Applications of the Conservation Laws

J. Gordon Leishman

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

25 Applications of the Conservation Laws

Introduction

The practical application of the conservation equations in fluid dynamics is best learned by studying and understanding exemplar problems. Remember that exemplars of the field are “Key examples chosen to be typical of designated levels of quality of competence.”^[1] Such exemplars in fluid dynamics include applications to flow through pipes and ducts, venturimeters used for flow measurement, Pitot tubes and Pitot-static systems for flow speed and airspeed measurement, understanding the essential performance and efficiency of propulsion systems such as jet engines, and the forces on airfoil sections and other bodies. After working through a few exemplars, the systematic solution process for applying the conservation equations will become apparent, enabling the tackling of new, more ambitious problems with greater confidence.

Some of these examples may appear at first to be generic engineering problems. However, they are important exemplars because they isolate the basic conservation principles in their simplest, most useful forms. Aerospace students must first learn to understand and solve these generic fluid-flow problems before they can expect to analyze more complex aerospace systems. The same principles govern wind tunnels, Pitot-static systems, aircraft fuel and hydraulic systems, environmental-control systems, engine inlets, nozzles, propulsion devices, and many other aerospace systems.

In general, all three conservation laws must be applied to solve practical problems in fluid dynamics, i.e., by using the governing equations to conserve mass, momentum, and energy. However, using all three equations is not always necessary. Conservation of mass is generally always required in problem-solving. Energy principles are often needed when pressure must be related to changes in velocity, elevation, heat transfer, or work, although pressure may also be determined from the momentum equation in appropriate problems. The momentum equation is necessary whenever forces are present. An auxiliary equation, the equation of state, may also be necessary for problems involving compressibility effects. The Bernoulli equation can be used as an alternative to the formal energy equation, but only for steady, incompressible, inviscid flows that involve no work or energy addition. This restriction on its use should always be kept in mind.

Learning Objectives

Recognize why generic fluid-flow exemplars, such as pipes, ducts, Venturis, siphons, and pipe bends, provide the foundation for analyzing more complex aerospace systems.
Apply the conservation of mass, momentum, and energy to determine flow properties in ducts, Venturis, wind tunnels, and related devices.
Understand the principles associated with the operation of Pitot tubes and Pitot-static systems, including the measurement of static, dynamic, and total pressures.
Use the integral form of the momentum equation to determine forces on fluids, pipe bends, airfoil sections, and other bodies.
Analyze the essential performance of propulsive systems, such as jet and rocket engines, using mass, momentum, and energy balances.

Flow in a Venturi

Consider the steady, incompressible flow (i.e., $\varrho$ is a constant) through a convergent-divergent nozzle, as shown in Figure 1, which is called a Venturi (with a capital V), after Giovanni Battista Venturi. The Venturi introduces the essential idea that pressure and velocity are coupled. If the flow is assumed to be incompressible, then the fluid can be either a liquid or a gas, such as air, but flowing at low velocities, i.e., all flow velocities must 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 referred to as the mouth, and the narrowest part is called the throat. According to Bernoulli’s principle, a pressure drop occurs at the throat of a Venturi tube, where the flow accelerates. This behavior can be advantageous in several practical applications, such as flow meters, carburetors, and wind tunnels.

The first step in analyzing flow through a Venturi is to draw a control surface around the Venturi, with $\vec{n} \, dS = d\vec{S}$ pointing outward from the control volume by convention, as shown in Figure 2. 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, for example, through 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 venturi’s walls, so the mass flow of fluid entering the venturi’s mouth per unit time is equal to the mass flow leaving at its exit. A one-dimensional flow assumption can significantly simplify the problem when cross-sectional area variation is relatively moderate. However, further consideration is needed of why a flow through a rapidly converging or expanding Venturi (or duct) may not be readily justifiable as a one-dimensional 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 a velocity $V_3$ at the exit of area ${A_3}$ . The fluid mass flow coming in is

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

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

Therefore, by using the continuity equation, then

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

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 equation states that the volume flow rate $Q$ through the Venturi is constant.

It can also be seen from the latter equation that, as the throat area decreases, the flow velocity must increase there to satisfy continuity, i.e., the mass and volume flow rates are constant at any cross-section. Conversely, if the throat area 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) $\begin{equation*} p + \frac{1}{2} \varrho \, V^2 = \mbox{\small constant} \end{equation*}$

or

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

giving

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

This form assumes that elevation changes are negligible between sections, so that the hydrostatic term $\varrho \, g \, z$ can be neglected.

The pressure difference $p_1 - p_2$ can be measured across the mouth and throat of the Venturi, which can be accomplished by drilling two small holes^[2] (i.e., pressure taps) perpendicular to the Venturi walls at locations 1 and 2. The pressures are then measured by connecting tubes from these holes to a differential pressure gauge or the two sides (legs) 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) through 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) $\begin{eqnarray*} p_{1} + \frac{1}{2} \varrho V_{1}^{2} & = & p_{2} + \frac{1}{2}\varrho V_{1}^{2} \left( \frac{A_1}{A_2} \right)^2 \\[6pt] p_{1} - p_{2} & = & \frac{1}{2} \varrho V_{1}^{2} \bigg( \left( \frac{A_1}{A_{2}} \right)^{2} - 1 \bigg) \end{eqnarray*}$

and so

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

This result applies to a converging section where $A_1 > A_2$ , so that the denominator remains positive.

Therefore, given the pressure difference $p_{1} - p_{2}$ and the Venturi geometry (the area ratio), 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 $\Delta h$ , using

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

where $\varrho_l$ is the density of the manometer liquid and $\varrho$ is the density of the flowing fluid. For gas flows, $\varrho_l \gg \varrho$ , so this result is often approximated as $p_{1}-p_{2} \simeq \varrho_l g \Delta h$ .

Applications of a Venturi

A Venturi has many applications in practical engineering. Its primary characteristic is that the flow velocity is higher in the Venturi’s throat. Consequently, the static pressure at the throat is lower than the upstream static pressure. Depending on the inlet conditions and area ratio, the throat pressure may also fall below atmospheric pressure, which is helpful for various purposes.

Flowmeter (Venturimeter)

As the previous analysis suggests, a Venturi forms the basis for a device used to measure 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, as shown in Figure 3. While venturimeters operate fundamentally on Bernoulli’s principle, which states that pressure decreases in the throat of the Venturi, they are usually factory-calibrated to account for losses, allowing volumetric flow rates to be more accurately related to the measured pressure drop across the throat. Four pressure taps are typically located at each section and are pneumatically averaged to provide a more consistent and accurate measurement of the static pressure drop between the entrance and the throat.

