Applications of the Conservation Laws

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.n19

21 Applications of the Conservation Laws

Introduction

Using the conservation equations of aerodynamics and fluid dynamics is best learned by studying and fully understanding exemplar problems. After a few problems, the systematic solution process will become apparent to the student, and new problems can be tackled more confidently. Generally speaking, all three conservation laws must be used to solve practical problems in fluid dynamics, i.e., using the governing equations for the conservation of mass, momentum, and energy, although not always. Energy principles will inevitably be needed whenever pressure is involved as an unknown in the problem. An auxiliary equation, the equation of state, may also be required for problems involving compressibility effects. The Bernoulli equation can be used instead of the formal energy equation, but only for steady, incompressible, inviscid flows without work or energy addition, and this critical restriction should be remembered.

Learning Objectives

Use the fundamental conservation laws of aerodynamics to calculate the flow properties through a venturi and the contraction and test section of a wind tunnel.
Understand the principles associated with the operation of a Pitot tube and how to measure static and dynamic pressure,
Use the conservation laws in the integral form to find the drag on a body.
Analyze the essential performance of a jet engine and an airplane propeller.

Flow in a Venturi

Consider the steady, incompressible flow (i.e., $\varrho$ is a constant) through a convergent-divergent nozzle, as shown below, called a venturi. If the flow is assumed to be incompressible, then the fluid could be a liquid, or it would have to be a gas such as air but flowing at low speed, i.e., all flow velocities need to be below a Mach number of 0.3 (which may not be known a priori).

The advantages of a pressure drop in the throat of a venturi are used in many practical engineering applications.

The intake to a venturi is called the mouth, and the narrowest part is called the throat. Following Bernoulli’s principle, there is a pressure drop at the throat of a venturi where the flow speeds up, which can be used to advantage in several practical applications such as flow meters, carburetors, wind tunnels, etc.

The first step in the analysis is to draw a control surface around the venturi with $\vec{n} \, dS = d\vec{S}$ pointing out of the control volume by convention, as shown below. Let the inlet conditions be at section 1, the throat at section 2, and the outlet at section 3. The cross-sectional areas in each case are assumed to be known, such as by measurement. The flow can be considered steady and radially axisymmetric, i.e., an appropriate one-dimensional flow assumption.

Flow model for the flow through a venturi.

The steady form of the continuity equation in integral form is

(1) $\begin{equation*} \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}$

In words: “What mass flow comes into the control volume per unit of time then leaves the control volume in the same time, i.e., no fluid mass accumulates inside the control volume.”

There is no flow over the walls of the venturi, so what fluid mass flow comes into the mouth of the venturi per unit of time also leaves at the exit of the venturi at the same time. If the cross-sectional area variation is relatively moderate, the problem can be simplified significantly with a one-dimensional flow assumption. However, also of further consideration is why a flow through a rapidly converging or expanding venturi (or duct) may not be as readily justifiable as a one-dimensional flow problem.

The flow enters the mouth of the venturi of area $A_1$ with an average velocity $V_1$ , has a velocity $V_2$ at the throat of area $A_2$ , and velocity $V_3$ at the exit of area $A_3$ . The fluid mass flow coming in is

(2) $\begin{equation*} \oiint_{\rm intake} \varrho \, \vec{V} \bigcdot d\vec{S} = -\varrho_1 A_1 V_1 \end{equation*}$

The minus sign appears because the flow is opposite to the direction of $d\vec{S}$ at the entrance face. At the exit face, then

(3) $\begin{equation*} \oiint_{\rm exit} \varrho \, \vec{V} \bigcdot d\vec{S} = \varrho_3 A_3 V_3 \end{equation*}$

Therefore, by using the continuity equation, then

(4) $\begin{equation*} \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = \oiint_{\rm intake} \varrho \, \vec{V} \bigcdot d\vec{S} + \oiint_{\rm exit} \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}$

which, for a one-dimensional flow, becomes

(5) $\begin{equation*} -\varrho_1 V_1 A_1 + \varrho_3 V_3 A_3 = 0 \end{equation*}$

If the flow is incompressible, $\varrho_1 = \varrho_2 = \varrho_3 = \varrho$ . Under these circumstances, the continuity equation can now be written as

(6) $\begin{equation*} -\varrho V_1 A_1 + \varrho V_3 A_3 = 0 \end{equation*}$

or

(7) $\begin{equation*} \varrho V_1 A_1 = \varrho V_3 A_3 = \overbigdot{m} \end{equation*}$

Also, it will be apparent that

(8) $\begin{equation*} \varrho V_1 A_1 = \varrho V_2 A_2 = \overbigdot{m} = \varrho Q \end{equation*}$

Canceling through the density gives

(9) $\begin{equation*} V_1 A_1 = V_2 A_2 = V_3 A_3 = Q \end{equation*}$

This statement states that the volume flow rate $Q$ through the venturi is constant.

