29 Internal Flows

Introduction

Aerospace engineers are often concerned with external fluid flows, such as those that develop over the outside surfaces of a body, including an airfoil, wing, or a complete flight vehicle. However, internal fluid flows are also crucial in many aerospace engineering applications. An internal flow can be defined as a flow in which its downstream development is confined or altered by all of the boundary surfaces inside which it flows. Internal flows may comprise gases, such as air, and liquids. The flow of liquids is often referred to as hydraulics, a field studied extensively and applied across all branches of engineering.

Such internal flows are encountered, but are not limited to those within engines and combustion chambers, engine intake systems, pneumatic and hydraulic systems, fuel systems, heat exchangers and cooling systems, air conditioning systems, fire suppression systems, and wind tunnels. Figure 1 below illustrates an example of internal flow within a jet engine. Velocity non-uniformities in the wakes behind each blade affect engine performance and noise generation.

Internal flow through part of a jet engine. The blade surfaces are colored by the magnitude of the static pressure, and the surrounding flow field is visualized using vorticity.

Learning Objectives

  • Appreciate the differences between external flows and internal flows, as well as understand why the action of viscous effects gives pressure losses for internal flows.
  • Distinguish between the effects of laminar and turbulent flow in pipes and ducts.
  • Calculate the frictional losses and pressure drops associated with the flows through pipes and ducts using the Colebrook-White equation and a Moody Chart.
  • Estimate the pumping power required to achieve specified mass flow rates and velocities in pipes and ducts.
  • Understand the design process of producing specific flow velocities in wind tunnels and calculating power requirements.

Internal Versus External Flows

An interesting consequence of all internal flows is that the associated surface boundary layers cannot develop freely, and the flows on each inner surface merge. In this regard, the boundary layer thickness and other properties will not be the same as those of a boundary layer formed over an external surface, as illustrated in Figure 2. This outcome is achieved because, for external flows, the boundaries or surfaces (often referred to as walls) introduce viscous effects into the flow that progressively diminish away from the wall. However, in internal flows, all parts of the walls of a duct, tube, or pipe constrain the flow development, so the entire cross-section of the flow is affected by the walls.

With an external flow, the effects diminish away from the wall; however, with an internal flow, all walls constrain the flow’s development.

The foregoing behavior of the surface boundary layers is one reason why external and internal flows must be carefully distinguished; i.e., all boundaries in an internal flow will interact. Because viscous losses (i.e., skin friction) are significant along all flow surfaces, in internal flows these effects manifest as substantial pressure drops along the interior surfaces where the fluid flows; the friction and the associated energy loss are irrecoverable and appear as heat. Evaluating these pressure drops is crucial for understanding internal flows and determining the engineering requirements for moving fluids through pipes and ducts.

Internal Flow Problems

Internal fluid flow problems occur in many aerospace systems, including fuel, hydraulic, and pneumatic systems. A rocket engine, as shown in the schematic in Figure 3, has numerous internal flows that require very high mass flow rates of fuel and oxidizer. These fluids must pass through several pumps, pipes, and valves, resulting in significant pressure losses. Fuel is also pre-circulated through channels or a jacket around the exhaust nozzle to keep it cool, where considerable pressure drops also occur. Therefore, turbopumps are required to deliver the fuel and oxidizer to the rocket engine’s combustion stage, a process that typically involves cryogenic liquid flow at extremely high mass flow rates.

Schematic showing the internal flow paths of fuel and oxidizer inside a rocket engine, which have to flow through pipes at high volumetric rates.

Aircraft hydraulic systems operate at relatively high pressures. Hence, they require high flow rates to operate the large actuators that power their landing gear, flaps, slats, flight control systems, and other components. Additionally, the hydraulic lines may extend through the structure over considerable distances, resulting in significant pressure drops. Aircraft typically have multiple hydraulic systems to provide redundancy in the event of failure, ensuring that the systems are entirely independent. Airliners, for example, typically have three parallel hydraulic systems, each comprising multiple engine-driven, electrically driven, air-driven, and hydraulic pumps.

In another example of an internal flow, consider a fuel delivery system on an airplane, where fuel must be pumped from one or more fuel tanks (e.g., located in the wings) to the engine(s), as illustrated in the schematic of Figure 4. The distribution of the fuel lines also involves considerable lengths, often with many twists and turns through the aircraft structure. Therefore, pressure drops, pumping power, and flow rates must be carefully considered to ensure that the engines receive the required fuel volumes at the required pressures. Redundancy in fuel delivery is achieved in this case through a mechanical (engine-driven) pump and a completely independent backup electrical pump.

In fuel delivery, even for a relatively simple aircraft, pressure drops, pumping power, and flow rates must be accounted for to ensure the fuel reaches the engines at the required pressure.

An interesting aerospace example of a combined external/internal flow interaction problem was encountered during the design of the X-43A, a hypersonic lifting-body research vehicle. A CFD solution for the flow is shown in Figure 5. At hypersonic Mach numbers, the shock waves bend back rather steeply and come close to the fuselage. Hence, the design used the lower side of the forward fuselage (external flow) as part of the engine intake to create the necessary sequence of shock waves that slow the air and increase pressure before it enters the engine’s intake (internal flow). Therefore, the external flow provides an upstream boundary condition for the internal flow through the engine, thereby ensuring proper engine operation.

The X-43A was developed to test a supersonic combustion ramjet (scramjet). The external flow around the forebody was integral to preconditioning the air intake, illustrating an interaction between external and internal flows.

Categories of Internal Flow Problems

In general, the types of internal flow problems that often arise for aerospace engineers to solve usually fall into these categories:

1. Determining the pressure drop (or the so-called head loss) when flows move through pipes of a given length and diameter for a required volume or mass flow rate.

2. Determining the volume or mass flow rate when the pipe length and diameter are known, given a specified (or allowable) pressure drop, which may be constrained for several reasons.

3. Finding the pumping power needed from a mechanical pump to obtain a certain exit pressure and/or flow rate for a pipe of a specific diameter over a certain length.

4. Determining the pipe diameter when the pipe length and flow or mass rate are known for a particular pressure drop. The pipe size required will also affect its weight and cost, both of which are significant considerations for aerospace applications.

5. Identifying the heating effects of pumping gases through ducts and pipes, especially at higher flow speeds. Frictional losses and gas compression generate heat, and heat dissipation may be required in some applications.

Even when most relevant information is known, not all of these problems can be solved directly. Design purposes sometimes require iterative solutions that start with initial estimates. However, several fundamental issues arise in determining pressure drops and calculating pumping power through pipes.

Fluid Dynamics of Internal Flows

Internal flows occur in ducts and pipes of many shapes and sizes, but flow through a circular pipe is often used as an example, as shown in Figure 6. Viscosity causes the fluid in adjacent layers to slow down gradually, thereby developing a velocity gradient in the pipe, i.e., an axisymmetric boundary layer. To compensate for this reduction in velocity, the fluid velocity at the pipe centerline must increase to maintain a constant net mass flow rate. Consequently, a significant velocity gradient eventually develops across the entire cross-section of the pipe. The primary manifestation of viscous friction on the interior walls is a pressure drop along the pipe length, which depends on the velocity profile, the volumetric or mass flow rate, and the fluid properties.

An internal flow through a pipe develops a velocity profile, a manifestation of viscous friction along the pipe walls, resulting in a pressure drop along the pipe length.

Determining the velocity profile and the corresponding mass (or volume) flow rate is a problem in the analysis of internal flows. For example, the average velocity V_{\rm av} at a downstream cross-section of a pipe or duct can be determined from the velocity profile using conservation of mass. With reference to Figure 7, the mass flow rate \overbigdot{m} is given by

(1)   \begin{equation*} \overbigdot{m} = \varrho \, V_{\rm av} \, A_c = \int_{A_c} \varrho \, u(r) \, dA_c \end{equation*}

where A_c is the pipe’s cross-sectional area, and {u} represents the velocity profile in the pipe as a function of the distance r from the centerline; notice that the profile is assumed to be axisymmetric.

The average flow velocity in a pipe is calculated by integrating the flow rate over the cross-section to determine the mass (or volume) flow rate and then dividing by the pipe’s cross-sectional area.

For the incompressible flow in a circular pipe of diameter D and radius R = D/2, then

(2)   \begin{equation*} V_{\rm av} = \frac{\displaystyle{\int_{A_c} \varrho u(r) \, dA_c}}{\varrho \, A_c} = \frac{\displaystyle{\int_{A_c} \varrho u(r) \, 2 \pi r dr }}{\varrho \pi R^2 } = \frac{2}{R^2} \int_0^R u(r) r dr \end{equation*}

Therefore, if the velocity profile {u(r)} is known (or the mass flow rate or volume flow rate for an incompressible fluid is given), then the average velocity in the pipe can be determined.

