Continuity in Cylindrical Coordinates Axisymmetric Flow

Burst Strength of RTP Pipeline

Qiang Bai , Yong Bai , in Subsea Pipeline Design, Analysis, and Installation, 2014

Coordinate Systems

A cylindrical coordinate system, as shown in Figure 27.3, is used for the analytical analysis. The coordinate axis r, θ, and z denote the radial, circumferential, and axial directions of RTP pipe, respectively. The local material coordinate system of the reinforced tape layers is designated as (L, T, r), where L is the wound direction, T is the direction perpendicular to the aramid wire in plane, and r is the normal direction, same as in the cylindrical coordinate system. The term α is the wound angle of reinforced layer, the angle between L direction and z direction.

Figure 27.3. Coordinate systems.

(For color version of this figure, the reader is referred to the online version of this book.)

The mathematical solutions are based on the model developed by Zheng et al. [4] for predicting the short-term burst pressure of PSP, by applying the 3D anisotropic elasticity and the maximum stress failure criterion, to calculate the short-term burst pressure of RTP pipe, except for the maximum strain failure criterion.

Provided that the interfaces between the fiber yarn and PE are perfectly bonded, the strain of the aramid wire and PE in the aramid-wound direction can be considered to be equal. Because Young's modulus of the aramid fiber is far greater than that of PE, the stresses in the aramid fiber are much greater than those in the PE. When the RTP is subjected to internal pressure, aramid wires first reach their strength limits and break, resulting in the RTP losing the reinforcement of the fibers yarn and bursting in the short term.

The detailed mathematical equations for the strains of each layers are found in the Bai et al. paper [12].

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Potential Flow

E.L. Houghton , ... Daniel T. Valentine , in Aerodynamics for Engineering Students (Sixth Edition), 2013

3.3.1 Cylindrical Coordinate System

The cylindrical coordinate system is illustrated in Fig. 3.20. The three coordinate surfaces are the planes z = constant and θ = constant, with the surface of the cylinder having radius r. For the Cartesian system, in contrast, all three coordinate surfaces are planes. As a consequence for the Cartesian system, the directions (x, y, z ) of the velocity components are fixed throughout the flow field. For the cylindrical coordinate system, though, only one of the directions (z ) is fixed throughout the flow field; the other two (r and θ) vary depending on the value of the angular coordinate θ. In this respect there is a certain similarity to the polar coordinates introduced earlier in the chapter. The velocity component q_r is always locally perpendicular to the cylindrical coordinate surface, and q_θ is always tangential to that surface. Once this elementary fact is properly understood, cylindrical and Cartesian coordinates are equally easy to use.

Figure 3.20. Cylindrical coordinates.

Like the relationships between velocity potential and velocity components derived for polar coordinates (see Section 3.1.3), the following relationships are obtained for cylindrical coordinates:

(3.60) $q_r=\frac{\partial \varphi }{\partial r},q_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta },q_z=\frac{\partial \varphi }{\partial z}$

An axisymmetric flow is defined as one for which the flow variables (i.e., velocity and pressure) do not vary with the angular coordinate θ. This would be so, for example, for a body of revolution about the z-axis, along which the oncoming flow is directed. For such an axisymmetric flow a stream function can be defined. The continuity equation for axisymmetric flow in cylindrical coordinates can be derived in a similar manner as for two-dimensional flow in polar coordinates (see Section 2.4.3); it takes the form

(3.61) $\frac{1}{r}\frac{\partial r{q}_r}{\partial r}+\frac{\partial q_z}{\partial z}=0$

The relationship between stream function and velocity component must be such as to satisfy Eq. (3.61); hence it can be seen that

(3.62) $q_r=-\frac{1}{r}\frac{\partial \psi }{\partial z},q_z=\frac{1}{r}\frac{\partial \psi }{\partial r}$

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Potential Flow

E.L. Houghton , ... Daniel T. Valentine , in Aerodynamics for Engineering Students (Seventh Edition), 2017

5.4.1 Cylindrical Coordinate System

The cylindrical coordinate system is illustrated in Fig. 5.27. The three coordinate surfaces are the planes z = constant and θ = constant and the surface of the cylinder having radius r. In contrast, for the Cartesian system all three coordinate surfaces are planes. As a consequence for the Cartesian system the directions $(x,y,z)$ of the velocity components, say, are fixed throughout the flow field. For the cylindrical coordinate system, though, only one of the directions (z) is fixed throughout the flow field; the other two (r and θ) vary throughout the flow field depending on the value of the angular coordinate θ. In this respect there is a certain similarity to the polar coordinates introduced earlier in the chapter. The velocity component $V_r$ is always locally perpendicular to the cylindrical coordinate surface and $V_{\theta }$ is always tangential to that surface. Once this elementary fact is properly understood cylindrical coordinates become as easy to use as the Cartesian system.

Figure 5.27. Components of the velocity V in cylindrical coordinates.

