30 Time-Dependent Flows

Introduction

In fluid dynamics and aerodynamics, a time-dependent or unsteady flow is one in which any of the properties (e.g., velocity, pressure, temperature, and density) at a given point in space vary with time. This behavior contrasts with steady flow, in which all 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 and must be considered in fields such as aeroelasticity and flutter, rotating machinery, including jet engines and helicopter rotors, mixing processes, hydraulic and pneumatic control systems, and combustion in air-breathing and rocket engines. Analyzing and predicting unsteady flow can be more challenging than steady flow because one or more relevant fluid properties vary with time. Although the inclusion of time may initially appear to complicate the analysis, exposure to representative exemplar problems helps develop familiarity and confidence in accounting for time-dependent effects using the usual conservation principles. For more complex unsteady flows, experimental techniques and computational fluid dynamics are frequently used to obtain practical engineering predictions.

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 flow is one in which the 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, at least one flow property at a point varies with time, as shown in Figure 1(b). In this case, the relevant time-dependent terms in the governing equations must be retained, i.e.,

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

 

Time histories of a flow property at a point illustrate the distinction between steady and unsteady flow.

Unsteady flow phenomena are encountered in many engineering applications. Examples include flows in turbomachinery and 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, denoted by St, is a dimensionless quantity used in fluid dynamics to characterize periodic or oscillatory flow phenomena. It is useful 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 angular 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 characterize the relative importance of unsteady aerodynamic effects 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 m/s, then L must be in m 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, quantifies 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 motion’s time history, 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 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 over 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 \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 discharge velocity depends on the instantaneous height of the liquid in the tank, h, as given by Torricelli’s law. 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 that when t = 0 then h = h_0, and when {h = 0} the tank is considered empty. Performing the integration gives

(17)   \begin{equation*} t = \frac{2\, A}{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}, then

(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. This result gives the ideal emptying time, assuming no flow losses through the drain or valve. In practice, flow resistance reduces the discharge velocity and increases the emptying time.

Check Your Understanding #2 – Emptying a tapered tank

A conical tank of height H and upper radius R is mounted with its narrow end downward. A small drain of area a is located at the bottom of the tank. The tank is initially filled with liquid to its upper edge. Assuming ideal discharge, determine the time required to empty the tank. Compare this time with that required to empty a cylindrical tank having the same initial liquid volume, liquid depth, and drain area. Which tank empties faster, and why?

Show solution/hide solution.

Let h be the instantaneous height of the liquid above the drain. From similar triangles, the radius of the liquid surface at height h is

    \[ r=\frac{R}{H} \, h \]

The horizontal cross-sectional area of the liquid is then

    \[ A(h)=\pi r^2 = \pi\frac{R^2}{H^2}h^2 \]

For an incremental change in liquid height, then

    \[ \frac{d{\cal V}}{dt} = A(h)\frac{dh}{dt} \]

The volume flow rate through the drain is

    \[ -\frac{d{\cal V}}{dt} = a\sqrt{2gh} \]

so

    \[ -\pi\frac{R^2}{H^2}h^2\frac{dh}{dt} = a\sqrt{2gh} \]

Separating the variables gives

    \[ dt = -\frac{\pi R^2}{aH^2\sqrt{2g}}h^{3/2}\,dh \]

Integrating from h=H at t=0 to h=0 at the emptying time t_{\rm cone} gives

    \[ t_{\rm cone} = \frac{\pi R^2}{aH^2\sqrt{2g}} \int_0^H h^{3/2}\,dh \]

Therefore,

    \[ t_{\rm cone} = \frac{\pi R^2}{aH^2\sqrt{2g}} \left(\frac{2}{5}H^{5/2}\right) \]

or

    \[ t_{\rm cone} = \frac{2\pi R^2}{5a} \sqrt{\frac{H}{2g}} \]

The initial volume of the conical tank is

    \[ {\cal V}_0 = \frac{1}{3}\pi R^2H \]

A cylindrical tank having the same volume and liquid depth must have a constant cross-sectional area A_c given by

    \[ A_cH = \frac{1}{3}\pi R^2H \]

and so

    \[ A_c = \frac{1}{3}\pi R^2 \qquad \text{and} \qquad R_c = \frac{R}{\sqrt{3}} \]

The emptying time for the cylindrical tank is

    \[ t_{\rm cyl} = \frac{2A_c}{a} \sqrt{\frac{H}{2g}} \]

which gives

    \[ t_{\rm cyl} = \frac{2\pi R^2}{3a} \sqrt{\frac{H}{2g}} \]

The ratio of the two emptying times is

    \[ \frac{t_{\rm cone}}{t_{\rm cyl}} = \frac{3}{5} \]

Hence, t_{\rm cone} = 0.60\,t_{\rm cyl}. For the conical tank, the cross-sectional area is smaller than that of the equal-volume cylindrical tank over the lower portion of the liquid depth and larger over the upper portion. Therefore, less of the conical tank’s volume is discharged during the slow, low-head part of the process, which gives a shorter total emptying time.