Entrance Length

At the beginning of any internal flow, there is a transition period as it organizes itself to become a fully developed internal flow. As shown in Figure 8, the entrance region or length in such a flow, L_e, is the length needed to transition to a fully developed internal flow. In the entrance region, the velocity profile transitions from the inlet state to a fully developed profile, in which the effects of the walls are felt across the entire cross-sectional area of the pipe, duct, or channel.

The flow into a pipe or duct takes time to reach a steady-state condition, but the viscous effects of the walls play a crucial role in determining the final velocity profile. Observe the pressure drop, or “loss,” as the flow develops along the pipe.

Therefore, a fully developed internal flow is defined as one in which the velocity profile remains constant along the flow direction, i.e., in the axial downstream direction. When the duct cross-section is constant, the velocity profile is the same at successive axial locations. If the pipe changes cross-section or turns a corner, it will take some distance for the flow to fully redevelop.

As expected, the entrance length depends on whether the flow is laminar or turbulent and on the roughness of the internal surface. For example, for the laminar flow in a smooth, straight cylindrical pipe of diameter D, the entrance length is known to be

(3)   \begin{equation*} \frac{L_e}{D} \approx 0.05 \, Re_{D} \end{equation*}

where Re_D is the Reynolds number based on the diameter of the pipe, i.e.,

(4)   \begin{equation*} Re_D = \frac{\varrho \, V_{\rm av} \, D}{\mu} = \frac{V_{\rm av} \, D}{\nu} \end{equation*}

For a fully developed turbulent flow, the entrance length is

(5)   \begin{equation*} \frac{L_e}{D} \approx 4.4 \, Re_D^{~1/6} \end{equation*}

This latter result indicates that turbulence-induced mixing significantly accelerates the transition to fully developed internal flow.

Under most practical conditions, the flow in a circular pipe is laminar for Re_{D} \le 2,300 and turbulent for Re_{D} > 4,000 and will be transitional and intermittent in between, i.e., not entirely one state or another. In practice, the lengths of pipes used are often many times the entrance length, so for many analyses, the flow may be assumed to be fully developed over most of the pipe. However, whether the fully developed flow is laminar or turbulent still depends primarily on the Reynolds number. Of course, these preceding results will change somewhat for different geometries or if the pipe has a bend, a turn, or an obstruction, such as a pump or a flow meter.

Hydraulic Diameter

For flows through non-circular pipes (for which many types are encountered in engineering practice), the Reynolds number is redefined based on the hydraulic diameter D_h, i.e.,

(6)   \begin{equation*} D_h = \dfrac{4 A_c}{p} \end{equation*}

where A_c is the cross-sectional area of the pipe, and p is its wetted perimeter. The hydraulic diameter is sometimes referred to as the hydraulic mean diameter. The basic idea is illustrated in Figure 9. Naturally, the hydraulic diameter is defined to reduce to the physical diameter for circular pipes, i.e., D_h = D.

The principle of hydraulic diameter is to find an equivalent circular pipe based on the wetted perimeter of the non-circular pipe.

For example, for a pipe with a rectangular cross-section of width a and height b, the area of the duct is A_c = ab, and the perimeter is p = 2a + 2b. Therefore, the equivalent hydraulic diameter for this cross-section is

(7)   \begin{equation*} D_h = \frac{4 A_c}{p} = \frac{4 ab}{2(a + b)} = \frac{2ab}{a+b} \end{equation*}

Notice that in a square duct where b = a, then D_h = a.

The corresponding Reynolds number will now be calculated by using the hydraulic diameter, i.e.,

(8)   \begin{equation*} { Re_{D_{h}} = \frac{\varrho \, V_{\rm av} \, D_h}{\mu} = \frac{V_{\rm av} \, D_h}{\nu} } \end{equation*}

However, the concept of hydraulic diameter is most effective for fully developed turbulent flow, i.e., for Re_{D_h} greater than approximately 4,000, where the effects of pipe geometry are less critical.

Laminar Flow in Pipes

Consider the analysis of the steady laminar flow of an incompressible fluid with constant properties in the fully developed region of a smooth straight pipe of circular cross-section. This problem, often referred to as a Hagen-Poiseuille flow, provides one of the few exact solutions for viscous internal flows. Such laminar flows are also referred to as Poiseuille flows, i.e., pressure-driven viscous flows. The approach is exceptionally instructive and serves as a prerequisite for understanding turbulent pipe flow, a topic that involves a semi-empirical analysis.

In fully developed laminar flow, the velocity profile does not change with downstream distance. The velocity profile {u(r)} remains unchanged with downstream distance, i.e., it is self-similar. It can also be assumed that there is no radial flow, thereby reducing the problem to a one-dimensional axisymmetric flow. In many cases, assuming steady flow is reasonable, but it must be justified on a problem-by-problem basis.

Flow Model

Consider an area (annular disk) of the flow at a given pipe cross-section. As shown in Figure 10, the forces acting on the annular disk arise from a balance of pressure and viscous (shear) forces. The area of the annular disk is 2 \pi r dr, and its perimeter is 2\pi r. It is assumed that the flow is fully developed.

Flow model for the flow in a circular pipe, as used to develop the exact laminar solution.

Force equilibrium of the annulus requires a balance of pressure and shear forces such that

(9)   \begin{equation*} p (2\pi r \, dr) - \left( p + \frac{dp}{dx} dx \right) (2\pi r \, dr) + \tau (2 \pi r \, dx) - \left( \tau + \frac{d\tau}{dr} dr \right) (2\pi (r + dr) \, dx) = 0 \end{equation*}

Expanding out the terms and canceling terms where possible gives

(10)   \begin{equation*} -\frac{dp}{dx} (2 \pi r \, dr \, dx) + \tau (2 \pi r \, dx) - \tau (2 \pi r \, dx) - \frac{d\tau}{dr} (2 \pi r \, dr \, dx) - \tau (2 \pi \, dr \, dx) = 0 \end{equation*}

Neglecting higher-order terms (i.e., those involving dr^2) and simplifying gives

(11)   \begin{equation*} -\frac{dp}{dx} (2 \pi r \, dr \, dx) - \frac{d\tau}{dr} (2 \pi r \, dr \, dx) - \tau (2 \pi \, dr \, dx) = 0 \end{equation*}

Dividing through by 2\pi \, dr \, dx gives

(12)   \begin{equation*} - r \frac{dp}{dx} - r \frac{d\tau}{dr} - \tau = 0 \end{equation*}

Rearranging leads to

(13)   \begin{equation*} \frac{d}{dr}\left( r \tau \right) = -\, r \frac{dp}{dx} \end{equation*}

or equivalently

(14)   \begin{equation*} \frac{1}{r} \frac{d}{dr}\left( r \tau \right) = -\frac{dp}{dx} \end{equation*}

Shear Stresses

The shear stress is given by Newton’s law of viscosity. Because the axial velocity decreases with increasing radial distance from the centerline, the wall-directed velocity gradient is negative. It is convenient to define the positive shear stress as resisting the motion, so

(15)   \begin{equation*} \tau = -\mu \left( \frac{du}{dr} \right) \end{equation*}

Substituting Newton’s law of viscosity into the preceding force balance gives

(16)   \begin{equation*} \frac{1}{r}\frac{d}{dr}\left( r \frac{du}{dr} \right) = \frac{1}{\mu}\frac{dp}{dx} \end{equation*}

or

(17)   \begin{equation*} \frac{dp}{dx} = \mu \left[ \frac{1}{r}\frac{d}{dr}\left( r \frac{du}{dr} \right) \right] \end{equation*}

The left-hand side of this equation is a function of the downstream distance x, while the right-hand side is a function of the radial coordinate r. The equality in Eq. 17 must hold for any value of r and x, so both sides must be equal to the same constant. Therefore, dp/dx is constant for fully developed flow in a straight pipe of constant cross-section.