In a similar way as the relationships between velocity potential and velocity components are derived for polar coordinates (see Section5.1.3 above), the following relationships are obtained for cylindrical coordinates

(5.59) $V_r=\frac{\partial \varphi }{\partial r},V_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta },V_z=\frac{\partial \varphi }{\partial z}.$

An axisymmetric flow is defined as one for which the flow variables, i.e. velocity and pressure, do not vary with the angular coordinate θ. This would be so, for example, for a body of revolution about the z axis with the oncoming flow directed along the z axis. For such an axisymmetric flow a stream function can be defined. The continuity equation for axisymmetric flow in cylindrical coordinates can be derived in a similar manner as it is for two-dimensional flow in polar coordinates (see Section2.5.3); it takes the form

(5.60) $\frac{1}{r}\frac{\partial r{V}_r}{\partial r}+\frac{\partial V_z}{\partial z}=0.$

The relationship between stream function and velocity component must be such as to satisfy Eq. (5.60); hence it can be seen that

(5.61) $V_r=-\frac{1}{r}\frac{\partial \psi }{\partial z},V_z=\frac{1}{r}\frac{\partial \psi }{\partial r}.$

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Observations of Winds, Storms, and Related Phenomena

Richard J. Doviak , Dušan S. Zrnić , in Doppler Radar and Weather Observations (Second Edition), 1993

9.2.1.1 COPLAN

The cylindrical coordinate system is illustrated in Fig. 9.3 where effects of the earth's curvature are ignored. The measured Doppler velocity needs to be corrected for the scatterers' terminal velocity w _t. In this chapter overbars, used in Section 5.2, are dropped; nevertheless, it is important to be aware of the reflectivity and I(r ₀, r ₁) weights. Thus the estimate of the radial component of air motion is

Fig. 9.3. Cylindrical coordinate system used for dual radar data analysis. The radars are located at the points 1 and 2, and a _r, a _s, a _α are the unit normals defining the direction of the three orthogonal velocity components. The cylinder axis is along the line connecting the radars, and r is the range from the axis to the data point.

(9.1) $v_{1,2}=v_{1,2}^{\text{h}}+w_{\text{t}}\sin {\theta }_{\text{e}1,2},$

where $v_{1,2}^{\text{h}}$ are the weighted mean hydrometeor velocities measured by radars 1 and 2, w _t is positive, and θ_e is the elevation angle. To estimate w _t for each resolution volume, we could use Eq. (8.79) or the empirical expression (Atlas et al., 1973)

(9.2) $w_t=2.65Z^{0.1114}{\left(\frac{{\gamma }_0}{\gamma }\right)}^{0.4}{\text{ms}}^{-1},$

where the parenthetical term is a correction [Eq. (8.4)] to account for the height-dependent air density γ. This relation represents, to within a standard error of 1 m s⁻¹, the experimental data of Joss and Waldvogel (1970) over a large range of Z (i.e., 1 ≤ Z ≤ 10⁵ mm⁶ m⁻³) and drop-size distributions for regions of liquid water; but large errors (up to several meters per second) in w _t estimates can be caused by using Eq. (9.2) for regions of hail. Terminal velocity relations for hail can be found using, for example, Eq. (8.6c). Usually there is little or no information to identify hail regions, and errors in vertical wind w can result. It has been shown, however, for typical arrangements of storms relative to the two-radar placement, that these errors are significantly smaller than errors in the estimate w _t (Doviak et al., 1976).

The estimated radial velocities v _1,2 of the air can be interpolated to uniformly spaced grid points in planes at an angle α to the horizontal surface containing the baseline. Interpolation filters the data and reduces the variance. Several interpolation schemes are possible, and a particularly simple and effective one employs the Cressman function (Cressman, 1959) W _i to weight data at points inside a volume centered on a grid point from which the distance to the ith datum is D_i .

(9.3) ${\text{W}}_{\text{i}}=(\begin{array}{lll}\left(R_{\text{i}}^2-D_i^2\right)/\left(R_{\text{i}}^2+D_i^2\right)\hfill & \text{for}\hfill & D_i\le R_{\text{i}},\hfill \\ 0\hfill & \text{for}\hfill & D_i>R_{\text{i}}.\hfill \end{array}$

R _i is the influence radius that determines the size of the interpolation volume. The shape of this volume, usually selected to be a sphere, is dictated by the functional dependence of R _i on the direction of a datum from the grid point. The cylindrical wind components w _r, w _s in the r, s plane (Fig. 9.3) are related to ${\overline{v}}_1,{\overline{v}}_2$ as

(9.4a) $w_{\text{r}}=\left[\left(s+d\right)r_1{\overline{v}}_1-\left(s-d\right)r_2{\overline{v}}_2\right]/2dr,$

(9.4b) $w_s=\left(r_2{\overline{v}}_2-r_1{\overline{v}}_1\right)/2d,$

where ${\overline{v}}_{1,2}$ are the interpolated Doppler velocities of air. Here the overbar signifies two averaging processes: (1) an I(r ₀, r ₁) and η weighted spatial average, and (2) an interpolated average of v _1,2 over several resolution volumes.

