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As seen in thermodynamics, there is a difference between system and control volume analysis. The former are used for a fixed mass of fluid, constantly moving, the latter are used for fluid flow through a defined boundary.

We hardly ever model a fixed mass of fluid, and as such we always use control volume analysis in fluid dynamics.

Mass Flow Rate

Mass flow rate, flow rate through a pipe, control volume analysis, fluid dynamics

Over a set time \delta t, a distance \delta x is travelled by any given fluid particles. Therefore, the swept volume over time \delta t is A \delta x. This volume has a mass, \rho A \delta x.

The mass flow rate is the derivative of this with respect to time:

\dot m = \rho A \frac{dx}{dt}

This can be re-written in terms of normal velocity, u:

    \[ \dot m = \rho u A \]

Mass flux is another description of flow, given as the mass flow rate per unit area:

    \[ \dot f = \frac{\dot m}{A} = \rho u \]

The velocity must be normal (perpendicular) to the area.

Normal vectors, perpendicular component of velocity

If the velocity is not normal to the area, \delta A, we need to find the normal component as the scalar product of the velocity and normal vectors:

\delta \dot m = \rho (\vec u . \vec n)\delta A

Writing \delta A as a vector:

\overrightarrow{\delta A} = \delta A \hat n

\delta \dot m = \rho \vec u . \overrightarrow{\delta A}

\delta \dot m = \rho \vec u . \overrightarrow{\delta A}

    \[ \dot m = \int_A \rho \vec u. \overrightarrow{dA} \]

This is the most important equation for mass flow rate. It will crop up in the derivations of everything that follows in this notes sheet.

Through a control volume, where there is inflow and outflow across the control surface (CS):

    \[ \dot m_{out} - \dot m_{in} = \int_{CS} \rho \vec u . \overarrowright{dA} \]

Outflow gives a positive value, inflow gives a negative value.

Volume Integrals

If certain properties through the control volume, such as density, are not constant, then we need to treat the total volume as an infinite number of infinitesimally small volumes, each with mass:

\delta m = \rho \delta V

    \[ m = \int_V \rho dV \]

When density is constant:

    \[ m = \rho \int_V dV = \rho V \]

Reynolds Transport Theorem

The Reynolds transport theorem is used to model the conservation of any given extensive property N. This could be any property: we will use it for mass, energy, and momentum.

In the Reynolds transport theorem, the specific form of the property is used, \eta:

    \[N=\eta m\]

Therefore, \rho \eta is the property per unit volume.

The theorem is:

    \[\frac{d}{dt} \int_{V_{sys(T)}} \eta \rho dV = \frac{d}{dt} \int_{CV} \eta \rho dV + \int_{CS} \eta \rho \vec u . \overrightarrow{dA} \]

  • The term on the left-hand side is the rate of change of amount of property N in the system at any time.
  • The first right-hand term is the rate of change of the amount of property N in the control volume overlapping with the system at the same time.
  • The second right-hand term is the net flow rate of property N out of the system.

See the derivation for the Reynolds Transport Theorem here.

Steady Flow

In steady flow, there is no net change in the amount of property N in the control volume, so the middle term disappears:

    \[\frac{d}{dt} \int_{V_{sys(T)}} \eta \rho dV = \int_{CS} \eta \rho \vec u . \overrightarrow{dA} \]

This does not mean that there is no change at all: some amount of property N could be created within the control volume, but if this same amount flows out of it, the total amount of N in the CV is fixed.

Conservation of Mass

The Reynolds transport theorem can be used to get to the equation for the conservation of mass in a control volume.

To do this, we set the property we wish to conserve, R, equal to m, meaning that \eta equals 1:

    \[\eta = \frac{R}{m} = \frac{m}{m} = 1\]

Applying this to the Reynolds Transport Theorem:

    \[\frac{d}{dt} \int_{V_{sys(T)}} \rho dV = \frac{d}{dt} \int_{CV} \rho dV + \int_{CS}\rho \vec u . \overrightarrow{dA} \]

The left hand term is the rate of change of mass in the system – since the mass is conserved, there is no change and this term must equal zero. This leads to the mass continuity equation:

    \[\frac{d}{dt} \int_{CV} \rho dV = -\int_{CS} \rho \vec u. \overrightarrow{dA}\]

  • The left-hand term represents the rate at which the control volumes gains mass
  • The right-hand term is the net rate of flow of mass across the control surface
    • Negative, as flow is into the control volume

Algebraically, this is written as:

    \[\frac{dm}{dt} = \sum \dot m_{in} - \sum \dot m_{out} \]

Where the sums are all the inflows and outflows added together.

Steady Flow

For steady flow, there is no net flow across a control surface, so the right-hand side of the continuity equation equals zero:

    \[\int_{CS} \rho \vec u . \overrightarrow{dA} = 0\]

Algebraically, this is generally written as:

    \[ \sum \dot m_{in} = \sum \dot m_{out}\]

For steady flow, the total flow in equals the total flow out.

