Reynolds Transport Theorem

Derivation

Reynolds Transport Theorem

D-314 Home PU-307 Threads matter · symmetry Depends on leibniz-integral-rule, divergence-theorem

Statement

Let \(\phi(\mathbf{x},t)\) be an intensive field (quantity per unit volume) transported by a flow with velocity \(\mathbf{v}(\mathbf{x},t)\), and let \(V(t)\) be a material volume — a region always composed of the same fluid particles — bounded by the material surface \(\partial V(t)\) with outward unit normal \(\hat{\mathbf{n}}\). The Reynolds Transport Theorem states that the total rate of change of the extensive quantity \(\Phi(t)=\int_{V(t)}\phi\,dV\) following the material equals the local rate of change accumulated over a fixed control volume plus the net flux of \(\phi\) carried across its boundary by the flow:

\[ \frac{d}{dt}\int_{V(t)}\phi\,dV \;=\; \int_{V}\frac{\partial \phi}{\partial t}\,dV \;+\; \oint_{\partial V}\phi\,(\mathbf{v}\cdot\hat{\mathbf{n}})\,dA. \]

Why it matters

Every conservation law of continuum mechanics — mass, momentum, energy, entropy — is naturally stated for a fixed lump of material (a system), because that is what Newton's and the thermodynamic laws describe. But we almost always want to compute using a fixed region of space (a control volume), because that is where our instruments and mesh points live. The Reynolds Transport Theorem is the exact bridge between the two viewpoints: it converts the Lagrangian material derivative \(\tfrac{d}{dt}\int_{V(t)}\) into an Eulerian expression involving only fields and their fixed-volume integrals.

Applied to \(\phi=\rho\) it yields the continuity equation; to \(\phi=\rho\mathbf{v}\) it yields the momentum (Cauchy / Navier–Stokes) equations; to \(\phi=\rho e\) the energy equation. The theorem is therefore the single kinematic identity from which the differential balance laws of fluid dynamics are all obtained, and its structure — local change plus boundary flux — recurs throughout physics.

Assumptions

The volume \(V(t)\) is material — its boundary moves everywhere with the local flow velocity \(\mathbf{v}\), so no fluid crosses \(\partial V(t)\).If the boundary moves with an independent velocity \(\mathbf{v}_b\ne\mathbf{v}\), the flux term uses \(\mathbf{v}_b\cdot\hat{\mathbf{n}}\) and the left side is no longer the material derivative of \(\Phi\); one gets the more general "moving control volume" form instead.

\(\phi(\mathbf{x},t)\) is continuously differentiable in space and time on \(\bar V(t)\).Without \(C^1\) smoothness the Leibniz rule and the divergence theorem used below do not apply pointwise, and shocks or contact discontinuities require the integral (weak / jump-condition) form instead.

The velocity field \(\mathbf{v}(\mathbf{x},t)\) is single-valued and the flow map \(\boldsymbol{\chi}(\mathbf{X},t)\) is a smooth, invertible diffeomorphism.If particle trajectories cross or the Jacobian \(J=\det(\partial\mathbf{x}/\partial\mathbf{X})\) vanishes, the material volume is no longer well defined and the change-of-variables step collapses.

The boundary \(\partial V(t)\) is piecewise smooth (Lipschitz), so a well-defined outward normal \(\hat{\mathbf{n}}\) exists almost everywhere.If dropped, the surface integral \(\oint_{\partial V}\phi\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA\) is not defined and the divergence theorem cannot be invoked to interconvert volume and surface forms.