A classic Venturi meter measures the mass flow rate or volumetric flow rate of a fluid through a pipe.

The volumetric flow rate can be obtained from the ideal fluid assumption derived previously, with a correction factor to account for any viscous losses, i.e.,

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

where $K_{\rm cal}$ is a calibration factor. The value of $K_{\rm cal}$ is typically close to 1, but it depends on the specific Venturi.

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 in Figure 4. 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 required suction pressure; therefore, venturis are no longer required.

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

From the continuity equation, the velocity at the throat, $V_t$ , for a given airspeed, ${V_{\infty}}$ , is

(19) $\begin{equation*} V_{t} = V_{\infty}\left(\frac{A_i}{A_t}\right) \end{equation*}$

where $A_i$ is the inlet area, and $A_t$ is the area of the throat. From the Bernoulli equation, then

(20) $\begin{equation*} p_{\infty} + \frac{1}{2}\varrho {V_{\infty}}^{2} = p_{t} + \frac{1}{2}\varrho {V_{t}}^{2} = p_{t} + \frac{1}{2}\varrho V_{\infty}^2\left(\frac{A_i}{A_t}\right)^2 \end{equation*}$

Therefore, the suction pressure available at the throat relative to static pressure is

(21) $\begin{equation*} p_{t} - p_{\infty} = \frac{1}{2}\varrho {V_{\infty}}^{2} \left[ 1 - \left(\frac{A_i}{A_t}\right)^2 \right] \end{equation*}$

which shows that the suction pressure is negative relative to the free-stream static pressure and that its magnitude increases with the square of the airspeed at a given density altitude.

Carburetor

A carburetor provides an air-fuel mixture to a piston (reciprocating) engine. A carburetor has a Venturi that mixes incoming air with fuel, as shown in the schematics of Figure 5. Fuel from the tank is routed to a fuel reservoir, where a float maintains the fuel level. The fuel delivery line opens into the Venturi in the carburetor’s throat, where the fuel is first atomized. The lower pressure in this region helps draw fuel into the airstream, mix it with the air, and vaporize it downstream of the throat before the resulting fuel-air mixture enters the cylinders for combustion. Metering the fuel requires precise adjustment to ensure that the correct air-fuel mixture is delivered to each cylinder, thereby maximizing engine power and minimizing hydrocarbon emissions.

Flow Speed in a Wind Tunnel

Wind tunnels are among the most direct aerospace applications of the conservation equations. Before forces and moments on a model can be interpreted, the test-section flow speed, density, and dynamic pressure must be known accurately. A low-speed wind tunnel is a large Venturi in which the airflow is driven by a fan connected to a motor drive. The wind tunnel fan is a large propeller that efficiently moves air through the test section, as shown in Figure 6. This is an open-return or Eiffel-type wind tunnel, while the other type is a closed-return or loop type.

The static pressure drop between the wind tunnel intake and test sections can be measured to determine the flow speed in the test section.

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 to satisfy the continuity condition. The model (e.g., a wing or a complete aircraft model) is placed in the test section, where its aerodynamic characteristics are measured and recorded. The flow then passes downstream into a diverging duct (diffuser), where, just before the fan, the area is ${A_3}$ , the velocity is $V_3$ , and the pressure is $p_3$ .

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

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

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

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

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

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

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

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

Using the continuity relation

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

gives

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

Rearranging, then

(28) $\begin{equation*} V_2^2 \left[ 1 - \left( \frac{A_2}{A_1} \right)^2 \right] = \frac{2}{\varrho} \left( p_1 - p_2 \right) \end{equation*}$

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

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

The area ratio $A_2/A_1$ is fixed for a given wind tunnel. Please remember that the density is a constant for an incompressible flow. Its value can be obtained from measurements of static pressure and temperature, along with the equation of state.

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 from the contraction to the test section. This latter technique is used in most low-speed wind tunnels to measure flow speed in the test section: the pressure drop between a point on the contraction and the test section is measured. This pressure drop is then related to the flow velocity in the test section via Bernoulli’s equation and subsequently verified 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, $K_{\rm cal}$ , that can be used to determine the flow speed more accurately, i.e., using

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

In practice, the calibration factor for a low-speed wind tunnel is close to 1.

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 at the throat is reduced to its vapor pressure point, $p_v$ , as illustrated in Figure 7. The resulting vapor and bubbles, known as cavitation, are analogous to boiling without adding heat. This phenomenon will limit the mass flow rate through the Venturi, preventing any further increase in mass flow rate at a given upstream inlet pressure. As the vapor bubbles travel downstream through the low-pressure region and into the diffuser, they eventually collapse as the static pressure recovers. The mass flow rate through the pipe is independent of the downstream pressure; therefore, the mass flow can be controlled by changing the inlet pressure, and hence the amount of cavitation in the throat. Cavitating venturimeters are often used as propellant mass flow limiters in rocket engines.

The cavitation number is usually expressed as

(31) $\begin{equation*} C_{\rm cav} = \dfrac{p - p_v}{\frac{1}{2} \varrho V_t^2} \end{equation*}$

where $V_t$ is the flow velocity at the throat. This is a form of pressure coefficient, so cavitation will occur as $C_{\rm cav} \rightarrow 0$ , i.e., when the local pressure approaches the vapor pressure, $p \rightarrow p_v$ . The volumetric flow rate through a Venturi is given by

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

where $p_e$ is the entrance pressure to the venturi, and $p_t$ is the pressure in the throat. The entrance area is $A_e$ , and the throat area is $A_t$ , both of which can be measured. If the pressure in the throat equals the vapor pressure, i.e., $p_t = p_v$ and $C_{\rm cav} = 0$ , then the flow rate will become cavitation-limited to a maximum value of

(33) $\begin{equation*} Q_{\rm max} = A_e \sqrt{ \frac{2( p_{e} - p_{v})}{\varrho \bigg( \left( \displaystyle{\frac{A_{e}}{A_{t}} } \right)^{2} - 1 \bigg) } } \end{equation*}$

Once this point is reached, further reductions in downstream pressure will be ineffective in increasing the flow rate for a given upstream pressure. However, the cavitation-limited flow rate can still be changed by changing the upstream pressure or the throat area. Vapor pressures, $p_v$ , for various liquids at different temperatures are available in online resources and in the table below. Knowing the vapor pressure also allows the Venturi to be designed to limit either the volumetric or the mass flow rate. For example, for given values of $Q_{\rm max}$ , $p_e$ , and $A_1$ , the throat area $A_t$ can be determined to limit the flow rate from the onset of cavitation.

