30 Time-Dependent Flows

Introduction

In fluid dynamics and aerodynamics, a time-dependent or unsteady flow is one in which the properties (e.g., velocity, pressure, temperature, and density) at a given point in space vary with time. This behavior contrasts with steady flow, where the fluid properties at a given point remain constant over time. Unsteady flows occur in many real-world situations, such as when a fluid is subjected to changes in operating conditions, when transient events begin, or when time-varying external forces act on flows or flow systems.

In aerospace engineering, unsteady flow effects are frequently encountered. For example, they are found and must be considered to predict fluid behavior in various fields, including aeroelasticity and flutter, rotating machinery (such as jet engines and helicopter rotors), mixing processes, hydraulic and pneumatic control systems, and combustion processes in air-breathing and rocket engines. Analyzing and predicting unsteady flow can be more challenging than steady flow because it requires careful consideration of all relevant time-dependent factors. Closed-form solutions to most unsteady flow problems are generally only realizable in cases with specific assumptions. Therefore, experimental techniques and computational fluid dynamics (CFD) are frequently employed to investigate unsteady flows in various engineering applications.

Learning Objectives

  • Understand how to differentiate between steady and unsteady flows and recognize situations in which unsteady flow properties are relevant to problem-solving.
  • Know about reduced frequency and Strouhal number, and why they are essential for characterizing unsteady flows and the degree of unsteadiness.
  • Be able to apply the conservation principles of fluid dynamics to solve some classic unsteady flow problems.

Classification of Time-Dependent Flows

Steady flows are characterized by properties that remain constant over time. A steady-state flow is a condition in which the macroscopic flow properties, such as velocity and pressure at a point, remain constant over time, as shown in Figure 1(a). Mathematically, for steady flows, then

(1)   \begin{equation*} \frac{\partial P}{\partial t} \equiv 0 \end{equation*}

where P is any property such as pressure, temperature, velocity, density, etc. However, in a time-dependent (unsteady) flow, the flow properties at a point vary with time, as shown in Figure 1(b). In this case, all of the unsteady terms in any equations used to describe the flow must be retained and solved, i.e.,

(2)   \begin{equation*} \frac{\partial P}{\partial t} \ne 0 \end{equation*}

 

Examples of laminar flow at a point in space illustrate the distinction between steady and unsteady flow.

Unsteady flow phenomena are encountered in many engineering applications. Examples include flows in turbomachinery and other internal-combustion engines, helicopter aerodynamics, wind turbines, and numerous aeroacoustics problems in which the creation of time-varying aerodynamic loads produces sound (noise). A turbulent flow is inherently unsteady because the instantaneous velocity, pressure, and other flow properties fluctuate with time. However, a turbulent flow can still be statistically steady. This means that its time-averaged quantities, such as the mean velocity and mean pressure at a point, do not change with time, even though the instantaneous fluctuations continue to vary. In other words, the fluctuations themselves are not steady, but their statistical measures, such as their mean-square values, may remain constant.

Figure 2 illustrates the difference between statistically steady and statistically unsteady turbulent flows. Turbulence enhances fluid mixing, promoting the transport of momentum, heat, and other properties. The irregular motion of fluid particles makes it difficult to predict the flow’s behavior over time; this randomness, or stochasticity, is a fundamental characteristic of turbulence. Statistically, a flow property P can be decomposed into a mean or average part, \overline{P}, and a fluctuating part, P', i.e.,

(3)   \begin{equation*} P = \overline{P} + P' \end{equation*}

where P' is the instantaneous departure from the mean value. For a statistically steady turbulent flow, the mean value \overline{P} is independent of time, while the fluctuation P' varies with time and has a zero mean over the averaging interval. This process is known as a Reynolds decomposition, a concept particularly helpful for modeling turbulent flows.

 

Examples of the turbulent flow at a point showing a statistically steady flow versus a statistically unsteady one.

One reason for distinguishing between steady and unsteady flows is that the former are often more tractable to analyze and predict. Eliminating time from the equations governing fluid-flow problems often yields significant simplification of both the equations and the mathematical and/or numerical techniques required to solve them. Characterizing steady, quasi-steady, or unsteady flows is often done using parameters such as the reduced frequency or the Strouhal number.

Strouhal Number

The Strouhal number, given the symbol St, is a dimensionless number used in fluid dynamics to characterize the behavior of oscillating or vibrating objects within a fluid flow. It is beneficial for analyzing phenomena such as vortex shedding and sound generation. The Strouhal number is defined as the product of the oscillation frequency and a characteristic length, divided by the reference velocity, such as V or the freestream velocity {V_{\infty}}. The equation is given by

(4)   \begin{equation*} St = \frac{f \, L}{V} \quad \text{or} \quad St = \frac{f \, L}{V_\infty} \end{equation*}

Notice that the physical frequency {\scriptstyle{f}} (in units of per second or Hz) is used to determine the Strouhal number rather than the circular frequency in the reduced frequency. The Strouhal number is named after the Czech physicist and engineer Vincenc Strouhal, who made significant contributions to the study of oscillating flows in the late 19th and early 20th centuries.