Derivation

\[ \mathbf{x}=\boldsymbol{\chi}(\mathbf{X},t),\qquad dV = J(\mathbf{X},t)\,dV_0,\qquad J=\det\!\left(\frac{\partial \mathbf{x}}{\partial \mathbf{X}}\right), \]

Pull the moving domain back to a fixed reference domain \(V_0\) via the flow map, so the time-dependence sits in the integrand rather than in the limits. \(\mathbf{X}\) labels the material particle at \(t=0\). A

\[ \frac{d}{dt}\int_{V(t)}\phi\,dV = \frac{d}{dt}\int_{V_0}\phi(\boldsymbol{\chi}(\mathbf{X},t),t)\,J\,dV_0 = \int_{V_0}\frac{d}{dt}\big(\phi\,J\big)\,dV_0. \]

The reference domain \(V_0\) is fixed, so differentiation and integration commute — this is exactly why we pulled back. The total derivative is now taken at fixed \(\mathbf{X}\), i.e. the material derivative \(\tfrac{D}{Dt}\). A

\[ \frac{dJ}{dt} = J\,(\nabla\cdot\mathbf{v}), \]

Euler's expansion formula. Differentiating \(J=\det(\partial x_i/\partial X_j)\) with Jacobi's formula gives \(\dot J = J\,\mathrm{tr}\!\big(F^{-1}\dot F\big)\); since \(\dot F_{ij}=\partial \dot x_i/\partial X_j\) and \(\dot{\mathbf x}=\mathbf v\), the trace collapses to \(\partial v_i/\partial x_i=\nabla\cdot\mathbf{v}\). C

\[ \frac{d}{dt}(\phi J) = \frac{D\phi}{Dt}\,J + \phi\,\frac{dJ}{dt} = \left(\frac{D\phi}{Dt} + \phi\,\nabla\cdot\mathbf{v}\right)J, \]

Product rule at fixed \(\mathbf{X}\), then substitute Euler's formula from Step 3. Here \(\dfrac{D\phi}{Dt}=\dfrac{\partial\phi}{\partial t}+\mathbf{v}\cdot\nabla\phi\) is the material derivative of the field. B

\[ \frac{d}{dt}\int_{V(t)}\phi\,dV = \int_{V(t)}\left(\frac{D\phi}{Dt} + \phi\,\nabla\cdot\mathbf{v}\right)dV. \]

Push the integral back to the current configuration \(V(t)\) using \(J\,dV_0=dV\). This is the theorem in material-derivative form. A

\[ \frac{D\phi}{Dt}+\phi\,\nabla\cdot\mathbf{v} = \frac{\partial\phi}{\partial t}+\mathbf{v}\cdot\nabla\phi+\phi\,\nabla\cdot\mathbf{v} = \frac{\partial\phi}{\partial t}+\nabla\cdot(\phi\,\mathbf{v}), \]

Expand the material derivative and recognise the product rule for divergence: \(\nabla\cdot(\phi\mathbf{v})=\mathbf{v}\cdot\nabla\phi+\phi\,\nabla\cdot\mathbf{v}\). Pure algebra — no new physics. B

\[ \frac{d}{dt}\int_{V(t)}\phi\,dV = \int_{V(t)}\frac{\partial\phi}{\partial t}\,dV + \int_{V(t)}\nabla\cdot(\phi\,\mathbf{v})\,dV. \]

Linearity of the integral splits the integrand. The first term is now the purely local (fixed-point) rate of change; the second is a divergence ripe for the divergence theorem. A

\[ \int_{V(t)}\nabla\cdot(\phi\,\mathbf{v})\,dV = \oint_{\partial V(t)}\phi\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA, \]

Divergence (Gauss) theorem applied to the vector field \(\phi\mathbf{v}\), legal because \(\phi\mathbf{v}\in C^1(\bar V)\) and \(\partial V\) is piecewise smooth (Assumptions 2–4). This converts the interior source into a boundary flux. B

\[ \boxed{\;\frac{d}{dt}\int_{V(t)}\phi\,dV = \int_{V}\frac{\partial\phi}{\partial t}\,dV + \oint_{\partial V}\phi\,(\mathbf{v}\cdot\hat{\mathbf{n}})\,dA\;} \]