**Vapor pressures of different liquids at a temperature of 25°C (77°F)**
Liquid	Vapor pressure (Pa)	Vapor pressure (lb/ft²)
Water	3,166.6	66.14
JET-A	~ 2,000	~ 41.77
100LL gasoline	~ 40,000	~ 835.42
Ethanol	5,853.7	122.26
Acetone	24,266.4	506.81
Diethyl ether	53,465.3	1,116.65
Methanol	16,931.5	353.62
Hexane	19,995	417.60
Chloroform	26,384.9	551.06
Carbon tetrachloride	12,125.7	253.25
Benzine	12,661.2	264.43

Siphon (Syphon)

A siphon, also spelled syphon, is a device used to transfer liquids from one level to another. It consists of a tube or hose bent into a U-shape, with one end placed in a liquid, such as water, and the other end set at a lower level, as shown in Figure 8. This arrangement allows water to flow from the tank to a lower level, relying on gravity and hydrostatic pressure. To initiate flow, the siphon tube must be filled with water. This can be done by evacuating air from the tube or by filling it with water and submerging one end in the tank. Once the flow starts, it will continue while the outlet remains below the liquid free surface and the siphon tube remains filled. The flow stops when the liquid level falls below the submerged inlet, the liquid column breaks, or the required elevation difference is no longer maintained.

The principle of a siphon is used to transfer a liquid from a container placed at a higher level to a lower level.

It is often claimed that a siphon 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 siphon gives

(34) $\begin{equation*} p_1 + \frac{1}{2} \varrho V_1^2 + \varrho \, g \, z_1 = p_2 + \frac{1}{2} \varrho V_2^2 + \varrho \, g \, z_2 \end{equation*}$

The change in atmospheric pressure between the elevations $z_1$ and $z_2$ is negligible, so it is reasonable to approximate $p_1 \approx p_2$ . Also, if the tank cross-sectional area is much greater than the cross-sectional area of the siphon tube, then the free-surface velocity is negligible, i.e., $V_1 \approx 0$ , and the free-surface elevation $z_1$ may be treated as approximately constant over a short time interval. Therefore, under these quasi-steady assumptions, the Bernoulli equation becomes

(35) $\begin{equation*} p_1 + 0 + \varrho \, g \, z_1 = p_2 + \frac{1}{2} \varrho V_2^2 + \varrho \, g \, z_2 \end{equation*}$

or

(36) $\begin{equation*} \frac{1}{2} V_2^2 = g \left( z_1 - z_2 \right) \end{equation*}$

Rearranging gives the outlet flow velocity as

(37) $\begin{equation*} V_2 = \sqrt{ 2 \, g \left( z_1 - z_2 \right)} = \sqrt{ 2 \, g \, h} \end{equation*}$

which is called Torricelli’s law.^[3] Notice that the flow velocity depends on the difference in the height of the water, ${h}$ , 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. Therefore, the siphon effect is produced by the hydrostatic pressure difference within the liquid, not the (slight) difference in atmospheric air pressure.

However, considering viscous effects (and hence losses) will change this result, as there will be a slight pressure drop along the length of the siphon tube. Calculating such internal pressure losses and flow velocity requires a viscous flow theory, which will be considered in a later chapter of this ebook.

Total, Static, & Dynamic Pressures

It must now be explained more precisely what the total and static pressures of a fluid mean. Consider a flow moving with velocity $V$ at a 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 of the random motion of 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 a direction 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 Figure 9, and the open end outside the flow (i.e., the reference end) of the manometer is at a lower static pressure than in the flow. In that case, a height difference $h_s$ will be observed on the manometer. The pressure is the static pressure relative to the ambient reference pressure, called the 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, causing it to stagnate; i.e., the flow velocity goes to zero at the tube entrance 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; this location is called a stagnation point. In this case, the pressure that is measured is called the total pressure, $p_T$ , which is the sum of the dynamic pressure and the static pressure, i.e.,

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

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

Finally, consider the case in which the reference pressure is 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

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

with an equivalent hydrostatic head of $h_d$ on the manometer.

Pitot Tube & Pitot-Static Tube

Pitot probes (or Pitot tubes), static probes, and Pitot-static probes are often used to measure flow velocities. For aerospace engineers, the Pitot-static system is one of the most important practical applications of Bernoulli’s equation. The measurement of airspeed, altitude, and rate of climb depends on pressure measurements, so understanding total, static, and dynamic pressure is essential to understanding how aircraft instruments work. 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 flow density at the probe insertion point is known. These three types of pressure probes are shown in Figure 10.

When a Pitot tube and static tube are combined into a single unit, it is called a Pitot-static tube (or probe). A “static tube” can be used to measure static pressure in a flow.

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 Figure 11. From Bernoulli’s equation it is known that $p+1/2\varrho V^{2} =$ constant along any given streamline, so that

(40) $\begin{equation*} p_{1} + \frac{1}{2} \varrho_{\infty} V_{1}^{2} = p_{2} + \frac{1}{2} \varrho_{\infty} V_{2}^{2} = p_T = p_{\infty} + \frac{1}{2} \varrho_{\infty} V_{\infty}^{2} \end{equation*}$

The difference in pressure between total pressure and static pressure can be measured 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 so

(41) $\begin{equation*} p_1 + \frac{1}{2} \varrho_{\infty} V_{1}^2 = p_{2} = p_T \quad \text{(total pressure at point 2)} \end{equation*}$

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

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

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

The preceding principles are used in airspeed measurement with an Air Speed Indicator (ASI), which is part of the aircraft’s pneumatic system for not only airspeed but also altitude and rate of climb, as shown in Figure 12. 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 and a separate static vent (usually placed 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 Pitot-static system on an airplane provides the reference pressures needed for the airspeed indicator and other pneumatic instruments. Notice that, in this case, a Pitot probe is used to measure total pressure, and a separate static port is used to measure static pressure.

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

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

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

Flow Speed in a Wind Tunnel (Take 2)

While it has been previously described that the flow speed in the test section of a wind tunnel can be obtained by measuring the static pressure drop between the inlet and the test section, an alternative is to use a Pitot probe, as shown in Figure 13. The Pitot probe measures the total pressure, $p_T$ , in the upstream section, or in any upstream section convenient for measurement. This approach tends to yield a more accurate measure of dynamic pressure because it is the difference between a higher and a lower pressure, rather than between two lower pressures of similar magnitude.

Another method of measuring the flow velocity in a wind tunnel is to use measurements of total pressure and static pressure.

