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Coolant mass flow rate per open flow area, measured in $\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. A decrease in total flow does not establish a decrease in mass flux when the open area also changes.

Take a reservoir with constant horizontal cross-sectional area $A$, one small orifice of area $a$, no inflow, and constant stipulated discharge coefficient $C_d>0$. Let the initial head be $H>0$. For constant mass density $\rho$, the stored mass above the outlet level is $\rho Ah$ and outgoing mass flow rate is $\rho Q$. Mass conservation cancels $\rho$ and gives the initial value problemThe downward speed of the free surface is $Q/A$, smaller than the ideal jet speed by the factor $C_{d}a/A$. Thus a small opening relative to the reservoir supports the negligible-surface-velocity assumption.

For $h>0$, separation of variables or the chain rule gives $d\sqrt{h}/dt=-k/2$. Integrating yieldsThe model reaches zero head in finite time even though volumetric flow rate tends to zero. Extending the squared expression beyond $T$ would make head grow again and violate the original ordinary differential equation. The zero-head continuation enforces no further outflow. For an elevated side opening, zero head means the surface has reached that opening, leaving fluid below it; $T$ is not the time to remove every drop from such a tank.

For an original synthetic example, stipulateThe initial volume above outlet level is $AH=0.0250{\mathrm{m}}^3=25.0\mathrm{L}$. The initial ideal jet speed is $2.215\mathrm{m}{\mathrm{s}}^{-1}$, the initial volumetric flow rate is $Q_0=0.1373\mathrm{L}{\mathrm{s}}^{-1}$, and the predicted zero-head time is $T=364.1\mathrm{s}$. These are consequences of stipulated parameters, not experimental measurements. At $t=T/2=182.1\mathrm{s}$, head is $H/4=0.0625\mathrm{m}$, stored volume is $6.25\mathrm{L}$, and volumetric flow rate is $Q_0/2=0.06866\mathrm{L}{\mathrm{s}}^{-1}$. Half the initial volume has already left at $t=T(1-1/\sqrt{2})=106.7\mathrm{s}$.

An independent mass conservation check integrates the outgoing volumetric flow rate:Multiply both sides by mass density to obtain discharged mass. A dimensional analysis check gives $[k]={\mathrm{m}}^{1/2}{\mathrm{s}}^{-1}$, $[Q]={\mathrm{m}}^3{\mathrm{s}}^{-1}$ and $[T]=\mathrm{s}$. The triangular area under $Q(t)$ explains why the initial-rate estimate $AH/Q_0$ is only half the actual model time. Doubling $A$ doubles $T$; doubling $a$ or $C_d$ halves $T$, within the same assumptions.

This finite-time result is an idealized prediction. The constant $C_d$ and quasi-steady approximation need not survive very small head or the initial startup. Appreciable viscosity in a narrow outlet tube requires a different pressure balance, as in Poiseuille flow; appreciable surface tension can prevent zero head altogether.

Heat-engine calorimetry measures heat transfers so that observed work can be compared with the correct energy input. Measuring motion alone cannot supply the denominator of thermal efficiency.

The institutional Laboratory 16: Stirling Engine describes a modern teaching arrangement: an ethanol lamp drives a Stirling engine coupled to an electrical output device, which supplies a resistor immersed in water in a calorimeter. The prescribed readouts include fuel mass consumed, the pressure-volume loop, electrical readings, and water temperature. The sequence separates fuel input, gas-cycle work, electrical output, and Joule heating of the water. Warming of the water is the final useful output; it is not the engine's rejected $Q_c$. The handout distinguishes warm-up from steady operation and asks students to compare the final warming with direct heating by the lamp. It gives a procedure, not a table of measured efficiencies. Its pp. 1–3 therefore establish an apparatus and measurement design without supplying a new experimental result.

For a proposed calibration, use a known input of electrical work to determine the effective heat capacity $C_{\mathrm{eff}}$ of the water, vessel and immersed parts; stir to make the measured temperature representative. A matched run with no input estimates exchange with the room. For constant $C_{\mathrm{eff}}$ and no phase transition, the resulting model iswhere $Q_{\mathrm{leak},\mathrm{out}}$ is the signed net heat lost to the room during the measurement. Here $E_{\mathrm{in}}$ is supplied electrical work when the resistor lies inside the calibration boundary; for a separate external heat input it is the received heat. These are proposed controls, not controls reported in that handout. Slow cooling, changing material properties and imperfectly matched room exchange limit this inference; the approach to thermal equilibrium and heat capacity matter as much as the thermometer reading.