Substitute Step 8 into Step 7. Because \(V(t)\) is material its instantaneous position coincides with a fixed control volume \(V\) at time \(t\), so on the right \(V\) and \(\partial V\) may be read as the fixed region the material momentarily occupies. A

Result

\[ \frac{d}{dt}\int_{V(t)}\phi\,dV = \underbrace{\int_{V}\frac{\partial\phi}{\partial t}\,dV}_{\text{local accumulation}} + \underbrace{\oint_{\partial V}\phi\,(\mathbf{v}\cdot\hat{\mathbf{n}})\,dA}_{\text{convective outflux}} \]

Reading. The rate at which the total amount of \(\phi\)-stuff carried by a chunk of moving material changes has two contributions. First, the stuff already inside the region can intensify or fade in place: that is \(\int_V \partial_t\phi\,dV\), measured as if the region were frozen in space. Second, as the material boundary sweeps through the field it carries \(\phi\) outward wherever \(\mathbf{v}\cdot\hat{\mathbf{n}}>0\) and inward wherever \(\mathbf{v}\cdot\hat{\mathbf{n}}<0\); the net of this is the surface flux \(\oint_{\partial V}\phi\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA\). The material change equals fixed-frame change plus what the flow drags across the border.

Units check. Let \([\phi]=Q\,\mathrm{m^{-3}}\) for any extensive quantity \(Q\). Then \(\Phi=\int\phi\,dV\) has units \(Q\), and the left side \(\tfrac{d\Phi}{dt}\) has \(Q\,\mathrm{s^{-1}}\). Right side, term one: \([\partial_t\phi]\,[dV]=(Q\,\mathrm{m^{-3}}\,\mathrm{s^{-1}})(\mathrm{m^3})=Q\,\mathrm{s^{-1}}\). Term two: \([\phi][v][dA]=(Q\,\mathrm{m^{-3}})(\mathrm{m\,s^{-1}})(\mathrm{m^2})=Q\,\mathrm{s^{-1}}\). All three agree. ✓

Limiting cases

Steady field (\(\partial_t\phi=0\)): the volume integral drops and \(\tfrac{d}{dt}\int_{V(t)}\phi\,dV=\oint_{\partial V}\phi\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA\). The material amount changes only because the boundary sweeps through a frozen field.
Uniform field, incompressible flow (\(\phi=\text{const}\), \(\nabla\cdot\mathbf{v}=0\)): \(\oint\phi\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA=\phi\oint\mathbf{v}\cdot\hat{\mathbf{n}}\,dA=\phi\int_V\nabla\cdot\mathbf{v}\,dV=0\), so \(\Phi\) is conserved — a material blob of incompressible fluid keeps constant volume.
\(\phi=\rho\) (mass density): the theorem gives \(\tfrac{d}{dt}\int_{V(t)}\rho\,dV=\int_V[\partial_t\rho+\nabla\cdot(\rho\mathbf{v})]\,dV\). Mass conservation forces the left side to zero for every \(V\), yielding the continuity equation \(\partial_t\rho+\nabla\cdot(\rho\mathbf{v})=0\).
Fixed control volume (\(\mathbf{v}\to\mathbf{0}\) on \(\partial V\), or equivalently a non-material fixed region): the flux term vanishes and \(\tfrac{d\Phi}{dt}=\int_V\partial_t\phi\,dV\) — the elementary Leibniz rule for a stationary domain.
One dimension: with \(V(t)=[a(t),b(t)]\), \(\dot a=v(a)\), \(\dot b=v(b)\), the theorem reduces to the Leibniz integral rule \(\tfrac{d}{dt}\int_{a}^{b}\phi\,dx=\int_a^b\partial_t\phi\,dx+\phi(b)\dot b-\phi(a)\dot a\).