From the Bernoulli equation, then

(44) $\begin{equation*} p_T = p_2 + \dfrac{1}{2} \varrho V_2^2 = p_{\infty} + \dfrac{1}{2} \varrho V_{\infty}^2 \end{equation*}$

so that

(45) $\begin{equation*} V_2 = V_{\infty} = \sqrt{ \frac{2 \left( p_T - p_{\infty} \right)}{\varrho_{\infty}} } \end{equation*}$

Again, the value of density, $\varrho_{\infty}$ , can be obtained from static pressure and temperature measurements in conjunction with the equation of state. The calibration would be verified by placing a Pitot probe in the test section to obtain any calibration factor, $K_{\rm cal}$ , i.e.,

(46) $\begin{equation*} V_{\infty} = K_{\rm cal} \sqrt{ \frac{2 \left( p_T - p_{\infty} \right)}{\varrho_{\infty}} } \end{equation*}$

where $K_{\rm cal}$ will be very close to unity.

Tesla Valve

Consider the branched circuit shown in Figure 14, which has a central main line and three loops. The question is: what happens to the flow as it moves through this unusual branched circuit? Two cases are to be examined. Case A depicts a flow entering from the right and exiting to the left. Case B is when the flow enters from the left and leaves to the right. The section title suggests that this situation involves valve-like action and that flow behavior varies with direction. Recall that a valve is a device used to regulate or limit the flow of a fluid. In understanding the flow, one immediately considers applying the principle of continuity, i.e., conservation of mass or volume, to the fluid entering and leaving the valve.

Consider a single branch loop of this system, as shown in Figure 15. In Case A, with flow from right to left, relatively little fluid passes around the branch loop compared with the flow through the main passage. Any fluid entering the loop rejoins the main flow farther downstream with a velocity component in the same general direction. By continuity, the flow rate entering the branch loop equals the flow rate leaving it, so the loop produces relatively little interference with the main flow.

Flow model for one loop of a Tesla valve. With the flow from left to right, the branch loop promotes backflow and stagnation, effectively limiting the volumetric flow rate.

In Case B, with flow from left to right, some of the flow splits at the first Y-branch, travels around the loop, and then rejoins the main passage at the second Y-branch with an upstream-directed component. The returning branch flow therefore opposes the local forward flow, producing regions of recirculation, backflow, and stagnation. The branch loop then acts as a fluidic restrictor. The flow entering the loop must still leave it, but the returning component does not contribute constructively to the downstream transport. Instead, it reduces the net forward flow that can be sustained for a given applied pressure difference.

The simple one-dimensional incompressible form of the continuity equation can be used to construct a first-order model of this directional behavior. Let the incoming volumetric flow rate before a loop be $Q$ . Suppose that a fraction $\scriptstyle f$ of the flow is diverted through the loop, while the remainder continues through the main passage. The corresponding flow rates are

(47) $\begin{equation*} Q_{\mathrm{loop}} = f \, Q \end{equation*}$

and

(48) $\begin{equation*} Q_{\mathrm{main}} = (1-f)Q \end{equation*}$

By continuity, the loop flow must return to the main passage. In the preferred direction, the returning flow has a downstream-directed component and rejoins the main flow with relatively little interference. In the restrictive direction, however, the returning flow has an upstream-directed component that opposes the local forward flow. If the full loop flow is idealized as returning directly against the main-passage flow, then the effective forward-flow component after one loop is

(49) $\begin{equation*} Q_1 = Q_{\mathrm{main}} - Q_{\mathrm{loop}} = (1-f)Q-fQ = (1-2f)Q \end{equation*}$

This result represents the net forward-flow component in the restrictive direction. It does not imply that fluid disappears from the system. For steady flow through the complete valve,

(50) $\begin{equation*} Q_{\mathrm{in}} = Q_{\mathrm{out}} \end{equation*}$

as required by continuity. Rather, the returning flow opposes the main flow and produces stagnation, recirculation, and additional resistance. Consequently, the inlet-to-outlet flow rate sustained by a given applied pressure difference is reduced.

For $n$ identical loops in series, a simple first-order model can be obtained by applying the same effective reduction factor at each stage, giving

(51) $\begin{equation*} Q_n = (1-2f)^n Q \end{equation*}$

where $Q$ is the relatively unimpeded reference flow rate and $Q_n$ is the modeled effective forward flow after $n$ restrictive loops. For example, if

(52) $\begin{equation*} f=\frac{1}{3} \end{equation*}$

then after three loops,

(53) $\begin{equation*} Q_3 = \left(1-\frac{2}{3}\right)^3Q = \left(\frac{1}{3}\right)^3Q = \frac{1}{27}Q \end{equation*}$

This model is intentionally idealized. It assumes that the same fraction $f$ is diverted at each loop and that the returning branch flow opposes the main flow directly. Continuity provides the flow-partition framework, but it does not determine the value of $f$ or guarantee that it remains constant from one loop to the next. Those quantities depend on the valve geometry, viscosity, Reynolds number, flow separation, momentum exchange, and pressure losses through the bends and junctions.

In practice, the performance of a Tesla valve is more complex than can be represented by this simple model, as shown in a viscous flow simulation here. Nevertheless, the branch-loop backflow and the resulting stagnation regions are readily apparent. In one direction, the side loops have relatively little effect, whereas in the other, they redirect part of the flow upstream and increase the resistance to forward motion. Therefore, a Tesla valve acts as a one-way fluidic restrictor that allows fluid to flow more easily in one direction than the other, with no moving parts.

Force on a Pipe Bend

The determination of flow rates, flow velocities, pressures, and forces on fluids flowing through pipes and channels is a typical application of the conservation laws in integral form. Although a pipe bend may seem far removed from aircraft or spacecraft, the same momentum-balance method is used to determine forces on ducts, inlets, nozzles, deflected jets, hydraulic lines, and propulsion-system components. The geometry is simple, but the force-balance logic is general.

Consider a flow through a pipe of circular cross-section that encounters a change in the area and height of the pipe as it passes through an elbow-type coupling, as shown in Figure 16. The pipe lies in a vertical plane, e.g., the ${x}$ – ${z}$ plane. Assume uniform flow properties across any cross-section and no losses, and that the fluid is water of density $\varrho_w$ .

Flow model for the flow through an elbow-type coupling.