The wind component w_α normal to the plane is obtained by introducing the continuity equation for air density γ,

(9.5a) $\partial \gamma /\partial t+\nabla ·\left(\gamma \text{v}\right)=0,$

where v is the vector wind. Unfortunately, this introduces another unknown. But since perturbations in air density are much smaller than the ambient density, it can be shown by using scale analysis for deep convection (Pielke, 1989) that the first-order velocity field is produced by the first-order density and pressure perturbations, and to this order the continuity equation is

(9.5b) $\nabla ·\left({\gamma }_0\text{v}\right)=0,$

where γ₀ is the ambient air density, which is assumed to depend only on height. This form of the continuity equation, first introduced by Ogura and Phillips (1962), is called anelastic because the energy conservation equation no longer has contribution from elastic energy (i.e., there are no sound waves; Ogura, 1963). In the cylindrical coordinate system this equation takes the form

(9.5c) $\frac{1}{r}\frac{\partial }{\partial r}\left(r{\text{γ}}_0{w}_r\right)+\frac{1}{r}\frac{\partial }{\partial \alpha }\left({\text{γ}}_0{w}_{\alpha }\right)+{\text{γ}}_0\frac{\partial }{\partial s}\left(w_s\right)=0.$

If the ambient atmosphere has constant potential temperature θ₀, γ₀ is given by Hess, 1959,

(9.5d) ${\text{γ}}_0\left(r,\alpha \right)={\text{γ}}_s{\left[\frac{{\theta }_0-\frac{g}{c_{\text{p}}}r\sin \alpha }{{\theta }_0}\right]}^{c_{\text{p}}M/R-1},$

where γ_s is air density at α = 0 (e.g., at the earth's surface), C _p = 1000 J kg⁻¹ K⁻¹ is the specific heat at constant pressure, g is the gravitational constant (9.8 m s⁻²), M is the mean molecular weight of air (0.029 kg mol⁻¹), and R is the universal gas constant (8.314 J mol⁻¹ K⁻¹). Estimates of θ₀ can be obtained from soundings of the environment within which the storm develops. Several techniques used to compute the vertical velocity are compared by Ray et al. (1980), and Ziegler et al. (1983) solve the anelastic form of the continuity equation using a variational formulation with downward integration to reduce errors in vertical velocity.

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Theoretical, Experimental, and Numerical Techniques

ZHAO-YUAN HAN , XIE-ZHEN YIN , in Handbook of Shock Waves, 2001

3.7.2.2.2.3 Three-Dimensional Equations for Shock Wave Propagation Through Nonuniform Flow Fields in a Cylindrical Coordinate System

The cylindrical coordinate system is illustrated in Fig. 3.7.21, and the unit vector for the normal to the shock wave and the flow ahead of the shock wave can be expressed as

FIGURE 3.7.21. The unit vectors in a cylindrical coordinate system.

(3.7.229) $\overrightarrow{\text{n}}=\text{cos}{\theta }_r{\overrightarrow{\text{e}}}_r+\text{cos}{\theta }_{\phi }{\overrightarrow{\text{e}}}_{\phi }+\text{cos}{\theta }_x{\overrightarrow{\text{e}}}_x$

(3.7.230) ${\overrightarrow{\tau }}_f=\text{cos}{\varepsilon }_r{\overrightarrow{\text{e}}}_r+\text{cos}{\theta }_{\phi }{\overrightarrow{\text{e}}}_{\phi }+\text{cos}{\theta }_x{\overrightarrow{\text{e}}}_x$

Neglecting the detailed derivation of the 3D equations, one can directly write the following equations in cylindrical coordinates as

(3.7.231) $\begin{array}{l}\frac{\partial }{\partial r}\left(\frac{(M\text{cos}{\theta }_r+mcos{\varepsilon }_r)r}{A^{\prime }L}\right)+\frac{\partial }{\partial \phi }\left(\frac{M\text{cos}{\theta }_{\phi }+m\text{cos}{\varepsilon }_{\phi }}{A^{\prime }L}\right)+\frac{\partial }{\partial x}\left(\frac{(Mcos{\theta }_x+m\text{cos}{\varepsilon }_x)r}{A^{\prime }L}\right)=0\\ \frac{\partial }{\partial x}\left(\frac{cos{\theta }_r}{La}\right)-\frac{\partial }{\partial r}\left(\frac{cos{\theta }_x}{La}\right)=0\\ \\ \frac{\partial }{\partial r}\left(\frac{rcos{\theta }_{\phi }}{La}\right)-\frac{\partial }{\partial \phi }\left(\frac{cos{\theta }_r}{La}\right)=0\\ {cos}^2{\theta }_r+{cos}^2{\theta }_{\phi }+{cos}^2{\theta }_x=1\\ \frac{1}{A^{\prime }}\left(\frac{\partial A^{\prime }}{\partial r}+{G^{\prime }}_1\frac{\partial A^{\prime }}{r\partial \phi }+{G^{\prime }}_2\frac{\partial A^{\prime }}{\partial x}\right)=-(E_r+{G^{\prime }}_1{E}_{\phi }+{G^{\prime }}_2{E}_x)\end{array}$

where

$\begin{array}{c}\begin{array}{cc}{E}_r=e\frac{\partial M}{\partial r}+\frac{h}{p}\frac{\partial p}{\partial r}+\frac{g}{a}\frac{\partial a}{\partial r},& E_{\phi }=e\end{array}\frac{\partial M}{r\partial \phi }+\frac{h}{p}\frac{\partial p}{r\partial \phi }+\frac{g}{a}\frac{\partial a}{r\partial \phi }\\ E_x=e\frac{\partial M}{\partial x}+\frac{h}{p}\frac{\partial p}{\partial x}+\frac{g}{a}\frac{\partial a}{\partial x},\\ \begin{array}{cc}{G^{\prime }}_1=\frac{M\text{cos}{\theta }_{\phi }+m\text{cos}{\varepsilon }_{\phi }}{M\text{cos}{\theta }_r+m\text{cos}{\varepsilon }_r},& {G^{\prime }}_1=\frac{M\text{cos}{\theta }_x+m\text{cos}{\varepsilon }_x}{M\text{cos}{\theta }_r+m\text{cos}{\varepsilon }_r}\end{array}\end{array}$