To measure the actual engine-input $Q_h$, one must instrument its hot-side transfer separately, or close a calibrated energy balance around a specified heater boundary. A heater's electrical work includes changes in its stored internal energy and heat transfer that bypasses the engine. At periodic operation a cold-side flowing-water measurement can instead estimate ${\dot{Q}}_c=\dot{m}{c}_p(T_{\mathrm{out}}-T_{\mathrm{in}})$, if mass flow rate $\dot{m}$ and specific heat capacity at constant pressure $c_p$ are known, with negligible storage and other heat exchange in that measurement boundary. It measures only the rejection captured by that stream. Include all remaining transfers before using the residual to infer ${\dot{Q}}_h=P_{\mathrm{out}}+{\dot{Q}}_c$. This measurement discipline makes a comparison with Carnot's theorem meaningful.

For constant gravity and downward rocket exhaust speed $u$, hovering requires rocket thrust $u\alpha =Mg$ with $\alpha =-\dot{M}$. Hence $M=M_0{e}^{-gt/u}$ and the required mass flow rate decays exponentially. A nonzero rocket dry mass ends the hover at finite time, even though this mass equation tends to zero only asymptotically.

An incompressible flow preserves the volume of each moving fluid element. Its local condition is zero velocity divergence, $\nabla \cdot \mathbf{u}=0$. Combined with mass conservation, it says that each parcel's mass density stays unchanged as it moves. This does not require every parcel to have the same mass density: an initially nonuniform distribution can be transported without compression. Nor does incompressibility require a constant velocity or constant pressure.

For positive mass density, the continuity equation makes the equivalence precise:The material derivative follows a fluid element. Thus $D\rho /Dt=0$ if and only if the velocity field is a divergence-free vector field. The distinction between following a parcel and observing a fixed position prevents confusing incompressibility with an unchanging density field. ^[ref]

For a rigid pipe with impermeable walls, no leaks and steady flow, mass conservation at constant mass density implies a constant volumetric flow rate:Here $\mathbf{n}$ points downstream, $A$ is the section's area and $\overline{U}$ is the mean normal velocity, not necessarily the centreline speed. No assumption of inviscid flow or a uniform velocity profile is needed. If unequal amounts entered and left a fixed segment, its mass would accumulate, contradicting steady flow.

For a worked example, take horizontal water flow with stipulated mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, inlet radius $R_1=0.010\mathrm{m}$, outlet radius $R_2=0.005\mathrm{m}$ and inlet mean velocity ${\overline{U}}_1=0.20\mathrm{m}{\mathrm{s}}^{-1}$. The circular areas giveThe radius halves, the area falls to one quarter, and the mean velocity quadruples. The mass flow rate is $\rho Q=0.02\pi \mathrm{kg}{\mathrm{s}}^{-1}$. The same amount of water passes both sections each second; narrowing changes its speed rather than compressing it. The Reynolds number and Bernoulli's principle discussions use this same example to examine viscous effects and the additional assumptions needed to calculate pressure.

Steady incompressible mass flow rate stays constant along a jet, while speed and cross-sectional area can change under gravity. At a horizontal apex the arriving speed is the initial horizontal component. The force on a target that stops this horizontal momentum is $\dot{m}{v}_{\mathrm{hit}}$, using the conserved nozzle mass flux. Substituting the nozzle area into $\rho A_{\mathrm{hit}}{v}_{\mathrm{hit}}^2$ without adjusting the area gives the wrong result.

A steady uniform incompressible jet has mass flow rate $\rho Au$, momentum flux $\rho A{u}^2$ and kinetic-energy flux $\rho A{u}^3/2$. These use the area and velocity at the same cross section. A nonuniform profile instead requires integrating powers of local speed; using mean speed in a cubic flux without a profile correction is generally wrong.

Consider an original model of a cold section enclosing a throttle and collecting region, with the precooling heat exchanger outside its boundary. The inlet has mass flow rate $\dot{m}>0$ and specific enthalpy $h_{\mathrm{in}}$. The outlets are equilibrium liquid and vapor at one fixed pressure, with specific enthalpies $h_l$ and $h_g$. Let $y$ be the outlet liquid mass fraction, and let $q=\dot{Q}/\dot{m}\ge 0$ be incoming unwanted heat per unit incoming mass. Assume steady flow, no shaft work, and negligible bulk kinetic energy and gravitational potential energy changes. The steady-flow energy balance givesHere $L$ is the latent heat per unit mass at the chosen outlet pressure. The mass flow rates of the outlets are $y\dot{m}$ and $(1-y)\dot{m}$. The formula describes a flowing cold section; a filling vessel with appreciable energy or mass accumulation requires storage terms.