It can also be seen from this latter equation that if the throat area decreases, then the flow velocity must increase there so that the continuity of the flow is satisfied, i.e., the mass and volume flow rate are constant. Conversely, if the area of the throat increases, the flow velocity must decrease. Moreover, from the previous discussion of Bernoulli’s equation, remember that when the velocity increases, the pressure decreases; conversely, when the velocity decreases, the pressure increases, i.e.,

(10) $\begin{equation*} p + \frac{1}{2} \varrho V^2 = \mbox{constant} \end{equation*}$

or

(11) $\begin{equation*} p_1 + \frac{1}{2} \varrho V_1^2 = p_2 + \frac{1}{2} \varrho V_2^2 \end{equation*}$

(12) $\begin{equation*} p_1 - p_2 = \frac{1}{2} \varrho \left( V_2^2 - V_1^2 \right) \end{equation*}$

The pressure difference $p_1 - p_2$ can be measured across the mouth and throat of the venturi, which can be accomplished by placing two small holes (i.e., pressure taps) drilled perpendicular to the venturi walls at locations 1 and 2. The pressure difference can be measured by connecting pressure tubes from these holes to a differential pressure gauge or both sides of a U-tube manometer; see here for a simple experimental demonstration of this process. This pressure difference can then be related to the unknown velocity (or flow rate) by calculation or direct calibration using measurements.

From the continuity equation, then

(13) $\begin{equation*} V_{2} = V_{1}\left(\frac{A_1}{A_2}\right) \end{equation*}$

and from Bernoulli’s equation

(14) $\begin{equation*} p_{1} + \frac{1}{2}\varrho {V_{1}}^{2} = p_{2} + \frac{1}{2}\varrho {V_{2}}^{2} \end{equation*}$

Substituting gives

(15) $\begin{eqnarray*} p_{1} + \frac{1}{2} \varrho V_{1}^{2} & = & p_{2} + \frac{1}{2}\varrho V_{1}^{2} \left( \frac{A_1}{A_2} \right)^2 \\ p_{1} - p_{2} & = & \frac{1}{2} \varrho V_{1}^{2} \bigg( \left( \frac{A_1}{A_{2}} \right)^{2} - 1 \bigg) \end{eqnarray*}$

and so

(16) $\begin{equation*} Q = A_1 \sqrt{ \frac{2( p_{1} - p_{2})}{\varrho \bigg( \left( \displaystyle{\frac{A_{1}}{A_{2}} } \right)^{2} - 1 \bigg) } } \end{equation*}$

Therefore, knowing the pressure difference $p_{1} - p_{2}$ and the geometry of the venturi (the area ratio), then the flow rate $Q$ can be determined. Finally, notice that the change in pressure can be related to a change in the hydrostatic head of the fluid $\Delta H$ , as shown in the previous figure, by using

(17) $\begin{equation*} p_{1} - p_{2} = \varrho_l \, g \, \Delta H \end{equation*}$

where $\varrho_l$ is the density of the liquid in the U-tube manometer.

Applications of a Venturi

A venturi finds many applications in practical engineering applications. Its primary characteristic is that the flow velocity is higher in the venturi’s throat. As a result, the pressure at the throat is lower than the atmospheric pressure, this pressure difference being useful for various purposes.

Flowmeter

As the previous analysis suggests, a venturi forms the basis for a device for measuring volume (or mass) flow rates. Such flowmeters, or venturimeters as they are often called, are commercially available in various sizes and capacities. They are widely used in the water, chemical, pharmaceutical, and oil and gas industries to measure fluid flow rates along ducts and pipes. While venturimeters operate fundamentally on Bernoulli’s principle that the pressure decreases in the throat of the venturi, they are usually calibrated at the factory so that volumetric flow rates can be accurately related to the measured pressure drop across the throat.

Suction Source for Flight Instruments

In early aircraft, a venturi (or sometimes a pair of venturi tubes) was mounted on the side of an aircraft’s fuselage, as shown in the photograph below. The venturi provided low (suction) pressure for air-driven gyroscopic instruments such as the artificial horizon and directional gyro. Modern light aircraft, however, are fitted with mechanical engine-driven vacuum pumps to provide the needed suction pressure.

A pair of venturi tubes mounted on the side of an early aircraft provided low (suction) pressure to drive gyroscopic flight instruments.

Carburetor

A carburetor has a venturi through which incoming air is mixed with fuel. The fuel delivery line opens into the venturi at the throat. The lower pressure in this region helps to draw fuel into the airstream and then mix it with the air downstream of the throat before the fuel-air mixture goes into the cylinders to be burned.

Wind-Tunnel

A low-speed wind tunnel is a large venturi where the airflow is driven by a fan connected to a motor drive. The wind tunnel fan is a large propeller, and it is specifically designed to move the air through the test section efficiently, as shown in the figure below.

