Control volume analysis¶

This module covers developing our governing equations (conservation of mass, momentum, and energy) by applying control volume analysis. It also introduces the concept of the stagnation reference state.

Our analysis requires us to make a series of assumptions about the fluid flow:

In general, flow is three-dimensional and unsteady, so $V = f(t, x, y, z)$
If we assume a flow is two-dimensional and unsteady, then $V = f(t, x, y)$
We can next assume the flow is unsteady and one-dimensional: $V = f(t, x)$
Finally, we can assume one-dimensional steaedy flow: $V = f(x)$

However, one-dimensional steady flow is not the same as unidirectional flow, since the overall flow direction can change.

Total derivative¶

For a control mass, the total/material/substantial derivative describes the time rate at which a property changes as we follow it around:

\[ \frac{D (\cdot)}{Dt} \text{ or } \frac{d(\cdot)}{dt} \]

The total derivative = convective derivative (changes due to movement of fluid to a new location) + local/partial derivative (changes with time at a given location). For example, the total derivative for pressure is

\[ \frac{dp}{dt} = \frac{\partial p}{\partial x} \frac{dx}{dt} + \frac{\partial p}{\partial y} \frac{dy}{dt} + \frac{\partial p}{\partial z} \frac{dz}{dt} + \frac{\partial p}{\partial t} \;, \]

where $\frac{dx}{dt}$, $\frac{dy}{dt}$, and $\frac{dz}{dt}$ are the components of flow velocity.

If fluid properties are independent of time, then we can assume steady flow and thus

\[ \frac{\partial (\cdot)}{\partial t} = 0 \]

Note

How good is the assumption of one-dimensional flow in a duct?

For laminar flow, the velocity forms a parabolic shape, with the peak in the center and zero at the walls. Thus, $\dot{m} = \rho A \frac{V_{\max}}{2} = \rho A V$, where V is the average velocity. Clearly, the velocity varies significantly at a given axial location, and so the assumption of one-dimensional flow is not good.

For turbulent flow, the larger inertial forces lead to more mixing, and so the velocity profile is much more uniform: $V_{\max} \approx V$, so the one-dimensional assumption is reasonable.

Since most applications in gas dynamics will be at high-enough Reynolds numbers for the flow to be fully turbulent, the one-dimensional assumption is pretty good!

Total derivative → control volume¶

With fluids, it is much more convenient to work with control volumes, so we need a way to convert from equations governing a control mass (written with a total derivative) to those that apply to control volumes.

We can work with any extensive property N, where

\[ N = \int \eta \, dm \equiv \iiint \rho \eta \, d \tilde{v} = \int_V \rho \eta \, d\tilde{v} \;, \]

where N is the total amount of the property in a given mass and $\eta$ is the amount per unit mass (the intensive version of this property). The overall change in the property N is then $\frac{dN}{dt}$.

Let’s follow a given control mass of fluid from time $t$ to time $t + \Delta t$, as it moves through space. The volume of space that the control mass occupied initially is our control volume, which the mass crosses as it moves. For a small time increment $\Delta t$, the volumes occupied by the fluid at time $t$ and time $t + \Delta t$ will overlap.

We can write the overall change in N with time as

\[\begin{split} \frac{dN}{dt} \equiv \lim_{\Delta t \to 0} \frac{ (\text{final value of } N)_{t+\Delta t} - (\text{initial value of } N)_t }{\Delta t} \\ \frac{dN}{dt} = \lim_{\Delta t \to 0} \frac{ (N_2 + N_3)_{t+\Delta t} - (N_1 + N_2)_t}{\Delta t} \;, \end{split}\]

where 1+2 represents the original volume and 2+3 represents the new region, and 2 represents the overlap between the two spaces.

Let’s focus on each of the terms present in that total derivative.