For a deliberately dimensionless example, choosePer unit incoming mass, one quarter leaves as liquid and three quarters as vapor. With zero incoming heat, the same inlet would give $y=0.30$. Increasing $q/L$ to $0.20$ reduces $y$ to $0.10$, and $q/L=0.30$ reaches the no-liquid boundary. These are chosen model values, not helium property data or a historical yield. In normalized enthalpy units with $h_l=0$ and $h_g=1$, conservation checks directly: $0.70+0.05=0.25\times 0+0.75\times 1$.

The admissible interval is $h_l\le h_{\mathrm{in}}+q\le h_g$, or $0\le y\le 1$. Outside it, the assumed equilibrium two-phase outlet fails: an enthalpy above $h_g$ requires a hotter single-vapor outlet under these assumptions, rather than a negative amount of liquid. Below $h_l$, a colder liquid or a different thermodynamic phase must be considered. The balance does not predict its own outlet pressure, temperature or helium property functions.

At fixed outlet pressure, hence fixed $L$, the derivatives are $\partial y/\partial h_{\mathrm{in}}=-1/L$ and $\partial y/\partial q=-1/L$. Lowering inlet specific enthalpy by regeneration and reducing incoming heat therefore buy the same increase in liquid mass fraction per unit energy per unit mass in this model. Adding any constant to all three specific enthalpies leaves $y$ unchanged. This independence of the arbitrary enthalpy reference makes the balance useful even when property tables choose different zero points.

Mass flux is the mass crossing a unit area per unit time. In a one-dimensional flow it is $\rho u$. Integrating its normal component over a surface gives the mass flow rate.

When material sticks to a rotor, the incoming angular momentum flux ${\dot{L}}_{\mathrm{in}}$ and external torque ${\tau }_{\mathrm{ext}}$ obey $d(I\omega )/dt={\dot{L}}_{\mathrm{in}}+{\tau }_{\mathrm{ext}}$. With zero incoming axial angular momentum and zero external axial torque, $I\omega$ is constant. For deposition at fixed radius $a$ at mass flow rate $\alpha$, $\dot{I}=\alpha a^2$ and the rotational kinetic energy satisfies $\dot{K}=-\alpha a^2{\omega }^2/2$. This loss is the irreversible cost of bringing the incoming material up to tangential speed; it does not violate conservation of angular momentum.

The momentum thickness weights the streamwise velocity deficit by the local velocity carrying it:Thus $\rho U^2\theta$ measures a momentum flux deficit per unit span relative to transporting the same local mass flow rate at $U$. It is not $\rho \int (U^2-u^2)dy$, which compares a different reference flux. A Blasius boundary layer has $\theta \simeq 0.664115\sqrt{\nu x/U}$; the ratio of displacement thickness to momentum thickness is about $2.591$.^[ref]

For the steady flow over an impermeable flat plate at constant $U$, integrating the boundary-layer equation and using mass conservation givesThis connects growth of the momentum thickness to local wall shear stress. Consequently the drag per unit span on one face up to $x=L$ is $D^{\prime }=\rho U^2\theta (L)$ in the ideal Blasius boundary layer with $\theta (0)=0$. Measured velocity profiles can therefore estimate integrated drag without differentiating noisy near-wall readings. The identity assumes no wall suction and zero outer pressure gradient.^[ref]

For a barotropic fluid, define $h(\rho )=\int dp/\rho$. Dotting the steady Euler equation with $\mathbf{u}$ givesThus the Bernoulli function $H=u^2/2+h+\psi$ is constant along each streamline. Different streamlines can have different constants; steady flow alone does not imply one spatially uniform $H$.

In a quasi-one-dimensional flow through area $A(z)$ with negligible gravity, the axial momentum equation and $dp=c_s^{2}d\rho$ giveMeanwhile the mass flow rate $\dot{M}=\rho uA$ is constant, so $d\ln \rho =-d\ln u-d\ln A$. Substitution proves the nozzle area-velocity relationA subsonic flow accelerates where $A$ decreases, whereas a supersonic flow accelerates where $A$ increases. A smooth accelerating transonic branch in a de Laval nozzle therefore crosses $u=c_s$ at a minimum-area throat. Suitable reservoir and downstream pressures are also required: a converging-diverging shape by itself does not exclude a wholly subsonic solution or a downstream shock.