The airflow enters the mouth of area $A_1$ at a flow velocity $V_1$ with pressure $p_1$ . The wind tunnel then contracts to a smaller area $A_2$ at the test section where the velocity has increased to $V_2$ . The velocity in the test section must increase if continuity is satisfied. The model (such as a wing or complete airplane model) is placed in the test section, where its aerodynamic characteristics are measured. The flow then passes downstream into a diverging duct called a diffuser, where just before the fan, the area is $A_3$ , the velocity reduces to $V_3$ , and the pressure increases to $p_3$ .

From the continuity equation, the air velocity in the test section is

(18) $\begin{equation*} V_2 = \left( \frac{A_1}{A_2} \right) V_1 \end{equation*}$

In turn, the velocity at the exit of the diffuser before the fan is

(19) $\begin{equation*} V_3 = \left( \frac{A_2}{A_3} \right) V_2 \end{equation*}$

The pressure at various locations in the wind tunnel is related to the velocity by Bernoulli’s equation

(20) $\begin{equation*} p_1 + \frac{1}{2} \varrho V_1^2 = p_2 + \frac{1}{2} \varrho V_2^2 = p_3 + \frac{1}{2} \varrho V_3^2 \end{equation*}$

The velocity in the working or test section (the most important quantity) can be related to the pressure drop across sections 1 and 2, i.e., the pressure drop between the mouth of the contraction section and the test section. From the Bernoulli equation, then

(21) $\begin{equation*} V_2^2 = \frac{2}{\varrho} \left( p_2 - p_1 \right) + V_1^2 \end{equation*}$

This means that

(22) $\begin{equation*} V_2^2 = \frac{2}{\varrho} \left( p_1 - p_2 \right) + \left( \frac{A_2}{A_1} \right)^2 V_2^2 \end{equation*}$

Solving for $V_2$ (the flow velocity in the test section) gives

(23) $\begin{equation*} V_{2} = \sqrt{ \frac{2\left(p_{1} - p_{2}\right)}{\varrho \bigg( 1 - \left( \displaystyle {\frac{A_{2}}{A_{1}} }\right)^{2} \bigg) } } \end{equation*}$

The area ratio $A_2/A_1$ is fixed for a given wind tunnel. Recall that for an incompressible flow, the density is a constant.

Therefore, the above equation can be used to determine the velocity in the test section by measuring the pressure drop from the intake to the contraction and the test section. This latter technique is used in most low-speed wind tunnels to measure the flow speed in the test section, i.e., the drop in pressure between some point on the contraction and the test section is measured. This pressure drop is then related to the flow velocity in the test section using Bernoulli’s equation and confirmed by calibration. In the calibration, a Pitot probe (see discussion below) is placed in the test section, and the pressure drop is measured for a range of flow speeds; any discrepancy leads to a calibration factor that can be used to determine the flow speed accurately.

Cavitating Venturi

Cavitating venturis can be used to control or limit the flow rate of a liquid through a pipe. The throat of a cavitating venturi is sized such that for a given differential pressure drop between the inlet and the throat, the pressure is reduced to its vapor pressure point. The resulting vapor bubbles that form, called cavitation, will then limit the mass flow rate through the venturi and prevent any additional increase in flow rate for a given upstream inlet pressure. The mass flow rate through the pipe is not dependent on the downstream pressure, so the mass flow can then be controlled by changing the inlet pressure and, hence, the amount of cavitation in the throat. Cavitating venturis are often used as propellant mass flow limiters in rocket motors.

Syphon

A syphon, also spelled siphon, is a device used to transfer liquids from one level to another. It consists of a tube or hose bent in a U-shape, with one end placed in some liquid, such as water, and the other set at a lower level, as shown in the figure below. This arrangement allows the water to flow from the tank to a lower level, which relies on the principle of gravity and hydrostatic pressure. To start the flow, the syphon tube must be filled with water, which can be done by suctioning the air out of the tube or by filling it with water and submerging one end in the tank. Once the flow starts, it will continue until the container is emptied.

It is often claimed that a syphon works on the principle of air pressure, but this is incorrect. To see why, the Bernoulli equation applied in the fluid between the tank and the outlet from the syphon gives

(24) $\begin{equation*} p_1 + \frac{1}{2} \varrho V_1^2 + \varrho \, g \, y_1 = p_2 + \frac{1}{2} \varrho V_2^2 + \varrho \, g \, y_2 \end{equation*}$

The change in air pressure between the levels at $y_1$ and $y_2$ is minimal, so it can be assumed that $p_1 = p_2$ . Also, if the fluid in the tank is assumed to be of much greater volume than the volume flow rate out of the syphon, then $y_2$ = constant. Therefore, under these assumptions, the Bernoulli equation becomes