The quantity

\[ \lim_{\Delta t \to 0} \frac{N_3 (t+\Delta t)}{\Delta t} \]

represents the amount of property $N$ lost by the fluid out of the control volume. (After all, region 3 is formed by the fluid moving out.) We can relate this to the fluid that crosses the surface of the control volume:

\[ \int_{S_{\text{out}}} \eta \rho \left( \vec{V} \cdot \hat{n} \right) \, dA \, \Delta t \approx \text{amount of $N$ in region 3} \;, \]

where $S_{\text{out}}$ is the area where fluid leaves the control volume, $\vec{V}$ is the velocity vector of the fluid, and $\hat{n}$ is the unit normal vector of the surface area. So, then

\[ \therefore \lim_{\Delta t \to 0} \frac{N_3 (t+\Delta t)}{\Delta t} = \int_{S_{\text{out}}} \eta \rho \left( \vec{V} \cdot \hat{n} \right) \, dA \]

and this represents the flux of $N$ out of the control volume.

We can do similar work for the flux of $N$ into the control volume:

\[ \lim_{\Delta t \to 0} \frac{N_1 (t)}{\Delta t} = \int_{S_{\text{in}}} \eta \rho \left( \vec{V} \cdot \hat{n}^{\prime} \right) \, dA \;, \]

where $\hat{n}^{\prime} = -\hat{n}$ is the unit normal vector pointing into the region.

Finally, the remaining terms are

(1)¶\[\begin{align} \lim_{\Delta t \to 0} \frac{ N_2 (t+\Delta t) - N_2 (t) }{\Delta t} &= \frac{\partial N_2}{\partial t}& \\ &= \frac{\partial N_{cv}}{\partial t}& \\ &= \frac{\partial}{\partial t} \int_{cv} \rho \eta \, d\tilde{v} \;,& \end{align}\]

because $N_2 \to N_{cv}$ as $\Delta t \to 0$.

Reynolds transport theorem¶

Now, we can rewrite the total derivative with these quantities based on the control volume:

\[ \frac{dN}{dt} = \frac{\partial}{\partial t} \int_{cv} \rho \eta \, d\tilde{v} + \int_{S_{\text{out}}} \eta \rho \left( \vec{V} \cdot \hat{n} \right) \, dA - \int_{S_{\text{in}}} \eta \rho \left( \vec{V} \cdot \hat{n}^{\prime} \right) \, dA \;, \]

but as $\Delta t \to 0$, and thus the region of overlap becomes the control volume itself, $\hat{n} = -\hat{n}^{\prime}$, and the areas where the fluid leave and enter the control volume are the control surface (CS), and so

(2)¶\[ \frac{dN}{dt} = \frac{\partial}{\partial t} \int_{CV} \rho \eta \, d\tilde{v} + \int_{CS} \eta \rho \left( \vec{V} \cdot \hat{n} \right) \, dA \;, \]

which is the Reynolds transport theorem. This expresses that the rate of change of an extensive property $N$ for a given mass as it moves around is equal to the rate of change of $N$ inside the control volume plus the net efflux (flow out - flow in) of $N$ from the control volume.

Conservation of mass¶

Conservation of mass states that, for a control mass, $\frac{d (\text{mass})}{dt} = 0$. We can apply this to a control volume by recognizing that $N = \text{mass}$ and $\eta = 1$. Then, applying this to Reynolds transport theorem (2) we get the continuity equation for a control volume:

\[ 0 = \frac{\partial}{\partial t} \int_{CV} \rho \, d\tilde{v} + \int_{CS} \rho (\vec{V} \cdot \hat{n}) \, dA \]

This applies to a general flow.