The flow rate, $Q$ , or the mass flow rate $\overbigdot{m}$ , is usually given, as well as the input and output areas, so that conservation of mass gives

(54) $\begin{equation*} \overbigdot{m} = \varrho_w A_1 V_1 = \varrho_w A_2 V_2 \implies Q = A_1 V_1 = A_2 V_2 \end{equation*}$

Therefore, the flow velocities are given by

(55) $\begin{equation*} V_1 = \dfrac{Q}{A_1} = \dfrac{\overbigdot{m}}{\varrho_w \, A_1} \quad \text{and} \quad V_2 = \dfrac{Q}{A_2} = \dfrac{\overbigdot{m}}{\varrho_w \, A_2} \end{equation*}$

Using the Bernoulli equation gives

(56) $\begin{equation*} p_1 + \frac{1}{2} \, \varrho_w \, V_1^2 + \varrho_w \, g \, z_1 = p_2 + \frac{1}{2} \, \varrho_w \, V_2^2 + \varrho_w \, g \, z_2 \end{equation*}$

If, for example, $p_1$ is a known value, then rearranging to solve for $p_2$ gives

(57) $\begin{equation*} p_2 = p_1 + \frac{1}{2} \varrho_w \, \left( V_1^2 - V_2^2 \right) + \varrho_w \, g \left( z_1 - z_2 \right) \end{equation*}$

The force on the pipe can be obtained by summing the net pressure forces and the time rate of change of momentum. It will be apparent that the pressure force and momentum change only in the ${x}$ direction. The one-dimensional form of the momentum equation becomes

(58) $\begin{equation*} F_{x} + (p_1 \, A_1 - p_2 \, A_2) = \overbigdot{m} V_2 - \overbigdot{m} V_1 = \overbigdot{m} \left( V_2 - V_1 \right) = \varrho_w \, Q \left( V_2 - V_1 \right) \end{equation*}$

where $F_x$ is the force on the fluid, so that

(59) $\begin{equation*} F_x = \left( p_2 \, A_2 - p_1 \, A_1 \right) + \overbigdot{m} \left( V_2 - V_1 \right) = ( p_2 \, A_2 - p_1 \, A_1) + \varrho_w \, Q \, \left( V_2 - V_1 \right) \end{equation*}$

Therefore, the reaction force on the pipe is $R_x = -F_x$ .

Jet Engine Performance

The thrust of a turbojet engine can be determined by applying conservation principles to a control volume surrounding the engine, as shown in Figure 17. The jet engine provides a direct aerospace application of the momentum equation. Thrust is produced because the engine changes the momentum of the flow passing through it, and the same control-volume logic applies to turbojets, turbofans, rockets, propellers, and rotors. 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 remains significant and must be accounted for.

A control volume approach for analyzing a turbojet engine, which works on the air to increase its momentum in the downstream direction; the thrust on the engine is directed opposite to the direction of the jet exhaust.

The mass flow of air into the engine will be

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

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

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

with $A_e$ as the exit area and $V_j$ as the exit or “jet” velocity. If the nozzle is approximately ideally expanded, so that $p_e \approx p_\infty$ , the pressure-thrust term is small and may be neglected, i.e.,

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

If the mass flow of the fuel (small) is neglected in comparison to the mass flow rate of air (high), then

(63) $\begin{equation*} T = \overbigdot{m}_{\rm air} V_j - \overbigdot{m}_{\rm air} V_{\infty} = \overbigdot{m} \left(V_j - V_{\infty} \right) \end{equation*}$

Using the principle of conservation of energy, the propulsive efficiency can be defined as

(64) $\begin{equation*} \eta_p = \frac{\mbox{\small Useful power produced}}{\mbox{\small Total power expended }} = \frac{T \, V_{\infty}}{ T \, V_{\infty} + \frac{1}{2} \overbigdot{m} \left( V_j - V_{\infty} \right)^2} \end{equation*}$

which shows that losses appear as a gain in kinetic energy in the downstream jet flow. Substituting for the thrust gives

(65) $\begin{equation*} \eta_p= \frac{\overbigdot{m} \left( V_j - V_{\infty} \right) V_{\infty}}{ \overbigdot{m} \left( V_j - V_{\infty} \right) V_{\infty} + \frac{1}{2} \overbigdot{m} \left( V_j - V_{\infty} \right)^2} = \frac{2}{1 + \left( \displaystyle{\frac{V_j}{V_{\infty} }}\right) } \end{equation*}$

where to be meaningful $V_j > V_{\infty}$ . Notice that a low value of $V_j$ for a given thrust and, hence, higher propulsive efficiency can be achieved by having a significantly high mass flow rate, $\overbigdot{m}$ , through the engine. Therefore, even with this relatively simple analysis, it can be concluded that it is more efficient to generate thrust by accelerating a large air volume (mass flow rate) at a lower $V_j$ than by accelerating a smaller air volume (mass flow rate) at a higher $V_j$ .

Drag Force

Consider a two-dimensional body placed inside a control volume bounded by the upstream plane AD, the downstream plane CB, and the upper and lower boundaries AB and CD. Let $V_{\infty}$ be the uniform upstream velocity and $V_2(y)$ be the streamwise velocity in the downstream wake. The reaction force exerted by the fluid on the body is $\vec{R}$ , whose ${x}$ -component is the drag per unit span, $D$ . Applying the ${x}$ -component of the momentum equation to this control volume gives

(66) $\begin{equation*} \iint_{\rm AD} p \, d\vec{S} = \iint_{\rm CB} p \, d\vec{S} \end{equation*}$

Therefore, the force on the fluid and drag on the body can be expressed only in terms of the streamwise velocity component $u$ , where ${u = V_{\infty}}$ upstream and $u = V_2$ downstream, i.e.,

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

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

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

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

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

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

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

Substituting back into the expression for $D$ gives

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

This latter expression gives the drag of a two-dimensional body in terms of ${V_{\infty}}$ and the flowfield properties $\varrho_2$ and ${V_2}$ , evaluated across a vertical plane 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

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

However, caution is needed, as this approach is effective only when flow separation or turbulence around the body is minimal. If there are significant wake losses because of viscous effects from turbulent eddies, the drag will be underestimated.

This latter technique is frequently employed in wind tunnel testing to measure forces on relatively streamlined bodies; the velocities downstream of the body or airfoil are measured using a wake rake. This rake is, in fact, an array of Pitot tubes used to measure dynamic pressure and flow velocity in the wake. The momentum deficit is measured at discrete points in the wake, so that the drag per unit span is

(73) $\begin{equation*} D \approx \varrho_{\infty} \sum_{i=1}^{N} u(x, y_i) \bigg( V_\infty - u(x, y_i) \bigg) \Delta y \end{equation*}$

where $u(x, y_i)$ is the streamwise velocity at the $i$ -th point in the wake, and $\Delta y$ is the uniform spacing between the measurement points. For nonuniform spacing, the summation must be modified to use the local interval widths. The values would be measured for each test point, e.g., different flow speeds and angles of attack.