(25) $\begin{equation*} p_1 + \frac{1}{2} \varrho V_1^2 + \varrho \, g \, y_1 = p_1 + 0 + \varrho \, g \, y_2 \end{equation*}$

or

(26) $\begin{equation*} \frac{1}{2} V_1^2 + g \, y_1 = g \, y_2 \end{equation*}$

Rearranging gives the outlet flow velocity as

(27) $\begin{equation*} V_1 = \sqrt{ 2 \, g \left( y_2 - y_1 \right)} = \sqrt{ 2 \, g \, h} \end{equation*}$

Notice that the flow velocity depends on the difference in height, $h$ , and acceleration under gravity, so the higher the height difference, the faster the water will flow. There is no significant change in atmospheric pressure between the two levels, so the syphon effect is produced by the hydrostatic pressure difference within the fluid, not atmospheric pressure. However, considering viscous effects (and hence losses) will change this result because there will be some pressure drop along the length of the syphon tube. Calculating such internal pressure losses and the flow velocity requires a viscous flow theory, which is considered in later chapters of this eBook.

Total, Static, & Dynamic Pressure

It must now be explained more precisely what a fluid’s total and static pressures mean. Consider a flow moving with velocity $V$ at pressure $p_{s}$ . If moving with the velocity of the airflow, then the pressure that will be felt is $p_{s}$ . This latter pressure is called static pressure and measures the effects produced by the random motion of the fluid molecules.

Now, if the boundary or “wall” of a flow is considered (e.g., the wall of a duct, pipe, or wind tunnel), and a small hole in the wall is drilled in perpendicular to the flow, then the pressure that is measured there would be the static pressure $p_{s}$ . Suppose a U-tube manometer is connected, as shown in the figure below, and the open end outside the flow (i.e., the reference end) of the manometer is at lower static pressure than in the flow. In that case, a difference in height $h_s$ will be shown on the manometer. The pressure is the static pressure between the flow and the ambient reference pressure, called gauge pressure. If the static pressures inside and outside are equal, then $h_s = 0$ .

Concepts of the measurement of static pressure, total pressure, and dynamic pressure.

Now consider a Pitot tube (after the hydraulic engineer, Henri Pitot) inserted into the flow, which is an open tube with its end facing into the flow. The fluid cannot flow through the tube so it will stagnate in the tube, i.e., the flow velocity will go to zero because there is no place for the fluid to flow. Hence, the streamline that infringes directly on the mouth of the Pitot tube will have zero velocity; hence, this location is called a stagnation point. In this case, the pressure that is measured is called the total pressure, $p_T$ , i.e., the sum of the dynamic pressure and the static pressure

(28) $\begin{equation*} p_T = p_s + \frac{1}{2} \varrho V^2 \end{equation*}$

where $\varrho$ is the density of the flowing fluid, with equivalent height on the manometer of $h_t$ .

Finally, consider the case where the reference pressure is now the static pressure in the flow. In this case, the pressure would be the dynamic pressure of the flow, i.e., the difference between the total pressure and the static pressure

(29) $\begin{equation*} p_s + \frac{1}{2} \varrho V^2 - p_s = \frac{1}{2} \varrho V^2 \end{equation*}$

with an equivalent height of $h_d$ on the manometer.

Pitot Tube & Pitot-Static Tube

Pitot probes (or tubes), static probes, and Pitot-static probes are often used to measure flow velocities. In each case, the principle is the same: pressure measurements can be used with the Bernoulli equation to calculate the flow velocity. The proviso is that the density of the flow where the probe is inserted is known. These three types of pressure probes are shown in the figure below.

When the Pitot tube and the static port(s) are combined into a single unit with concentric tubes, it is called a Pitot-static tube (or probe). A “static tube” is rarely used to measure static pressure.

For the analysis of a Pitot tube, consider point 1 upstream of the Pitot tube and point 2 at the entrance to the tube, as shown in the figure below. From Bernoulli’s equation it is known that $p+1/2\varrho V^{2} =$ constant along any given streamline, so that

(30) $\begin{equation*} p_{1} + \frac{1}{2} \varrho V_{1}^{2} = p_{2} + \frac{1}{2} \varrho V_{2}^{2} \end{equation*}$

The difference in pressure between the total pressure measured using a Pitot tube and the static pressure measured using a pressure tap is the dynamic pressure, which can be recorded on a suitably calibrated pressure gauge.

But at point 2 at the entrance to the Pitot tube, the fluid is brought to rest so $V_{2} = 0$ (i.e., it is a stagnation point), and

(31) $\begin{equation*} p_1 + \frac{1}{2} \varrho V_{1}^2 = p_{2} = p_T =\mbox{Total pressure at point 2} \end{equation*}$

The static pressure $p_1 = p_{s}$ must be measured to obtain the flow velocity from this latter expression. Like the venturi problem, this static pressure can be measured with a separate static vent or with static ports on the outer side of a concentric Pitot-static tube. Solving for $V_{1}$ gives

(32) $\begin{equation*} V_{1} = \sqrt{\frac{ 2\left(p_{T} - p_{s}\right)}{\varrho}} \end{equation*}$

Therefore, the upstream flow velocity can be measured by measuring the difference between a flow’s total and static pressure. Again, this pressure difference can be measured using a differential gauge or a manometer.