Solution to the Governing Equation

Equation 17 can be solved by integrating it twice. First, multiply by r and integrate once to give

(18)   \begin{equation*} r \frac{du}{dr} = \frac{1}{2\mu} \frac{dp}{dx} r^2 + C_1 \end{equation*}

or

(19)   \begin{equation*} \frac{du}{dr} = \frac{1}{2\mu} \frac{dp}{dx} r + \frac{C_1}{r} \end{equation*}

To ensure a physically finite solution at the centerline (r = 0), the term \dfrac{C_1}{r} must vanish, which requires C_1 = 0. Therefore,

(20)   \begin{equation*} \frac{du}{dr} = \frac{1}{2\mu} \frac{dp}{dx} r \end{equation*}

Integrating again gives

(21)   \begin{equation*} u(r) = \frac{1}{4\mu} \frac{dp}{dx} r^2 + C_2 \end{equation*}

Applying the no-slip condition at the wall, u(R) = 0, gives

(22)   \begin{equation*} C_2 = -\frac{1}{4\mu} \frac{dp}{dx} R^2 \end{equation*}

Because the pressure decreases in the direction of the flow, \dfrac{dp}{dx} < 0, so the velocity is positive. It is often more convenient to write the result in terms of the pressure drop, giving

(23)   \begin{equation*} u(r) = \frac{1}{4\mu} \left( -\frac{dp}{dx} \right) \left( R^2 - r^2 \right) \end{equation*}

Volumetric and Average Flow Rates

The volumetric flow rate is

(24)   \begin{equation*} Q = \int_0^R u(r) \, 2\pi r \, dr = 2\pi \int_0^R u(r) \, r \, dr \end{equation*}

so substituting u(r) and evaluating gives

(25)   \begin{equation*} Q = -\frac{\pi R^4}{8\mu} \frac{dp}{dx} \quad \text{or} \quad \frac{dp}{dx} = -\frac{8\mu Q}{\pi R^4} \end{equation*}

Therefore, the final velocity profile is

(26)   \begin{equation*} u(r) = \frac{2Q}{\pi R^4} \left( R^2 - r^2 \right) = \frac{2Q}{\pi R^2} \left( 1 - \frac{r^2}{R^2} \right) \end{equation*}

This is the classic Hagen-Poiseuille velocity profile for fully developed laminar flow in a circular pipe.

For the preceding velocity profile, the average velocity can be determined using

(27)   \begin{equation*} V_{\rm av} = \frac{2}{R^2} \int_0^R u(r) r \, dr \end{equation*}

Substituting the velocity profile and evaluating gives

(28)   \begin{equation*} V_{\rm av} = \frac{2}{R^2} \int_0^R u(r)\, r \, dr = \frac{R^2}{8\mu} \left( -\frac{dp}{dx} \right) \end{equation*}

Substituting for the pressure gradient in terms of V_{\rm av} gives

(29)   \begin{equation*} u(r) = 2 V_{\rm av} \left( 1 - \frac{r^2}{R^2} \right) \end{equation*}

The centerline, or maximum, velocity is obtained at r = 0, giving

(30)   \begin{equation*} V_{\rm max} = u(0) = 2 V_{\rm av} \end{equation*}

and, in terms of the pressure gradient,

(31)   \begin{equation*} V_{\rm max} = \frac{R^2}{4 \mu} \left( -\frac{dp}{dx} \right) \end{equation*}

However, this result applies only to this specific case.

Quantifying the Pressure Drop

A quantity of interest in the analysis of internal flows is the pressure drop \Delta p, mainly because this drop is directly related to the power requirements of the pump to maintain a certain flow rate through the pipe. For fully developed laminar flow through a straight pipe of constant circular cross-section, the pressure gradient dp/dx is constant. Therefore, integrating from location 1, where the pressure is p_1, to another point at a distance L downstream, where the pressure is p_2, gives

(32)   \begin{equation*} \frac{dp}{dx} = \frac{p_2 - p_1}{L} = -\frac{\Delta p}{L} \end{equation*}

where \Delta p = p_1 - p_2. Because V_{\rm av} is given by

(33)   \begin{equation*} V_{\rm av} = \frac{R^2}{8 \mu} \left( -\frac{dp}{dx} \right) \end{equation*}

then

(34)   \begin{equation*} \Delta p = p_1 - p_2 = \frac{8 \mu \, L \, V_{\rm av}}{R^2} = \frac{32 \mu \, L \, V_{\rm av}}{D^2} \end{equation*}

Of course, this pressure drop is a direct consequence of viscous effects, thereby representing an irreversible loss of mechanical energy.

Notice that for a given length of pipe, the pressure drop is proportional to the fluid’s viscosity and flow speed, so the faster a given fluid moves through a pipe or duct, the more significant the pressure drop will be. The pressure drop can also be reduced by using a larger diameter pipe. However, weight and cost are often substantial concerns in aerospace systems, where a trade-off exists between the weight and cost of a larger pipe to reduce losses versus a bigger pump to overcome existing losses. In practice, this pressure drop also varies with pipe cross-sections, surface finishes, and other factors.

The same pressure loss for fully developed pipe flow can be expressed using the Darcy-Weisbach form, i.e.,

(35)   \begin{equation*} \Delta p_L = \frac{1}{2} \varrho \, V_{\rm av}^2 \, f \left( \frac{L}{D} \right) \end{equation*}

where {\scriptstyle{f}} is called the friction factor. For a fully developed laminar flow in a smooth circular pipe, the friction factor is

(36)   \begin{equation*} f = \frac{64}{Re_D} \end{equation*}

where the Reynolds number is defined as

(37)   \begin{equation*} Re_D = \frac{\varrho \, V_{\rm av} \, D}{\mu} \end{equation*}

Therefore, for fully developed laminar flow in a circular pipe, the friction factor depends only on the Reynolds number. Friction factors for fully developed laminar flow in pipes with different cross-sections, such as square, oval, and triangular, are available in various sources. However, these shapes are less commonly encountered in aerospace fluid systems.

Head Loss

The head loss, h_L, is frequently used in hydraulics to quantify the magnitude of the frictional pressure loss, \Delta p_L. It represents the equivalent hydrostatic height required to overcome the pressure drop, so that

(38)   \begin{equation*} h_L = \frac{\Delta p_L}{\varrho \, g} \end{equation*}

Substituting for \Delta p_L gives

(39)   \begin{equation*} h_L = f \left( \frac{L}{D} \right) \frac{V_{\rm av}^2}{2 g} \end{equation*}

After the pressure or head loss is known, the required power P to overcome the pressure loss can be determined from

(40)   \begin{equation*} P = Q \, \Delta p_L = Q \, \varrho \, g \, h_L = \overbigdot{m} \, g \, h_L \end{equation*}

where Q is the volume flow rate and \overbigdot{m} = \varrho Q is the mass flow rate.

Poiseuille’s Law

The average velocity for laminar flow in a horizontal pipe (i.e., no gravitational hydrostatic pressure contributions) is

(41)   \begin{equation*} V_{\rm av} = \frac{\Delta p_L \, R^2}{8 \mu \, L} = \frac{\Delta p_L \, D^2}{32 \mu \, L} \end{equation*}

So the volume flow rate for laminar flow through the pipe becomes

(42)   \begin{equation*} Q = \dfrac{\pi D^4 \, \Delta p_L}{128 \mu \, L} \end{equation*}

which is known as Poiseuille’s law.[1] In terms of the pressure drop, then

(43)   \begin{equation*} \Delta p_L = \dfrac{128 \mu \, L \, Q}{\pi \, D^4} \end{equation*}

This result indicates that, for a specified flow rate, the pressure drop (and the required pumping power to overcome losses) is proportional to the pipe length and the fluid viscosity, but inversely proportional to the pipe diameter to the fourth power. This strong dependence shows that even modest increases in pipe diameter can significantly reduce pumping power. However, this approach may not always be practical (or feasible) for other reasons, including higher weight and cost.

Historical Context – Poiseuille’s Original Equation

Poiseuille did not originally derive the equation now known as Poiseuille’s law, which is now written as

    \[ \small{Q = \frac{\Delta p \, \pi \, D^4}{128 \, \mu \, L}} \]

where Q is the volumetric flow rate, \Delta p is the pressure drop along the pipe, D is the pipe diameter, \mu is the dynamic viscosity of the fluid, and L is the pipe length. Instead, based on his many meticulous experiments, Poiseuille’s equation for the flow rate in terms of pressure drop was written as

    \[ \small{Q = \frac{K'' \, \Delta p \, D^4}{L}} \]

where \small{K''} = 1836.7 \left(1 + 0.033678 T + 0.00022099 T^2 \right). Poiseuille did not mention viscosity in his work, only indicating that the value of \small{K''} depended on temperature. At that time, the understanding of viscosity was tentative, although Navier and Stokes later (independently) rationalized this physical property. The derivation of Poiseuille’s law from the Navier-Stokes equations came later. Poiseuille’s so-called constant, \small{K''}, was shown to be

    \[ \small{K'' = \frac{\pi}{128 \, \mu}} \]

Measurements of the values of the dynamic viscosity \mu (for water) were later found to match Poiseuille’s results, as expressed by

    \[ {\small \mu = \frac{\pi}{128 \, K''}} \]

Check Your Understanding #1 – Using Poiseuille’s Law

How many small-diameter pipes of diameter D_1 and length L does it take to discharge the same volume flow rate as a bigger pipe of the same length if the bigger pipe is twice the diameter of the smaller pipes, i.e., D_2 = 2 D_1 \,?