Lift Force

Consider now how to measure the lift on the body, which is the force in the $y$ direction.The lift on the C.V. is calculated as the net vertical force resulting from the pressure difference between the surface AB (the ceiling) and CD (the floor), using the normal components of the pressure forces, i.e.,

(74) $\begin{equation*} L = \iint_{_{\text{fl}}} p \, dS - \iint_{_{\text{ce}}} p \, dS = \iint_{\text{CD}} p \, dS - \iint_{\text{AB}} p \, dS \end{equation*}$

as illustrated in Figure 18, where $p_{_{\text{ce}}}(x)$ is the measured pressure distribution (using pressure taps) along the ceiling, and $p_{_{\text{fl}}}(x)$ is the pressure distribution along the floor. Again, while the floor and ceiling are streamlines to the flow (so there is no flow normal to these surfaces), the pressure can still vary along their lengths because of streamwise acceleration and deceleration induced by the presence of the body.

Lift on a body can be determined from the pressure distributions on the floor and ceiling of the wind tunnel.

The sectional lift per unit span is obtained using a one-dimensional pressure integral along the length of the control volume, i.e.,

(75) $\begin{equation*} L = \int_{x_1}^{x_2} \bigg( p_{\text{fl}}(x) - p_{\text{ce}}(x) \bigg) dx \end{equation*}$

where $x_1$ is located at the start of the test section and $x_2$ is at or near the end. Each pressure tap is connected to a pressure-measurement system to measure the local static pressure. This formulation assumes that viscous shear forces on the control surfaces are negligible, so that the net vertical force is determined solely by the pressure distribution.

Because the pressure is measured at discrete points along the wind tunnel walls, the lift per unit span is approximated by

(76) $\begin{equation*} L \approx \sum_{i=1}^{N} \bigg( p_{\text{fl}}(x_i) - p_{\text{ce}}(x_i) \bigg) \Delta x \end{equation*}$

where $x_i$ represents the measurement locations along the wind tunnel and $\Delta x$ is the spacing between measurement points; $\Delta x$ can be non-uniform. This summation represents the integral of the pressure difference between the floor and ceiling, i.e., the net vertical force per unit span on the control volume.

Historical Context: NACA and Abbott & Von Doenhoff

The integral method was extensively used by NACA researchers, notably Ira Abbott and Albert von Doenhoff, whose influential 1940s work was published as “Theory of Wing Sections & Summary of Airfoil Data.” Their approach inferred forces from test-section wall-pressure distributions and wake surveys, thereby eliminating the need for direct surface measurements, such as pressures on each airfoil. This method enabled the testing and comparison of more than 200 airfoil sections, significantly benefiting the aeronautics industry by providing a comprehensive catalog of airfoil shapes for selecting to meet specific aerodynamic requirements.

Firefighting Drone

It has been proposed that a quad-rotor drone could fight a high-rise fire by supporting a free-hanging hose supplied by a ground-based pump. At first glance, the idea appears interesting and plausible. A ground-based pump provides the pressure needed to raise the water to the drone, so it may seem that the drone needs only to position the nozzle and direct the water toward the fire. However, the feasibility of this concept also depends on the structural loads transmitted through the hose, the reaction force produced by the nozzle, the effects of wind and flow transients, and the pumping power required to lift and accelerate the water.

Consider a hovering drone supporting the upper end of a firefighting hose supplied by a ground-based pump, as shown in Figure 19. The hose has an internal diameter $D$ and extends through a vertical distance $L$ . If the building has $p$ stories and a representative story height is 3 m, then

(77) $\begin{equation*} L = 3 \, p \end{equation*}$

A proposed firefighting drone could be helpful, but does it make physical sense? The answer lies in the conservation laws of fluid mechanics.

The mass of water contained in the vertical portion of the hose is

(78) $\begin{equation*} m_w = \varrho_w \, A \, L = \varrho_w \left( \frac{\pi D^2}{4} \right)L \end{equation*}$

and its weight is

(79) $\begin{equation*} W_w = \varrho_w \, g \, A \, L \end{equation*}$

However, this water weight is not necessarily transmitted directly to the drone as an additional suspended dead load. For steady flow in a vertical hose of constant area, the pressure difference between its lower and upper ends provides the force needed to balance the weight of the water column. Neglecting friction for the moment, a force balance on the water gives

(80) $\begin{equation*} p_{\rm lower}A - p_{\rm upper}A - \varrho_w \, g \, A \, L = 0 \end{equation*}$

or

(81) $\begin{equation*} p_{\rm lower} - p_{\rm upper} = \varrho_w \, g \, L \end{equation*}$

Therefore, the structural load transmitted to the drone cannot be obtained simply by adding the weight of all the water contained in the hose. It depends on the pressure forces at the fittings, the support conditions at the lower end of the hose, the hose tension, the dry hose weight, the nozzle reaction, and dynamic loads. A complete determination would require a free-body analysis of the hose, fittings, nozzle, and their attachment points.

The dry hose may still impose a substantial suspended load. If the dry hose mass per unit length is $\mu_h$ and the entire suspended hose weight is carried at its upper attachment, then an upper-bound estimate of this contribution is

(82) $\begin{equation*} W_h = \mu_h \, g \, L \end{equation*}$

For example, if $\mu_h = 1.0~\mbox{kg/m}$ and $L = 45~\mbox{m}$ , then

(83) $\begin{equation*} W_h = 1.0 \times 9.81 \times 45 = 441~\mbox{N} \end{equation*}$

which corresponds to an equivalent mass of approximately 45 kg before accounting for the nozzle, fittings, hose tension caused by pressure forces, wind loading, and flow transients.

The next issue is the reaction force generated by the water discharged from the nozzle. For a control volume drawn around the nozzle, conservation of linear momentum gives

(84) $\begin{equation*} \vec{F}_{\text{on fluid}} = \overbigdot{m} \left( \vec{V}_{\text{out}} - \vec{V}_{\text{in}} \right) + \int_A (p-p_\infty) \,d\vec{S} \end{equation*}$

where $\overbigdot{m}$ is the mass flow rate, $\vec{V}_{\text{in}}$ and $\vec{V}_{\text{out}}$ are the inlet and outlet velocities, and the surface integral represents the pressure-force contribution.

For a one-dimensional nozzle aligned with the jet, the reaction transmitted through the nozzle depends on the inlet pressure, inlet and outlet areas, and the corresponding velocities. The outlet pressure term vanishes when the nozzle discharges to the atmosphere and gauge pressure is used, but the inlet pressure force generally remains important. The exact nozzle reaction must therefore be determined from the complete momentum balance.

