Conservation of mass, momentum, and energy

Start with a pipe region

A fixed horizontal pipe section is 0.50 m long. At each end its area is 0.004 m² and the assigned section speed is 2 m/s toward the right. The inlet absolute pressure is 140000 Pa; the outlet absolute pressure is 100000 Pa. Fluid continuously enters and leaves, while the flow at each fixed position is steady. If the entering and leaving mass rates are equal, what balances the pressure push? Where does the pressure-driven energy transfer go?

Supply a single-phase liquid with constant density ρ = 1000 kg/m³. Density is mass per volume; area times speed gives volume crossing an end per second. The pipe walls are rigid and stationary. We prescribe uniform section profiles, no leaks, no axial body force, no heat transfer and no pump or turbine. The pressure drop is an input, not a result of a friction law. This is an illustrative balance model rather than a resolved wall-layer calculation.

First follow the fixed-boundary sketch, tank storage table, force comparison and energy bars. Units, standards, and physical constants and Algebra, functions, graphs, and logarithms provide useful preparation. Newton’s laws and free-body diagrams, Momentum, impulse, and collisions, and Work, kinetic energy, potential energy, and power supply the earlier mechanical balances. The necessary fluid terms are defined here. Vectors and vector spaces and Integrals and multivariable calculus are optional return paths for the final vector and surface-integral forms.

Choose matter or a region

A material system contains the same identified matter as time passes. A marked group of fluid parcels may move and deform, but membership stays fixed. In this classical nonreacting model, its mass is constant. Its momentum changes under the net external force. Its total energy changes through heat supplied and work done on or by that matter. Conservation does not mean that every chosen subsystem has zero force or constant kinetic energy.

A control volume is a chosen region whose boundary is called the control surface. For our pipe it is the fixed fluid space between two end cuts and the inner walls. Fluid crosses the end cuts. The matter inside at a later time is generally different, so the material-system laws cannot be applied by ignoring that exchange.

For each quantity, ask two questions: how fast does it change inside the region, and how fast does it cross the boundary? The first is storage; the second is transport or flux. A control-volume balance relates these rates to the underlying material-system law. Reynolds’ transport theorem is the general bridge; its compact statement comes at the end.

Choose x toward the right and the control-surface unit normal n outward. The inlet normal points left and the outlet normal right. Fluid velocity v points right at both ends, but v·n is negative at the inlet and positive at the outlet. This sign counts crossing direction, rather than reversing the fluid’s actual velocity.

Count mass crossing and storage

Let m_CV be the fluid mass inside the region. Mass balance is dm_CV/dt = ṁ_in − ṁ_out for one inlet and one outlet, with each listed rate a positive magnitude. A dot indicates a rate per time; mass flow rate ṁ has units kg/s. Volume flow rate Q_v has units m³/s. For uniform density and section speed V normal to a port, Q_v = AV and ṁ = ρAV.

For the supplied pipe, Q_v = (0.004 m²)(2 m/s) = 0.008 m³/s and ṁ = 8 kg/s at each end. The net mass storage rate is zero. That statement allows substantial through-flow; it does not say that the fluid is stationary.

Compare a fixed tank with horizontal area 0.05 m², initial liquid volume 0.01 m³ and capacity 0.04 m³. Prescribe inflow 0.003 m³/s and outflow 0.001 m³/s over the six-second comparison. The mass rates are 3 kg/s in and 1 kg/s out, so mass grows at 2 kg/s. With vertical walls, the level grows at (0.002 m³/s)/(0.05 m²) = 0.04 m/s. These rates are imposed; no outlet pressure or discharge law has been supplied.

The first figure keeps the tank boundary fixed and distinguishes entering fluid, leaving fluid and stored liquid. Its height graph has a real time axis and one common height scale. The tank calculation concerns mass and level only. It supplies no transient momentum, temperature or mixing-energy prediction.

Separate steady flow from equal speed

Steady means that fields at fixed positions do not change with time. It does not mean zero velocity, or the same velocity at every position. A fluid parcel can accelerate while travelling through a steady converging region, because the velocity varies along its path.

For a steady single-inlet, single-outlet control volume, mass balance gives ρ₁A₁V₁ = ρ₂A₂V₂. Only if density is the same at both sections can this be reduced to A₁V₁ = A₂V₂. Compressible flow can have equal mass flow rates but unequal volume flow rates.

Keep the pipe’s inlet area and speed, but let a second fixed horizontal region converge to outlet area 0.001 m². With the same assigned liquid density, V₂ = (0.004 m²)(2 m/s)/(0.001 m²) = 8 m/s. Both ends still carry 8 kg/s. Continuity determines this relation; by itself it supplies neither the required pressure difference nor the wall force.

