sqrt{2gh}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    v=2(9.81)(2)v=\sqrt{2(9.81)(2)}

  3. Evaluate the expression; the result uses the units shown.

    Result=6.264184 m s−1\mathrm{Result}=6.264184\ {\rm m\,s}^{-1}

Interpretation. A discharge coefficient is required for a real opening.

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Example 08. Horizontal Bernoulli pressure drop

Definitions & inputs. Water,v1=1 m/s,v2=3 m/s,no loss.

  1. Choose the governing model and isolate the requested quantity.

    p1−p2=ρ(v22−v12)/2p_1-p_2=\rho(v_2^2-v_1^2)/2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Δp=1000(9−1)/2\Delta p=1000(9-1)/2

  3. Evaluate the expression; the result uses the units shown.

    Result=4000 Pa\mathrm{Result}=4000\ {\rm Pa}

Interpretation. Static pressure falls where the flow speeds up.

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Example 09. Reynolds number

Definitions & inputs. ρ=1000 kg/m³,v=.1 m/s,D=.01 m,μ=.001 Pa s.

  1. Choose the governing model and isolate the requested quantity.

    Re=ρvD/μRe=\rho vD/\mu

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Re=1000(0.1)(0.01)/0.001Re=1000(0.1)(0.01)/0.001

  3. Evaluate the expression; the result uses the units shown.

    Result=1000 \mathrm{Result}=1000\ {}

Interpretation. This is within the usual laminar circular-pipe regime for smooth disturbances.

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Example 10. Kinematic viscosity

Definitions & inputs. μ=.001 Pa s,ρ=1000 kg/m³.

  1. Choose the governing model and isolate the requested quantity.

    ν=μ/ρ\nu=\mu/\rho

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ν=0.001/1000\nu=0.001/1000

  3. Evaluate the expression; the result uses the units shown.

    Result=1×10−6 m2s−1\mathrm{Result}=1\times10^{-6}\ {\rm m}^2{\rm s}^{-1}

Interpretation. Kinematic viscosity controls viscous momentum diffusion.

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Example 11. Laminar Darcy factor

Definitions & inputs. Re=1000.

  1. Choose the governing model and isolate the requested quantity.

    fD=64/Ref_D=64/Re

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    fD=64/1000f_D=64/1000

  3. Evaluate the expression; the result uses the units shown.

    Result=0.064 \mathrm{Result}=0.064\ {}

Interpretation. The Fanning factor would be four times smaller.

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Example 12. Poiseuille flow

Definitions & inputs. R=.001 m,L=1 m,μ=.001 Pa s,Δp=100 Pa.

  1. Choose the governing model and isolate the requested quantity.

    Q=πR4Δp/(8μL)Q=\pi R^4\Delta p/(8\mu L)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Q=π(10−3)4(100)/(8⋅10−3)Q=\pi(10^{-3})^4(100)/(8\cdot10^{-3})

  3. Evaluate the expression; the result uses the units shown.

    Result=3.926991×10−8 m3s−1\mathrm{Result}=3.926991\times10^{-8}\ {\rm m}^3{\rm s}^{-1}

Interpretation. The implied Reynolds number is small, consistent with laminar flow.

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Example 13. Laminar peak velocity

Definitions & inputs. Mean velocity .2 m/s,fully developed circular pipe.

  1. Choose the governing model and isolate the requested quantity.

    umax=2vmeanu_{max}=2v_{mean}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    umax=2(0.2)u_{max}=2(0.2)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.4 m s−1\mathrm{Result}=0.4\ {\rm m\,s}^{-1}

Interpretation. The profile is parabolic under these assumptions.

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Example 14. Pipe wall shear

Definitions & inputs. Δp=100 Pa,D=.01 m,L=2 m.

  1. Choose the governing model and isolate the requested quantity.

    τw=ΔpD/(4L)\tau_w=\Delta pD/(4L)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    τw=100(0.01)/8\tau_w=100(0.01)/8

  3. Evaluate the expression; the result uses the units shown.

    Result=0.125 Pa\mathrm{Result}=0.125\ {\rm Pa}

Interpretation. This follows an axial force balance in fully developed flow.

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Example 15. Darcy head loss

Definitions & inputs. fD=.02,L=10 m,D=.1 m,v=2 m/s.

  1. Choose the governing model and isolate the requested quantity.

    hf=fD(L/D)v2/(2g)h_f=f_D(L/D)v^2/(2g)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    hf=0.02(100)(4)/(2⋅9.81)h_f=0.02(100)(4)/(2\cdot9.81)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.4077472 m\mathrm{Result}=0.4077472\ {\rm m}

Interpretation. The supplied friction factor must be appropriate for roughness and Reynolds number.

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Example 16. Pump shaft power

Definitions & inputs. Δp=100 kPa,Q=.01 m³/s,efficiency .7.

  1. Choose the governing model and isolate the requested quantity.

    P=ΔpQ/ηP=\Delta pQ/\eta

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    P=105(0.01)/0.7P=10^5(0.01)/0.7

  3. Evaluate the expression; the result uses the units shown.

    Result=1428.571 W\mathrm{Result}=1428.571\ {\rm W}

Interpretation. Hydraulic power is 1000 W; shaft power includes loss.

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Example 17. Drag force

Definitions & inputs. ρ=1.2 kg/m³,v=20 m/s,A=.5 m²,Cd=1.

  1. Choose the governing model and isolate the requested quantity.

    FD=CDρv2A/2F_D=C_D\rho v^2A/2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    FD=1(1.2)(400)(0.5)/2F_D=1(1.2)(400)(0.5)/2

  3. Evaluate the expression; the result uses the units shown.

    Result=120 N\mathrm{Result}=120\ {\rm N}

Interpretation. Cd is an input specific to the body and flow.

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Example 18. Mach number

Definitions & inputs. v=170 m/s,a=340 m/s.

  1. Choose the governing model and isolate the requested quantity.

    Ma=v/aMa=v/a

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Ma=170/340Ma=170/340

  3. Evaluate the expression; the result uses the units shown.

    Result=0.5 \mathrm{Result}=0.5\ {}

Interpretation. Compressibility may be significant at this Mach number.

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Example 19. Ideal-gas sound speed

Definitions & inputs. γ=1.4,Rs=287 J/(kg K),T=300 K.

  1. Choose the governing model and isolate the requested quantity.

    a=γRsTa=\sqrt{\gamma R_sT}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    a=1.4(287)(300)a=\sqrt{1.4(287)(300)}

  3. Evaluate the expression; the result uses the units shown.

    Result=347.1887 m s−1\mathrm{Result}=347.1887\ {\rm m\,s}^{-1}

Interpretation. Thermodynamic composition and temperature set γ and Rs.

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Example 20. Capillary rise

Definitions & inputs. Surface tension .072 N/m,contact angle 0°,radius .001 m,ρ=1000 kg/m³.

  1. Choose the governing model and isolate the requested quantity.

    h=2γcos⁡θ/(ρgr)h=2\gamma\cos\theta/(\rho gr)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    h=2(0.072)/[1000(9.81)(0.001)]h=2(0.072)/[1000(9.81)(0.001)]

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0146789 m\mathrm{Result}=0.0146789\ {\rm m}

Interpretation. The tube is narrow enough for capillarity and the meniscus contact angle is prescribed.

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Symbols and units

Each derivation and problem defines its own symbols and inputs. Symbols may be reused with different meanings in other subjects. Keep units consistent, retain sufficient precision during calculation, and apply the stated validity limits.