For axisymmetric flows, that is, ∂/∂φ = 0, and without a circumferential flow, that is, θ_φ = π/2 and ɛ_φ = π/2 letting θ _x = θ, ɛ _x = ɛ and θ _r = π/2 – θ, ɛ _r = π/2 – ɛ equations (3.7.231) can be simplified to read

(3.7.232) $\begin{array}{l}\frac{\partial }{\partial r}\left[\frac{(M\text{sin}\theta +m\text{sin}\varepsilon )r}{(M+m\text{cos}(\theta -\varepsilon ))A^{\prime }}\right]+\frac{\partial }{\partial x}\left[\frac{(M\text{cos}\theta +m\text{cos})r}{(M+m\text{cos}(\theta -\varepsilon ))A^{\prime }}\right]=0\\ \frac{\partial }{\partial x}\left[\frac{sin\theta }{(M+m\text{cos}(\theta -\varepsilon ))a}\right]-\frac{\partial }{\partial r}\left[\frac{\text{cos}\theta }{(M+m\text{cos}(\theta -\varepsilon ))a}\right]=0\\ \frac{1}{A^{\prime }}\frac{\partial A^{\prime }}{\partial r}+G^{\prime }\frac{1}{A^{\prime }}\frac{\partial A^{\prime }}{\partial x}=-(E_r+G^{\prime }{E}_x)\end{array}$

where

$\begin{array}{l}{E}_r=e\frac{\partial M}{\partial r}+\frac{h}{p}\frac{\partial p}{\partial r}+\frac{g}{a}\frac{\partial a}{\partial r},\\ E_x=e\frac{\partial M}{\partial x}+\frac{h}{p}\frac{\partial p}{\partial x}+\frac{g}{a}\frac{\partial a}{\partial x}\\ G^{\prime }=\frac{Msin\theta +msin\varepsilon }{M\text{cos}\theta +m\text{cos}\varepsilon }\end{array}$

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Formulation and Solution Procedure

Singiresu S. Rao , in The Finite Element Method in Engineering (Sixth Edition), 2018

13.3 Governing Equation for Three-Dimensional Bodies

Consider a small element of material in a solid body as shown in Fig. 13.1. The element is in the shape of a rectangular parallelepiped with sides dx, dy, and dz. The energy balance equation can be stated as follows [13.1]:

Figure 13.1. An element in Cartesian coordinates.

(13.7) $\begin{array}{llll}\text{Heat}\text{inflow}& \text{Heat}\text{generated}& \text{Heat}\text{outflow}& \text{Change}\text{in}\\ \text{during}\text{time}\text{d}t& +\text{by}\text{internal}& =\text{during}\text{d}t& +\text{internal}\\ & \text{sources}\text{during}\text{d}t& & \text{energy}\text{during}\text{d}t\end{array}$

With the help of rate equations, Eq. (13.7) can be expressed as

(13.8) $(q_x+q_y+q_z)\text{d}t+\dot{q}\text{d}x\text{d}y\text{d}z\text{d}t=(q_{x+\text{d}x}+q_{y+\text{d}y}+q_{z+\text{d}z})\text{d}t+\rho c\text{d}T\text{d}x\text{d}y\text{d}z$

where

(13.9) $\begin{array}{c}{q}_x=\text{heat}\text{inflow}\text{rate}\text{into}\text{the}\text{face}\text{located}\text{at}x\\ =-k_x{A}_x\frac{\partial T}{\partial x}=-k_x\frac{\partial T}{\partial x}\text{d}y\text{d}z\end{array}$

(13.10) $\begin{array}{c}{q}_{x+\text{d}x}=\text{heat}\text{outflow}\text{rate}\text{from}\text{the}\text{face}\text{located}\text{at}x+\text{d}x\\ =q{|}_{x+\text{d}x}\approx q_x+\frac{\partial q_x}{\partial x}\text{d}x\\ =-k_x{A}_x\frac{\partial T}{\partial x}-\frac{\partial }{\partial x}\left(k_x{A}_x\frac{\partial T}{\partial x}\right)\text{d}x\\ =-k_x\frac{\partial T}{\partial x}\text{d}y\text{d}z-\frac{\partial }{\partial x}\left(k_x\frac{\partial T}{\partial x}\right)\text{d}x\text{d}y\text{d}z\end{array}$

k _x is the thermal conductivity of the material in x direction, A _x is the area normal to the x direction through which heat flows   =   dy dz, T is the temperature, $\dot{q}$ is the rate of heat generated per unit volume (per unit time), $\rho$ is the density of the material, and c is the specific heat of the material. By substituting Eqs. (13.9) and (13.10) and similar expressions for q _y , q _y+dy , q _z , and q _z+dz into Eq. (13.8) and dividing each term by dx dy dz dt, we obtain