Uniform section profiles are a separate simplification. An average speed is enough for volume flow at constant density, but momentum transport depends on squared speed and kinetic-energy transport on cubed speed within a section. A nonuniform profile therefore needs its own integrals or correction factors. The final optional section and last practice explain this distinction.

Balance force with momentum transport

Linear momentum per unit mass is the velocity vector. For a fixed control volume, external force equals momentum storage rate plus net outward momentum flux. In our steady straight-through cases, the x balance becomes ΣF_x = ṁV₂ − ṁV₁. Each term ṁV has force units, kg·m/s² = N, but represents a rate of transported momentum, rather than a separate contact force.

Pressure pushes inward on the control surface. With outward normal n, its traction is −p n. The inlet cut therefore receives +p₁A₁ along x; the outlet receives −p₂A₂. Add the actual wall-on-fluid resultant R_x and any axial body force. Use the same chosen fluid region for every term. Here gravity has no axial component.

For the pipe, inlet pressure supplies +560 N and outlet pressure supplies −400 N. The net pressure force is +160 N. The signed momentum fluxes are −16 N at entry and +16 N at exit, so their sum is zero. Momentum balance gives 560 N − 400 N + R_x = 0, or R_x = −160 N. The wall force opposes the pressure push even though mass and momentum storage are steady.

An unsteady case requires the momentum stored inside as well. Equal mass inflow and outflow do not force that momentum-storage term to vanish: velocities within a fixed region can change while its total mass stays constant. Our two steady examples omit storage because their prescribed fields are time independent, not because a fluid momentum balance always lacks it.

Compare a converging momentum balance

For the converging region, retain the assigned absolute port pressures 140000 Pa and 100000 Pa. The inlet pressure force remains +560 N, but the smaller outlet area gives −100 N. The net pressure force is therefore +460 N. Momentum leaves at 64 N and enters with signed flux −16 N, giving net outward flux +48 N.

The actual wall-on-fluid force is R_x = 48 N − 460 N = −412 N. Adding all three contact contributions verifies the balance: 560 N − 100 N − 412 N = 48 N. The net force changes the momentum of matter passing through the region, while the fixed-region storage remains zero.

The fluid’s contact reaction on the inner walls has opposite sign. That reaction alone is not automatically the net load on a mounting: outside pressure, solid connections and other mechanical forces may also act. No mounting geometry or design-load calculation has been supplied.

All forces in the comparison use absolute pressure on the cuts. Subtracting a common reference from those cut pressures while keeping the actual wall force unchanged would alter the equation when port areas differ. A consistently gauge-adjusted wall resultant is a different quantity; its relation is stated in the optional return section. The force figure separates outward normals, actual flow direction, momentum transport and wall contact.

Include pressure flow work in energy

Total specific energy in this nonreacting fluid model is e_int + V²/2 + gz, measured in J/kg. Internal energy e_int concerns the material’s thermal and microscopic state; V²/2 is kinetic energy relative to the chosen inertial frame; gz is gravitational potential energy with upward height z. Total stored energy in the control volume integrates all these contributions over its fluid mass.

Energy also crosses by heat and work. Take heat rate Q̇_heat positive into the fluid and shaft or device work rate Ẇ_s positive out. A pump supplying work has negative Ẇ_s with this convention; a turbine extracting it has positive Ẇ_s. Stationary rigid walls in our stated model supply no mechanical boundary work. A nonzero wall force does not therefore become a shaft-power term obtained by multiplying that force by the bulk flow speed.

Flow through a cut also involves pressure work: pressure times volume flow is pQ_v. Incoming surroundings push fluid in, supplying pressure work; outgoing fluid pushes its surroundings away. Combine this flow work with internal energy per mass to define specific enthalpy h_f = e_int + p/ρ. Its units are J/kg. Enthalpy is an accounting combination, not another independent energy contribution to add after internal energy and pressure work have already been counted.

For one steady inlet and outlet with uniform properties, Q̇_heat − Ẇ_s = ṁ[(h_f,2 − h_f,1) + (V₂² − V₁²)/2 + g(z₂ − z₁)]. More generally, stored total energy changes when heat, work and incoming-minus-outgoing energy flux do not balance. Our steady formula assumes no additional moving-boundary or shear-work transfer; those terms must be retained in models that have them.

Account for the pipe’s pressure power

For the pipe, pressure flow power is (140000 Pa)(0.008 m³/s) = 1120 W in and (100000 Pa)(0.008 m³/s) = 800 W out. Their difference is 320 W. Kinetic-energy flux is ṁV²/2 = 16 W at each end, and elevations are equal. With no heat or shaft work, the internal-energy flux must increase by 320 W.