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 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. Let the initial density and pressure be \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}} = \overbigdot{m}_{\rm in} \end{equation*}

Assuming uniform mixing of the air, 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*}

Emptying an Air Tank

Consider the complementary problem of air discharging from a rigid tank of volume {\cal V}, which is the reverse of the flow direction shown in Figure 6. The tank initially contains air at pressure p_0 and temperature T_0, and the air is discharged (by opening a valve) into the surroundings at pressure p_a. Assume that the air inside the tank remains well mixed and that the discharge process is sufficiently slow for the temperature to remain constant at T_0. Unlike the preceding filling problem, the mass flow rate leaving the tank is not assumed to remain constant.

As the tank pressure approaches ambient pressure, the pressure difference decreases, reducing the discharge rate. A simple model for this behavior is

(27)   \begin{equation*} \overbigdot{m}_{\rm out}=K\left(p-p_a\right) \end{equation*}

where K is a flow-conductance coefficient for the valve or opening. The value and units of K depend on the particular valve and the units used for pressure and mass flow rate. The conservation of mass for the tank gives

(28)   \begin{equation*} \frac{dM}{dt}=-\overbigdot{m}_{\rm out} \end{equation*}

where M is the instantaneous mass of air remaining in the tank. For an ideal gas, then

(29)   \begin{equation*} M=\frac{p \, {\cal V}}{R \, T_0} \end{equation*}

Because the tank’s volume and temperature are assumed constant, then

(30)   \begin{equation*} \frac{dM}{dt} = \frac{{\cal V}}{R \, T_0}\frac{dp}{dt} \end{equation*}

Substituting the assumed discharge relation gives

(31)   \begin{equation*} \frac{{\cal V}}{R \, T_0}\frac{dp}{dt} = -K\left(p-p_a\right) \end{equation*}

or

(32)   \begin{equation*} \frac{dp}{p-p_a} = -\frac{K \, R \, T_0}{\cal V}\,dt \end{equation*}

Integrating from p=p_0 at t=0 to the pressure p at time t gives

(33)   \begin{equation*} \int_{p_0}^{p}\frac{dp}{p-p_a} = -\frac{K \, R \, T_0}{\cal V}\int_0^t dt \end{equation*}

and so

(34)   \begin{equation*} \ln\left(\frac{p-p_a}{p_0-p_a}\right) = -\frac{K \, R \, T_0}{\cal V}t \end{equation*}

The pressure in the tank is then

(35)   \begin{equation*} p = p_a+ \left(p_0-p_a\right) \exp\left(-\frac{K \, R \, T_0}{\cal V}t\right) \end{equation*}

Defining the characteristic emptying time as

(36)   \begin{equation*} \tau=\frac{\cal V}{K \, R \, T_0} \end{equation*}

allows the pressure history to be written as

(37)   \begin{equation*} p = p_a+ \left(p_0-p_a\right) \exp\left(-\frac{t}{\tau}\right) \end{equation*}

This result shows that the pressure difference between the tank and its surroundings decreases exponentially with time. At t=0, the tank pressure is p_0, while at sufficiently large times p\rightarrow p_a. The corresponding mass flow rate is

(38)   \begin{equation*} \overbigdot{m}_{\rm out} = K\left(p_0-p_a\right) \exp\left(-\frac{t}{\tau}\right) \end{equation*}

so the discharge rate is greatest when the valve is first opened and decreases continuously as the tank pressure approaches the ambient pressure.

The time required for the tank pressure to reach any specified value p can be obtained by rearranging the pressure equation, giving

(39)   \begin{equation*} t = \tau \ln\left(\frac{p_0-p_a}{p-p_a}\right) \end{equation*}

The tank pressure approaches the ambient pressure asymptotically, so the idealized model predicts that exact pressure equalization requires an infinite time. In practice, the tank may be considered depressurized when its pressure has decreased to a specified value that is sufficiently close to ambient pressure. Notice that the assumed relation \overbigdot{m}_{\rm out}=K(p-p_a) is a simplified model of the valve flow. Actual gas discharge may involve other effects when the pressure difference is large. However, the present model illustrates how coupling the discharge rate to the instantaneous tank pressure produces a genuine unsteady response.