For steady one-dimensional flow we can simplify this further. First, $\frac{\partial}{\partial t} = 0$, and for a given inlet/outlet we can write

\[ \int \rho (\vec{V} \cdot \hat{n}) \, dA = \rho \vec{V} \cdot \hat{n} \int dA = \rho V A \]

Thus, for a steady one-dimensional flow, conservation of mass says that

\[ 0 = \sum \rho V A \]

over all inlets and outlets. For flow with a single inlet and single outlet, this simplifies to

\[ \dot{m} = \rho A V = \text{constant} \]

Alternately, we can express conservation of mass by differentiating this expression:

\[\begin{split} d (\rho A V) = 0 = A V \, d\rho + \rho V \, dA + \rho A \, dV \\ \frac{d \rho}{\rho} + \frac{dA}{A} + \frac{dV}{V} = 0 \, \end{split}\]

which is a useful expression for flow through a duct, channel, or streamline. Also, it shows us that if $dA = 0$ (i.e., the area is constant), then increasing flow velocity means decreasing flow density, and vice versa.

Conservation of energy¶

For a control mass, conservation of energy (i.e., the first law of thermodynamics) says

\[\begin{split} Q = W + \Delta E \\ \frac{\delta Q}{dt} = \frac{\delta W}{dt} + \frac{dE}{dt} \end{split}\]

where the first two terms are the instantaneous rates of heat and work transfer between the system and its surroundings, and $\frac{dE}{dt}$ is the total derivative of energy. In this case, $N = E$, and $\eta = e = u + \frac{V^2}{2} + gz$, so applying Reynolds transport theorem (2) gives:

\[\begin{split} \frac{dE}{dt} = \frac{\partial}{\partial t} \int_{CV} e \rho \, d\tilde{v} + \int_{CS} e \rho (\vec{V} \cdot \hat{n}) \, dA \\ \rightarrow \frac{\delta Q}{dt} = \frac{\delta W}{dt} + \frac{\partial}{\partial t} \int_{CV} e \rho \, d\tilde{v} + \int_{CS} e \rho (\vec{V} \cdot \hat{n}) \, dA \end{split}\]

For one-dimensional flow, we can simplify the integral over the control surface to consider just the inlets and outlets separately:

\[ \int_{CS} e \rho (\vec{V} \cdot \hat{n}) \, dA = \sum \dot{m} e \;, \]

where contributions are positive where fluid leaves the control volume (outlet) and negative where fluid enters the control volume (inlet).

Example: pressure-volume and shaft work¶

Consider a system with pressure-volume work from a piston and shaft work. We should choose the control surface appropriately so that there is no fluid motion at the boundary, except where

fluid enters and leaves the control volume, and
a device (e.g., shaft) crosses the boundary.

There is no pressure/shear work along the sidewalls, since the fluid velocity at the wall is zero.

The pressure work is then

\[ \delta W' = \vec{F} \cdot d\vec{x} = p A \, dx = p A V \, dt \]

and the rate of this work is

\[ \frac{\delta W'}{dt} = p A V = \dot{m} p v \;, \]

which represents the work done to push the fluid out of the control volume. The total work is then

\[ \frac{\delta W}{dt} = \frac{\delta W_s}{dt} + \sum \dot{m} p v \]

and we can rewrite conservation of energy as

\[ \frac{\delta Q}{dt} = \frac{\delta W_s}{dt} + \frac{\partial}{\partial t} \int_{CV} e \rho \, d\tilde{v} + \sum \dot{m} (e + pv) \]

For steady one-dimensional flow, this simplifies to

\[ \frac{\delta Q}{dt} = \frac{\delta W_s}{dt} + \sum \dot{m} (e + pv) \;, \]

and for a single inlet, single outlet duct, where $\dot{m}_{\text{in}} = \dot{m}_{\text{out}} = \dot{m}$, we have

\[ q = w_s + \left( e + p v \right)_{\text{out}} - \left( e + p v \right)_{\text{in}} \]

where $q = \frac{1}{\dot{m}} \frac{\delta Q}{dt}$ and $w_s = \frac{1}{\dot{m}} \frac{\delta W_s}{dt}$.