The Strouhal number is often used to analyze various fluid-flow phenomena, such as vortex shedding behind cylinders and other bluff bodies (i.e., non-streamlined bodies); see Figure 3. It helps characterize the vortex-shedding frequency relative to the fluid flow velocity and the object’s size. For different flow types and body shapes, characteristic Strouhal number ranges often correspond to specific flow regimes and behaviors. For a bluff body, once periodic shedding is established, the shedding frequency is related to the flow speed and characteristic length through the Strouhal number. If the body is flexible or elastically mounted, the shedding may also couple with a structural natural frequency, producing vortex-induced vibration.

 

The frequency of vortex shedding from a bluff body, such as a circular cylinder, is often characterized by the Strouhal number.

Check Your Understanding #1 – Calculating the value of the Strouhal number

A circular cylinder with a diameter {d} of 0.1 m is immersed in a fluid flow with a velocity V of 1.0 m/s. The shedding frequency {\scriptstyle{f}} of vortex shedding behind the cylinder is measured to be 2 Hz. What is the Strouhal number in this case?

Show solution/hide solution.

The Strouhal number is defined by

    \[ St = \frac{f \, L}{V} \]

Using the numerical values gives

    \[ St = \frac{ 2.0 \times 0.1}{1.0} = 0.2 \]

For a circular cylinder, a Strouhal number near 0.2 is commonly associated with periodic vortex shedding over a broad range of Reynolds numbers. However, the onset of vortex shedding is governed primarily by the Reynolds number, not by the Strouhal number alone. Therefore, a measured value of St \approx 0.2 indicates that the shedding frequency is consistent with the usual cylinder-shedding behavior, and vortex-induced vibrations may be possible if the shedding frequency couples with a structural natural frequency. The specific Strouhal number can vary with Reynolds number, body shape, surface roughness, and other flow conditions.

Reduced Frequency

Another nondimensional parameter used to categorize whether a flow is steady or unsteady is the reduced frequency k. The reduced frequency is defined, in general, as

(5)   \begin{equation*} { k = \frac{\omega \, L}{V} } \end{equation*}

where \omega is a characteristic angular frequency of the unsteady flow, in radians per second, L is a characteristic length scale, and V is a reference flow velocity. Dimensionally consistent units must be used to obtain the correct numerical value for the reduced frequency; for example, if V is in ft/s, then L must be in ft and \omega in rad/s. For a wing or airfoil, such as one oscillating in angle of attack, the reduced frequency is often defined in terms of its semi-chord, i.e., L = c/2, and the freestream velocity, {V_{\infty}}, i.e., in this case, it is defined as

(6)   \begin{equation*} { k = \dfrac{\omega \, c}{2 \, V_{\infty}} } \end{equation*}

Flows with characteristic reduced frequencies of about k \gtrsim 0.05 may exhibit significant unsteady aerodynamic effects, so the unsteady terms in the governing equations may need to be retained. Such problems are more challenging because local flow properties and aerodynamic forces can depend on the prior time history of the motion, not just on the instantaneous angle of attack or flow condition. Compressibility effects, however, are not determined by reduced frequency alone; they also depend on the Mach number and on the relative acoustic time scales of the problem.

The reduced frequency, k, provides a measure of the importance of unsteady aerodynamic effects. When k \ll 1, the flow behaves quasi-steadily, and the aerodynamic forces at any instant are approximately the same as those for steady flow at the same instantaneous conditions. However, as k increases, the aerodynamic forces depend on the time history of the motion, leading to phase differences between the motion and the aerodynamic response that are critical in problems such as flutter, gust response, and rotor aerodynamics.

Time-Dependent Fluid Flows

Time-dependent flows are often more challenging to handle in fluid dynamics, but the same principles of conservation of mass, momentum, and energy still apply. Studying exemplars is an excellent way to learn how to solve problems involving unsteady flows. Several classic examples in fluid dynamics can be used to establish the principles of solution when time is included as an additional dimension.

Torricelli’s Law

Torricelli’s law relates the velocity of a liquid flowing out of an orifice to the height of the liquid above the orifice’s level, as shown in Figure 4. The derivation of this law assumes ideal conditions where the liquid is quasi-steady, incompressible, and inviscid. Although the problem is fundamentally unsteady, the time rates of change are sufficiently small that the flow can be treated as quasi-steady. Therefore, the Bernoulli equation is applicable, as it describes the instantaneous energy balance between points in the flow.