The preceding principles are used in airspeed measurement using an airspeed indicator, which is part of the pneumatic system used on an aircraft for not only airspeed measurement but also altitude and rate of climb, as shown in the figure below. An airspeed indicator is a suitably calibrated differential pressure gauge. The dynamic pressure source needed for an airspeed indicator is either a Pitot tube or a Pitot-static tube. Notice, in this case, the use of a Pitot probe to measure the total pressure probe and a separate static vent (somewhere on the side of the fuselage) to measure the reference pressure. A heater prevents the Pitot probe from malfunctioning in icing conditions by preventing ice accumulation on the exposed probe.

A typical pitot-static system on an airplane provides the reference pressures needed for the airspeed indicator and other pneumatic instruments. Notice, in this case, the use of a pitot probe to measure total pressure and a separate static vent to measure the static reference pressure.

More often than not, airplanes use Pitot tubes (to measure total pressure $p_T$ ) with static taps (to measure $p_s$ ) placed at some other point on the airframe, as previously shown. In this case, then the airspeed $V_{\infty}$ will be

(33) $\begin{equation*} V_{\infty} = \sqrt{\frac{ 2\left(p_{T} - p_{s}\right)}{\varrho_{\infty}}} \end{equation*}$

In either case, the purpose is the same: to obtain a measurement of the aircraft’s airspeed. This information is then provided to the pilot on an airspeed indicator. Therefore, an airspeed indicator is just a dynamic pressure gauge calibrated in units of speed; usually, units of nautical miles per hour (kts) are used, but sometimes miles per hour (mph). An online simulator of the Pitot-static system is helpful in understanding how it works and what happens under various failure scenarios.

Jet Engine Performance

The thrust of a turbojet engine can be determined using conservation principles applied to a control volume surrounding the engine. There is a mass flow into the engine from the air and an additional mass flow from the fuel. Fuel is much denser than air, so although the volume flow of fuel may be relatively low, its mass flow is still significant and needs to be accounted for.

Control volume approach for analyzing a turbojet engine, which works on the air to increase its momentum in the downstream direction and produces a reaction force, which is the thrust, directed in the upstream direction.

The mass flow of air into the engine will be

(34) $\begin{equation*} \overbigdot{m}_{\rm air} = \varrho_{\infty} V_{\infty} A_i \end{equation*}$

where $A_i$ is the inlet area, and the mass flow rate of fuel is $\overbigdot{m}_{\rm fuel}$ . Therefore, the thrust is

(35) $\begin{equation*} T = (\overbigdot{m}_{\rm air} + \overbigdot{m}_{\rm fuel} ) V_e - \overbigdot{m}_{\rm air} V_{\infty} + ( p_e - p_{\infty} ) A_e ) \end{equation*}$

with $A_e$ as the exit area and $V_e$ as the exit or “jet” velocity. The pressure term in Eq. 35 is relatively small compared to the change in momentum of the flow and so may be neglected, i.e.,

(36) $\begin{equation*} T = (\overbigdot{m}_{\rm air} + \overbigdot{m}_{\rm fuel} ) V_e - \overbigdot{m}_{\rm air} V_{\infty} \end{equation*}$

Drag on a Two-Dimensional body

The conservation laws can be used to find the drag of a two-dimensional body in terms of fluid properties. Consider the physical problem of a body immersed in a fluid flow. The fluid will generally exert pressure and viscous forces on the body, which are the sources of the resultant force, $R$ , and the drag, $D$ . Physically, a wake with a lower flow velocity will form downstream of the body. The objective is to use the equations of motion of the fluid (in this case, continuity and momentum) to calculate the drag force on the body.

In this case, the upstream and downstream sections labeled 1 and 2 and the top and bottom sides bound the control volume. The top and bottom sides of the control volume are considered far enough away from the body that they are in the free stream and, therefore, are streamlines to the flow, i.e., there will be no flow over these boundaries. Remember that $d\vec{S}$ points out of the control volume by convention. The upstream or free-stream (undisturbed) velocity is $V_1$ = constant = $V_{\infty}$ . Because of viscosity, a wake of reduced velocity or retarded flow exists behind the body; this gives a wake region a velocity profile in which $V_2 < V_\infty$ .

Consider the forces on the fluid in the control volume. They stem from two sources:

1. The net effect of the pressure distribution over the control volume, i.e., over the surface ABCD, then

(37) $\begin{equation*} \mbox{Net pressure force} = -\iint_{ABCD}p \, d\vec{S} \end{equation*}$

2. The reaction of the surface forces on the body, i.e., those forces produced over the inner part of the control volume as created by the body, i.e., $-R$ .