For example, for a system with one inlet, one outlet, shaft work, and heat transfer, conservation of energy simplifies down to the usual

\[\begin{split} q = w_s + \left( e_2 + p_2 v_2 \right) - \left( e_1 + p_1 v_1 \right) \\ h_1 + \frac{V_1^2}{2} + g z_1 + q = h_2 + \frac{V_2^2}{2} + g z_2 + w_s \end{split}\]

Entropy¶

We can also develop a useful expression that relates pressure, energy, and entropy. Recall that $\Delta S \equiv \int \frac{\delta Q_R}{T}$.

We can express changes in entropy as $dS = dS_e + dS_i$, separating into changes caused by actual heat transfer ($dS_e$) and changes caused by irreversibilities ($dS_i$). By definition,

\[ dS_e = \frac{\delta Q}{T} \;, \]

where $dS_e = 0$ for adiabatic processes. Then, $dS_i \geq 0$, and this comes from irreversible effects (e.g., temperature or pressure gradients, friction). For a reversible process, $dS_i = 0$.

Let’s consider a cyclic integral of entropy (meaning, integrate over a cycle):

\[ \oint dS = \oint dS_e + \oint dS_i = 0 \;, \]

by definition, since entropy is a property and we are integrating over a cycle (which ends back at the original state). But,

\[\begin{split} \oint dS_i \geq 0 \\ 0 = \oint dS_e + (\geq 0) \\ \therefore \oint \frac{\delta Q}{T} \leq 0 \\ \end{split}\]

which is the inequality of Clausius.

Pressure-energy equation¶

We can obtain a pressure-energy equation by integrating our property relation into conservation of energy, taking advantage of this approach to considering entropy:

\[\begin{split} T \, ds = dh - v \, dp \\ T \, ds_e + T \, ds_i = dh - \frac{dp}{\rho} \\ dh = T \, ds_e + T \, ds_i + \frac{dp}{\rho} \end{split}\]

Incorporate this into the energy equation:

\[\begin{split} \delta q = \delta w_s + \left( T \, ds_e + T \, ds_i + \frac{dp}{\rho} \right) + \frac{dV^2}{2} + g \, dz \\ \frac{dp}{\rho} + \frac{dV^2}{2} + g \, dz + \delta w_s + T \, ds_i = 0 \;, \end{split}\]

which is the pressure-energy equation.

Note

We can obtain Euler’s equation from the pressure-energy equation by applying it to a reversible process with no shaft work:

\[ \frac{dp}{\rho} + \frac{dV^2}{2} + g \, dz = 0 \]

Conservation of momentum¶

For a control mass, conservation of momentum comes from Newton’s second law:

\[ \sum \vec{F} = \frac{ d (\vec{\text{momentum}})}{dt} \;. \]

We can apply Reynolds transport theorem (2) to momentum, where $\vec{N}$ is momentum and $\eta = \vec{V}$:

\[ \frac{ d (\vec{\text{momentum}})}{dt} = \frac{\partial}{\partial t} \int_{CV} \vec{V} \rho \, d\tilde{v} + \int_{CS} \vec{V} \rho (\vec{V} \cdot \hat{n} ) \, dA \]

For one-dimensional flow, the area integral simplifies to

\[ \int_{CS} \vec{V} \rho (\vec{V} \cdot \hat{n} ) \, dA = \sum \dot{m} \vec{V} \;, \]

and for steady, one-dimensional flow, we can write conservation of momentum as

\[\begin{split} \sum \vec{F} = \sum \dot{m} \vec{V} \\ \sum \vec{F} = \dot{m} \left( \vec{V}_{\text{out}} - \vec{V}_{\text{in}} \right) \;, \end{split}\]

where the latter form applies if we have one inlet and one outlet.

For example, consider flow through a duct, which includes friction at the wall. We can focus on the $x$-component of the momentum equation:

\[\begin{split} \sum F_x = \dot{m} \left( V_{\text{out}, x} - V_{\text{in}, x} \right) \\ \rightarrow p_1 A_1 - p_2 A_2 - F_f = \dot{m} (V_2 - V_1) \;. \end{split}\]

Gas Dynamics notes