Pressure Lag in a Pitot-Static System

Pitot-static instruments and pressure transducers are connected to their pressure sources through tubing and internal passages. Under steady conditions, the air in these passages is stagnant. When the source pressure changes, however, a small transient flow occurs as air is redistributed within the pressure line and instrument cavity. This flow continues only until the instrument pressure adjusts to the new source pressure. As a result, the measured pressure may lag behind the actual pressure.

Consider a pressure source at pressure p_d(t) connected through a narrow tube to an instrument cavity of fixed volume {\cal V}_i; see Figure 7. Let the pressure inside the instrument (e.g., a pressure transducer) be p_i(t). Assume that the air in the instrument cavity remains well mixed and that the process is sufficiently slow to be approximately isothermal at temperature T_0. For a small pressure difference, the mass flow rate through the pressure line can be approximated as being proportional to the pressure difference, i.e.,

(40)   \begin{equation*} \overbigdot{m} = K \left( p_d - p_i \right) \end{equation*}

where K is a flow-conductance coefficient for the tubing and internal passages. The mass of air in the pressure transducer cavity is

(41)   \begin{equation*} M_i = \frac{p_i \, {\cal V}_i}{R \, T_0} \end{equation*}

The lag between a pressure source and the measurement point, such as a pressure transducer, can introduce pneumatic time delays.

Because the cavity volume and temperature are constant, then

(42)   \begin{equation*} \frac{dM_i}{dt} = \frac{{\cal V}_i}{R \, T_0} \frac{dp_i}{dt} \end{equation*}

Conservation of mass requires that the rate of increase of mass in the pressure transducer equal the mass flow rate entering through the pressure line. Therefore,

(43)   \begin{equation*} \frac{{\cal V}_i}{R \, T_0} \frac{dp_i}{dt} = K \left( p_d - p_i \right) \end{equation*}

Rearranging gives

(44)   \begin{equation*} \tau \frac{dp_i}{dt} + p_i = p_d \end{equation*}

where

(45)   \begin{equation*} \tau = \frac{{\cal V}_i}{K \, R \, T_0} \end{equation*}

is the time constant of the pressure-line system. A larger pressure transducer volume or greater resistance in the pressure line produces a larger time constant and a slower pressure response.

Suppose that the source pressure changes suddenly from p_0 to a new constant value p_1 at t = 0. The pressure transducer initially remains at p_i(0) = p_0. For t > 0, Eq. 44 becomes

(46)   \begin{equation*} \tau \frac{dp_i}{dt} + p_i = p_1 \end{equation*}

or

(47)   \begin{equation*} \frac{dp_i}{p_1 - p_i} = \frac{dt}{\tau} \end{equation*}

Integrating from p_i = p_0 at t = 0 to the pressure p_i at time t gives

(48)   \begin{equation*} p_i = p_1 + \left( p_0 - p_1 \right) \exp \left( -\frac{t}{\tau} \right) \end{equation*}

This result shows that the measured pressure approaches the new source pressure exponentially rather than changing instantaneously. The fraction of the total pressure change measured at time t is

(49)   \begin{equation*} \frac{p_i - p_0}{p_1 - p_0} = 1 - \exp \left( -\frac{t}{\tau} \right) \end{equation*}

After one time constant, then

(50)   \begin{equation*} \frac{p_i - p_0}{p_1 - p_0} = 1 - e^{-1} \approx 0.632 \end{equation*}

so the pressure transducer has reached approximately 63.2% of its response. After three time constants, the response is approximately 95% complete, and after five time constants, it is more than 99% complete.

Pressure lag can affect pneumatic measurements of airspeed, altitude, and vertical speed whenever the external pressure changes rapidly. Long or narrow pressure lines, leaks, moisture, and restrictions can increase the response time and produce larger transient measurement errors. The model developed here represents the pressure system as a first-order response and is most appropriate only for small pressure differences.