Breaks when

Discontinuous fields (shocks, contact surfaces, phase boundaries). Across a shock \(\phi\) and \(\mathbf{v}\) jump, \(\partial_t\phi\) contains a Dirac delta, and the divergence theorem in Step 8 fails pointwise. One must split \(V(t)\) at the moving discontinuity \(\Sigma(t)\) and add a surface term \(\int_\Sigma[\![\phi(\mathbf{v}-\mathbf{v}_\Sigma)\cdot\hat{\mathbf{n}}]\!]\,dA\), giving the Rankine–Hugoniot jump conditions.
Non-material or arbitrary-Lagrangian boundary. If \(\partial V\) moves with \(\mathbf{v}_b\ne\mathbf{v}\) (a deforming mesh, a growing bubble, a fixed pipe with the boundary held still), the derivation's Step 1 change of variables no longer tracks fluid particles; the correct flux uses \(\mathbf{v}_b\cdot\hat{\mathbf{n}}\) and the left side is \(\tfrac{d}{dt}\int_{V_b(t)}\), not the material derivative of \(\Phi\).
Singular or multivalued flow map. Where trajectories cross or focus (caustics, cavitation, vanishing Jacobian \(J\to 0\)), the pullback in Steps 1–5 is not invertible and "the same material particles" ceases to be a meaningful set. The theorem must be restricted to regions where the flow map stays a diffeomorphism.
Relativistic / non-Galilean kinematics. The split into a fixed-frame \(\partial_t\) plus a convective flux presumes an absolute time and a 3-velocity; at speeds near \(c\) one instead uses the covariant divergence \(\nabla_\mu T^{\mu\nu}=0\) on a spacetime volume, and the 3+1 flux decomposition is frame-dependent.

Failure modes

Pulling \(\tfrac{d}{dt}\) inside as \(\partial_t\) directly. Writing \(\tfrac{d}{dt}\int_{V(t)}\phi\,dV=\int_{V(t)}\partial_t\phi\,dV\) forgets the moving boundary entirely; it drops the whole flux term and only holds for a stationary domain.
Using \(\tfrac{D\phi}{Dt}\) where \(\partial_t\phi\) belongs. After converting to the fixed-volume form the correct integrand is the partial time derivative \(\partial_t\phi\); students who carry over the material derivative double-count the convective term \(\mathbf{v}\cdot\nabla\phi\).
Forgetting the \(\phi\,\nabla\cdot\mathbf{v}\) term for compressible flow. Assuming \(\nabla\cdot\mathbf{v}=0\) collapses the material-derivative form incorrectly whenever the fluid actually compresses (gases, high-speed flow), corrupting the continuity and energy balances.
Sign error on the outward normal. Taking \(\hat{\mathbf{n}}\) inward flips the flux term, turning outflow into inflow and reversing the sign of the entire boundary contribution.
Confusing \(\mathbf{v}\) (fluid velocity) with \(\mathbf{v}_b\) (boundary velocity). For a material volume they coincide, but writing the general form and then silently setting them equal — or unequal — where the problem demands otherwise gives the wrong control-volume equation.
Treating \(\phi\) as the extensive quantity. \(\phi\) is the density (per unit volume); the extensive quantity is \(\Phi=\int\phi\,dV\). Plugging total mass \(m\) where \(\rho\) belongs (or vice versa) breaks the units immediately.

Discussion

The theorem is best understood as a generalisation of the Leibniz integral rule from one dimension to three, with the flow supplying the boundary velocity. In 1D, differentiating \(\int_{a(t)}^{b(t)}\phi\,dx\) produces an interior term \(\int\partial_t\phi\,dx\) and two endpoint terms \(\phi\dot b-\phi\dot a\); the endpoints are a zero-dimensional "surface" whose motion carries integrand value across. In 3D the endpoints fatten into the closed surface \(\partial V\), the endpoint velocities become \(\mathbf{v}\cdot\hat{\mathbf{n}}\), and the two boundary terms merge into a single flux integral. Nothing new is assumed physically — only the calculus of moving domains — which is why the result is called a transport theorem: it is pure kinematics, agnostic to what \(\phi\) represents.