(13.11) $\frac{\partial }{\partial x}\left(k_x\frac{\partial T}{\partial x}\right)+\frac{\partial }{\partial y}\left(k_y\frac{\partial T}{\partial y}\right)+\frac{\partial }{\partial z}\left(k_z\frac{\partial T}{\partial z}\right)+\dot{q}=\rho c\frac{\partial T}{\partial t}$

Eq. (13.11) is the differential equation governing the heat conduction in an orthotropic solid body. If the thermal conductivities in x, y, and z directions are assumed to be the same, k _x   = k _y   = k _z   = k  =   constant, Eq. (13.11) can be written as

(13.12) $\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}+\frac{{\partial }^{2}T}{\partial z^2}+\frac{\dot{q}}{k}=\frac{1}{\alpha }\frac{\partial T}{\partial t}$

where the constant $\alpha =\left(k/\rho c\right)$ is called the thermal diffusivity. Eq. (13.12) is the heat conduction equation that governs the temperature distribution and the conduction heat flow in a solid having uniform material properties (isotropic body). If heat sources are absent in the body, Eq. (13.12) reduces to the Fourier equation

(13.13) $\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}+\frac{{\partial }^{2}T}{\partial z^2}=\frac{1}{\alpha }\frac{\partial T}{\partial t}$

If the body is in a steady state (with heat sources present), Eq. (13.12) becomes the Poisson's equation

(13.14) $\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}+\frac{{\partial }^{2}T}{\partial z^2}+\frac{\dot{q}}{k}=0$

If the body is in a steady state without any heat sources, Eq. (13.12) reduces to the Laplace equation

(13.15) $\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}+\frac{{\partial }^{2}T}{\partial z^2}=0$

13.3.1 Governing Equation in Cylindrical Coordinate System

If a cylindrical coordinate system (with r, $\varphi$ , z coordinates) is used instead of the Cartesian x, y, z system, Eq. (13.12) takes the form

(13.16) $\frac{{\partial }^{2}T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}T}{\partial {\varphi }^2}+\frac{{\partial }^{2}T}{\partial z^2}+\frac{\dot{q}}{k}=\frac{1}{\alpha }\frac{\partial T}{\partial t}$

This equation can be derived by taking the element of the body as shown in Fig. 13.2.

Figure 13.2. An element in cylindrical coordinates.

13.3.2 Governing Equation in Spherical Coordinate System

By considering an element of the body in a spherical r, $\varphi$ , $\psi$ coordinate system as indicated in Fig. 13.3, the general heat conduction Eq. (13.12) becomes

Figure 13.3. An element in spherical coordinates.

(13.17) $\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial T}{\partial r}\right)+\frac{1}{r^2·\sin \varphi }·\frac{\partial }{\partial \varphi }\left(\sin \varphi ·\frac{\partial T}{\partial \varphi }\right)+\frac{1}{r^2·{\sin }^2\varphi }·\frac{{\partial }^{2}T}{\partial {\psi }^2}+\frac{\dot{q}}{k}=\frac{1}{\alpha }\frac{\partial T}{\partial t}$

13.3.3 Boundary and Initial Conditions

Since the differential equation—Eq. (13.11) or (13.12)—is second order, two boundary conditions need to be specified. The possible boundary conditions are

(13.18) $T\left(x,y,z,t\right)=T_0\text{for}t>0\text{on}{S}_1$

(13.19) $k_x·\frac{\partial T}{\partial x}·l_x+k_y·\frac{\partial T}{\partial y}·l_y+k_z·\frac{\partial T}{\partial z}·l_z+q_0=0\text{for}t>0\text{on}{S}_2$

(13.20) $k_x·\frac{\partial T}{\partial x}·l_x+k_y·\frac{\partial T}{\partial y}·l_y+k_z·\frac{\partial T}{\partial z}·l_z+h\left(T-T_{\infty }\right)=0\text{for}t>0\text{on}{S}_3$

where q ₀ is the boundary heat flux, h is the convection heat transfer coefficient, T _∞ is the surrounding temperature, and l _x , l _y , l _z are the direction cosines of the outward drawn normal to the boundary.

Eq. (13.18) indicates that the temperature is specified as T ₀ (as T ₀(t) in an unsteady state problem) on the surface S ₁. This boundary condition is applicable, for example, when the surface is in contact with a melting solid or a boiling liquid. There will be heat transfer in both these cases and the surface remains at the temperature of the phase change process. Eq. (13.19) represents the existence of a fixed or constant heat flux q ₀ at the surface S ₂. The equation basically states that the heat flux q ₀ is related to the temperature gradient at the surface by Fourier's law. This boundary condition is realized when a thin film or patch electric heater is attached or bonded to the surface. A special case of this boundary condition corresponds to the perfectly insulated or adiabatic surface for which the temperature gradient is zero. Eq. (13.20) denotes the existence of convection heat transfer (heating or cooling of the body) at the surface S ₃ of the body. Eq. (13.20) represents energy balance at the surface. This boundary condition is realized when cold (or hot) air flows around the hot (or cold) surface. The air may be blown by a fan to increase the convection heat transfer. The boundary condition stated in Eq. (13.18) is known as the Dirichlet condition and those stated in Eqs. (13.19) and (13.20) are called Neumann conditions.

