# 20 Applications of the Conservation Laws

# Introduction

Using conservation equations of aerodynamics and fluid dynamics is best learned by studying and fully understanding exemplar problems. After a few simple examples, the systematic process of solving such problems will become apparent to the student, and new problems can then be tackled more confidently. Generally speaking, all three conservation laws must be used to solve practical problems in fluid dynamics, i.e., the use of 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 needed 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 important restriction should be remembered.

Objectives of this Lesson

- Use the basic conservation laws of aerodynamics to calculate the flow properties through a venturi, as well as through 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 integral form to find the drag on a body.
- Analyze the basic performance of a jet engine and an airplane propeller.

# Flow in a Venturi

Consider the steady, incompressible flow (i.e., is a constant) through a convergent-divergent nozzle, as shown below, which is 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.

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 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 assumed to be steady and radially axisymmetric, i.e., an appropriate one-dimensional flow assumption.

The steady form of the continuity equation in integral form is

(1)

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

In this case, 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. If the cross-sectional area variation is relatively moderate, then 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 with an average velocity , has a velocity at the throat of area , and velocity at the exit of area . The fluid mass flow coming in is

(2)

noting the minus sign appears because the flow is in the opposite direction to the direction of at the entrance face. At the exit face then

(3)

(4)

which for a one-dimensional flow this just becomes

(5)

If the flow is incompressible as stated, then . Under these circumstances, the continuity equation can now be written as

(6)

or

(7)

Also, it will be apparent that

(8)

Cancelling through the density gives

(9)

which is simply a statement that the volume flow rate through the venturi is constant.

(10)

(11)

and from Bernoulli’s equation

(12)

Substituting gives

(13)

and so

(14)

(15)

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

# Applications of a Venturi

## 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

On 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 was used to provide low (suction) pressure for air-driven gyroscopic instruments such as the artificial horizon and directional gyro. Modern light aircraft, however, are fitted with mechanic engine-driven vacuum pumps to provide the needed suction pressure.

## Carburetor

## Wind-Tunnel

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

(16)

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

(17)

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

(18)

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

(19)

This means that

(20)

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

(21)

The area ratio is fixed for a given wind tunnel. Recall that for an incompressible flow then 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.

# Total, Static, & Dynamic Pressure

It must now be explained more precisely what is meant by the total and static pressures in a fluid. Consider a flow moving with velocity at pressure . If moving with the velocity of the airflow, then the pressure that will be felt is . This latter pressure is called static pressure and is a measure of 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 . If 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, then a difference height will be shown on the manometer. The pressure is the static pressure between that in the flow and the reference ambient pressure, called gauge pressure. If the static pressures inside and outside are equal, then .

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 that 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, and hence this location is called a *stagnation point*. In this case, the pressure that is measured is called the total pressure, , i.e., the sum of the dynamic pressure and the static pressure

(22)

where is the density of the flowing fluid, with equivalent height on the manometer of .

Finally, consider the case shown 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

(23)

with an equivalent height of on the manometer.

# Pitot Tube & Pitot-Static Tube

(24)

*stagnation point*) and

(25)

(26)

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

(27)

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 really 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).

# Jet Engine Performance

The mass flow of air into the engine will be

(28)

where is the inlet area and the mass flow rate of fuel is . Therefore, the thrust is

(29)

with as the exit area and as the exit or “jet” velocity. The pressure term in Eq. 29 is relatively small compared to the change in momentum of the flow and so may be neglected.

# Drag on a Two-Dimensional 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 points out of the control volume by convention. The upstream or free-stream (undisturbed) velocity is = constant = . Because of the action of viscosity, a wake of reduced velocity or retarded flow exists behind the body, this gives a wake region a velocity profile in which .

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

(30)

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., .

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

Therefore, the total force on the fluid volume is

(31)

and from the conservation of momentum equation then

(32)

For steady flow , so that this latter equation is simplified to

(33)

Notice that this is a vector equation.

To find the drag, the -component of this latter equation is needed, noting that and are the inflow and outflow velocities in the -direction and that the -component of is the drag on the section, i.e., 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

(34)

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

(35)

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

(36)

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

(37)

Multiplying by in this case (which is a constant) gives

(38)

Substituting back into the expression for gives

(39)

This latter expression gives the drag of a two-dimensional body in terms of and the flow field properties and across a vertical dimension downstream of the body.

Notice that is the velocity deficiency or decrement at a given station (or section) in the downstream wake. Also, is the mass flux, so the product 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 = constant, then things can be simplified even further to get

(40)

Therefore, it has been shown that by applying the momentum equation in its integral form, this 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 hence 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 is, then there are losses in the wake that do not show up just as a loss in the momentum of the flow, so 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 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 points out of the control volume by convention. The upstream or free-stream (undisturbed) velocity is and the pressure there is , and 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.

The flow is entrained and accelerated into the propeller, so at the plane of the propeller, the flow velocity is plus an increment , i.e., the velocity there is . The static pressure will change there too. The propeller does work on the air to increase its kinetic energy, so by continuity considerations, in the slipstream (downstream behind the propeller), the velocity there is . In the slipstream, the static pressure will also recover to ambient conditions, i.e., the pressure downstream will be .

Conservation of mass requires constant mass flow through the control volume. The areas of at the upstream and downstream parts of the streamtube (stations 1 and 2) are not known, but the area of the propeller disk, , is known ( where is the diameter of the propeller, so the mass flow rate through the propeller (and hence through the boundaries of the control volume) is

(41)

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.,

(42)

where the direction of is downstream, and the thrust on the propeller, , is in the opposite direction (pointing left in the figure) so that

(43)

At this point, the application of 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.,

(44)

so that the power required for the propeller to produce the thrust is

(45)

Notice that the term on the left-hand side of the above equation can be written as the sum of the term, which is the work done to move the propeller forward (this is the useful work). The term is the work done on the air or the i*nduced* power loss; notice that 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. 43 into Eq. 45 gives

(46)

or simplifying gives

(47)

which gives the relationship between and , i.e.,

(48)

For the next step, take Eq. 43 and substitute in Eq. 41 (mass flow rate) and the connection between and () to get

(49)

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

(50)

or

(51)

In fact, the ratio has a special name and is called the propeller disk loading. It can be seen that latter equation a quadratic in which can be solved to get

(52)

for which there must be two roots, i.e., from the 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

(53)

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

(54)

Notice again that the useful work is and the second term is an induced loss, i.e., a loss that is irrecoverable.

Notice that for the limiting case where goes to zero then the power (i.e., the power required in the static thrust condition) is

(55)

The efficiency of the propeller can also be derived. The useful power is so the efficiency of the propeller, , would be

(56)

# Summary & Closure

Many fluid flow problems can be analyzed by 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 obviously the earliest 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

- 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 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 in action.
- This Pitot Static System Simulator allows you visualize the Pitot static system on an airplane under varying atmospheric conditions, as well as what happens when parts of the system become blocked.