The two equivalent forms carry different intuitions. The material-derivative form \(\int_{V}(\tfrac{D\phi}{Dt}+\phi\nabla\cdot\mathbf{v})\,dV\) is Lagrangian in spirit: it tracks how the density seen by a moving particle changes, plus a dilation term because the parcel's own volume stretches at rate \(\nabla\cdot\mathbf{v}\). The flux form \(\int_V\partial_t\phi\,dV+\oint_{\partial V}\phi\mathbf{v}\cdot\hat{\mathbf{n}}\,dA\) is Eulerian: everything is evaluated on a fixed region, with the flow's effect banished entirely to the boundary. The divergence theorem is precisely the hinge that swings between them, which is why RTT presupposes Gauss's theorem as a lemma.

Physically, the theorem is the engine that converts integral conservation laws into differential field equations. Because \(V(t)\) is arbitrary, any statement "\(\tfrac{d}{dt}\int_{V(t)}\phi\,dV = \int_V S\,dV\)" (source \(S\)) forces the integrands to match pointwise: \(\partial_t\phi+\nabla\cdot(\phi\mathbf{v})=S\). This "for all volumes, therefore pointwise" localisation argument, powered by RTT, is the standard route to the continuity, momentum, and energy PDEs. Coupled with a constitutive law it closes the equations of motion of a continuum.

At a deeper level RTT is the classical shadow of a general fact about the Lie derivative of a differential form advected by a flow. Writing the extensive quantity as the integral of a top-degree form \(\omega=\phi\,dV\) over a material chain, one has \(\tfrac{d}{dt}\int_{V(t)}\omega=\int_{V(t)}\mathcal{L}_\mathbf{v}\omega\), and Cartan's magic formula \(\mathcal{L}_\mathbf{v}=d\,\iota_\mathbf{v}+\iota_\mathbf{v}\,d\) splits \(\mathcal{L}_\mathbf{v}\omega\) into an exact piece (the boundary flux, via Stokes) and an interior piece (the local rate). The flux/material duality is thus Cartan's formula plus Stokes' theorem; the divergence-theorem step above is exactly the \(d\,\iota_\mathbf{v}\) branch. This viewpoint generalises RTT verbatim to curved manifolds, moving surfaces (Thomas's surface transport theorem), and relativistic spacetime, where the naïve 3-velocity split loses its meaning.

Common misconceptions. RTT is not itself a conservation law — it contains no physics beyond kinematics and is true for any smooth field, conserved or not. The conservation content enters only when you assert that \(\tfrac{d}{dt}\int_{V(t)}\phi\,dV\) equals a specific source (zero for mass, applied force for momentum). A second misconception is that the flux term measures fluid "leaving the volume": for a material volume no fluid ever crosses \(\partial V\) — the flux term measures the field value swept across the fixed region that the material momentarily occupies, which is a statement about the reference control volume, not about particles escaping.

Worked examples

Example 1 — Mass in an expanding spherical parcel. A gas parcel is a sphere of radius \(R(t)\) whose surface moves radially outward with the fluid at speed \(\dot R\). The density is uniform in space but decreasing in time, \(\rho(t)=\rho_0\,e^{-\alpha t}\). Find \(\tfrac{d}{dt}\int_{V(t)}\rho\,dV\) at \(t=0\) and confirm it against the direct mass \(m(t)\).

\[ \frac{d}{dt}\int_{V(t)}\rho\,dV = \int_V \frac{\partial\rho}{\partial t}\,dV + \oint_{\partial V}\rho\,(\mathbf{v}\cdot\hat{\mathbf{n}})\,dA. \]

Apply RTT with \(\phi=\rho\). A

\[ \int_V\frac{\partial\rho}{\partial t}\,dV = \left(-\alpha\rho_0 e^{-\alpha t}\right)\cdot\frac{4}{3}\pi R^3,\qquad \oint_{\partial V}\rho\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA = \rho\,\dot R\cdot 4\pi R^2. \]