Divide by 8 kg/s to obtain e_int,2 − e_int,1 = 40 J/kg. Supply the simple liquid property model e_int = c(T − 293 K), with c = 4200 J/(kg·K), independent of pressure at the assigned constant density. The inlet temperature is 293 K, so ΔT = (40 J/kg)/c = 1/105 K ≈ 0.009524 K. This is a derived illustrative value, not a measured temperature or material-accuracy claim.

The pressure-driven mechanical contribution becomes internal energy through the dissipative flow represented by the supplied states. It has not disappeared from total energy, and no heat crosses our adiabatic boundary. Heat transfer and internal conversion are different processes. The stationary-wall force is part of momentum accounting, while the pressure flow work appears at the moving fluid across the end cuts.

Check the complete accounting with the chosen internal-energy reference: inlet energy flux is 1120 W + 16 W + 0 = 1136 W. Outlet energy flux is 800 W + 16 W + 320 W = 1136 W. Mechanical-energy loss terminology describes a conversion or transfer relative to the mechanical subset; it does not mean destruction of total energy.

Compare energy conversion in the converging case

The converging region has the same volume flow and pressure difference, so the pressure flow power still decreases by 320 W. Its kinetic-energy flux rises from 16 W at 2 m/s to 256 W at 8 m/s. The increase is 240 W, leaving 80 W of increased internal-energy flux under the same zero-heat, zero-shaft-work assumptions.

Per unit mass, the kinetic-energy increase is (8² − 2²) m²/s²/2 = 30 J/kg. Pressure work supplies 40 J/kg, so internal energy increases by 10 J/kg. The property model gives ΔT = 1/420 K ≈ 0.002381 K. Outlet specific enthalpy is 110 J/kg and kinetic energy is 32 J/kg; together they equal the inlet 140 J/kg + 2 J/kg. No Bernoulli assumption is needed for this total-energy balance.

The third figure compares the same 320 W pressure-flow decrease in both cases. In the straight pipe it all appears as increased internal-energy flux; in the converging region, 240 W appears as kinetic-energy increase and 80 W as internal-energy increase. The bar segments share one power scale and are differences between section fluxes. They are not stored energy in joules or a record of the local conversion rate at each position.

Different heat transfer, device work, elevation, density or section profiles change these results. A dissipative flow need not preserve a mechanical-only Bernoulli quantity. Euler’s equation and Bernoulli’s principle is the neighbouring lesson for that conditional relation; the present lesson keeps the full control-volume balances.

Optional return: transport and section profiles

Return after the finite balances and figures for the integral structure. For the instant being considered, select the material system occupying the chosen region. For a material-system property B and its specific value b per unit mass, Reynolds’ transport theorem for a fixed region is dB_system/dt = d/dt∫_CVρb dV + ∮_CSρb(v·n) dA. The first integral stores the property; the second counts its signed outward transport. Take b = 1 for mass, b = v for momentum, or b = e_int + |v|²/2 + gz for total energy. The physical material-system law supplies the left side.

Mass balance gives d/dt∫_CVρ dV + ∮_CSρv·n dA = 0. For a sufficiently smooth field, the divergence theorem and an arbitrary fixed region yield ∂ρ/∂t + ∇·(ρv) = 0. Constant material density reduces this to ∇·v = 0. That incompressibility condition is distinct from steady flow; density and velocity need their own stated assumptions.

Momentum gives ΣF_ext = d/dt∫_CVρv dV + ∮_CSρv(v·n) dA. Pressure contributes −∮p n dA; shear traction and body forces supply their own terms. The velocities in momentum and kinetic energy are inertial-frame velocities. For a moving or deforming control surface, crossing is instead counted by (v − v_boundary)·n. Changing the boundary does not justify replacing the transported inertial momentum by relative velocity without also changing the force/frame formulation.

At a straight port, let w be the local positive through-flow speed and V̄ = (∫_A w dA)/A its mean. For constant density, momentum flux magnitude is ρ∫_A w² dA = βρAV̄², with β = (∫_A w² dA)/(AV̄²). Kinetic-energy flux is (ρ/2)∫_A w³ dA = α(ρAV̄³/2), with α = (∫_A w³ dA)/(AV̄³). Our uniform supplied profiles give α = β = 1. Density variations or flow not normal to a port require the full corresponding surface integrals.

The energy flux form combines pressure work with advected internal energy. Stored energy itself uses e_int + |v|²/2 + gz, rather than adding p/ρ a second time. Heat, device work and any extra boundary work are transfers, with signs defined by the chosen region. For unsteady flow, keep dE_CV/dt; an equal mass rate alone cannot remove changing thermal or kinetic storage.