Check Your Understanding #3 – Oscillatory response of a pressure measuring system

A pressure transducer is connected to a pressure source through a pressure line with a time constant of \tau = 0.20 s. The source pressure varies harmonically according to

    \[ p_d(t) = \overline{p} + p_1 \sin(\omega t) \]

Starting with

    \[ \tau \frac{dp_i}{dt} + p_i = p_d(t) \]

assume that, after the initial transient has decayed, the indicated pressure has the form

    \[ p_i(t) = \overline{p} + p_{i_1} \sin(\omega t - \phi) \]

Derive expressions for the amplitude ratio p_{i_1}/p_1 and the phase lag \phi. Then consider the specific source pressure

    \[ p_d(t) = 100 + 4\sin(2\pi t) \quad \text{kPa} \]

Determine the amplitude of the indicated pressure oscillation, the phase lag relative to the source pressure, and an equation for the indicated pressure.

Show solution/hide solution.

The assumed indicated pressure is

    \[ p_i(t) = \overline{p} + p_{i_1} \sin(\omega t - \phi) \]

and its time derivative is

    \[ \frac{dp_i}{dt} = \omega \, p_{i_1}\cos(\omega t-\phi) \]

Substituting into the pressure-line equation gives

    \[ \tau\, \omega \, p_{i_1}\cos(\omega t-\phi) + p_{i_1}\sin(\omega t-\phi) = p_1\sin(\omega t) \]

Using the trigonometric identities for \sin(\omega t-\phi) and \cos(\omega t-\phi) and equating the coefficients of \sin(\omega t) and \cos(\omega t) gives

    \[ p_{i_1} \left( \cos\phi+\omega \, \tau\sin\phi \right) = p_1 \]

and

    \[ p_{i_1} \left( \omega \, \tau\cos\phi-\sin\phi \right) = 0 \]

The second equation gives

    \[ \tan\phi=\omega \, \tau \]

and so

    \[ \phi=\tan^{-1}(\omega \, \tau) \]

The amplitude ratio is then

    \[ \frac{p_{i_1}}{p_1} = \frac{1}{\sqrt{1+\left(\omega\tau\right)^2}} \]

For the specified pressure variation, p_1 = 4 kPa and \omega = 2\pi rad/s, then \omega \, \tau = 2\pi\times 0.20 = 1.257. The amplitude ratio is

    \[ \frac{p_{i_1}}{p_1} = \frac{1}{\sqrt{1+\left(1.257\right)^2}} \approx 0.623 \]

and hence

    \[ p_{i_1} = 0.623\times4 \approx 2.49~\mbox{kPa} \]

The phase lag is

    \[ \phi = \tan^{-1}(1.257) \approx 51.5^\circ \]

The indicated pressure is

    \[ p_i(t) = 100 + 2.49\sin\left(2\pi t-51.5^\circ\right) \quad \text{kPa} \]

The pressure-line system reduces the oscillation amplitude from 4 kPa to approximately 2.49 kPa and causes the indicated pressure to lag the source pressure by approximately 51.5^\circ.

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 8. 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. When a flowing fluid experiences a sudden change in velocity, its momentum must change rapidly, producing a pressure wave that propagates through the system. 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 liquid compression and elastic strain in the pipe wall, with some of it 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, whereas for an elastic pipe, it is reduced by the pipe wall’s compliance. When the compression wave reaches a constant-pressure boundary, such as a large reservoir, it reflects as a pressure-reducing wave that propagates downstream. Repeated wave propagation and reflection produce pressure and velocity oscillations that gradually decay because of friction and other losses. The repeated propagation and reflection of the pressure waves can produce the characteristic thumping or hammering sound that may be felt or heard in the piping system.

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 9. 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

(51)   \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. 51 can also be written as

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

There are two primary cases of interest when predicting water hammer effects: 1. Gradual closure of the valve. 2. Sudden closure of the valve. In either case, the pipe’s elasticity affects the wave propagation speed and must be accounted for when appropriate. 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

(53)   \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 for an approximately linear reduction in flow velocity, the pressure increase may be estimated by

(54)   \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

(55)   \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

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

and the equivalent pressure head is

(57)   \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 the wave speed, a, is required to determine whether the valve closure is gradual or sudden and to calculate the pressure rise for sudden closure. If the pipe is rigid, then

(58)   \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

(59)   \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

(60)   \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

(61)   \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 #4 – 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 depends on its density and elastic properties. For a liquid, it is approximately

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

where E_b is the bulk modulus and \varrho is the density. Although liquids are denser than gases, they are also much less compressible and have much higher bulk moduli. The increase in bulk modulus outweighs the increase in density, giving liquids a higher speed of sound than gases. Solids can transmit both longitudinal and shear waves, and their wave speeds likewise depend on the appropriate elastic moduli and density.