Remember that the flow will exert pressure and shear forces on the body, resulting in a resultant overall force $R$ . This means that by Newton’s third law, the airfoil will exert a force $-R$ on the fluid within the control volume.

Therefore, the total force on the fluid volume is

(38) $\begin{equation*} \mbox{Net force} = -\iint_{ABCD} p \, d\vec{S} - R \end{equation*}$

and from the conservation of momentum equation, then

(39) $\begin{equation*} -\iint_{ABCD} p \, d\vec{S} - R = \frac{d}{dt}\oiiint_{\cal{V}} \varrho\vec{V} d {\cal{V}} + \iint_{S} \left(\varrho \, \vec{V} \bigcdot d\vec{S} \right) \vec{V} \end{equation*}$

For steady flow $\partial/ \partial t = 0$ , so that this latter equation is simplified to

(40) $\begin{equation*} R = -\iint_{S} \left(\varrho\vec{V}\bigcdot d\vec{S}\right) \vec{V} -\iint_{ABCD} p \, d\vec{S} \end{equation*}$

Notice that this is a vector equation.

To find the drag, the $x$ -component of this latter equation is needed, noting that $V_{\infty}$ and $V_2$ are the inflow and outflow velocities in the $x$ -direction and that the $x$ -component of $R$ is the drag on the section, i.e., $D$ per unit span. The pressures far away from the body are at ambient conditions, i.e., around the entire control volume, then the pressure forces are

(41) $\begin{equation*} \iint_{ABCD} p \, d\vec{S} = 0 \end{equation*}$

Therefore, the force on the fluid and drag on the body can be expressed only in terms of the $u$ velocity profiles upstream and downstream, i.e.,

(42) $\begin{equation*} D = -\iint_{S} \left(\varrho\vec{V}\bigcdot d\vec{S} \right) u \end{equation*}$

and so this integral becomes a one-dimensional integral (noting carefully the signs on the components), i.e.,

(43) $\begin{equation*} D = -\bigg( - \int_{D}^{A} \varrho_{\infty} V_{\infty}^{2} dy + \int_{C}^{B}\varrho_{2}V_{2}^{2}dy \bigg) = \int_{D}^{A}\varrho_{\infty}V_{\infty}^{2}dy - \int_{C}^{B}\varrho_{2} V_{2}^{2}dy \end{equation*}$

The continuity equation can also be applied to the control volume, i.e., based on the principle of conservation of mass, then

(44) $\begin{equation*} \int_{D}^{A}\varrho_{\infty}V_{\infty}dy = \int_{C}^{B} \varrho_{2} V_{2} dy \end{equation*}$

Multiplying by $V_{\infty}$ in this case (which is a constant) gives

(45) $\begin{equation*} \int_{D}^{A}\varrho_{\infty}V_{\infty}^2 dy = \int_{C}^{B} \varrho_{2} V_{2} V_{\infty} dy \end{equation*}$

Substituting back into the expression for $D$ gives

(46) $\begin{eqnarray*} D & = & \int_{D}^{A}\varrho_{\infty}V_{\infty}^{2}dy - \int_{C}^{B}\varrho_{2} V_{2}^{2}dy \nonumber \\[6pt] & = & \int_{C}^{B} \varrho_{2} V_{2} V_{\infty}dy - \int_{C}^{B}\varrho_{2}V_{2}^{2}dy = \int_{C}^{B}\varrho_{2}V_{2} \left(V_{\infty} - V_{2}\right)dy \end{eqnarray*}$

This latter expression gives the drag of a two-dimensional body in terms of $V_{\infty}$ and the flow field properties $\varrho_2$ and $V_2$ across a vertical dimension downstream of the body.

Notice that $V_{\infty} - V_2$ is the velocity deficiency or decrement at a given station (or section) in the downstream wake. Also, $\varrho_{2}V_{2}$ is the mass flux, so the product $\varrho_{2}V_{2}(V_{\infty}-V_{2})$ gives the decrement in momentum. The integral of this expression leads to the total decrement in momentum behind the body and, hence, the drag of the section.

If $\varrho$ = constant, then things can be simplified even further to get

(47) $\begin{equation*} D=\varrho\int_{C}^{B}V_{2}\left(V_{\infty}-V_{2}\right) dy \end{equation*}$

Therefore, it has been shown that applying the momentum equation in its integral form can be related to the drag on a body to the flow field properties upstream and downstream.

This latter technique is frequently employed in wind tunnel testing for measuring forces on relatively streamlined bodies; the velocities downstream of the body or airfoil are measured with what is known as a wake rake. This rake is, in fact, an array of Pitot tubes, which are used to measure the dynamic pressure and the flow velocities in the wake. However, some caution is appropriate because this approach works well only without significant flow separation and turbulence from the body. If there are losses in the wake that do not show up, just as a loss in the momentum of the flow, the drag will be underestimated.