Show solution/hide solution.

Poiseuille’s law can be expressed in terms of flow rate as

    \[ Q = \dfrac{\pi \, D^4 \, \Delta p_L}{128 \mu \, L} = \dfrac{\pi \, R^4 \, \Delta p_L}{8 \mu \, L} \]

Assume that the same pressure drop acts across each pipe, so that \Delta p_L is constant. For the larger pipe, the flow rate is

    \[ Q_2 = \dfrac{\pi \, D_2^4 \, \Delta p_L}{128 \mu \, L} \]

For N smaller pipes in parallel, the total flow rate is

    \[ Q_1 = N \, \dfrac{\pi \, D_1^4 \, \Delta p_L}{128 \mu \, L} \]

For the same total flow rate, Q_1 = Q_2, so

    \[ \dfrac{\pi \, D_2^4 \, \Delta p_L}{128 \mu \, L} = N \, \dfrac{\pi \, D_1^4 \, \Delta p_L}{128 \mu \, L} \]

Solving for N gives

    \[ N = \dfrac{D_2^4}{D_1^4} = \dfrac{2^4 D_1^4}{D_1^4} = 16 \]

Therefore, sixteen smaller pipes are needed to achieve the same volumetric flow rate as one pipe with twice the diameter, assuming the same length, fluid, and pressure drop. This result shows why a larger pipe can significantly reduce pressure losses, because the flow rate in laminar pipe flow is proportional to D^4. See the demonstration here.

Recall that the Poiseuille equation (often called the Hagen-Poiseuille equation) is derived under specific assumptions, i.e., laminar, steady, and incompressible flow. When these assumptions are violated, as in turbulent flow and under geometric conditions such as those found in wide or short pipes, the Poiseuille equation may no longer accurately describe the flow behavior. A rule of thumb is that the ratio of the pipe length to diameter should satisfy \dfrac{L}{D} \gtrsim \dfrac{Re_D}{48} for the assumption of the Poiseuille law to be valid.

Check Your Understanding #2 – Calculating the pressure drop for laminar flow in a pipe

The design of the fuel delivery system requires a flow through a smooth pipe  200 m long and 15 mm in diameter. The required fuel flow rate is 125 kg hr-1. The fluid properties of the fuel are given as \varrho = 800 kg m-3 and \mu = 0.00164 kg m-1 s-1. All entrance effects should be disregarded. Calculate: 1. The pressure drop along the length of the pipe. 2. The pump’s pressure requirements (in terms of head). 3. The pumping power requirements.

Show solution/hide solution.

The cross-sectional area of the pipe is

    \[ A_c = \frac{\pi D^2}{4} = \frac{\pi \times 0.015^2}{4} = 0.0001767~\mbox{ m${^{2}}$} \]

The mass flow rate \overbigdot{m} is given as 125~kg hr^{-1} = 0.0347~kg s^{-1}, i.e.,

    \[ \overbigdot{m} = \varrho \, A_c \, V_{\rm av} = 0.0347~\mbox{ kg s$^{-1}$} \]

The average flow velocity in the pipe is

    \[ V_{\rm av} = \frac{\overbigdot{m}}{\varrho \, A_c} = \frac{0.0347}{800.0 \times 0.0001767} = 0.246~\mbox{m s$^{-1}$} \]

The Reynolds number based on pipe diameter is

    \[ Re_D = \frac{\varrho V_{\rm av} \, D}{\mu} = \frac{ 800.0 \times 0.246 \times 0.015}{0.00164} = 1,797 \]

Notice that this Reynolds number is in the laminar regime, so the friction factor {\scriptstyle{f}} is given by

    \[ f = \frac{64}{Re_D} = \frac{64}{1,797} = 0.0356 \]

The pressure drop \Delta p is given by

    \[ \Delta p = \frac{1}{2} \varrho \, V_{\rm av}^2 \, f \left( \frac{L}{D} \right) \]

and inserting the given values leads to

    \[ \Delta p = \frac{1}{2} \times 800.0 \times 0.246^2 \times 0.0356 \times \frac{200}{0.015} = 11,490.0~\mbox{ Pa} = 11.46~\mbox{kPa} \]

The equivalent head loss, {h}, will be

    \[ h = \frac{\Delta p}{\varrho \, g} = \frac{11,490.0}{800.0 \times 9.81} = 1.46~\mbox{ m} \]

Finally, the pumping power P can be determined from

    \[ P = \frac{\overbigdot{m} \, \Delta p}{\varrho} = \frac{0.0347 \times 11,490.0}{800.0} = 0.498~\mbox{W} \]

Turbulent Flows in Pipes

Unlike laminar-flow expressions, expressions for losses in pipes or ducts with turbulent flow are based on analysis and measurements and provide empirical or semi-empirical relationships. In fact, because of their importance in many industrial applications, systematic experiments have been conducted on pipe flows to measure pressure losses across a range of flow rates, Reynolds numbers, and surface-roughness levels.

Representative relative velocity profiles for fully developed laminar and turbulent flows are shown in Figure 11. Remember that the fully laminar velocity profile is parabolic in laminar flow. However, the profile shape is much “fuller” in turbulent flow because of mixing between fluid layers, resulting in a more uniform centerline velocity and a sharper velocity drop near the pipe wall. This characteristic is similar to a turbulent boundary-layer profile as it develops over an external surface.

The fully established velocity profiles differ for fully laminar and fully turbulent pipe flows.

Any irregularity or “roughness” on the pipe surface, as shown in Figure 12, will disturb the boundary-layer development and increase wall stresses. Therefore, the friction factor in turbulent pipe flow depends on internal surface roughness; pipes with greater roughness yield higher pressure drops. The roughness, \epsilon, is a property of the pipe material.

The interior roughness of a pipe can be quantified in terms of an average roughness height \epsilon.

It should be appreciated that “roughness” is a relative concept. In practice, roughness becomes significant only when the roughness elements reach the thickness of the viscous sublayer, although this layer is relatively thin. Many extruded plastic pipes are generally considered hydrodynamically smooth. However, most other surfaces are rough to some degree, so this roughness causes the boundary layer to become turbulent more quickly, i.e., after a shorter downstream development distance.

The pressure loss with a turbulent flow is still given by

(44)   \begin{equation*} \Delta p_L = \frac{1}{2} \varrho \, V_{\rm av}^2 \, f \left( \frac{L}{D} \right) \end{equation*}

but now, the value of {\scriptstyle{f}} cannot be obtained from theoretical analysis. Indeed, the friction factor in fully developed turbulent pipe flows (sometimes called the Darcy-Weisbach friction factor or just the Darcy friction factor) depends on the Reynolds number and the relative roughness height { \epsilon/D }, which is the ratio of the mean height of roughness of the pipe \epsilon to the pipe diameter D, i.e., for the friction factor then

(45)   \begin{equation*} f = \mbox{\small function} \! \left( Re_D, \frac{\epsilon}{D} \right) \end{equation*}

which follows from dimensional analysis, although the specific functional relationship must be determined from experiment.

Many important results on pressure drops and friction factors for turbulent pipe flow were obtained from experiments with smooth pipes, commercial pipes, and artificially roughened surfaces. In controlled rough-pipe experiments, particles of known size were bonded to the inner surfaces of the pipes to produce a measurable roughness height. Ludwig Prandtl and his students conducted influential experiments of this type in the 1930s, and subsequent generations of hydraulic engineers have extended the measurements to many pipe materials and flow conditions. The resulting friction factor {\scriptstyle{f}} is calculated from the flow rate measurements and the pressure drop.

Pipe friction factors can be presented in tabular, graphical, or functional forms, obtained by fitting a curve to measured data. Commercially available pipes are evaluated based on equivalent roughness values, enabling engineers to perform accurate calculations for design and installation. However, the pipe’s relative roughness may increase with daily use from corrosion or abrasion (depending on the fluid), thereby altering the friction factor over time. Good engineering practice would always account for such effects, ensuring that the required pressure at the end of the pipe can be maintained for its expected service life.