 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 10. 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

(62)   \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

(63)   \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,

(64)   \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

(65)   \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 momentum flux of the exhaust and any pressure thrust, i.e.,

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

where V_e is the nozzle exit velocity, 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 incorporated into an effective exhaust velocity, so that

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

where V_e is now interpreted as the effective exhaust velocity. Therefore, the acceleration of the spacecraft, which is not constant because of its continuously decreasing mass, is given by

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

Therefore,

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

Integrating the equation gives

(70)   \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 #5 – 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. An airfoil may experience unsteady motion because of control inputs, atmospheric disturbances, structural vibration, rotor motion, or aeroelastic effects. In such cases, the aerodynamic forces may vary not only because the angle of attack changes, but also because the airfoil has a finite pitch rate.

Consider a two-dimensional thin airfoil in a uniform flow of velocity {V_{\infty}} undergoing small-amplitude harmonic pitching motion about its quarter-chord. The quarter-chord point is assumed fixed, while the airfoil rotates about this point. Let the angle of attack vary as

(71)   \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. The airfoil therefore pitches periodically between angles of attack \alpha_0-\alpha_1 and \alpha_0+\alpha_1.

If the pitching motion is sufficiently slow, the flow may be treated as quasi-steady. Under this approximation, the aerodynamic forces at each instant are assumed to be nearly the same as those obtained from steady-flow theory at the same instantaneous angle of attack. For a thin symmetric airfoil, or for an angle of attack measured from the zero-lift angle, thin airfoil theory gives

(72)   \begin{equation*} C_L(t) = 2\pi \, \alpha(t) \end{equation*}

and so

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

This result accounts only for the instantaneous geometric angle of attack. However, when the airfoil pitches, points along the chord also move vertically relative to the freestream. This motion changes the local direction of the relative airflow and produces an additional effective angle of attack.

For pitching about the quarter-chord, the pitch-rate contribution may be represented using the vertical velocity at the three-quarter-chord point. This point lies a distance c/2 behind the pitching axis. For small angular motions, its vertical velocity is approximately

(74)   \begin{equation*} V_p = \frac{c}{2}\frac{d\alpha}{dt} \end{equation*}

The corresponding change in the direction of the relative airflow is approximately

(75)   \begin{equation*} \Delta\alpha_p = \frac{V_p}{V_{\infty}} = \frac{c}{2V_{\infty}}\frac{d\alpha}{dt} \end{equation*}

The effective angle of attack is then

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

The sign of the pitch-rate term follows from the adopted sign convention. With positive nose-up pitching taken as positive, a positive pitch rate increases the effective angle of attack in the present formulation.

Using the thin-airfoil lift relation gives

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

Because

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

then

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

The relative importance of the pitch-rate contribution is determined by the reduced frequency

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

The reduced frequency compares the time scale of the airfoil motion with the time required for the freestream to travel a distance equal to the airfoil semi-chord. Small values of k correspond to relatively slow pitching motion, while larger values indicate that the airfoil motion occurs more rapidly relative to the convective time scale of the flow.

Using the reduced frequency gives

(81)   \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 first term, 2\pi\alpha_0, is the mean lift coefficient. The oscillatory part is

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

The term proportional to \sin(\omega t) results from the variation of the geometric angle of attack. The term proportional to k\cos(\omega t) results from the pitch rate. Because the pitch rate is greatest when the airfoil passes through its mean angle of attack and is zero at the maximum and minimum angles, the two contributions are out of phase.

The oscillatory lift coefficient can also be written as a single harmonic function

(83)   \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

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

and the amplitude of the oscillatory lift coefficient is

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

This result shows how the pitch-rate contribution changes both the amplitude and phase of the lift response. When k\ll1, the pitch-rate term is small, and the lift varies approximately in phase with the angle of attack. 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.

Phase relationships are important in problems such as aeroelastic flutter, rotor aerodynamics, and the prediction of loads on oscillating lifting surfaces. The present model provides a useful first approximation of the effects of pitching motion. For higher reduced frequencies, a more complete unsteady aerodynamic theory is required to accurately predict the lift amplitude and phase.

Summary & Closure

Unsteady flows, characterized by temporal variations in fluid properties, occur throughout fluid dynamics and are particularly important in many engineering and scientific applications. 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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