Propeller Performance

Another classic application of the conservation laws in integral form is determining the essential performance characteristics of a propeller. The physical problem is that the propeller does work on the air as it passes through the propeller disk (the propeller applies a force to the air in the downstream direction). So, it changes the momentum and kinetic energy of the air. As a result, the force on the propeller, which is produced because of a pressure difference between one side of the propeller disk and the other, is in the opposite direction to the force on the fluid, i.e., the thrust $T$ is directed upstream.

The flow model for this propeller problem is shown in the figure below. The upstream and downstream sections labeled 1 and 2 bound the control volume. Remember that when the integral form of the conservation equations is used, then $d\vec{S}$ points out of the control volume by convention. The upstream or free-stream (undisturbed) velocity is $V_{\infty}$ , and the pressure there is $p_{\infty}$ . It will be assumed it is uniform (a reasonable assumption unless the propeller is affected by a wing or another part of the airframe). It will also be assumed that a one-dimensional, steady flow applies throughout so that the flow velocities only change with downstream distance.

Control volume used for the analysis of a propeller flow.

The flow is entrained and accelerated into the propeller, so at the plane of the propeller, the flow velocity is $V_{\infty}$ plus an increment $v$ , i.e., the velocity there is $V_{\infty} + v$ . The static pressure will change there too. The propeller works on the air to increase its momentum and kinetic energy, so the velocity is $V_{\infty} + w$ downstream. In the slipstream, the static pressure will also recover to ambient conditions, i.e., the pressure downstream will be $p_{\infty}$ .

Conservation of mass requires constant mass flow through the control volume. The areas of the upstream and downstream parts of the streamtube (stations 1 and 2) are unknown, but the area of the propeller disk, $A$ , is known ( $A = \pi D^2/4$ where $D$ is the diameter of the propeller, so the mass flow rate through the propeller (and hence through the boundaries of the control volume) is

(48) $\begin{equation*} \overbigdot{m} = \varrho A (V_{\infty} + v ) \end{equation*}$

Conservation of momentum requires that the net change in momentum of the fluid (as applied by the propeller) is equal to the force on the fluid, i.e.,

(49) $\begin{equation*} F_{\rm fluid} = -\overbigdot{m} V_{\infty} + \overbigdot{m} (V_{\infty} + w) = -\overbigdot{m} V_{\infty} + \overbigdot{m} V_{\infty} + \overbigdot{m} w = \overbigdot{m} w \end{equation*}$

where the direction of $F_{\rm fluid}$ is downstream, and the thrust on the propeller, $T$ , is in the opposite direction (pointing left in the figure) so that

(50) $\begin{equation*} T = \overbigdot{m} w \end{equation*}$

At this point, the conservation of energy could be applied. Conservation of energy tells us that the work done on the propeller (to move it forward) plus the work done on the air (to create the aerodynamic force) must be equal to the gain in kinetic energy of the slipstream as it passes through the propeller, i.e.,

(51) $\begin{eqnarray*} T (V_{\infty} + v ) & = & \frac{1}{2} \overbigdot{m} (V_{\infty} + w)^2 - \frac{1}{2} \overbigdot{m} V_{\infty}^2 \nonumber \\[6pt] & = & \frac{1}{2} \overbigdot{m} (V_{\infty}^2 + 2 V_{\infty} w + w^2) - \frac{1}{2} \overbigdot{m} V_{\infty}^2 \nonumber \\[6pt] & = & \frac{1}{2} \overbigdot{m} ( 2 V_{\infty} w + w^2) \end{eqnarray*}$

so that the power required $P$ for the propeller to produce the thrust $T$ is

(52) $\begin{equation*} P = T (V_{\infty} + v ) = \frac{1}{2} \overbigdot{m} ( 2 V_{\infty} w + w^2) \end{equation*}$

Notice that the $T (V_{\infty} + v)$ term on the left-hand side of the above equation can be written as the sum of the $T V_{\infty}$ term, which is the work done to move the propeller forward (this is the useful work). The $T v$ term is the work done on the air or the induced power loss; the induced power is an irrecoverable power loss and so constitutes a loss in efficiency of the propeller in producing useful work.