For constant density (incompressible flow) and uniform velocity normal to the surface:

    \[\dot m = \rho u A\]

    \[ \sum_{in} \rho u A = \sum_{out} \rho u A\]

Constant Density

In this case, the left-hand term of the continuity equation equals zero:

    \[\frac{d}{dt} \int_{CV} \rho dV = 0 = -\int_{CS} \rho \vec u.\overrightarrow{dA}\]

    \[\rho \int_{CS} \vec u.\overrightarrow{dA}=0\]

    \[\int_{CS} \vec u.\overrightarrow{dA}=0\]

This represents the volumetric flow rate through the control surface. Algebraically, it is written as:

    \[\sum_{in} uA = \sum_{out} uA\]

This only applies when the control volume only contains fluid: it is full (not being filled/drained).

Conservation of Momentum

From Newton’s second law, we know that the sum of the forces is equal to the derivative of the momentum of a system:

    \[\sum \vec F = m \vec a = m \frac{d \vec u}{dt} = \frac{d}{dt} (m \vec u) \]

Rearranging this in terms of a volume integral and density, instead of mass:

    \[\sum \vec F = \frac{d}{dt} \int_{V_{sys(t)}} \vec u \rho dV\]

This is clearly the left hand side of the Reynolds Transport Theorem, where \eta = \vec u.

Swapping the right-hand side of the Reynolds transport theorem for the left-hand side gives the equation for the conservation of momentum in a control volume:

    \[\sum \vec F = \frac{d}{dt} \int_{CV} \vec u \rho dV + \int_{CS} \vec u \rho \vec u .\overrightarrow{dA}\]

Steady Flow

In steady flow, there is no change with respect to time. Therefore, the first term on the right-hand side becomes zero:

    \[\sum \vec F_x = \int_{CS} u_x \rho \vec u. \overrightarrow{dA}\]

    \[\sum \vec F_y = \int_{CS} v_y \rho \vec u. \overrightarrow{dA}\]

  • This can be interpreted to mean the sum of forces in a particular direction is equal to the momentum outflow minus the momentum inflow in that direction.

As seen in the notes sheet of forces in fluids, all fluids consist of body and surface forces.

  • All body forces in the control volume (e.g., weight) must be accounted for.
  • Only the surface forces at the control volume boundary, however, must be taken account for here (as internal forces cancel out).
  • These could be atmospheric pressure or reaction forces.

Algebraic Formulation

The integrated form of the conservation of momentum equation above is not all that useful. Instead, if velocity is uniform at an inlet or outlet, this inlet or outlet can be looked at individually.

The surface integral for the whole control volume therefore becomes an area integral for just that inlet/outlet:

    \[\int_A u\rho \vec u.\overrightarrow{dA} = u\int_A \rho \vec u.\overrightarrow{dA} = u \dot m\]

  • (See mass flow rate above)

Therefore, the conservation of momentum can be written in terms of the sum of all outlet momentums minus those at the inlets:

    \[\sum F_x = \sum_{x, out} \dot m u - \sum_{x, in} \dot m u\]

    \[\sum F_y = \sum_{y, out} \dot m u - \sum_{y, in} \dot m u\]

For one inlet and one outlet:

    \[\sum F_x = (\dot mu)_{x,out} - (\dot mu)_{x,in}\]

Using conservation of mass:

    \[\dot m_{in} = \dot m_{out} = \dot m\]

    \[\sum F_x = \dot m(u_{out} - u_{in})_x\]

    \[\sum F_y = \dot m(u_{out}-u_{in})_y\]

  • Mass flow rate is given as \dot m = \rho uA where u is the normal velocity
  • If the velocity is given as a surface integral: \dot m = \int_A \vec u . \overrightarrow{dA}
  • Sign Convention: outflow is positive, inflow is negative
  • Reynolds transport theorem can be used to model the conservation of a specific property, \eta in a control volume:
    • \frac{d}{dt} \int_{V_{sys(t)}} \eta \rho dV = \frac{d}{dt} \int_{CV} \eta \rho dV + \int_{CS} \eta \rho \vec u . \overrightarrow{dA}
    • The first term is the instantaneous rate of change of \eta in the system
    • The second term is the instantaneous rate of change of \eta in the control volume
    • The third term is the net flow rate of \eta out of the system
  • The conservation of mass states that the total mass flow rate in must equal the total mass flow rate out: \frac{dm}{dt} = \sum \dot m_{in} - \sum \dot m_{out}
  • The conservation of momentum allows us to calculate resultant forces from fluids:
    • \sum \vec F = \frac{d}{dt} \int_{CV} \vec u \rho dV + \int_{CS} \vec u \rho \vec u .\overrightarrow{dA}
Conservation Fluids Mass Momentum Reynolds Transport Theorem

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