Torricelli’s law relates the velocity of a fluid flowing out of an orifice to the hydrostatic head of the fluid above the orifice.

The application of Bernoulli’s equation between levels 1 and 2 gives

(7)   \begin{equation*} p_1 + \frac{1}{2} \varrho \, V_1^2 + \varrho \, g \, h_1 = p_2 + \frac{1}{2} \varrho \, V_2^2 + \varrho \, g \, h_2 \end{equation*}

where p_1 is the pressure at the surface, and p_2 is the pressure at the discharge from the orifice. The fluid is a liquid, so \varrho = constant. It is assumed that there are no losses from the discharge through the orifice. Assuming p_1 = p_2 (i.e., both are equal to atmospheric pressure, p_{\rm atm}), and V_1 \approx 0, the Bernoulli equation simplifies to

(8)   \begin{equation*} \varrho \, g ( h_1 - h_2 ) = \frac{1}{2} \varrho \, V^2 \end{equation*}

Solving for the discharge or “jet” velocity, V_2, gives

(9)   \begin{equation*} V_2 = \sqrt{2 \, g \, \Delta h} \end{equation*}

This result shows that the discharge velocity is proportional to the square root of the difference in hydrostatic height or “head” between the free surface at h_1 and the orifice at h_2.

To find the horizontal distance, R, reached by a jet from any one of the orifices requires the time, t, it takes for the jet to reach the ground, i.e.,

(10)   \begin{equation*} t = \sqrt{\frac{2 H}{g}} \end{equation*}

where H is the height of the orifice above the ground. Therefore, the horizontal distance reached by the jet, R, is given by

(11)   \begin{equation*} R = V t = \sqrt{2 \, g \, \Delta h} \, \sqrt{\frac{2 \, H}{g}} = 2\sqrt{\Delta h \, H} \end{equation*}

This latter result shows that the distance traveled by the fluid jet depends on the hydrostatic head of fluid above the orifice, \Delta h, and the height above the ground, H.

Time for Liquid to Drain from a Tank

Consider a tank with a constant cross-sectional area, A, that discharges liquid through a drain valve at the bottom, as shown in Figure 5. For an ideal discharge with no losses, the discharge velocity, {V_d}, varies with the height, {h}, of the fluid level above the drain according to the relationship V_d = \sqrt{ 2 g h}. Notice that the height, {h}, will decrease with time as the liquid leaves the tank. For a real drain or orifice, this velocity is usually corrected using a discharge coefficient.

Illustration of a time-dependent flow problem where a fluid continuously discharges from a drain.

The application of the continuity equation gives

(12)   \begin{equation*} -\frac{d{\cal{V}}}{dt} = a \, V_d = 0 \end{equation*}

where a is the area of liquid discharge from the drain. The volume of liquid \cal{V} in the tank for any height {h} is

(13)   \begin{equation*} {\cal{V}} = A h \end{equation*}

where A is the cross-sectional area of the tank. It is given that

(14)   \begin{equation*} V_d = \sqrt{ 2 g h} \end{equation*}

and this flow rate depends on the instantaneous height of the liquid in the tank, h, which, as previously derived, is given by Torricelli’s theorem. Therefore, using the conservation of mass gives

(15)   \begin{equation*} \frac{dh}{dt} = - \frac{a}{A} \sqrt{ 2 g h} \end{equation*}

Separating the variables and integrating them gives

(16)   \begin{equation*} t = \int_{0}^{t} dt = - \frac{A}{a} \int_{h_0}^0 \frac{dh}{\sqrt{ 2 g h}} = - \frac{A}{a \sqrt{ 2 g}} \int_{h_0}^0 \frac{dh}{\sqrt{h}} = \frac{A}{a \sqrt{ 2 g}} \int_0^{h_0} \frac{dh}{\sqrt{h}} \end{equation*}

where the limits of integration are: At t = 0, then h = h_0, and when {h = 0}, the tank is considered empty. Performing the integration gives

(17)   \begin{equation*} t = \frac{2A}{a \sqrt{ 2 \, g}} \sqrt{h_0} \end{equation*}

If both the tank cross-section and the drain outlet are circular, then in terms of the tank diameter, D, and the outlet diameter, {d},

(18)   \begin{equation*} \frac{A}{a} = \frac{D^2}{d^2} \end{equation*}

Therefore, the ideal time required to empty the tank will be

(19)   \begin{equation*} t = \frac{2D^2}{d^2} \sqrt{\frac{h_0}{2 \, g}} \end{equation*}

where h_0 is the initial height of the liquid in the tank when the drain is first opened. For a real drain or valve, the discharge coefficient would increase the emptying time above this ideal value.