A useful first-order estimate based only on the discharged jet momentum is

(85) $\begin{equation*} F_{\text{jet}} \approx \overbigdot{m} \, V_j = \varrho_w \, Q \, V_j \end{equation*}$

where $Q$ is the volumetric flow rate and $V_j$ is the jet exit speed. For a representative flow rate of

(86) $\begin{equation*} Q = 0.010~\mbox{m$^3$/s} \end{equation*}$

the mass flow rate is

(87) $\begin{equation*} \overbigdot{m} = \varrho_w Q = 1,000 \times 0.010 = 10.0~\mbox{kg/s} \end{equation*}$

For representative jet speeds of $V_j = 25$ – $40~\mbox{m/s}$ , the jet-momentum estimate gives

(88) $\begin{equation*} F_{\text{jet}} \approx 250\text{--}400~\mbox{N} \end{equation*}$

Because the jet is directed primarily horizontally toward the fire, the drone must produce a horizontal thrust component to oppose this reaction while maintaining sufficient vertical thrust to support its own weight and any load transmitted through the hose and fittings. If $T_v$ denotes the total required vertical thrust and $F_h$ denotes the horizontal nozzle and hose load, then the required rotor thrust magnitude is

(89) $\begin{equation*} T = \sqrt{T_v^2+F_h^2} \end{equation*}$

and the thrust-vector tilt angle from vertical is

(90) $\begin{equation*} \theta = \tan^{-1} \left( \frac{F_h}{T_v} \right) \end{equation*}$

The actual values of $T_v$ and $F_h$ require a structural and momentum balance for the complete hose-nozzle system.

The ground-based pump must provide the pressure and power needed to raise the water and produce the required jet velocity. For steady vertical flow, the ideal pressure rise associated with the elevation change is

(91) $\begin{equation*} \Delta p_{\text{static}} = \varrho_w \, g \, L \end{equation*}$

For $L = 45~\mbox{m}$ ,

(92) $\begin{equation*} \Delta p_{\text{static}} = 1,000 \times 9.81 \times 45 = 4.41\times10^5~\mbox{Pa} = 441~\mbox{kPa} \end{equation*}$

This pressure difference balances the weight of the vertical water column and also represents an energy requirement that must be supplied by the pump. The corresponding hydraulic power is

(93) $\begin{equation*} P_{\text{static}} = \Delta p_{\text{static}} \, Q \end{equation*}$

For

*** QuickLaTeX cannot compile formula:
Q = 0.010~\mbox{m

*** Error message:
File ended while scanning use of \mbox .

^3

*** QuickLaTeX cannot compile formula:
/s}

*** Error message:
Extra }, or forgotten $.
leading text: $/s}

,

(94) $\begin{equation*} P_{\text{static}} = 4.41\times10^5 \times 0.010 = 4.41\times10^3~\mbox{W} = 4.41~\mbox{kW} \end{equation*}$

Assuming a representative pump efficiency of $\eta_p = 0.70$ , the required shaft power associated with the elevation change alone is

(95) $\begin{equation*} P_{\text{shaft,static}} = \frac{P_{\text{static}}}{\eta_p} = \frac{4.41}{0.70} \approx 6.3~\mbox{kW} \end{equation*}$

The pump must also supply the kinetic energy of the discharged jet. Neglecting inlet velocity and internal nozzle losses, the ideal pressure drop needed to form a jet of speed $V_j$ is

(96) $\begin{equation*} \Delta p_{\text{nozzle}} \approx \frac{1}{2} \varrho_w V_j^2 \end{equation*}$

and the corresponding hydraulic jet power is

(97) $\begin{equation*} P_{\text{jet}} = \Delta p_{\text{nozzle}}Q = \frac{1}{2} \varrho_w Q V_j^2 \end{equation*}$

For $V_j = 25$ – $40~\mbox{m/s}$ , these values are approximately

(98) $\begin{equation*} \Delta p_{\text{nozzle}} \approx 0.31\text{--}0.80~\mbox{MPa} \end{equation*}$

and

(99) $\begin{equation*} P_{\text{jet}} \approx 3.1\text{--}8.0~\mbox{kW} \end{equation*}$

The corresponding shaft power associated with producing the jet is approximately

(100) $\begin{equation*} P_{\text{shaft,jet}} \approx 4.4\text{--}11.4~\mbox{kW} \end{equation*}$

for a pump efficiency of 70%.

Even before hose friction, bends, fittings, and other losses are included, the total pump shaft power required to overcome the elevation change and produce the jet is therefore of the order

(101) $\begin{equation*} P_{\text{shaft,total}} \approx 10\text{--}18~\mbox{kW} \end{equation*}$

or approximately 13–24 hp per drone.

Therefore, the feasibility of a hose-supported firefighting drone cannot be assessed merely by calculating the mass of water contained in the hose and assigning its full weight to the drone. The hydrostatic pressure gradient supports the vertical water column, while the drone load depends on the dry hose weight, pressure forces at the fittings, hose support conditions, nozzle reaction, wind loading, and dynamic effects. Even so, the required hose loads, horizontal nozzle reaction, control authority, and pumping power remain substantial and would require careful system-level engineering analysis.

Compressible Flow in a Converging-Diverging Nozzle

In many applications of a Venturi, the flow speeds remain low enough that density changes are negligible, and the incompressible form of Bernoulli’s equation applies. However, if the pressure ratio across a converging-diverging duct becomes sufficiently large, the flow can become appreciably compressible, and the throat may reach the sonic condition, i.e., ${M = 1}$ . Under these circumstances, the device is more properly regarded as a converging-diverging nozzle, as shown in Figure 20, rather than a Venturi used purely as a flow meter.

This distinction is important in aerospace engineering because a converging-diverging nozzle is the essential device in rocket engines that converts the thermal energy and pressure of the combustion gases into directed kinetic energy. In a rocket engine, high-pressure, high-temperature gas is produced in the combustion chamber. The gas then accelerates through the converging section of the nozzle, reaches sonic speed at the throat, and, if the pressure ratio is sufficiently large, expands to supersonic speed in the diverging section. The resulting high-speed exhaust jet produces thrust by increasing the downstream momentum of the flow, while the reaction force propels the vehicle in the opposite direction.

Compressible flow through a nozzle can be analyzed using the standard isentropic equations of fluid dynamics.

