Exercises — Reynolds transport theorem

They are wrong. That term is the surface flux — the net rate at which the property is carried across the boundary by moving fluid. Plain-word names:

• Left, $\frac{d{B}_{sys}}{dt}$ = "the rate of change following the original lump of matter" (Lagrangian).
• Storage, $\frac{d}{dt}{\int }_{CV}\rho bdV$ = "how fast the amount of stuff sitting inside the fixed box is changing".
• Surface, ${\oint }_{CS}\rho b(\vec{v}\cdot \hat{n})dA$ = "net outflow of stuff through the walls" (out minus in).

The colleague confused inside change (storage) with crossing the wall (surface).

$b=\frac{dB}{dm}$ = property per unit mass.

• (a) Mass: $B=m$, so $b=\frac{dm}{dm}=1$.
• (b) Momentum: $B=m\vec{v}$, so $b=\vec{v}$.
• (c) Energy: $B=E$, so $b=e$ (specific energy, energy per kilogram).

Set up RTT with $b=1$ (mass): $\frac{d{m}_{sys}}{dt}=0$, and steady flow kills storage: $0=\underset{=0\text{ (steady)}}{\underbrace{\frac{d}{dt}{\int }_{CV}\rho dV}}+{\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA.$ Evaluate the surface integral over the two ports. At the inlet the fluid flows in, opposite to the outward normal ${\hat{n}}_1$, so $\vec{v}\cdot {\hat{n}}_1=-v_1$. At the outlet fluid flows out, along ${\hat{n}}_2$, so $\vec{v}\cdot {\hat{n}}_2=+v_2$. The pipe walls contribute nothing ($\vec{v}\cdot \hat{n}=0$ there). Thus: $0=\rho (-v_1)A_1+\rho (+v_2)A_2\Rightarrow v_2=\frac{A_1}{A_2}{v}_1=\frac{0.02}{0.008}(3)=7.5\text{m/s}.$ Mass flow: $\dot{m}=\rho A_1{v}_1=1000(0.02)(3)=\boxed{60\text{kg/s}}$ (same at both ports, as it must be).

$b=1$, but now NOT steady — the tank content is growing, so keep the storage term: $0=\frac{d}{dt}{\int }_{CV}\rho dV+{\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA.$ Only one port (inflow), $\vec{v}\cdot \hat{n}=-v$, so the surface term $=-{\dot{m}}_{in}=-4$. Density constant, pull it out: $\frac{d}{dt}\int \rho dV=\rho \frac{dV}{dt}$. Then: $0=\rho \frac{dV}{dt}-{\dot{m}}_{in}\Rightarrow \frac{dV}{dt}=\frac{{\dot{m}}_{in}}{\rho }=\frac{4}{1000}=\boxed{0.004{\text{m}}^3/\text{s}}.$

Use RTT with $b=\vec{v}$ (momentum), steady ⇒ no storage: $\sum \vec{F}={\oint }_{CS}\rho \vec{v}(\vec{v}\cdot \hat{n})dA.$ Mass flow: $\dot{m}=\rho Av=1000(1.5\times 10^{-3})(20)=30\text{kg/s}$.

Inlet ($\vec{v}=(20,0)$, flows in so $\vec{v}\cdot \hat{n}=-v$): contribution $=\rho \vec{v}(\vec{v}\cdot \hat{n})A=\dot{m}(-1)(20,0)=(-600,0)$.

Outlet ($\vec{v}=(0,20)$, flows out so $\vec{v}\cdot \hat{n}=+v$): contribution $=\dot{m}(+1)(0,20)=(0,+600)$.

Sum the flux (this equals $\sum \vec{F}$ on the water): $\sum F_x=-600+0=-600\text{N},\sum F_y=0+600=+600\text{N}.$ So the vane pushes the water with $\boxed{F_x=-600\text{N},F_y=+600\text{N}}$ (i.e. leftward-and-up to remove its rightward motion and give it upward motion). Magnitude $\sqrt{600^2+600^2}\approx 848.5\text{N}$.

The integrand is $\rho \vec{v}(\vec{v}\cdot \hat{n})$. Two things carry sign:

• $\vec{v}$ itself is $+x$ (the momentum being carried), factor $+20$.
• $(\vec{v}\cdot \hat{n})$ is $-v$ at an inlet (velocity opposes the outward normal), factor $-20$.