\(\rho\) is spatially uniform so it factors out; on the sphere \(\mathbf{v}\cdot\hat{\mathbf{n}}=\dot R\) is constant, and \(\oint dA=4\pi R^2\). B

\[ \frac{dm}{dt}= \frac{4}{3}\pi R^3\big(-\alpha\rho\big) + 4\pi R^2\,\rho\,\dot R = \rho\Big(4\pi R^2\dot R - \tfrac{4}{3}\pi R^3\alpha\Big). \]

Symbolic sum of the two terms. A

\[ \rho_0=1.2\ \mathrm{kg\,m^{-3}},\ \alpha=0.5\ \mathrm{s^{-1}},\ R=2\ \mathrm{m},\ \dot R=0.3\ \mathrm{m\,s^{-1}},\ t=0. \]

Insert numbers only now; at \(t=0\), \(\rho=\rho_0=1.2\ \mathrm{kg\,m^{-3}}\). A

\[ \frac{dm}{dt}=1.2\big(4\pi(2)^2(0.3) - \tfrac{4}{3}\pi(2)^3(0.5)\big)=1.2\big(15.08 - 16.76\big)\ \mathrm{kg\,s^{-1}}. \]

Evaluate: \(4\pi(4)(0.3)=15.08\); \(\tfrac43\pi(8)(0.5)=16.76\). A

\[ \left.\frac{dm}{dt}\right|_{t=0}\approx -2.0\ \mathrm{kg\,s^{-1}}. \]

Reading. The parcel's mass is falling: dilution (\(-16.76\)) outpaces the extra volume the expanding boundary sweeps in (\(+15.08\)). Cross-check: \(m(t)=\rho_0 e^{-\alpha t}\cdot\tfrac43\pi R(t)^3\); with \(R(t)=2+0.3t\), \(\dot m(0)=\tfrac43\pi\rho_0[-\alpha R^3+3R^2\dot R]=\tfrac43\pi(1.2)[-0.5(8)+3(4)(0.3)]=\tfrac43\pi(1.2)(-0.4)\approx-2.0\ \mathrm{kg\,s^{-1}}\). ✓ Units. \(\mathrm{kg\,m^{-3}}\cdot\mathrm{m^2}\cdot\mathrm{m\,s^{-1}}=\mathrm{kg\,s^{-1}}\). ✓

Example 2 — Derive the continuity equation from RTT. Mass is conserved: the mass of a material volume is constant, \(\tfrac{d}{dt}\int_{V(t)}\rho\,dV=0\) for every such volume. Obtain the differential form.

\[ 0=\frac{d}{dt}\int_{V(t)}\rho\,dV = \int_V\frac{\partial\rho}{\partial t}\,dV + \oint_{\partial V}\rho\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA. \]

RTT with \(\phi=\rho\), left side zero by mass conservation. A

\[ \oint_{\partial V}\rho\,\mathbf{v}\cdot\hat{\mathbf{n}}\,dA = \int_V\nabla\cdot(\rho\mathbf{v})\,dV \;\Rightarrow\; \int_V\left[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})\right]dV=0. \]

Divergence theorem converts the flux back to a volume integral; combine under one integral. B

\[ \frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})=0. \]

The integral vanishes for every volume \(V\); a continuous integrand with zero integral over all subregions must vanish pointwise (localisation). B

\[ \text{Check, incompressible: } \rho=1000\ \mathrm{kg\,m^{-3}}\text{ const}\ \Rightarrow\ \nabla\cdot\mathbf{v}=0. \]

Numerical sanity: constant \(\rho\) kills \(\partial_t\rho\) and factors out, leaving the incompressibility condition. A

\[ \boxed{\ \frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})=0\ } \]