Furthermore, the differential equation, Eq. (13.11) or (13.12), is first-order in time t, and hence it requires one initial condition. The commonly used initial condition is

(13.21) $T\left(x,y,z,t=0\right)={\overline{T}}_0\left(x,y,z\right)\text{in}V$

where V is the domain (or volume) of the solid body and ${\overline{T}}_0$ is the specified temperature distribution at time zero.

Example 13.1

A 20-cm thick wall of an industrial furnace is constructed using fireclay bricks that have a thermal conductivity of k  =   2   W/m °C. During steady state operation, the furnace wall has a temperature of 800 °C on the inside and 300 °C on the outside. If one of the walls of the furnace has a surface area of 2   m² (with 20-cm thickness), find the rate of heat transfer and rate of heat loss through the wall.

Solution

Assuming that the heat loss is due to conduction only, the rate of heat transfer (heat flowing through a unit surface area of the wall) is given by

$q=k\frac{dT}{dx}\approx k\frac{\Delta T}{\Delta x}=2.0\left(\frac{800-300}{0.2}\right)=5000\text{W}/{\text{m}}^2$

The rate of heat loss can be determined as follows:

$\begin{array}{c}\text{Rate}\text{of}\text{heat}\text{loss}=\text{Rate}\text{of}\text{heat}\text{transfer}\times \text{Surface}\text{area}\text{of}\text{the}\text{wall}\text{through}\text{which}\text{heat}\text{flows}\\ =\left(5000\right)\left(2\right)=\mathrm{10,000}\text{W}\end{array}$

Example 13.2

A metal pipe of 10 cm outer diameter carrying steam passes through a room. The walls and the air in the room are at a temperature of 20 °C while the outer surface of the pipe is at a temperature of 250 °C. If the heat transfer coefficient for free convection from the pipe to the air is h  =   20   W/m² °C find the rate of heat loss from the pipe.

Solution

The convection heat loss from the pipe of unit length (l  =   1   m) is given by

(E.1) $q=hPl\left(T-T_{\infty }\right)$

where P is the perimeter of the pipe cross section, l is the length of the pipe, T is the temperature of the outer surface of the pipe, and $T_{\infty }$ is the temperature of the air. Using the data given, Eq. (E.1) gives:

Heat loss per 1-m length of pipe (q)   =   20 {π (0.1)} (1) (250–20)   =   1445.1360   W.

Example 13.3

A metal pipe of 10-cm outer diameter carrying steam passes through a room. The walls and the air in the room are at a temperature of 20 °C while the outer surface of the pipe is at a temperature of 250 °C. If the emissivity of the outer surface of the pipe is 0.75, determine the rate of heat loss from the pipe to the surrounding air and walls of the room by radiation.

Solution

The radiation heat loss (q) from the pipe, of unit length (l  =   1   m), is given by

(E.1) $q=\sigma \epsilon A\left(T^4-T_{\infty }^4\right)$

where σ  =   5.7   ×   10–8   W/m² °K⁴, ε   =   0.75, A  =   surface area from which radiation heat transfer occurs   = Pl  = π Dl  = π(0.1) (1)   =   0.1 π  m², T  =   250 °C   =   523 °K, and $T_{\infty }$   =   20 °C   =   293 °K. Thus, Eq. (E.1) gives

$\text{Heat}\text{loss}\text{per}\text{unit}\text{length}\text{of}\text{pipe}\left(q\right)=5.7\times {10}^{-8}\left(0.75\right)\left(0.1\pi \right)({523}^4-{293}^4)=905.8496\text{W}$

Example 13.4

Compute the thermal diffusivity of the following building materials with given values of thermal conductivity (k), density ( $\rho$ ), and specific heat (c _p ):

Material	Thermal Conductivity, k W/m °C	Density, $\rho$ Kg/m3	Specific Heat, c p J/kg °C
Plywood	0.12	540	1215
Brick	0.70	1920	835
Gypsum or plaster board	0.15	900	1080

Solution

The thermal diffusivity (α) is defined as

(E.1) $\alpha =\frac{k}{\rho c_p}$

From the known data, the thermal diffusivities of the indicated materials can be determined from Eq. (E.1) as follows:

$\text{Playwood}:\alpha =\frac{0.12}{\left(540\right)\left(1215\right)}=183\times {10}^{-9}{\text{m}}^2/\text{s}$

$\text{Brick}:\alpha =\frac{0.70}{\left(1920\right)\left(835\right)}=437\times {10}^{-9}{\text{m}}^2/\text{s}$

$\text{Gypsum}\text{or}\text{plaster}\text{board}:\alpha =\frac{0.15}{\left(900\right)\left(1080\right)}=154\times {10}^{-9}{\text{m}}^2/\text{s}$

Example 13.5

In an automobile radiator, the coolant (assumed to be water) enters at 170 °F and leaves at 130 °F with a flow rate of 10   gallons per minute. If the surrounding air is at 75 °F, determine the surface area of the radiator. Assume the following data: Convection heat transfer coefficient   =   12 BTU/(ft²-hr °F), Specific heat of water at 140 °F   =   1.0245 BTU/lb_m °F, Density of water at 140 °F: 61.1794 lb_m/ft³.