For the straight-through force convention, a gauge-adjusted wall resultant obeys R_x,g = R_x + p_ref(A₁ − A₂). At p_ref = 100000 Pa, the converging example gives −112 N for this adjusted quantity, while actual wall contact remains −412 N. The difference compensates the reference removed from unequal cut areas. It is not a new physical contact force or an independently determined net mount load.

Use these balances to find compatible totals before seeking a detailed velocity, pressure or temperature field. Constitutive laws, boundary conditions and geometry are needed to predict those fields or a pressure drop. The adjacent Euler/Bernoulli topic and later viscous-flow lessons add those conditions; they do not replace conservation.

Worked examples

Try it yourself

1. In a separate steady mass-only gas example, inlet density is 2 kg/m³, area 0.01 m² and speed 3 m/s. Outlet density is 1 kg/m³ and area 0.005 m². Find outlet speed and both volume flow rates. Must the volume flow rates be equal?

   Show the answer and reasoning

   Mass flow is 0.06 kg/s. Outlet speed is 12 m/s. Volume flows are 0.03 m³/s in and 0.06 m³/s out; they need not be equal.

   Use ρ₁A₁V₁ = ρ₂A₂V₂. The less dense outlet fluid needs twice the volume per second to carry the same mass. This mass-only calculation supplies no gas pressure, heat, work or temperature relation.

2. In a separate tank comparison, halve the prescribed inflow to 0.0015 m³/s, retain outflow 0.001 m³/s, initial volume 0.01 m³, area 0.05 m² and density 1000 kg/m³. Find mass storage rate and level after 4 s.

   Show the answer and reasoning

   Mass storage is +0.5 kg/s. Level after 4 s is 0.24 m, with mass 12 kg.

   Net volume addition is 0.0005 m³/s, so 4 s adds 0.002 m³. Total volume 0.012 m³ gives mass 12 kg and height 0.012/0.05 m = 0.24 m. Both rates are imposed separately; halving a real inlet would not guarantee an unchanged outlet without its own control or discharge model.

3. For the converging case, subtract reference 100000 Pa from both port pressures. The cut-pressure force is then +160 N. Is the resulting −112 N wall term the same as the actual −412 N wall contact?

   Show the answer and reasoning

   No. −112 N is the consistently gauge-adjusted wall resultant; actual wall contact remains −412 N in the absolute-pressure balance.

   The unequal cut areas lose p_ref(A₁ − A₂) = (100000 Pa)(0.004 − 0.001) m² = 300 N of pressure force when shifted to gauge values. The wall term must increase by the same 300 N to preserve the +48 N momentum-flux balance. A net mounting load requires the actual external solid-force accounting too.

4. Keep the straight pipe’s supplied pressure and velocity states, but prescribe heat rate −320 W into the fluid, with no device work. Under the same liquid property model, what outlet temperature follows?

   Show the answer and reasoning

   The outlet temperature is 293 K, equal to the inlet temperature.

   The 320 W net pressure flow work is now balanced by 320 W of heat leaving through the boundary. Kinetic and potential fluxes still do not change, so the internal-energy rise is zero. A stationary wall can transfer heat even though it supplies no mechanical shaft work. The heat rate is prescribed; no heat-transfer coefficient has been inferred.

5. At a separate port of total area 0.004 m² and density 1000 kg/m³, one half-area has through-flow speed 1 m/s and the other 3 m/s. Find mean speed, mass flow, momentum-flux magnitude and kinetic-energy flux. Compare with a uniform profile at the same mean speed.

   Show the answer and reasoning

   Mean speed is 2 m/s and mass flow 8 kg/s. Momentum flux is 20 N and kinetic-energy flux 28 W, versus 16 N and 16 W for the uniform profile. The correction factors are β = 1.25 and α = 1.75.

   Each half has area 0.002 m². Volume flow is 0.002(1 + 3) m³/s = 0.008 m³/s. Sum momentum rates: 1000 × 0.002(1² + 3²) N = 20 N. Sum kinetic-energy rates: (1000/2) × 0.002(1³ + 3³) W = 28 W. Divide by the corresponding uniform values to obtain β and α. This port-only comparison does not by itself determine a complete pipe pressure drop or wall force.

What to carry forward

• A material system follows fixed matter. A control volume follows a region and must account for matter crossing its boundary.
• Mass storage equals inflow minus outflow. Steady through-flow can have zero storage and nonzero velocity.
• External force balances momentum storage plus signed outward momentum flux; equal mass rates do not imply equal momentum fluxes.
• Enthalpy combines internal energy and pressure flow work. Count heat, device work, kinetic energy and potential energy with one consistent sign convention.
• Mechanical-energy conversion to internal energy does not destroy total energy and does not require heat to cross the boundary.
• Constant density, steady fields and uniform section profiles are different assumptions. Change the appropriate balance when one fails.