Their product is negative: momentum $+x$ is entering, and "entering" is written as a negative outflow. Physically, momentum flooding in through the inlet is momentum the CV gains, so to keep the system's momentum steady the vane must supply an equal-and-opposite force. The negative sign is exactly what later flips (via $\sum F=\text{outflow}-\text{inflow}$) to give the reaction force.

Step 1 — continuity ($b=1$) gives $v_2$: $v_2=\frac{A_1}{A_2}{v}_1=\frac{0.01}{0.002}(2)=10\text{m/s},\dot{m}=\rho A_1{v}_1=1000(0.01)(2)=20\text{kg/s}.$ Step 2 — momentum ($b=\vec{v}$), steady. The $x$-momentum equation: $\sum F_x={\oint }_{CS}\rho v_x(\vec{v}\cdot \hat{n})dA=\dot{m}{v}_2-\dot{m}{v}_1=\dot{m}(v_2-v_1).$ (Outlet $+\dot{m}{v}_2$; inlet $-\dot{m}{v}_1$ because inflow carries the minus.) $\sum F_x=20(10-2)=160\text{N}.$ Step 3 — list the real forces making up $\sum F_x$: pressure on the inlet face pushes fluid in the $+x$ direction: $+p_1{A}_1$. Pressure on the exit face is atmospheric gauge $=0$: $+p_2{A}_2=0$. Plus the wall reaction $R_x$: $\sum F_x=p_1{A}_1+p_2{A}_2+R_x=50000(0.01)+0+R_x=500+R_x.$ Step 4 — solve: $500+R_x=160\Rightarrow R_x=160-500=\boxed{-340\text{N}}$. The nozzle pulls the water back ($-x$) by 340 N; by reaction the water pulls the nozzle forward with 340 N (this is the thrust you feel holding a hose nozzle).

Key idea: for a moving CV the surface term uses the relative velocity ${\vec{v}}_r=\vec{v}-{\vec{v}}_{CS}$, because only motion relative to the surface actually crosses it. $v_r=v-v_{CS}=15-5=10\text{m/s}.$ Mass flow across the moving surface uses $v_r$: ${\dot{m}}_r=\rho A{v}_r=1000(2\times 10^{-3})(10)=20\text{kg/s}.$ Momentum RTT (relative), steady in the cart frame, $b={\vec{v}}_r$: $\sum F_x=\oint \rho v_{r,x}({\vec{v}}_r\cdot \hat{n})dA.$ Inlet: relative velocity $+10$, crosses inward $({\vec{v}}_r\cdot \hat{n}=-v_r)$ ⇒ term $={\dot{m}}_r(-1)(+10)=-200$. Outlet: deflected $180^{\circ }$, relative velocity $-10$, crosses outward $(+v_r)$ ⇒ term $={\dot{m}}_r(+1)(-10)=-200$. $\sum F_x=-200+(-200)=\boxed{-400\text{N}}.$ The vane pushes the water in $-x$ with 400 N; the water pushes the cart forward (+x) with 400 N.

Start ($b=1$, fixed CV): $0=\frac{\partial }{\partial t}{\int }_{CV}\rho dV+{\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA.$ (Time-derivative moves inside as $\partial /\partial t$ since the CV is fixed.) Convert the surface flux to a volume integral with the Divergence theorem, ${\oint }_{CS}\vec{F}\cdot \hat{n}dA={\int }_{CV}\nabla \cdot \vec{F}dV$, using $\vec{F}=\rho \vec{v}$: ${\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA={\int }_{CV}\nabla \cdot (\rho \vec{v})dV.$ Combine into one volume integral: ${\int }_{CV}\left[\frac{\partial \rho }{\partial t}+\nabla \cdot (\rho \vec{v})\right]dV=0.$ This holds for every choice of CV, however small, so the bracket must vanish pointwise: $\boxed{\frac{\partial \rho }{\partial t}+\nabla \cdot (\rho \vec{v})=0.}$ This is exactly the point analogue, and it links back to the Material derivative and Continuity equation notes.

L1 :::: naming terms, choosing $b$ (Recognition) L2 :::: plugging numbers, steady vs unsteady storage (Application) L3 :::: vector momentum, inflow-minus signs (Analysis) L4 :::: continuity + momentum + pressure forces combined (Synthesis) L5 :::: moving-CV relative velocity, differential form via divergence theorem (Mastery)