Filling an Air Tank

Consider now a rigid tank of volume {\cal{V}} with air pumped at a constant mass flow rate, as shown in Figure 6. In this problem, the effect of compressibility must be accounted for. Still, it can be assumed that the process is slow enough to be approximately isothermal, so that heat generated during compression is transferred through the tank wall to the surroundings.

Filling an air tank is a process that involves unsteady flow and compressibility effects.

This is an unsteady flow problem because air is pumped into a fixed volume, so mass conservation requires that the air density increases with time. The initial density and pressure are \varrho_0 and p_0, respectively. The general form of the continuity equation is

(20)   \begin{equation*} \frac{\partial}{\partial t}\oiiint \varrho \ d {\cal{V}} + \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

In this case

(21)   \begin{equation*} \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = -\overbigdot{m}_{\rm in} \end{equation*}

and so

(22)   \begin{equation*} \frac{\partial}{\partial t}\oiiint \varrho \, d{\cal{V}} = {\cal{V}}\frac{d \varrho}{dt} \end{equation*}

Assume uniform mixing of air, so the density within the volume is uniform. Therefore,

(23)   \begin{equation*} {\cal{V}} \frac{d \varrho}{dt} = \overbigdot{m}_{\rm in} \end{equation*}

Integrating with respect to time gives

(24)   \begin{equation*} \int_{\varrho_0}^{\varrho} d \varrho = \frac{ \overbigdot{m}_{\rm in}}{{\cal{V}}} \int_{t_0}^{t} dt \end{equation*}

so that

(25)   \begin{equation*} \varrho = \varrho_0 + \frac{ \overbigdot{m}_{\rm in}}{{\cal{V}}} \left( t - t_0 \right) \end{equation*}

which shows that, under the assumed constant inlet mass flow rate, the air density increases linearly with time. The corresponding pressure can be obtained from the equation of state, i.e., p = \varrho \, R \, T. If the process is also assumed to be isothermal, then T = T_0, and so

(26)   \begin{equation*} p = \varrho_0 \, R \, T_0 + \frac{ R \, T_0 \, \overbigdot{m}_{\rm in}}{{\cal{V}}} \left( t - t_0 \right) = p_0 + \frac{R \, T_0 \, \overbigdot{m}_{\rm in}}{{\cal{V}}} \left( t - t_0 \right) \end{equation*}

Hydraulic Shock

Hydraulic shock, also known as “water hammer” or “hydraulic hammer,” is a time-dependent flow phenomenon that occurs in fluid systems, such as hydraulic and fuel systems, when flow velocity changes suddenly. This abrupt change can generate strong pressure waves that propagate through the fluid, causing transient pressure surges within the system, as illustrated in Figure 7. Water hammer is common in fluid flow systems where valves and regulators are rapidly opened or closed. Rapid, time-dependent pressure changes can generate loud banging or knocking sounds, often audible throughout the piping system.

Hydraulic shock, also known as “water hammer,” involves the generation of large pressure pulses in a flowing fluid after the rapid closure of a valve.

The fundamental cause of water hammer is the fluid’s inertia and the need to conserve energy when the flow suddenly stops. When a fluid is in motion and experiences a sudden change in velocity, its momentum must change rapidly. If a valve is closed suddenly, the fluid velocity at the valve drops to zero, causing a local pressure rise and initiating a compression wave that propagates through the pipe. The associated energy is stored partly as compression of the liquid and elastic strain of the pipe wall, with some energy eventually dissipated by friction and other losses.

This process generates a pressure wave that propagates upstream through the pipe. This pressure wave moves through the pipe at the pressure-wave speed, a, often called the wave speed. For a rigid pipe, this speed is close to the acoustic speed in the liquid, but for an elastic pipe, it is reduced by the compliance of the pipe wall. When the pressure wave reaches a point where the fluid is stagnant, such as at a reservoir, the reflected pressure wave propagates downstream, causing continued oscillations until the wave’s energy is dissipated through friction. The increase in pressure causes a “hammer” effect, i.e., a thumping effect, which can be felt and/or heard by an observer.

The resulting pressure pulses can be several times higher than the normal operating pressure. The change in pressure from the water hammer depends on several factors, including the speed at which the valve is closed, the type of fluid, the length of the pipe, its elastic properties, and how the pipe is mounted, such as whether it is supported at its ends and/or clamped along its length. Prolonged or severe water hammer can cause structural damage to pipes, fittings, and other system components. The phenomenon can cause fatigue, leaks, and pipe bursts, and can accelerate wear and tear on valves and pumps, thereby increasing maintenance requirements. Preventing or mitigating water hammer effects involves implementing engineering solutions and using devices like surge tanks, air chambers, and water hammer arrestors, which are designed to absorb excess pressure and prevent damage to the system.

