11.6 Control-Volume Analysis: Continuity, Momentum & Energy

System Versus Control Volume

A system is a fixed quantity of matter. Newton's laws and thermodynamic laws are naturally stated for systems, because they track the same particles.

A control volume is a fixed region in space through which fluid flows. This is what an engineer actually wants, because nobody follows individual fluid particles through a pump; they want to know what enters and leaves the casing.

The Reynolds transport theorem bridges the two. For any extensive property $B$ with intensive counterpart $b$ per unit mass:

$\frac{dB_{\text{sys}}}{dt} = \underbrace{\frac{\partial}{\partial t}\int_{CV} b\rho\,d\forall}_{\text{rate of storage inside}} + \underbrace{\int_{CS} b\rho\,(\mathbf{V}\cdot \mathbf{n})\,dA}_{\text{net flux out across the surface}}$

Substituting $b = 1$, $b = \mathbf{V}$ and $b = e$ produces the continuity, momentum and energy equations respectively. Everything in this section follows from that one statement.

Continuity: Conservation of Mass

Setting $B = m$ and $b = 1$, with $dm_{\text{sys}}/dt = 0$:

$\frac{\partial}{\partial t}\int_{CV}\rho\,d\forall + \int_{CS}\rho(\mathbf{V}\cdot\mathbf{n})\,dA = 0$

For steady flow the storage term vanishes, leaving

$\sum \dot{m}_{\text{out}} = \sum \dot{m}_{\text{in}}, \qquad \dot m = \rho A V$

For steady incompressible flow the density cancels:

$A_1V_1 = A_2V_2 = Q$

This is why a nozzle accelerates flow: halving the area doubles the velocity.

Differential form

Applying the same argument to an infinitesimal element gives the differential continuity equation:

$\frac{\partial \rho}{\partial t} + \frac{\partial(\rho u)}{\partial x} + \frac{\partial(\rho v)}{\partial y} + \frac{\partial(\rho w)}{\partial z} = 0$

which for incompressible flow reduces to the divergence-free condition

$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0$

A velocity field is a possible incompressible flow only if it satisfies this — a standard objective item where a field is given and the candidate must test it.

Linear Momentum

Setting $b = \mathbf{V}$ and using Newton's second law for the system:

$\sum \mathbf{F} = \frac{\partial}{\partial t}\int_{CV}\rho\mathbf{V}\,d\forall + \int_{CS}\mathbf{V}\rho(\mathbf{V}\cdot\mathbf{n})\,dA$

For steady flow with one inlet and one outlet:

$\boxed{\sum \mathbf{F} = \dot{m}\left(\mathbf{V}_{\text{out}} - \mathbf{V}_{\text{in}}\right)}$

The forces on the left include pressure forces on the inlet and outlet faces, body forces such as weight, and the reaction from the solid boundary, which is usually what is being sought.

Working procedure

1. Draw the control volume, cutting through the inlet and outlet.
2. Mark all forces acting on the fluid inside it.
3. Choose axes and a sign convention, and stay with it.
4. Write the momentum equation separately for each component.
5. The force on the solid boundary is equal and opposite to the force on the fluid.

Step 4 is where most errors occur — momentum is a vector, and treating it as a scalar loses the direction information that is usually the point of the question.

Application 1: Force on a Pipe Bend

A horizontal pipe bend turns flow through angle $\theta$. Taking the inlet along $x$ and applying the momentum equation to each component:

$F_x = p_1A_1 - p_2A_2\cos\theta - \dot m\left(V_2\cos\theta - V_1\right)$

$F_y = -p_2A_2\sin\theta - \dot m\,V_2\sin\theta$

The resultant force on the bend is $\sqrt{F_x^2 + F_y^2}$, acting opposite to the force on the fluid.

Worked example. Water at 300 kPa gauge flows at 0.1 m$^3$/s through a 90-degree bend of constant 100 mm diameter.

Area $= \pi(0.05)^2 = 7.854\times10^{-3}$ m$^2$, so $V = 0.1/7.854\times10^{-3} = 12.73$ m/s and $\dot m = 100$ kg/s.

With $\theta = 90^\circ$, taking pressure as unchanged for a constant-area bend:

$F_x = pA + \dot m V_1 = 300000(7.854\times10^{-3}) + 100(12.73) = 2356 + 1273 = 3629 \text{ N}$

$F_y = -\left[pA + \dot m V_2\right] = -3629 \text{ N}$

Resultant $= 3629\sqrt2 = 5132$ N at 45 degrees to the inlet. This is why large pipe bends need thrust blocks — and note that the pressure term dominates the momentum term here, a point often overlooked.

Application 2: Jet on a Vane

The hemispherical result — that reversing a jet gives twice the force of stopping it — is the physical basis of the Pelton wheel bucket shape, and the reason buckets are split with a central ridge and turned through slightly less than 180 degrees so the deflected water clears the following bucket.

For a single moving flat plate, the power is $F u = \rho A(V-u)^2u$, maximised when $u = V/3$. For a series of vanes on a wheel, maximum efficiency occurs at $u = V/2$ — the standard result for impulse turbines.

Angular Momentum: Euler's Turbomachine Equation

Applying the moment-of-momentum form of the transport theorem to a rotor gives

$T = \dot m\left(r_2 V_{w2} - r_1 V_{w1}\right)$

and multiplying by angular velocity $\omega$ gives the power, or per unit weight of fluid the Euler head:

$H_e = \frac{u_2V_{w2} - u_1V_{w1}}{g}$

Every pump, fan, compressor and turbine analysis in the turbomachinery topic descends from this one control-volume result.

Energy Equation

Setting $b = e$, the energy per unit mass, and applying the first law of thermodynamics to the control volume gives the steady flow energy equation:

$\dot Q - \dot W_s = \dot m\left[\left(h_2 - h_1\right) + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1)\right]$

For an incompressible fluid with no shaft work, no heat transfer and no friction, this collapses to Bernoulli's equation:

$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + z_2$

With losses and machines included, the practical working form is

$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} + z_1 + h_{\text{pump}} = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_{\text{turbine}} + h_{L}$

Which Equation to Use

This choice is the practical skill the topic teaches.

A critical distinction: the momentum equation does not require the flow to be frictionless, because internal losses are internal forces that cancel. Bernoulli's equation does. So for a sudden expansion, where losses are significant, momentum gives the correct force while Bernoulli would not — and indeed applying momentum and energy together to a sudden expansion is precisely how the Borda-Carnot head loss $ (V_1-V_2)^2/2g$ is derived.