Colebrook-White Equation

The most widely used equation for determining the friction factor in turbulent pipe flows is the Colebrook-White equation. This equation has no closed-form theoretical derivation and is based on curve fitting to experimental measurements. The equation is given by

(46)   \begin{equation*} \frac{1}{\sqrt{f}} = -2 \log_{10}\left( \displaystyle{ \frac{\epsilon}{3.7 \, D_h}} + \frac{2.51}{Re_{D_h} \, \sqrt{f}} \right) \end{equation*}

where \epsilon is the pipe roughness, D_h is the hydraulic diameter, and Re_{D_h} is the Reynolds number based on D_h. Notice that the Colebrook-White equation is an implicit function of {\scriptstyle{f}} (i.e., the value of {\scriptstyle{f}} appears on both the left- and right-hand sides), requiring an iterative numerical solution. The equation is readily solved numerically and converges rapidly in practice.

Moody Chart

Most available results for turbulent flow through common types of rough pipes have been summarized in the form of a Moody Chart, an example of which is shown in Figure 13. Engineers can use this chart to calculate the friction factor for a given internal flow along a pipe. The friction factor is then used to calculate the pressure loss, pumping power, and other related quantities. Lewis Moody solved the Colebrook-White equation to create this type of chart, combining the dimensional terms of roughness, \epsilon, and pipe diameter, D_h, into a non-dimensional relative roughness, \epsilon/D_h, together with the Reynolds number to determine the friction factor.

A standard Moody chart for estimating the Darcy-Weisbach friction factor in pipes and ducts. It is essential to recognize that careful interpolation on the chart will be required to find the needed values.

On the Moody Chart, the friction factor, {\scriptstyle{f}}, is on the left-hand ordinate, and the Reynolds number based on diameter (or hydraulic diameter), Re_{D_h}, is on the abscissa. Notice that the chart has a log-log scale. Curves of constant values of relative roughness, \epsilon/D_h, are then plotted, which are represented by the blue lines shown on the chart. Notice that the right side of the graph is not on a scale. Online calculator versions of the chart are also available; however, some may use approximations to the Colebrook-White equation rather than iteratively solving the actual equation.

Although the Moody Chart was initially developed for flow through circular pipes, it can also be applied to non-circular pipes by replacing the circular diameter, D, with the equivalent hydraulic diameter, D_h, as previously discussed. Experience suggests that this approach is sufficiently accurate for calculating the pressure drop along non-circular pipes when the flow is fully turbulent. Remember that the Reynolds number based on diameter or hydraulic diameter is calculated using

(47)   \begin{equation*} Re_{D_h} = \frac{\varrho \, V_{\rm av} \, D_h}{\mu} = \frac{V_{\rm av} \, D_h}{\nu} \end{equation*}

with the fluid properties being represented in terms of density, \varrho, and viscosity, \mu, or the kinematic viscosity, \nu (=\mu / \varrho). The average flow velocity, V_{\rm av}, can be calculated from the volume flow rate, Q, and the area of the cross-section, A_c, i.e.

(48)   \begin{equation*} V_{\rm av} = \frac{Q}{A_c} \end{equation*}

When using the Moody chart, if two of the three parameters are known, i.e., the friction factor, {\scriptstyle{f}}, the Reynolds number, Re_{D_h}, and the relative roughness, \epsilon/D_h, then the third parameter may be determined using the chart. However, in practice, the Reynolds number and relative roughness are usually known or calculated, so the friction factor is the desired output obtained from the chart.

MATLAB code to create the Moody chart using the Colebrook-White equation

Show code/hide code.

% Moody Chart in MATLAB

clear; clc;

% Define the Reynolds number range
Re = logspace(3, 8, 500); % Reynolds number from 10^3 to 10^8

% Define the relative roughness values (epsilon/D)
rel_roughness = [0, 0.00001, 0.0001, 0.001, 0.01, 0.02, 0.05];

% Preallocate friction factor array
friction_factors = NaN(length(rel_roughness), length(Re));

% Calculate friction factors using the Colebrook-White equation
for i = 1:length(rel_roughness)
for j = 1:length(Re)

if Re(j) <= 2300

% Laminar flow
friction_factors(i, j) = 64/Re(j);

elseif Re(j) >= 4000

% Turbulent flow using Colebrook-White equation
f = 0.02; % Initial guess for f

while true
f_new = 1 / (-2*log10((rel_roughness(i)/3.7) + …
(2.51/(Re(j)*sqrt(f)))))^2;

if abs(f_new – f) < 1e-6
break;
else
f = f_new;
end
end

friction_factors(i, j) = f_new;

end
end
end

% Plot the Moody chart
figure;
loglog(Re, friction_factors, ‘LineWidth’, 2);
hold on;

% Laminar flow line
Re_laminar = Re(Re <= 2300);
friction_laminar = 64 ./ Re_laminar;
loglog(Re_laminar, friction_laminar, ‘k–‘, ‘LineWidth’, 2);

% Annotate plot
grid on;
xlabel(‘Reynolds Number, Re’);
ylabel(‘Friction Factor, f’);
title(‘Moody Chart’);
legend(arrayfun(@(x) sprintf(‘\\epsilon/D = %g’, x), rel_roughness, …
‘UniformOutput’, false), ‘Location’, ‘Best’);

set(gca, ‘XScale’, ‘log’, ‘YScale’, ‘log’);
xlim([1e3 1e8]);
ylim([1e-3 1]);

% Add text annotations
text(2e3, 0.08, ‘Laminar Flow’, ‘FontSize’, 12, ‘Rotation’, -45);
text(1e7, 0.008, ‘Turbulent Flow’, ‘FontSize’, 12);

% Add grid lines for guidance
set(gca, ‘MinorGridLineStyle’, ‘-‘, ‘GridAlpha’, 0.5, ‘MinorGridAlpha’, 0.5);

hold off;

Interpretation

The following points should be understood when using and interpreting the curves on the Moody chart:

1. The laminar flow line represents a theoretical lower bound of the friction factor for low Reynolds numbers, i.e.,

(49)   \begin{equation*} f = \frac{64}{Re_D} \end{equation*}

This result applies only to fully developed laminar flow in smooth pipes. Nevertheless, it shows that the friction factor decreases as the Reynolds number increases and is independent of surface roughness.

2. The shaded area indicates a critical (transition) zone. This is where the flow begins to develop turbulence but remains unsteady and intermittent. The friction factor can vary significantly in this region, indicating that the flow may be either laminar or turbulent. The empirical data here are limited in scope and show considerable scatter. Fortunately, it is uncommon to have to use this part of the chart.

3. The friction factors increase with surface roughness for any given Reynolds number, particularly at higher Reynolds numbers. Notice that even entirely smooth pipes containing turbulent flow, such as plastic and glass, for which \epsilon/D \rightarrow 0, will still have a finite friction factor and corresponding pressure drop, although these effects diminish with increasing Reynolds number.

4. At higher Reynolds numbers, referred to as the fully turbulent or hydraulically rough regime, the friction factors become nearly independent of the Reynolds number. This behavior occurs because the roughness elements protrude through the viscous sublayer, so viscous scaling no longer controls the wall resistance. In this limiting case, the Colebrook-White equation reduces to the fully rough asymptote, i.e.,

(50)   \begin{equation*} \frac{1}{\sqrt{f}} \approx -2 \log_{10}\!\left(\frac{\epsilon}{3.7\,D_h}\right) \end{equation*}

5. In the fully rough regime, the friction factor becomes effectively independent of Reynolds number for a given value of relative roughness, \epsilon/D. In this case, the friction factor curves become nearly horizontal on the Moody chart. The Reynolds number at which this behavior occurs depends on the relative roughness, with smoother pipes requiring higher Reynolds numbers to reach the fully rough regime.

Reading a Moody Chart

Reading a Moody chart is not difficult, but it does require some practice, as it is a log-log chart that requires graphical interpolation. While the friction factor can be calculated numerically using the Colebrook-White equation, it is still essential to understand the visual interpretation of the physical behavior and the process of using the chart, as shown in the example in Figure 14. Reading the chart to three decimal places is usually sufficient. Notice that some published Moody charts, such as those found on the internet, may give slightly different friction factor values because they are based on approximations to the Colebrook-White equation.

How to read a Moody chart. Remember, it is a log-log scale, so some care is required when interpreting the results.

In this case, it is assumed that the information needed to calculate the Reynolds number and the relative roughness is available, and the desired output is the friction factor, {\scriptstyle{f}}.