Consider the analysis of this problem. A liquid stored in a tank flows steadily through a pipe of length L, as shown in Figure 8. At the time t = 0, the valve at the downstream end is quickly closed, producing the classic pressure pulse of a water hammer. Water hammer transients in long, slender pipes are usually modeled as one-dimensional axial wave propagation problems using cross-sectionally averaged flow quantities. Radial motion of the pipe wall and the fluid is not resolved explicitly in this simple model, but its effect is accounted for through the pressure-wave speed.

 

Unsteady effects, known as water hammer, are produced when a valve is closed in a pipeline.

Nikolay Joukowsky laid the foundation of water-hammer theory [1] where the pressure amplitude, \Delta p, in the liquid in the pipe is related to the change in flow velocity, {\Delta V}, using

(27)   \begin{equation*} \Delta p = \pm \, \varrho \, a \, \Delta V \end{equation*}

where a is the wave propagation speed and \varrho is the fluid density. This equation is commonly called the Joukowsky equation. The sign depends on the chosen coordinate direction and on whether the disturbance corresponds to a pressure rise or a pressure drop. For a sudden valve closure that reduces the local flow speed by V_0, the pressure rise is usually written as \Delta p = \varrho \, a \, V_0. Because pressure head is often used in the field of hydraulics, Eq. 27 can also be written as

(28)   \begin{equation*} \Delta h = \frac{\Delta p}{\varrho \, g} = \pm \left( \frac{ a \, \Delta V}{g} \right) \end{equation*}

There are three cases of interest when it comes to predicting water hammer effects, all of which involve the wave propagation speed:

  1. Gradual closure of the valve.
  2. Sudden closure of the valve.
  3. Closure of the valve with allowance for the elasticity of the pipe, which affects the wave propagation speed.

The transit time, also known as the reflection time, for the pressure wave to propagate from the valve to the tank and then back to the valve, is

(29)   \begin{equation*} t_t = \frac{2 L}{a} \end{equation*}

If the valve closure time t_c is less than this reflection time, i.e., if t_c < 2L/a, then the closure is considered sudden. If t_c > 2L/a, then the closure is considered gradual.

If t_c > 2L/a, which corresponds to a gradual closure of the valve, then the approximate increase in pressure from the water hammer is given by

(30)   \begin{equation*} \Delta p = \frac{2 \, \varrho \, L \, V_0}{t_c} \end{equation*}

where V_0 is the average flow velocity inside the pipe before the valve closes. The equivalent pressure head of the water hammer is

(31)   \begin{equation*} \Delta h = \frac{\Delta p}{\varrho \, g} = \frac{2 \, \varrho \, L \, V_0}{\varrho \, g \, t_c} = \frac{2 \, L \, V_0}{g \, t_c} \end{equation*}

If t_c < 2L/a, which is a sudden closure of the valve, then the increase in pressure from the water hammer is given by

(32)   \begin{equation*} \Delta p = \varrho \, a \, V_0 \end{equation*}

and the equivalent pressure head is

(33)   \begin{equation*} \Delta h = \frac{a \, V_0}{g} \end{equation*}

where V_0 represents the average flow velocity of the fluid inside the pipe before the valve is closed. This velocity is critical to determining the magnitude of the pressure increase associated with the water hammer effect. It reflects the kinetic energy of the moving fluid that is converted into pressure when the flow is suddenly stopped.

Note that, in both cases, the speed of sound, a, is required. If the pipe is rigid, then

(34)   \begin{equation*} a = \sqrt{ \dfrac{E_b}{\varrho} } \end{equation*}

where E_b is called the bulk modulus of the liquid, for which values for various liquids are available. This latter equation is often referred to as the Newton-Laplace equation.

If the pipe is elastic, which most will be to a lesser or greater degree, then

(35)   \begin{equation*} a = \sqrt{ \frac{E_c}{\varrho} } \end{equation*}

where {{E_c}} is called the effective bulk modulus of the liquid in the pipe. The value of {{E_c}} can be obtained using

(36)   \begin{equation*} \frac{1}{E_c} = \frac{1}{E_b} + \frac{D \, K}{E_p \, e} \end{equation*}

where {E_p} is the modulus of elasticity of the pipe, D is the pipe diameter, and {e} is the wall thickness of the pipe. The value of K depends on the exact manner in which the pipe is mounted and anchored; typically, K \approx 1.0. The modulus of elasticity of the pipe is a material property for which values are also widely available.