For the stipulated constant-$K$ polytropic flow, $c_s^2=\gamma K{\rho }^{\gamma -1}$. At the throat $u_m=c_{s,m}$, soThe specific enthalpy is $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)=c_s^2/(\gamma -1)$. Evaluating $H$ at the throat and substituting $\rho =\dot{M}/(Au)$ elsewhere givesOn the outward supersonic branch, $A\sim C{r}^2$ lets the gas expand and its specific enthalpy tend to zero for $\gamma >1$. ConsequentlyFor $\gamma =5/3$, $u_{\infty }=2c_{s,m}$. Also $T\propto {\rho }^{\gamma -1}\propto r^{-2(\gamma -1)}$. These are the adiabatic nozzle terminal speed and expansion laws for a lossless escaping branch; a subsonic branch could instead approach finite pressure with $u\propto A^{-1}$. A pressure-shaped astrophysical jet is being approximated by a nozzle here: the assumed geometry does not by itself explain the confinement or account for magnetic forces or relativistic motion in an actual active galactic nucleus.

In an adiabatic outflow without additional heating, the reservoir's specific enthalpy supplies both escaping kinetic energy and the work against gravity. For a monatomic ideal gas, $\gamma =5/3$ and $h=c_s^2/(\gamma -1)=3c_s^2/2$, where $c_s$ is the reservoir's adiabatic sound speed. The Bernoulli equation, measured from a reservoir initially at rest at radius $r$, givesThe maximum terminal speed uses all available specific enthalpy, so $h_{\infty }\to 0$ andA freely escaping adiabatic flow needs a nonnegative radicand; equality is marginal escape. If the radicand is negative, this reservoir cannot supply an unheated escaping outflow. The initial rest condition is a stagnation-reservoir approximation: an exactly stationary finite-density surface cannot simultaneously carry a nonzero steady mass flow rate. Further heating, radiation forces or an imposed confining pressure would change the terminal-energy budget.

An accretion disk forms when infalling material has enough angular momentum to orbit before it can reach the central object. In a thin disc dominated by a central mass $M$, the orbital motion is approximately Keplerian rotation:Vertical hydrostatic equilibrium gives the disk scale height $H\sim c_s/{\Omega }_K$; a geometrically thin disc has $H/R\ll 1$. The surface density is $\Sigma =\int \rho dz$, and a steady inward mass flow rate is $\dot{M}=-2\pi R\Sigma v_R>0$.

Moving inward requires losing specific angular momentum. Internal stresses transport angular momentum outward; much of the mass can then accrete while an outer fraction spreads to larger radius. Molecular viscosity alone is generally inadequate for rapid astrophysical accretion. Turbulent and magnetic stresses, including those generated by the magnetorotational instability, provide effective transport. A common parametrization is $\nu =\alpha c_{s}H$ for an effective kinematic viscosity, with $\alpha$ a dimensionless stress parameter.

For a thin Keplerian accretion disk, the evolution equation isThis viscous evolution of an accretion disk occurs on the viscous timescale $t_{\nu }\sim R^2/\nu$, much longer than an orbital period in a thin disc with moderate effective stress.

The orbital specific energy is $-GM/(2R)$, so inward motion releases binding energy. In a steady, optically thick, radiatively efficient disc with zero torque at its inner edge, the emergent flux from each face isThe Stefan–Boltzmann law identifies the local effective temperature. Far from the inner boundary, $T_{\mathrm{eff}}\propto R^{-3/4}$, so the emitted spectrum combines hotter inner and cooler outer annuli. Integrating both faces gives $L_{\mathrm{disc}}=GM\dot{M}/(2R_{\mathrm{in}})$ under these assumptions. This is not a universal spectrum or efficiency: optical depth, inner torque, energy advection and outflows can change the radiation. The essential connection is outward angular-momentum transport, inward accretion, and release of binding energy.

Hydrostatic equilibrium with self-gravity. A resting fluid obeys $\nabla p=-\rho \nabla \Psi$. Spherical symmetry and the Poisson equation for Newtonian gravity giveEliminate the Newtonian gravitational potential derivative to obtainEquivalently, the shell theorem gives ${\Psi }^{\prime }=GM(r)/r^2$ and $M^{\prime }(r)=4\pi r^2\rho$, which leads to the same hydrostatic equilibrium equation.