  1. Determine the Reynolds number based on the average flow velocity, V_{\rm av}, and fluid properties (\varrho and \mu or \nu) using the hydraulic diameter, D_h, of the pipe, i.e.,

        \[ Re_{D_h} = \frac{\varrho \, V_{\rm av} \, D_h}{\mu} = \frac{V_{\rm av} \, D_h}{\nu} \]

    where the average flow velocity is determined from the flow rate and cross-sectional area, i.e.,

        \[ V_{\rm av} = \frac{Q}{A_c} \]

  2. Calculate the value of the relative roughness from the actual dimensional roughness, \epsilon, and calculated hydraulic diameter, D_h, i.e.,

        \[ \mbox{\small Relative roughness} = \frac{\epsilon}{D_h} \]

    The roughness value, \epsilon, is often specified for industrial piping or can be found in standard tables.

  3. Find the appropriate (blue) curve for the calculated value of relative roughness on the Moody diagram, recognizing that only a finite number of curves are shown, so interpolation on the logarithmic scales will likely be required.
  4. Find where the friction factor curve (or interpolated curve) intersects the Reynolds number (at the red dot).
  5. Estimate the friction factor by reading the value on the vertical axis corresponding to the intersection of the Reynolds number and relative roughness curve.

Finally, some caution is warranted when using the Moody chart. First, as with all engineering calculations, it is crucial to ensure that the correct units are used consistently when calculating non-dimensional quantities such as the Reynolds number and relative roughness. Ensure that the Reynolds number is calculated based on the pipe diameter (or hydraulic diameter), not its length. Second, because of the numerous lines on the chart, careful interpolation is required. While the chart provides good engineering estimates, greater accuracy may require solving the Colebrook-White equation directly or using an accepted approximation.

Approximations for the Friction Factor

The von Kármán rough pipe law, also known as the von Kármán-Prandtl roughness law, is an empirical relationship used to estimate the friction factor for fully developed turbulent flow in the hydraulically rough regime. In this limiting case, the friction factor becomes essentially independent of Reynolds number and depends only on the relative roughness, i.e.,

(51)   \begin{equation*} \frac{1}{\sqrt{f}} = -2.0 \, \log_{10}\left( \frac{\epsilon}{3.7 D_h} \right) \end{equation*}

This result is the fully rough asymptote of the Colebrook-White equation.

The Swamee-Jain equation, also known as the Swamee-Jain friction factor equation, provides an explicit approximation for the Darcy friction factor, i.e.,

(52)   \begin{equation*} f = \frac{0.25}{\left[\log_{10}\left( \displaystyle{ \frac{\epsilon}{3.7 D_h} + \frac{5.74}{Re_{D_h}^{\,0.9}}} \right)\right]^2} \end{equation*}

Roughness Values

In most cases, roughness values cover a range and will depend on the specific surface, as shown in the table below. As manufactured, roughness values will be on the lower end of the scale; however, all pipe types may develop roughness over time due to oxidation or corrosion. It is essential to consult the relevant standards and specifications, or obtain specific roughness values, for the pipe(s) used in a given application.

Roughness factors for different types of internal surfaces. In most cases, roughness values cover a range.
Pipe Material Roughness, ε (mm)
Smooth Plastic (PVC) 0.001 – 0.01
Glass 0.001 – 0.01
Smooth Metal (e.g., SS) 0.001 – 0.03
Concrete 0.2 – 1.5
Galvanized Iron 0.15 – 0.5
Commercial Steel 0.045 – 0.09
Riveted Steel 0.9 – 1.2
Corrugated Metal 3.0 – 8.0
Cast Iron (new) 0.15 – 0.5
Cast Iron (old) 0.6 – 3.0
Smooth Copper 0.001 – 0.01
Polyethylene (PE) 0.001 – 0.01
Polyvinyl Chloride (PVC) 0.001 – 0.01
Fiberglass 0.01 – 0.03
Stainless Steel (welded) 0.025 – 0.05
Stainless Steel (seamless) 0.02 – 0.045
Aluminum 0.05 – 0.1
Brass 0.02 – 0.05
PVC (Corrugated) 1.5 – 3.0
HDPE (Corrugated) 0.4 – 1.5
Ductile Iron 0.025 – 0.05
Polypropylene 0.01 – 0.03
ABS 0.02 – 0.06

Check Your Understanding #3 – Calculating the pressure drop for turbulent flow in a pipe using the Moody chart

Oil with a density \varrho_{\rm oil} = 900 kg/m{^{3}} and kinematic viscosity \nu_{\rm oil} = 10^{-5}m{^{2}}/s, flows at 0.2 m{^{3}}/s through a 500 m length of 300 mm diameter cast iron pipe. The average roughness of the pipe’s surface is 0.26 mm. Calculate: 1. The average flow velocity in the pipe. 2. The Reynolds number of the flow. 3. Is the flow in the pipe laminar or turbulent? 4. The pressure drop and head loss along the pipe. 5. The minimum hydraulic power needed to overcome the pressure loss.

Show solution/hide solution.
  1. The average flow velocity is calculated from the volume flow rate, i.e.,

        \[ V_{\rm av} = \frac{Q}{A } = \frac{4Q}{\pi D^2} = \frac{4 \times 0.2}{\pi \times 0.3^2} = 2.83~\mbox{ m/s} \]

  2. The Reynolds number of the flow in the pipe is

        \[ Re_D = \frac{\varrho_{\rm oil} \, V_{\rm av} \, D}{\mu_{\rm oil} } =\frac{V_{\rm av} \, D}{\nu_{\rm oil}} = \frac{2.83\times 0.3 }{10^{-5}} = 84,900 \]

  3. The flow will be turbulent because the Reynolds number is greater than 4,000, and the Moody chart will be needed to find the friction factor {\scriptstyle{f}}.
  4. To find the pressure drop and use the Moody chart, the relative surface roughness is needed, which is

        \[ \frac{\epsilon}{D} = \frac{0.26}{300} = 0.00087 \]

    From the Moody chart for a Reynolds number of 84,900 and a relative roughness of 0.00087 (using interpolation), then f \approx 0.022. Notice: You may get a slightly different friction factor depending on the Moody chart you use. Therefore, the pressure loss over the length of the pipe is

        \[ \Delta p = \frac{1}{2} \varrho_{\rm oil} \, V_{\rm av}^2 \, f \left( \frac{L}{D}\right) = 0.5\times 900\times 2.83^2 \times 0.022\times \frac{500}{0.3} = 132.15~\mbox{ kPa} \]

    The corresponding head loss over this pipe is

        \[ h = \frac{\Delta p}{\varrho_{\rm oil} \, g} = \frac{132.15\times 10^3}{900\times 9.81} = 14.97~\mbox{m} \]

  5. The ideal hydraulic or pumping power required to overcome the pressure loss is

        \[ P_{\rm hyd} = Q \, \Delta p = 0.2 \times 132.15 \times 10^3 = 26.43~\mbox{kW} \]

    The actual pump shaft power would exceed this value and depend on the pump efficiency.

Pressure Drop Over a Tapered Pipe

There is no simple general closed-form Darcy-Weisbach expression for the pressure drop along a tapered pipe; an example of such a shape is shown in Figure 15. The main issue is that the flow velocity, hydraulic diameter, Reynolds number, and friction factor vary continuously along the pipe. Therefore, numerical solutions are often obtained by dividing the pipe into short segments and summing the local losses. In fully turbulent flow at higher Reynolds numbers, where the friction factor is only weakly dependent on Reynolds number, an approximate estimate can be obtained using a representative hydraulic diameter and flow velocity.

Calculating the pressure drop over a tapered pipe is possible under certain assumptions, including fully turbulent flow at higher Reynolds numbers.

In this case, an approximate average hydraulic diameter D_{h,\rm avg} may be defined as

(53)   \begin{equation*} D_{h,\rm avg} = \frac{1}{2} \left( \frac{4ab}{2a + 2b} + \frac{4cd}{2c + 2d} \right) = \frac{ab}{a + b} + \frac{cd}{c + d} \end{equation*}

The average flow velocity, V_{\rm av}, based on continuity considerations, can be approximated from the inlet and exit conditions, i.e.,

(54)   \begin{equation*} V_{\rm av} \approx \frac{Q}{A_{\rm avg}} \end{equation*}

where, for this estimate, a representative cross-sectional area may be taken as the arithmetic average of the inlet and exit areas, i.e.,

(55)   \begin{equation*} A_{\rm avg} = \frac{1}{2}(ab + cd) \end{equation*}

The Reynolds number based on the average hydraulic diameter and average flow velocity can then be obtained, i.e.,

(56)   \begin{equation*} Re_{D_{h,\rm avg}} = \frac{\varrho \, V_{\rm av} \, D_{h,\rm avg}}{\mu} = \frac{V_{\rm av} \, D_{h,\rm avg}}{\nu} \end{equation*}

Then the Moody chart can be used with the relative roughness to obtain the friction factor, {\scriptstyle{f}}, or the Swamee-Jain equation can also be used, i.e.,

(57)   \begin{equation*} f = \frac{0.25}{\left[\log_{10}\left( \displaystyle{ \frac{\epsilon}{3.7 D_{h,\rm avg}} + \frac{5.74}{Re_{D_{h,\rm avg}}^{\,0.9}}} \right)\right]^2} \end{equation*}

Therefore, the pressure loss over the length of the tapered pipe may be estimated as

(58)   \begin{equation*} \Delta p_L = \frac{1}{2} \varrho \, V_{\rm av}^2 \, f \left( \frac{L}{D_{h,\rm avg}} \right) \end{equation*}

These pressure-drop estimates are sufficiently accurate for many practical pipe and duct calculations when the assumptions used to define the average velocity, hydraulic diameter, and friction factor are appropriate.