Finally, consider the valve’s sudden closure when the pipe’s elasticity is taken into account. In this case

(37)   \begin{equation*} \Delta p = \varrho \, a \, V_0 = \varrho \sqrt{ \frac{E_c}{\varrho} } \, V_0 = V_0 \sqrt{ \frac{\varrho}{ \displaystyle{ \left( \frac{1}{E_b} + \frac{D \, K}{E_p \, e} \right) }} } \end{equation*}

Therefore, the system’s elasticity mitigates the pressure effects of the water hammer.

Check Your Understanding #2 – Determining hydraulic shock pressure in a pipeline

A hydraulic fluid with a density \varrho = 850 \, \text{kg/m}^3 flows through a titanium pipe in an aircraft’s hydraulic system. The pipe has a diameter of D = 0.05 \, \text{m} and a wall thickness of e = 0.005 \, \text{m}. The bulk modulus of the hydraulic fluid is E_b = 1.5 \times 10^9 \, \text{Pa}, and the modulus of elasticity of titanium is E_{\text{Ti}} = 116 \times 10^9 \, \text{Pa}. The hydraulic fluid has an average velocity V_0 = 3 \, \text{m/s} before a valve is suddenly closed in the system. Assume K = 1. Calculate the pressure increase \Delta p in the pipe from the hydraulic shock, or water hammer, effect.

Show solution/hide solution.

To calculate the pressure increase, we first need the wave speed {a}, which depends on the effective bulk modulus {E_c}, i.e.,

    \[ a = \sqrt{\frac{E_c}{\varrho}} \]

where E_c is calculated using

    \[ \frac{1}{E_c} = \frac{1}{E_b} + \frac{D \, K}{E_{\text{Ti}} \, e} \]

Substituting the given values into the equation for E_c gives

    \[ \frac{1}{E_c} = \frac{1}{1.5 \times 10^9} + \frac{0.05 \times 1}{116 \times 10^9 \times 0.005} \approx 7.53 \times 10^{-10} \, \text{Pa}^{-1} \]

and so

    \[ E_c \approx 1.33 \times 10^9 \, \text{Pa} \]

The wave speed, a, can now be calculated, i.e.,

    \[ a = \sqrt{\frac{1.33 \times 10^9}{850}} \approx 1,250 \, \text{m/s} \]

The pressure increase \Delta p from the water hammer effect for sudden valve closure is given by

    \[ \Delta p = \varrho \, a \, \Delta V = \varrho \, a \, V_0 \]

Substituting the known values gives the change in pressure as

    \[ \Delta p = 850 \times 1,250 \times 3 \approx 3.19 \times 10^6 \, \text{Pa} = 3.19 \, \text{MPa} \]

Why is the speed of sound so much higher in a liquid than in air?

The speed of sound in a medium, whether it’s a gas, liquid, or solid, depends on the properties of that medium. Sound travels faster in liquids and solids than in gases because the molecules in these substances are much closer together, allowing for the more rapid transmission of mechanical waves. The speed of sound in a medium is determined by its density and elastic properties, specifically its bulk modulus. The bulk modulus quantifies a substance’s resistance to volume changes under pressure. Liquids and solids generally have higher bulk moduli than gases, making them less compressible. In a liquid, the molecules are closer together than in a gas, and they can still transmit mechanical waves. This close molecular packing and the resistance to compression contribute to a higher sound speed in liquids than in gases.

 Producing Jet Thrust

A monopropellant thruster is a basic rocket engine, often used for satellite propulsion and control, that stores the propellant under pressure in a pressurized tank. The propellant is released over time by opening the valve, allowing it to flow over a catalyst bed. This process causes thermal decomposition and energy release from the propellant, as shown in Figure 9. The flow then expands through a nozzle to produce an exit velocity and, hence, a thrust resulting from the time rate of change of the momentum of the expanding gases. The idea is that the thrust produces a controlled change in the satellite’s velocity. Depending on the burn’s direction and timing, this velocity change can be used for orbit raising, orbit lowering, circularization, station-keeping, attitude control, or other maneuvers.

Jet thrust can be produced by expelling a high-velocity flow from a pressurized tank.

This is a time-varying flow problem because propellant mass is continuously discharged from a tank, thereby reducing the propellant mass within the control volume. The most general form of the continuity equation is

(38)   \begin{equation*} \frac{\partial}{\partial t}\oiiint \varrho \, d {\cal{V}} + \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

In this case, it can be reduced to

(39)   \begin{equation*} \frac{d M_p}{d t} + \overbigdot{m}_p = 0 \end{equation*}

where M_p is the mass of the propellant remaining in the tank and \overbigdot{m}_p is the positive propellant mass flow rate leaving the tank. Therefore,

(40)   \begin{equation*} \frac{d M_p}{d t} = -\overbigdot{m}_p \end{equation*}

which shows that the propellant mass decreases with time.