Solution

Assuming steady state, the heat lost through water must be equal to the heat gain by the surrounding air. Thus

(E.1) $m_w{c}_p(T_{in}-T_{out})=h{A}_{rad}(T_w-T_{\infty })$

where $m_w$ denotes the mass flow rate of water given by

$m_w=\text{volume}\text{flow}\text{rate}\times \text{density}\text{of}\text{water}=\frac{10\left(0.1337\right)}{60}\left(61\text{.}1794\right)=1.3631\text{lbm/s}$

Using the relation: 1   gallon   =   231 in³  =   0.1337   ft³. Eq. (E.1) gives, for 1   h,

(E.2) $(1.3631\times 3600)\left(1.0245\right)(170-130)=12A_{\text{rad}}(150-75)$

where the average temperature of water is used for $T_w$ as

(E.3) $T_w=\frac{T_{\text{IN}}+t_{\text{OUT}}}{2}=\frac{170+130}{2}=150°F$

The solution of Eq. (E.2) gives the surface area of the radiator as $A_{\text{rad}}=223.4393{\text{ft}}^2$ .

Note

The large area required for the radiator surface area is accommodated within the available space of the radiator by using a very large number of fins, each fin designed to provide some of the heat exchanger surface area.

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Onset of Rayleigh-Bénard Convection in Porous Bodies

P.A. Tyvand , in Transport Phenomena in Porous Media II, 2002

4.5.1 Genera] linearized equations in cylindrical coordinates

We introduce the cylindrical coordinate system ( x, r, ϕ) for a porous cylinder with axis along the x-direction. The transformations between Cartesian and cylindrical coordinates are as follows:

(4.64) $\left(x,y,z\right)=\left(x,r\cos \varphi ,r\sin \varphi \right)$

and the velocity components are denoted by (u, v_r , ν_ϕ ).

After eliminating the pressure, the governing equations (4.7) – (4.9) can be expressed as follows:

(4.65) $\frac{\partial }{\partial r}\left(r{v}_{\varphi }\right)-\frac{\partial v_r}{\partial \varphi }=Ra\left(r\frac{\partial \mathrm{\Theta }}{\partial r}\cos \varphi -\frac{\partial \mathrm{\Theta }}{\partial \varphi }\sin \varphi \right)\text{,}$

(4.66) $\frac{1}{r}\frac{\partial u}{\partial \varphi }-\frac{\partial v_{\varphi }}{\partial x}=-Ra\cos \varphi \frac{\partial \mathrm{\Theta }}{\partial x}\text{,}$

(4.67) $\frac{\partial v_r}{\partial x}-\frac{\partial u}{\partial r}=Ra\sin \varphi \frac{\partial \mathrm{\Theta }}{\partial x}\text{,}$

(4.68) $\frac{\partial u}{\partial x}+\frac{1}{r}\frac{\partial }{\partial r}\left(r{v}_r\right)+\frac{1}{r}\frac{\partial v_{\varphi }}{\partial \varphi }=0\text{,}$

(4.69) $v_r\sin \varphi +v_{\varphi }\cos \varphi +\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \mathrm{\Theta }}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2\mathrm{\Theta }}{\partial {\varphi }^2}+\frac{{\partial }^2\mathrm{\Theta }}{\partial x^2}=0\text{.}$

In order to solve the challenging problem of zero perturbation temperature along all boundaries, including the end walls, one would have to start with these general equations.

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Torsion in Structures

Chai H. Yoo , Sung C. Lee , in Stability of Structures, 2011

5.5.1 Elastic Torque

Based on the cylindrical coordinate system shown in Fig. 5-6, the equation for the membrane is given by

Figure 5-6.

(a) $z=z_0\frac{r^2}{R^2}$

Hence, the equation of the dome under the membrane is

(b) $z=z_0-z_0\frac{r^2}{R^2}=z_0\left(1-\frac{r^2}{R^2}\right)$

$dV=rd\theta dzdr$

$V={{\int }_0^{2\pi }{\int }_0^R{\int }_0^{z_0}}^{\left(1-\frac{r^2}{R^2}\right)}dzdrrd\theta =z_0{\int }_0^{2\pi }d\theta {\int }_0^R\left(r-\frac{r^3}{R^2}\right)dr=\frac{z_0{R}^2}{4}{\int }_0^{2\pi }d\theta =\frac{z_0{R}^2}{4}2\pi =\frac{\pi R^2{z}_0}{2}$

(c) $M_{ze}=T_e=2V=\pi R^2{z}_0$

(d) $\frac{dz}{dr}{|}_{r=R}={\tau }_{\max }=\frac{2z_0}{R}=\tan \alpha$

Equilibrium

$\sum F_z=0\Rightarrow q\pi R^2=F(2\pi R)\sin \alpha =2\pi RF\frac{2z_0}{R}=4\pi z_{0}F$

$\frac{q}{F}=\frac{4z_0}{R^2}\left(\mathrm{c}\mathrm{f}.\frac{q}{F}=2G{\theta }^{\prime }=\frac{4z_0}{R^2}\right)$

(e) $z_0=\frac{G{\theta }^{\prime }{R}^2}{2}$

Substituting Eq. (e) into Eq. (c), one gets

(f) $M_{ze}=\pi R^2\frac{G{\theta }^{\prime }{R}^2}{2}=\frac{\pi R^{4}G{\theta }^{\prime }}{2}$