For steady, one-dimensional flow, the continuity equation still requires that the mass flow rate be constant, i.e.,

(102) $\begin{equation*} \overbigdot{m} = \varrho \, A \, V = \mbox{\small constant} \end{equation*}$

However, $\varrho$ is no longer constant, and so the volumetric flow rate $Q = A \, V$ is not constant along the nozzle. This behavior differs fundamentally from the incompressible Venturi, where the density is assumed constant and the volume flow rate remains the same at each section.

For adiabatic flow with no shaft work, the appropriate energy relation is expressed in terms of stagnation (total) enthalpy,

(103) $\begin{equation*} h_0 = h + \frac{V^2}{2} = \mbox{\small constant} \end{equation*}$

For a perfect gas with constant specific heats, $h = c_p T$ , so this may be written in terms of the stagnation temperature $T_0$ as

(104) $\begin{equation*} T_0 = T \left( 1 + \frac{\gamma - 1}{2} \, M^2 \right) \end{equation*}$

where $M = V/a$ is the Mach number and $a = \sqrt{\gamma \, R \, T}$ is the local speed of sound. This equation shows how thermal energy is converted into kinetic energy as the gas accelerates through the nozzle. In a rocket nozzle, this conversion is the fundamental mechanism by which the combustion chamber gas is accelerated to produce thrust.

If the flow is isentropic, so that no shocks are present and viscous dissipation is negligible, then the stagnation pressure is also constant, and the static-to-stagnation relations are

(105) $\begin{equation*} \frac{p_0}{p} = \left( 1 + \frac{\gamma - 1}{2} \, M^2 \right)^{\frac{\gamma}{\gamma - 1}} \qquad \text{and} \qquad \frac{\varrho_0}{\varrho} = \left( 1 + \frac{\gamma - 1}{2} \, M^2 \right)^{\frac{1}{\gamma - 1}} \end{equation*}$

A key result for compressible nozzle flow is the area-Mach number relation, i.e.,

(106) $\begin{equation*} \frac{A}{A^*} = \frac{1}{M} \left[ \frac{2}{\gamma + 1} \left( 1 + \frac{\gamma - 1}{2} \, M^2 \right) \right]^{\frac{\gamma + 1}{2(\gamma - 1)}} \end{equation*}$

where $A^*$ is the critical area at which ${M = 1}$ . This relation implies that a given area ratio $A/A^*$ generally corresponds to two possible solutions, one subsonic and one supersonic. The converging part of the nozzle accelerates subsonic flow toward the throat, while the diverging part accelerates supersonic flow downstream of the throat. This is why a rocket nozzle must have a throat followed by a diverging section if supersonic exhaust is to be produced.

As back pressure decreases, the Mach number at the throat increases. When the throat reaches the sonic condition ${M = 1}$ , the flow is said to be choked. Under choked conditions, the mass flow rate reaches a maximum for the given upstream stagnation state $(p_0, T_0)$ , and further reductions in back pressure do not increase $\overbigdot{m}$ . The choked mass flow rate is

(107) $\begin{equation*} \overbigdot{m}_{\max} = A^* p_0 \sqrt{\frac{\gamma}{R \, T_0}} \left( \frac{2}{\gamma + 1} \right)^{\frac{\gamma + 1}{2(\gamma - 1)}} \end{equation*}$

This result is especially important in rocket propulsion because, once the nozzle is choked, the mass flow rate is set primarily by the chamber stagnation pressure, chamber stagnation temperature, gas properties, and throat area. The downstream pressure can affect the expansion process and the pressure thrust, but it does not directly increase the choked mass flow rate.

If the back pressure is low enough, the flow accelerates to supersonic speed in the diverging section. If the back pressure is not low enough to support fully supersonic flow, a normal shock may stand in the diverging section, causing an abrupt transition from supersonic to subsonic flow and a substantial loss of stagnation pressure. Depending on the back pressure, a converging-diverging nozzle may therefore operate in one of several regimes: entirely subsonic flow, choked flow with a normal shock in the diverging section, or fully supersonic expansion to the exit.

For a rocket engine, the useful operating condition is usually a choked throat with supersonic expansion through the diverging section. The nozzle exit pressure may be higher than, lower than, or equal to the ambient pressure, depending on the altitude and nozzle expansion ratio. When the exit pressure equals the ambient pressure, the nozzle is said to be ideally expanded. If the exit pressure is higher than ambient pressure, the nozzle is underexpanded; if it is lower, it is overexpanded. These effects change the pressure contribution to thrust, but the nozzle’s essential purpose remains the same: to convert the chamber gas’s energy into a high-speed exhaust jet.

Summary & Closure

Many fluid-flow problems can be analyzed using the continuity and momentum equations in integral form, together with the energy equation, often in the surrogate form of the Bernoulli equation. In most practical problems, all three conservation principles may be needed, along with the complete energy equation and the equation of state when compressibility effects are present. The momentum equation is always required when calculating forces, while the energy equation is needed whenever pressure, velocity, elevation, heat transfer, or work interactions must be related.

Problems that can be assumed to be steady, inviscid, and incompressible are the easiest to understand and predict, and several such examples have been discussed in this chapter. These examples show how conservation of mass determines flow rates and velocities, how Bernoulli’s equation relates pressure changes to changes in speed or elevation, and how the momentum equation can be used to determine forces on fluids and solid boundaries. The same principles also provide useful first-order interpretations of devices such as Venturis, Pitot tubes, wind tunnels, siphons, Tesla valves, pipe bends, and propulsion systems.

The experience gained from solving these exemplar fluid-flow problems builds confidence in applying the conservation equations to more complex flows. Real flows may include viscosity, turbulence, separation, heat transfer, shaft work, unsteadiness, and compressibility effects, but the same conservation principles remain the foundation for their analysis. The challenge is to decide which assumptions are justified, which terms must be retained, and which measurements or auxiliary relations are needed to close the problem.

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?
Why does a balloon filled with helium deflate more quickly than a balloon filled with air?
If the drag on a body can be measured using a momentum integral method, how can the lift on the body be measured?

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.

Sadler, D. R., "Interpretations of Criteria-Based Assessment and Grading in Higher Education." Assessment and Evaluation in Higher Education, 30 (2), 2005, pp. 175–194. ↵
. The diameter of such holes should be less than 0.5 mm or 20-thousandths of an inch. ↵
Evangelista Torricelli's original derivation can be found in the second book "De motu aquarum" of his "Opera Geometrica." See also: A. Malcherek, "History of the Torricelli Principle and a New Outflow Theory," Journal of Hydraulic Engineering, 142 (11), 1–7, 2016. ↵

License

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

Introduction to Aerospace Flight Vehicles Copyright © 2022–2026 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