Let the initial mass of the spacecraft be {M_0}. Conservation of mass gives the current mass of the spacecraft, M, at time t later, as the propellant is discharged as

(41)   \begin{equation*} M = M_0 - \overbigdot{m}_p \, t \end{equation*}

where \overbigdot{m}_p = constant. The thrust force from the propulsion system is given by the time rate of change of momentum of the exhaust flow, with any pressure thrust included in an effective exhaust velocity, i.e.,

(42)   \begin{equation*} T = \overbigdot{m}_p V_e \end{equation*}

where V_e is interpreted as the effective exhaust velocity. More generally, if the nozzle exit pressure differs from the ambient pressure, then

(43)   \begin{equation*} T = \overbigdot{m}_p V_e + \left( p_e - p_a \right) A_e \end{equation*}

where p_e is the nozzle exit pressure, p_a is the ambient pressure, and A_e is the nozzle exit area. For the present simplified analysis, the pressure-thrust contribution is included in V_e. Therefore, the acceleration of the spacecraft, which is not constant because of its continuously decreasing mass, is given by

(44)   \begin{equation*} a = \frac{T}{M} = \frac{\overbigdot{m}_p \, V_e}{M} = \frac{dV}{dt} \end{equation*}

Therefore,

(45)   \begin{equation*} \frac{dV}{dt} = \frac{\overbigdot{m}_p \, V_e}{M_0 -\overbigdot{m}_p \, t} \end{equation*}

Integrating the equation gives

(46)   \begin{equation*} \Delta V = V_e \ln \left( \frac{M_0}{M_0 -\overbigdot{m}_p \, t} \right) \end{equation*}

which has a specific name, the rocket equation.

Check Your Understanding #3 – Discharge of a monopropellant thruster

An orbiting satellite is propelled using a monopropellant thruster. The satellite has an initial mass of 5,000 kg, including its propellant mass. To give a slight corrective boost to its orbit, it opens the control valve and ejects propellant at a constant rate of 0.1 kg/s, with an effective exit velocity of 2,000 m/s. Assume the satellite operates in a vacuum and that, during this short burn, the incremental velocity change can be calculated from the thrust alone. Determine the thrust, the change in velocity of the satellite, and the new mass after 90 seconds from the start of the propulsive burn.

Show solution/hide solution.

The thrust produced can be determined from the time rate of change of momentum of the exhaust gases, i.e.,

    \[ T = \overbigdot{m}_p \, V_e = 0.1 \times 2,000 = 200~\mbox{N} \]

After 90 seconds of thrusting, i.e., t = 90, then using the rocket equation gives

    \[ \Delta V = 2,000 \ln \left( \frac{5,000}{5,000 - 0.1 \times 90.0 } \right) = 3.6~\mbox{m/s} \]

The final mass after the burn will be

    \[ M = M_0 -\overbigdot{m}_p \, t = 5,000 - 0.1 \times 90 = 4,991~\mbox{kg} \]

Unsteady Lift on a Pitching Airfoil

Time-dependent or unsteady flow plays an important role in determining the aerodynamic forces on airfoils and other lifting surfaces. Consider a two-dimensional thin airfoil in a uniform flow of velocity {V_{\infty}} undergoing small-amplitude harmonic pitching motion. Let the angle of attack vary as

(47)   \begin{equation*} \alpha(t) = \alpha_0 + \alpha_1 \sin(\omega t) \end{equation*}

where \alpha_0 is the mean angle of attack, \alpha_1 is the pitching amplitude in radians, and \omega is the angular frequency of the motion. Under quasi-steady conditions, i.e., when the reduced frequency k \ll 1, the instantaneous lift coefficient for a thin symmetric airfoil, or for an angle of attack measured from the zero-lift angle, can be approximated from thin airfoil theory as

(48)   \begin{equation*} C_L(t) = 2\pi \, \alpha(t) = 2\pi \left( \alpha_0 + \alpha_1 \sin(\omega t) \right) \end{equation*}

Consider a two-dimensional thin airfoil in a uniform flow of velocity {V_{\infty}} undergoing small-amplitude harmonic pitching motion about its quarter-chord. Let the angle of attack vary as

(49)   \begin{equation*} \alpha(t) = \alpha_0 + \alpha_1 \sin(\omega t) \end{equation*}

A first approximate correction for unsteady aerodynamic effects can be introduced by defining an effective angle of attack that includes a pitch-rate contribution. Using the semi-chord, c/2, as the characteristic length gives

(50)   \begin{equation*} \alpha_{\rm eff}(t) = \alpha(t) + \frac{c}{2 V_{\infty}} \frac{d\alpha}{dt} \end{equation*}

so that

(51)   \begin{equation*} C_L(t) = 2\pi \, \alpha_{\rm eff}(t) = 2\pi \left( \alpha + \frac{c}{2 V_{\infty}} \frac{d\alpha}{dt} \right) \end{equation*}