A uniform-density liquid. With constant $\rho$, $M(r)=4\pi \rho r^3/3$. There is no central point mass, so the interior gravitational acceleration magnitude is $4\pi G\rho r/3$. ThusUse $p(R)=0$ at a vacuum boundary, neglecting surface tension. The uniform-density planetary pressure profile isThe central value can also be writtenIt has units pascals and decreases smoothly from its maximum at the centre to zero at the surface.

An isothermal gas. Here $c$ is the constant isothermal sound speed, with $p=c^2\rho$. Try $p=A/r^2$, so $\rho =A/(c^2{r}^2)$. The two sides of the equilibrium equation becomeThus $A=c^4/(2\pi G)$, giving the singular isothermal sphere,This does not describe a finite gaseous sphere in vacuum. Both pressure and mass density diverge at the centre, and $p(R)>0$ for every finite outer radius $R$, so it cannot meet $p=0$ there. Extending it to infinity removes the finite-radius pressure jump but makes the total mass infinite. It is useful as an idealized profile over an intermediate radial range, or as a truncated model with nonzero external pressure. The vacuum boundary obstruction for an isothermal sphere also applies to the regular, nonsingular isothermal solution; removing the central singularity does not produce a finite vacuum surface.

Steady outflow. In the wind calculation, take the central mass $M$ to be fixed and neglect the wind's own self-gravity. Constant mass flow rate and the isothermal equation of state giveThe steady radial Euler equation isSubstituting the density derivative givesThis is the isothermal Parker wind equation. A smooth sonic point requires both sides to vanish when $u=c$, soMaintaining constant temperature during expansion generally requires heat input; it is an isothermal model, not the adiabatic blast-wave model of Question 5.

The escaping branch at large radius. Integrate the wind equation, using dimensionless logarithm arguments:Set $x=r/r_s$ and $w=u^2/c^2$. ThenFor the accelerating transonic branch, $w(1)=1$, which fixes $C_0=-3$. At $x\gg 1$ on its supersonic branch, $w\to \infty$ and $\ln w$ is smaller than $w$. ConsequentlyA first correction makes the ordering explicit:This is the large-radius asymptotics of an isothermal Parker wind. The leading expression also holds for other supersonic branches with finite $C_0$.

The source implicitly selects an escaping wind. Positive steady outward flow alone does not determine that branch: a subsonic isothermal breeze also satisfies the differential equation. Its $w\ll 1$ limit gives $w\sim e^{-C_0}{x}^{-4}$, hence $u\propto r^{-2}$ and $\rho \to {\rho }_{\infty }>0$. Such a breeze needs nonzero pressure at infinity. Selecting a regular outflow from a subsonic base into vanishing ambient pressure instead gives the accelerating transonic wind used above.

The asymptotic speed has the same units as $c$; the logarithm is dimensionless. Its unlimited growth is a formal feature of indefinite isothermal heating and a prescribed point-mass potential. Real heating, surrounding pressure, and eventually the neglected wind mass limit that extrapolation.

Use the ideal one-dimensional pipe model: constant density, steady inviscid flow, approximately uniform endpoint speeds $u_1,u_2$, and equal body-force potential at the two ends. Conservation of mass gives the constant volumetric flow rate $q=u_1{S}_1=u_2{S}_2$, and the mass flow rate is $m=\rho q$. The steady Bernoulli equation givesFor flow from a wider to a narrower section, $S_1>S_2$ and $\Delta p>0$. Solving for the positive mass throughput gives the Venturi effect formulaThe formula requires these hydraulic assumptions; steady incompressibility alone does not impose uniform endpoint profiles or equal potential. If $V_1\ne V_2$, the same derivation replaces $\Delta p$ by $\Delta p+\rho (V_1-V_2)$. Viscous losses would also modify the pressure balance. The equal-area case is singular in this pressure-based determination: in the ideal equal-potential model its pressure difference must vanish and does not determine the throughput.

Continue the same-rod-number and same-height comparison from part(a), and assume unchanged pump speed, similar inlet conditions and single-phase forced flow for the normal-operation hydraulic argument.

Core power density increases. The active core volume decreases while total thermal power stays fixed. The fuel's own average volumetric power and the mean fuel linear heat rate do not increase merely from bringing the same rods closer together: fuel volume and rod count are unchanged.