Reading. RTT plus "mass of a material blob is fixed" yields the continuity equation with no further physics — a textbook instance of the localisation route from integral to differential balance. Units. \([\partial_t\rho]=\mathrm{kg\,m^{-3}\,s^{-1}}\); \([\nabla\cdot(\rho\mathbf{v})]=\mathrm{m^{-1}}\cdot\mathrm{kg\,m^{-3}}\cdot\mathrm{m\,s^{-1}}=\mathrm{kg\,m^{-3}\,s^{-1}}\). ✓

Problems

(A) One-dimensional Leibniz check. For \(V(t)=[0,\,L(t)]\) with \(L(t)=1+2t\), boundary velocity \(v(L)=\dot L\), and \(\phi(x,t)=x\,t\), compute \(\tfrac{d}{dt}\int_0^{L(t)}\phi\,dx\) at \(t=1\) both by RTT and by direct integration.
Solution
Direct: \(\int_0^{L}xt\,dx=\tfrac{t L^2}{2}\). With \(L=1+2t\): \(F(t)=\tfrac{t(1+2t)^2}{2}\). \(F'(t)=\tfrac12[(1+2t)^2 + t\cdot 2(1+2t)(2)]=\tfrac12(1+2t)[(1+2t)+4t]=\tfrac12(1+2t)(1+6t)\). At \(t=1\): \(\tfrac12(3)(7)=10.5\). RTT: local term \(\int_0^L\partial_t\phi\,dx=\int_0^L x\,dx=\tfrac{L^2}{2}=\tfrac{9}{2}=4.5\); flux term \(\phi(L,t)\dot L=(L t)(2)=(3\cdot1)(2)=6\). Sum \(=4.5+6=10.5\). ✓ Both give \(\boxed{10.5}\).
(B) Expanding cube, uniform field. A material cube has side \(a(t)=1+0.1t\) m expanding uniformly with the flow (\(\mathbf{v}=0.1\,\mathbf{x}/|\text{edge}|\) so each face moves at \(\dot a/2\) outward... use \(\mathbf{v}\cdot\hat{\mathbf n}=\dot a/2\) on each face). The scalar \(\phi=5\) (const, uniform). Find \(\tfrac{d}{dt}\int_{V}\phi\,dV\) at \(t=0\).
Solution
\(\phi\) constant and uniform \(\Rightarrow\) local term \(=0\). Flux \(=\phi\oint\mathbf{v}\cdot\hat{\mathbf n}\,dA\). A cube of side \(a\) has 6 faces of area \(a^2\), each with outward normal velocity \(\dot a/2\): \(\oint\mathbf v\cdot\hat{\mathbf n}\,dA=6a^2(\dot a/2)=3a^2\dot a\). At \(t=0\): \(a=1\), \(\dot a=0.1\), so \(=3(1)(0.1)=0.3\ \mathrm{m^3 s^{-1}}\). Thus \(\tfrac{d}{dt}\int\phi\,dV=5(0.3)=\boxed{1.5\ \text{(units of }\phi)\cdot\mathrm{m^3 s^{-1}}}\). Check via \(V=a^3\): \(\tfrac{d}{dt}(\phi a^3)=\phi\,3a^2\dot a=5(3)(1)(0.1)=1.5\). ✓
(C) Momentum flux term. Water (\(\rho=1000\ \mathrm{kg\,m^{-3}}\)) flows steadily through a fixed control volume; at an inlet disk of area \(A=0.02\ \mathrm{m^2}\) the velocity is \(u=3\ \mathrm{m\,s^{-1}}\) normal to the disk and directed inward. Using the RTT-derived momentum flux \(\oint\rho\mathbf{v}(\mathbf{v}\cdot\hat{\mathbf n})\,dA\), find the \(x\)-momentum flux carried through the inlet.
Solution