Because

(52)   \begin{equation*} \frac{d\alpha}{dt} = \omega \alpha_1 \cos(\omega t) \end{equation*}

then

(53)   \begin{equation*} C_L(t) = 2\pi \alpha_0 + 2\pi \alpha_1 \left[ \sin(\omega t) + \frac{\omega c}{2 V_{\infty}} \cos(\omega t) \right] \end{equation*}

Using the definition of reduced frequency, i.e.,

(54)   \begin{equation*} k = \frac{\omega c}{2 V_{\infty}} \end{equation*}

gives

(55)   \begin{equation*} C_L(t) = 2\pi \alpha_0 + 2\pi \alpha_1 \left( \sin(\omega t) + k \cos(\omega t) \right) \end{equation*}

The oscillatory part of the lift coefficient may therefore be written as

(56)   \begin{equation*} C_{L_1}(t) = 2\pi \alpha_1 \left[ \sin(\omega t) + k \cos(\omega t) \right] \end{equation*}

or, equivalently,

(57)   \begin{equation*} C_{L_1}(t) = 2\pi \alpha_1 \sqrt{1+k^2} \, \sin \left( \omega t + \phi \right) \end{equation*}

where the phase angle is

(58)   \begin{equation*} \phi = \tan^{-1} k \end{equation*}

and the amplitude of the oscillatory lift coefficient is

(59)   \begin{equation*} C_{L_1,{\rm max}} = 2\pi \alpha_1 \sqrt{1+k^2} \end{equation*}

This result shows explicitly how the unsteady contribution scales with the reduced frequency. The term proportional to \sin(\omega t) is the quasi-steady lift, while the term proportional to k \cos(\omega t) is associated with the pitch-rate effect. As k increases in this simplified model, the amplitude of the lift oscillation increases and the lift response leads the pitching motion by the phase angle \phi = \tan^{-1} k. These effects illustrate why phase relationships are important in aeroelastic phenomena such as flutter and in predicting aerodynamic loads on oscillating lifting surfaces. For larger values of reduced frequency, however, a more complete unsteady aerodynamic theory, such as Theodorsen’s theory for attached incompressible flow, is required to accurately predict the lift amplitude and phase.

Summary & Closure

Unsteady flows, characterized by temporal fluctuations in fluid properties, are a fundamental aspect of fluid dynamics and are particularly relevant across many engineering and scientific disciplines. Unlike steady flows, unsteady flows are more representative of real-world scenarios where conditions are rarely constant. Studying unsteady flows is essential for designing and optimizing advanced aerodynamic systems, propulsion technologies, and other applications where time-dependent behavior is critical. The practical implications of unsteady flows extend beyond fundamental fluid dynamics, impacting areas such as turbulence modeling, wave propagation, and the transient behavior of fluid systems. Addressing these complexities requires sophisticated mathematical models and computational methods capable of capturing transient phenomena. An essential step in analyzing unsteady flows is their classification, often achieved using dimensionless parameters such as the reduced frequency, which helps quantify the degree of unsteadiness relative to the system’s inherent time scales. Throughout this chapter, various classic examples have illustrated the principles of solving simple unsteady flow problems. This understanding is crucial for developing accurate predictions and practical solutions in applications where unsteady flows are a key consideration.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • Can you provide real-world examples in which unsteady flow is crucial in fluid systems or engineering applications?
  • Why is reduced frequency an essential parameter for analyzing fluid flow oscillatory or vibratory motion?
  • In what scenarios might the Strouhal number be a relevant parameter to consider?
  • How does turbulence contribute to the complexity of fluid dynamics, and why is it often associated with unsteadiness?
  • Discuss the practical implications of unsteady flows in engineering design. How might engineers account for unsteady conditions in the design of systems like pipelines, aircraft, or water turbines?
  • How might unsteady flows impact the efficiency and performance of propulsion systems, such as those in aircraft or marine vehicles?

 

Other Useful Online Resources

To learn more about unsteady flows, take a look at some of these online resources:

  • A good video explaining the differences between steady and unsteady flows.
  • An explanation of unsteady flows, including the effects of thermodynamics.
  • An explanation of the conservation of mass in unsteady flow.
  • A video presentation on the application of the continuity equation to unsteady flows.
  1. Joukowsky, N., “Über den hydraulischen Stoß in Wasserleitungsröhren.” (“On the hydraulic hammer in water supply pipes”) Mémoires de l'Académie Impériale des Sciences de St.-Pétersbourg, Série 8, 9(5), 1-71, 1900 (in German). ↵

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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.

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