It is easy to see that by substituting Eq. 50 into Eq. 52 gives

(53) $\begin{equation*} (V_{\infty} + v ) w = \frac{1}{2} ( 2 V_{\infty} w + w^2) \end{equation*}$

or simplifying gives

(54) $\begin{equation*} 2 V_{\infty}w + 2 v w = 2 V_{\infty} w + w^2 \end{equation*}$

which gives the relationship between $v$ and $w$ , i.e.,

(55) $\begin{equation*} w = 2 v \end{equation*}$

For the next step, take Eq. 50 and substitute in Eq. 48 (mass flow rate) and the connection between $v$ and $w$ ( $w = 2 v$ ) to get

(56) $\begin{equation*} T = \varrho A (V_{\infty} + v ) w = 2 \varrho A (V_{\infty} + v ) v \end{equation*}$

so the induced velocity in the plane of the propeller can be solved for in terms of the thrust, i.e.,

(57) $\begin{equation*} (V_{\infty} + v ) v = \frac{T}{2 \varrho A} \end{equation*}$

or

(58) $\begin{equation*} v^2 + V_{\infty} v - \frac{T}{2 \varrho A} = 0 \end{equation*}$

The ratio $T/A$ has a special name called the propeller disk loading. It can be seen that the latter equation is quadratic in $v$ , which can be solved to get

(59) $\begin{equation*} v = \frac{1}{2} \bigg( -V_{\infty} \pm \sqrt{V_{\infty}^2 + \frac{2T}{\varrho A}} \bigg) \end{equation*}$

for which there must be two roots, i.e., from the $\pm$ term. Only one can be a physical root (the other one violates the assumed flow model where the flow is assumed to be through the propeller from left to right), which is

(60) $\begin{equation*} v = \frac{1}{2} \bigg( -V_{\infty} + \sqrt{V_{\infty}^2 + \frac{2T}{\varrho A}} \bigg) \end{equation*}$

Therefore, the power required to drive the propeller and produce thrust becomes

(61) $\begin{equation*} P = T (V_{\infty} + v ) = T V_{\infty} + T v = T V_{\infty} + \frac{T}{2} \bigg( -V_{\infty} + \sqrt{V_{\infty}^2 + \frac{2T}{\varrho A}} \bigg) \end{equation*}$

Notice again that the useful work is $T V_{\infty}$ , and the second term is an induced loss, i.e., an irrecoverable loss.

For the limiting case where $V_{\infty}$ goes to zero, then the power (i.e., the power required in the static thrust condition) is

(62) $\begin{equation*} P = \frac{ T^{3/2}}{\sqrt{2 \varrho A}} \end{equation*}$

The efficiency of the propeller can also be derived from the foregoing analysis. The useful power is $T V_{\infty}$ so the efficiency of the propeller, $\eta$ , would be

(63) $\begin{equation*} \eta = \frac{T V_{\infty}}{T (V_{\infty} + v )} = \frac{T V_{\infty}}{T V_{\infty} \sqrt{1 + \displaystyle{\frac{2T}{\varrho A V_{\infty}^2}}}} \end{equation*}$

which says that the propeller becomes more efficient at higher airspeeds where the induced velocity becomes a smaller fraction of the total velocity through the propeller. In practice, some additional profile power, say $P_0$ , will be required to overcome the drag of the blades so that the efficiency can be written as

(64) $\begin{equation*} \eta = \frac{T V_{\infty}}{T (V_{\infty} + v ) + P_0} \end{equation*}$

confirming that the efficiency of the propeller at higher airspeeds is dictated by the profile losses, i.e.,

(65) $\begin{equation*} \eta \rightarrow \frac{1}{1 + P_0/T V_{\infty}} \end{equation*}$

Summary & Closure

Many fluid flow problems can be analyzed using the continuity and momentum equations in the integral form and the energy equation in the surrogate form of the Bernoulli equation. All three conservation principles will be needed in most practical problems, along with the complete energy equation and the equation of state if compressibility effects are involved. Problems that can be assumed to be steady, inviscid, and incompressible are the easiest to understand and predict, and to this end, some examples have been given. The experience gained in solving more straightforward fluid flow problems lends confidence in analyzing more complex problems, such as those involving unsteadiness and compressibility effects.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

Why is the flow through a rapidly converging or expanding venturi (or duct) more difficult to justify as one-dimensional?
What happens when an aircraft flies at higher Mach numbers? Does Bernoulli’s equation apply?
Who was Henri Pitot? What did he do besides invent the Pitot tube?
Using a drinking straw and a ruler, explain how you would measure the flow velocity in a water channel.
What happens in the throat of a venturi as the flow speed becomes subsonic and then supersonic?
Think about some aeronautical problems that cannot be tackled using the conservation principles in integral form.
It is proposed that the drag of a circular cylinder is to be measured in a wind tunnel using the momentum deficiency technique. What is the concern here, and why?

Additional Online Resources

Explore some of these additional online resources to help understand the application of the conservation laws when applied to simple fluid flow problems:

Explore this interactive simulation of flow through a pipe.
Video of worked example problems using the conservation laws.
View this video for a simple experimental demonstration of a U-tube manometer.
This Pitot Static System Simulator allows you to visualize the Pitot static system on an airplane under varying atmospheric conditions and what happens when system parts become blocked.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles Copyright © 2022, 2023 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n19