At the inlet \(\hat{\mathbf n}\) is outward but flow is inward, so \(\mathbf v\cdot\hat{\mathbf n}=-u\). Flow is along \(+x\): \(\mathbf v=u\,\hat{\mathbf x}\). Flux of \(x\)-momentum \(=\rho\,u_x(\mathbf v\cdot\hat{\mathbf n})A=\rho\,(u)(-u)A=-\rho u^2 A\). Symbolic: \(-\rho u^2 A\). Numbers: \(-(1000)(3^2)(0.02)=-(1000)(9)(0.02)=-180\ \mathrm{N}\). The negative sign means momentum enters the volume at \(180\ \mathrm N\) worth of flux. \(\boxed{-180\ \mathrm{kg\,m\,s^{-2}}=-180\ \mathrm N}\). Units: \(\mathrm{kg\,m^{-3}}(\mathrm{m\,s^{-1}})^2\mathrm{m^2}=\mathrm{kg\,m\,s^{-2}}\). ✓
(D) Non-material (fixed) control volume. A field \(\phi(x,t)=\phi_0\cos(kx-\omega t)\) fills a fixed box \(V=[0,\ell]\times[0,w]\times[0,h]\) (boundary velocity zero, but fluid velocity \(\mathbf v=U\hat{\mathbf x}\) nonzero). Write both the material rate over the momentarily-coincident material volume and the fixed-box rate \(\int_V\partial_t\phi\,dV\), and state which term distinguishes them.
Solution
Fixed-box rate: \(\int_V\partial_t\phi\,dV=\int_V \phi_0\omega\sin(kx-\omega t)\,dV\). The material rate over the coincident material volume adds the convective flux \(\oint_{\partial V}\phi\,\mathbf v\cdot\hat{\mathbf n}\,dA\); with \(\mathbf v=U\hat{\mathbf x}\) only the two \(x\)-faces contribute (\(\hat{\mathbf n}=\pm\hat{\mathbf x}\)): \(Uwh[\phi(\ell,t)-\phi(0,t)]=Uwh\,\phi_0[\cos(k\ell-\omega t)-\cos(-\omega t)]\). The distinguishing term is exactly this convective flux \(\oint\phi\,\mathbf v\cdot\hat{\mathbf n}\,dA\) — present for the material volume, absent for the stationary box. \(\boxed{\text{difference}=Uwh\,\phi_0[\cos(k\ell-\omega t)-\cos(\omega t)]}\) (using \(\cos(-\theta)=\cos\theta\)).
(C) Compressible dilation. A material parcel of volume \(V_0=0.5\ \mathrm{m^3}\) sits in a flow with uniform divergence \(\nabla\cdot\mathbf v=0.4\ \mathrm{s^{-1}}\). The transported density is \(\phi=\rho=2\ \mathrm{kg\,m^{-3}}\), momentarily uniform, with material rate \(\tfrac{D\rho}{Dt}=-0.8\ \mathrm{kg\,m^{-3}\,s^{-1}}\). Using the material-derivative form of RTT, find \(\tfrac{d}{dt}\int_{V(t)}\rho\,dV\) and interpret the sign.
Solution
Material form: \(\tfrac{d}{dt}\int_{V(t)}\rho\,dV=\int_V(\tfrac{D\rho}{Dt}+\rho\nabla\cdot\mathbf v)\,dV\). Integrand uniform: \(=(\tfrac{D\rho}{Dt}+\rho\nabla\cdot\mathbf v)V_0\). Numbers: \((-0.8 + 2(0.4))(0.5)=(-0.8+0.8)(0.5)=0\). \(\boxed{0\ \mathrm{kg\,s^{-1}}}\). Interpretation: the parcel's density is dropping at exactly the rate its volume expands (\(\dot V/V=\nabla\cdot\mathbf v\)), so total mass is conserved — consistent with continuity \(\tfrac{D\rho}{Dt}+\rho\nabla\cdot\mathbf v=0\). The parcel is neither gaining nor losing mass. Units: \(\mathrm{kg\,m^{-3}\,s^{-1}}\cdot\mathrm{m^3}=\mathrm{kg\,s^{-1}}\). ✓