Pumps

Many internal flow problems involve the use or design of various types of pumps to move fluid from one place to another. Such devices include piston, centrifugal, gear, vane, and axial-flow pumps for liquids, as well as fans, blowers, and compressors for gases. For aerospace applications, pumps must be lightweight for the required flow rate and pressure rise, reliable, and capable of withstanding harsh operating environments, including large temperature swings and vibration. Pumps obtained “off-the-shelf” are usually specified by their volumetric flow rate and the pressure rise (or head) they can produce. It is essential to note the units in which these quantities are given, which typically are not base units; e.g., flow rates may be quoted in liters/hr, gallons/minute, or other units.

In general, pumps compensate for pressure losses so that fluid can be transported from one location to another, as shown in Figure 16. While all pumps increase the fluid pressure, they do so with varying efficiencies and operating characteristics depending on the fluid and application. Pressure drops depend on several factors, including the flow rate, Reynolds number, and friction factor.

Pumps are usually located in-line with the overall flow and are used to overcome pressure losses and keep the fluid moving through the pipe from one end to the other.

Types of Pumps

Figure 17 shows four common types of fluid pumps: a gear pump, a lobe pump, a vane pump, and a centrifugal pump. A fifth type, the diaphragm pump, is shown separately in Figure 18. In each case, the pump must be driven by a power source. This source could be an accessory drive through a belt or gear system from an engine or propulsion system, or a direct drive from a separate source, such as an electric motor. Such pumps are usually inserted in series into a pipeline and connected using flanges sealed by gaskets.

There are many different types of pumps, but in each case, their purpose is to move fluid through a pipe to overcome pressure losses.

Gear Pumps

Gear pumps are positive-displacement pumps that use two meshing gears to transfer fluids by creating chambers between the gear teeth and the pump casing. They come in various types, including external and internal configurations, offering specific advantages and applications. Gear pumps are valued for their simplicity, reliability, and ability to handle a wide range of viscosities, making them suitable for applications in hydraulics, lubrication, and fuel transfer. Lubrication systems in engines often use gear pumps because of their simplicity, low cost, and good reliability.

Lobe Pumps

Lobe pumps are positive-displacement pumps characterized by two or more lobes that rotate synchronously within a casing, creating chambers that transport fluid from the inlet to the outlet. Unlike gear pumps, lobe pumps use smooth lobes rather than gear teeth, resulting in smoother fluid handling and lower shear. They are widely used because they can handle more viscous fluids, such as oils. They are reliable but tend to be more expensive than gear pumps.

Vane Pumps

Vane pumps are positive-displacement pumps that use rotating vanes inside a cylindrical housing to move fluid from the inlet to the outlet. These pumps typically feature an off-center rotor carrying the vanes. The vanes slide in and out of their slots while maintaining close contact with the housing. Vane pumps deliver a smooth, consistent flow, making them suitable for applications requiring precise, steady fluid delivery, such as hydraulic and pneumatic systems. They offer relatively quiet operation and can handle a range of viscosities, though they exhibit reduced efficiency with more viscous fluids. Furthermore, the vanes wear over time and require periodic maintenance to remain functional.

Centrifugal Pumps

Centrifugal pumps use an impeller to increase the fluid’s kinetic energy and pressure. Such pumps consist of an impeller enclosed within a casing that propels the fluid outward from the center of rotation. The fluid is introduced through a pipe at the center of the pump. As the fluid moves through the pump, it gains velocity and pressure, transferring fluid from the inlet to the outlet. Centrifugal pumps are widely used in applications requiring high flow rates. While efficient, they are generally better suited to low-viscosity liquids than to highly viscous liquids.

Diaphragm Pumps

Figure 18 shows a CFD simulation of the internal flow through a diaphragm pump or membrane pump. In this type of pump, common in fuel delivery systems, the diaphragm is flexed up and down by a motor or other actuator so that, as the chamber volume increases, the pressure decreases, drawing fluid from one end of the pipe into the chamber. When the chamber volume decreases, the pressure rises, forcing the fluid out through the other pipe. Notice that a pair of check valves, or one-way valves, prevents reverse flow of fluid through the pump.

An excellent example of an internal flow is this diaphragm pump, which draws the fluid from a pipe into a chamber and forces it out through the other pipe.

Pressure Drops Over Bends & Corners

When an internal fluid flow encounters a bend or corner, it experiences an additional pressure loss because the flow direction must change, as shown in Figure 19. Several factors contribute to this loss, including streamline curvature, secondary flows, changes in velocity distribution, wall friction, and possible flow separation. As the fluid moves through the bend, centrifugal effects create a pressure gradient across the cross-section, with the higher pressure generally occurring on the outside of the bend. If the corner is sharp, flow separation and higher pressure losses can be expected.

Flow around bends and corners incurs additional pressure losses.

The magnitude of the pressure drop around a bend or corner depends on several factors, including bend geometry, flow rate, fluid viscosity, Reynolds number, and surface roughness. Many studies have been conducted to determine the losses in bends, including by NIST. One way of assessing such pressure losses is by using a local loss coefficient K_L, for which values are published for various corners, bends, fittings, and valves. The local pressure loss, \Delta p_L, is given by

(59)   \begin{equation*} \Delta p_L = \frac{1}{2} \varrho \, V_{\rm av}^2 \, K_L = \varrho \, g \, h_L \end{equation*}

where h_L is the equivalent head loss. Therefore,

(60)   \begin{equation*} h_L = K_L \left( \frac{V_{\rm av}^2}{2 \, g} \right) \end{equation*}

where V_{\rm av}^2/(2g) is the velocity head. Representative values of K_L depend strongly on the bend radius, cross-sectional shape, Reynolds number, and details of the geometry. Sharp corners or square elbows generally produce much larger losses than smooth, large-radius bends, so tabulated values should be used for the specific fitting or bend geometry being analyzed.

Many hydraulic, pneumatic, and other internal fluid-flow systems consist of straight sections, turns, fittings, valves, and branches. The total pressure loss is the sum of all the losses incurred by each component of the system, i.e.,

(61)   \begin{equation*} \Delta p_{\rm total} = \Delta p_{1} + \Delta p_{2} + \Delta p_{3} + \cdots \end{equation*}

Today, computational fluid dynamics (CFD) can be used to predict and analyze the pressure drop for a given internal flow system. In general, the pressure drop around a corner can be minimized by using smooth bends with large radii, which reduce the likelihood of flow separation. Additionally, properly designed flow-control devices, such as corner vanes or turning vanes, can reduce pressure losses.

Pressure Losses in Wind Tunnels

One key challenge in wind tunnel design is determining the required fan or motor power to generate a desired test-section velocity or dynamic pressure. This depends on accurately estimating pressure losses throughout the tunnel circuit. Because wind tunnels contain ducts of varying shapes, areas, and transition pieces, the flow moving through them experiences friction and other losses, particularly at higher Reynolds numbers. Additional pressure losses arise from turning vanes at circuit corners, typically cascades of thin airfoil-shaped plates that help redirect flow but introduce some resistance. Careful estimation of these cumulative losses is essential to ensure the tunnel meets its performance targets.

In the conventional approach to wind tunnel design, the circuit losses are estimated to support the initial sizing of the fan and motor by breaking the tunnel circuit into its primary parts:

  1. Cylindrical sections, or equivalent duct sections.
  2. Corners and turning vanes.
  3. Expanding sections, i.e., diffusers.
  4. Contracting sections, i.e., nozzles.
  5. Turbulence screens.
  6. Heat exchangers.
  7. Other miscellaneous parts.