Coolant mass flow rate normally decreases. Tighter passages reduce flow area and hydraulic diameter, increasing the core's hydraulic resistance. The unchanged centrifugal pumps do not enforce constant flow. The pump operating point moves to the intersection of the same pump curve with the steeper system-resistance curve, normally giving smaller flow at greater pump head.^[ref]

The core pressure drop normally increases for the usual pump curve. Its resistance at any given flow is higher. At the new, smaller flow, unchanged external piping and component losses are smaller, whereas a conventional falling pump head-versus-flow curve gives greater available head. The core takes the difference. This conclusion uses the stated pump-curve assumptions; holding flow fixed or holding pump head fixed would be different models. The exact new flow and pressure drop need both system and pump characteristics.

Fuel temperature would normally tend to rise, especially downstream, if the loss of flow increases coolant heating without a compensating improvement in film heat transfer. At fixed inlet state,shows that smaller total mass flow rate raises the bulk coolant temperature rise. At unchanged local fuel linear heat rate, the fuel-to-surface conductive contribution is approximately unchanged if material properties and gap conditions are held fixed. Its boundary temperature and the coolant film drop can nevertheless change.

The local coolant mass flux is $G=\dot{m}/A_f$. Both total flow and flow area decrease, so the sign of the change in $G$, the heat transfer coefficient and the fuel temperature cannot be deduced from total flow alone. Changes in power peaking and inlet conditions add further qualifications to the usual hotter-fuel tendency.

The minimum departure from nucleate boiling ratio would commonly decrease, implying reduced boiling margin, but its sign is conditional. The defining local ratio isThe numerator is the critical heat flux for the local conditions, and the denominator is actual rod-surface heat flux.^[ref] The mean surface heat loading remains approximately unchanged for the same rods and total power. Greater coolant enthalpy rise tends to reduce margin, while changes in $G$, geometry, pressure and local power determine the actual critical heat flux. A geometry-appropriate correlation or experiment is needed for a definite MDNBR change; unchanged hardware does not supply it.

Let $\mu$ be the instantaneous total mass of the rocket, including its remaining rocket propellant. During a short ejection it changes by $d\mu <0$. The expelled mass $-d\mu$ has velocity $v-v_0$ to first order in a fixed inertial frame. With no external force, momentum conservation readsDiscarding the second-order product gives $\mu dv=-v_{0}d\mu$. Integrate from initial total mass $m_0+M$ to rocket dry mass $m_0$:For constant mass flow rate $\alpha =-\dot{\mu }>0$, the same rocket equation gives $\mu (t)=m_0+M-\alpha t$ and $v(t)=v_0\ln [(m_0+M)/\mu (t)]$ until $t=M/\alpha$. The final result depends on the rocket mass ratio, not on how rapidly the fuel is ejected. These are Newtonian velocities; a very large rocket mass ratio would require a relativistic treatment even if the exhaust is slow in the instantaneous rocket frame.

Take the rocket exhaust speed $u$ relative to the hovering ship, with exhaust directed vertically downward. If $\alpha =-\dot{M}>0$ is the mass flow rate, the upward rocket thrust is $u\alpha$. For example, over a short interval $dt$, the expelled mass $\alpha dt$ acquires downward momentum $-u\alpha dt$, giving the opposite recoil to the ship. Balancing its weight requiresSolving this differential equation with $M(0)=M_0$ yields the hovering rocket mass flowThe required mass flow rate decreases because the mass to support decreases. These expressions apply while fuel is available and $g,u$ remain constant. If the rocket dry mass is $M_d>0$, hovering ends at $t=(u/g)\log (M_0/M_d)$; the mathematical limit $M\to 0$ is not a model of an indefinitely fuelled physical ship.

Let $V(t)$ have a boundary moving with velocity ${\mathbf{v}}_b$, and let $\mathbf{n}$ be its outward unit normal vector. For a smooth scalar field $q$, the Reynolds transport theorem isIn the last expression, extend ${\mathbf{v}}_b$ smoothly into $V(t)$ and use the divergence theorem. For a material volume, its boundary moves with the fluid velocity $\mathbf{u}$, so ${\mathbf{v}}_b=\mathbf{u}$. Equivalently, the Reynolds transport theorem for a material volume readswhere $D/Dt$ is the material derivative.

For an arbitrary moving region, mass conservation includes the relative mass flow rate across its boundary:Substitution of the Reynolds transport theorem and the divergence theorem givesSince the region is arbitrary, a continuous integrand must vanish: if it had a positive or negative value at a point, a sufficiently small surrounding region would have an integral of the same sign. This gives the continuity equation. With constant positive mass density $\rho$, it reduces toThus the flow is incompressible. For a material volume, the same argument starts directly from conservation of its total mass, with zero relative boundary flux.