In each of these sections, energy is dissipated in the form of a static pressure drop \Delta p_L, which can be expressed using a dimensionless local loss coefficient

(62)   \begin{equation*} K_L = \frac{\Delta p_L}{q} \end{equation*}

where q is the local dynamic pressure. For corners, bends, and turning vanes, the loss coefficient K_L is typically determined empirically, as previously discussed. This loss is referenced to the test-section values, denoted by subscript 0, using

(63)   \begin{equation*} K_0 = K_L \left( \frac{q}{q_0} \right) \end{equation*}

For incompressible flow, this relationship can also be written as

(64)   \begin{equation*} K_0 = K_L \left( \frac{V}{V_0} \right)^2 \end{equation*}

where V is the local average velocity in the section and V_0 is the test-section velocity. The local velocity depends on the area ratio through continuity. Therefore, for incompressible flow,

(65)   \begin{equation*} V = V_0 \left( \frac{A_0}{A} \right) \end{equation*}

so that

(66)   \begin{equation*} K_0 = K_L \left( \frac{A_0}{A} \right)^2 \end{equation*}

where A is the local cross-sectional area and A_0 is the test-section area.

The hydraulic diameter may be used when estimating frictional losses in non-circular duct sections, i.e.,

(67)   \begin{equation*} D_h = \frac{4A}{P_w} \end{equation*}

where A is the cross-sectional area and P_w is the wetted perimeter of the section. However, the hydraulic diameter is used primarily to form Reynolds numbers and estimate friction factors. It does not, by itself, imply a fourth-power scaling of the loss coefficient for turbulent wind-tunnel circuit losses.

The next step is to express the rate of energy loss, or power loss, \Delta P, in terms of the test-section conditions. For a given section, this gives

(68)   \begin{equation*} \Delta P = \Delta p_L \, Q \end{equation*}

where Q = A_0 V_0 is the volume flow rate through the tunnel. Because

(69)   \begin{equation*} \Delta p_L = K_0 q_0 = K_0 \left( \frac{1}{2} \varrho V_0^2 \right) \end{equation*}

then

(70)   \begin{equation*} \Delta P = K_0 \left( \frac{1}{2} \varrho A_0 V_0^3 \right) \end{equation*}

The total circuit power loss is therefore

(71)   \begin{equation*} \sum \Delta P = \left( \sum K_0 \right) \left( \frac{1}{2} \varrho A_0 V_0^3 \right) \end{equation*}

The so-called energy ratio, ER_t, can then be defined as

(72)   \begin{equation*} \mbox{\small Energy~ratio} = \frac{\mbox{\small Power at test section}}{\displaystyle{ \sum~ \mbox{\small Circuit power losses}}} = ER_t \end{equation*}

so that

(73)   \begin{equation*} ER_t = \frac{\frac{1}{2} \varrho \, A_0 \, V_0^3}{\displaystyle{\sum K_0 \, \frac{1}{2} \varrho \, A_0 \, V_0^3}} = \frac{1}{\displaystyle{ \sum K_0}} \end{equation*}

This formulation shows that minimizing \sum K_0, including corner and turning-vane losses, improves the tunnel’s efficiency and reduces the fan power requirement.

The energy ratio, ER_t, represents the efficiency of a wind tunnel circuit and is inversely related to the total circuit losses. For a well-designed closed-return tunnel, ER_t typically ranges from 4 to 7. Lower pressure losses mean greater efficiency and lower power demand from the fan and motor. Major losses often arise in diffuser sections, corners, turning vanes, turbulence screens, and heat exchangers, making their design critical. Calculating the loss coefficients, K_0, for each section involves applying standard aerodynamic relationships for turbulent flow through ducts, fittings, vanes, screens, and other elements. Losses within the fan or motor are generally excluded from ER_t to isolate the tunnel circuit efficiency. Estimating these losses is necessary to determine the fan power required to achieve the target test-section velocity. Because some losses cannot be accurately estimated before construction, wind tunnel designs typically include power margins to ensure the specifications are met.

Check Your Understanding #4 – Calculating the power required to drive a low-speed wind tunnel

Estimate the minimum motor power required for a low-speed wind tunnel with a maximum flow speed of 230 mph in the test section. The test section area is 22.5 ft{^{2}}, and the energy ratio of ER_t for the tunnel is 5.2. The fan efficiency \eta_f is 74%, and the motor efficiency \eta_m is 90%.

Show solution/hide solution.

The specification of the energy ratio gives the cumulative losses in the tunnel, so the first step is to find the power in the flow through the test section, i.e.,

    \[ \mbox{\small Power} = \frac{1}{2} \varrho \, A_0 \, V_0^3 \]

The value of A_0 is 22.5 ft{^{2}} and V_0 = 230 mph = 337.33 ft/s. Assume MSL ISA for the air density, so \varrho = 0.002378 slug ft{^{-3}}. This gives

    \[ \mbox{\small Power} = \frac{1}{2} \varrho \, A_0 \, V_0^3 = 0.5 (0.002378) (22.5) (337.33)^3 = 1.0269 \times 10^6~\mbox{ft lb s$^{-1}$} \]

Therefore, the circuit loss power that must be supplied to the air by the fan is

    \[ P_{\rm air} = \frac{\mbox{\small Power at test section}}{ER_t} = \frac{1.0269 \times 10^6}{5.2 \times 550} = 359.1~\mbox{hp} \]

Taking into account the fan and motor efficiencies, the minimum motor input power required is

    \[ P_{\rm req} = \frac{P_{\rm air}}{\eta_f \, \eta_m} = \frac{359.1}{(0.74)(0.90)} = 539.1~\mbox{hp} \]

This latter result would only be valid for an empty test section. To achieve the same flow speed with an article in the test section, more power would be required to overcome the drag and the blockage caused by the article. This value is generally unknown a priori. However, it is usually considered reasonable to add a margin of power to overcome a winged test article with a wing span of 0.8 of the tunnel diameter or width, an aspect ratio of 5, and a C_D of 1.0. Furthermore, to account for the potential diversity of test articles, not all of which will be streamlined shapes, a margin of 50% more power may be needed. The final estimated rated motor power for the tunnel would be about 900–1,000 hp.

Summary & Closure

Internal flows are encountered in various aerospace applications, such as engine air intakes, fuel systems, hydraulic systems, air conditioning systems, and wind tunnels, where fluids flow through pipes and ducts. A significant consideration for internal flows is that frictional effects from viscosity produce pressure drops along the length of the pipe or duct. These pressure drops require a power source to pump the fluid through the pipe or duct, and the required power depends on the flow velocity and internal surface finish. Because many practical internal flows are turbulent, a Moody chart or an equivalent friction-factor correlation is often used to determine the friction factors and estimate the resulting pressure drops for design purposes. Internal flows are critical within the combustion chambers of jet and rocket engines, where combustion produces high-temperature, high-pressure gases to produce thrust. Aircraft and spacecraft have environmental control systems to maintain comfortable conditions for passengers and crew, and these systems also require the analysis of internal flows. Understanding and optimizing internal flows in aerospace applications are essential for improving efficiency, safety, and overall performance.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • For an aircraft’s hydraulic system, discuss some potential design trades in operating the hydraulic system with smaller pipes and higher pressure versus larger pipes and lower pressure.
  • What is the definition of internal flow in fluid dynamics? How does internal flow differ from external flow, and what are some examples of each?
  • Consider the fuel and oxidizer system for a rocket engine, which requires high mass flow rates of cryogenic liquids. What are the internal flow issues to consider regarding the delivery of the fuel and oxidizer to the combustion chamber in this case?
  • The corner or turning vanes in a wind tunnel are a large source of losses. Consider the engineering steps you might take to calculate and minimize such losses.
  • Explain the significance of the Reynolds number in internal flows. How does it affect the flow characteristics? Compare the flow regimes associated with low and high Reynolds numbers in internal flows.
  • Explain the major differences between flow in circular pipes and non-circular conduits. How do these differences impact flow characteristics?

 

Other Useful Online Resources

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

  1. See: "The History of Poiseuille's Law," by S. P. Sutera and R. Skalak, Annual Review of Fluid Mechanics, Volume 25, 1993. According to them, the term "Poiseuille's law" for Equation 42 appears in a publication by Hagenbach (1860), who, after giving the derivation, generously suggested calling it "Poiseuille's law". Jacobson (1860) also refers to Equation 42 as Poiseuille's law. See: Hagenbach, E. 1860, "Über die Bestimmung der Zähigkeit einer Flüssigkeit durch den Ausfluss aus Röhren," Poggendorff's Annalen der Physik und Chemie, Vol. 108, pp. 38–426, and Jacobson, H. 1860, "Beiträge zur Häemodynamik," Archives of Anatomy and Physiology, Vol. 80, pp. 80–113. ↵

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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.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n26

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