Recalling the polar moment of inertia (J) or the St. Venant torsional constant ( $K_T$ ) of a solid circle is $\pi R^4/2$ , the elastic twisting moment of a circular shaft is given by

(5.5.1) $M_{ze}=GJ{\theta }^{\prime }=G{K}_T{\theta }^{\prime }$

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Evaluation of Lennard-Jones Potential Fields

Duangkamon Baowan , ... Ngamta Thamwattana , in Modelling and Mechanics of Carbon-Based Nanostructured Materials, 2017

3.4.1 Cylindrical Coordinate System (r, θ, z)

We employ the usual cylindrical coordinate system ( r, θ, z). As usual, θ is measured in an anticlockwise direction from the positive x axis, as shown in Fig. 3.10, and therefore to map from cylindrical to Cartesian coordinates, we have

Fig. 3.10. A general point P described by cylindrical coordinates (r, θ, z).

(3.26) $\begin{array}{l}\hfill x=rcos\theta ,y=rsin\theta ,\end{array}$

and the z-coordinate is unchanged. The inverse mapping is given by

$\begin{array}{l}\hfill r=\sqrt{x^2+y^2},\theta =arctan\frac{y}{x}.\end{array}$

The vector line element in cylindrical coordinates is given by

$\begin{array}{l}\hfill d\mathbf{s}=dr\widehat{\mathbf{r}}+rd\theta \widehat{\theta }+dz\widehat{\mathbf{z}},\end{array}$

where $\widehat{\mathbf{r}}$ , $\widehat{\theta }$ , and $\widehat{\mathbf{z}}$ denote the unit vectors for the cylindrical coordinate system. Similarly, the vector area element is given by

$\begin{array}{l}\hfill d\mathbf{S}=rd\theta dz\widehat{\mathbf{r}}.\end{array}$

In this book, we often consider integrals of the Lennard-Jones potential function Φ(ρ) over the surface of a cylinder of radius b of the form

$\begin{array}{l}\hfill I=b{\int }_{z_1}^{z_2}{\int }_{-\pi }^{\pi }\frac{1}{{\rho }^{2n}}d\theta dz,\end{array}$

where ρ is some function of the geometric parameters of the problem, which will generally also be a function of θ and z.

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Direct Numerical Simulation of Rotating Turbulent Flows Through Concentric Annuli

M. Okamoto , N. Shima , in Engineering Turbulence Modelling and Experiments 6, 2005

NUMERICAL PROCEDURE AND FUNDAMENTAL EQUATION

The flow configuration and cylindrical coordinate system are given in Figure 1. The numerical scheme of the present DNS is the second-order center difference and interpolation, and the pressure Poisson equation is solved by the direct method with the fast Fourier transformation. The time advancement is done by the second-order Adams-Bashforth method. The computational domain size is 14δ  ×   2δ  ×   2π in the axial, radial and azimuthal directions, respectively. The variables are nondimensionalized by the radial half width δ and the axial global friction velocity u_τg defined by ${\mathit{u}}_{\mathit{\tau g}}=\sqrt{-\mathit{\delta dP}/\mathit{dx}}$ and the corresponding intrinsic dimensionless parameters in the present flow are the Reynolds number

Figure 1:. Flow configuration and coordinate system.

(1) $\mathrm{Re}=\frac{\mathit{\delta }{\mathit{u}}_{\mathit{\tau g}}}{\mathit{v}},$

the rotation number by the rotating wall velocity W ₀

(2) $\mathit{N}=\frac{{\mathit{W}}_0}{{{\mathit{u}}_{\mathit{\tau }}}_{\mathit{g}}},$

and the radius ratios

(3) $\mathit{\alpha }=\frac{{\mathit{R}}_{\mathit{in}}}{{\mathit{R}}_{\mathit{out}}}.$

In the present DNS, the fixed Re is set to 150, N is 0, 5 and 10 and α is 0.05, 0.1, 0.2 and 0.5. The computational mesh is listed in TABLE 1. In order to resolve all essential scales of the turbulence motion, the axial grid for the inner-wall rotation case and azimuthal one become sufficiently fine. The axial grid-resolution Δx ⁺ near the inner wall, which is normalized by the local friction velocity, is 6.2   ~   26, while Δx ⁺ near the outer wall is 3.9   ~   16. The radial one Δr ⁺ near the inner wall is 0.39   ~   0.70, while that near the outer wall is about 0.36. The azimuthal one (R_in Δθ) ⁺ is 1.0   ~   9.0 and (R_out Δθ)⁺ is 14.1   ~   18.1.

TABLE 1. Computational Mesh

Outer-Wall Rotation					Inner-Wall Rotation
α	N	x	r	θ	α	N	x	r	θ
0.05	0	128	128	256	0.05	0	128	128	256
	5	128	128	256		5	256	128	256
	10	128	128	256		10	512	128	256
0.1	0	128	128	256	0.1	0	128	128	256
	5	128	128	256		5	256	128	256
	10	128	128	256		10	512	128	256
0.2	0	128	128	256	0.2	0	128	128	256
	5	128	128	256		5	256	128	256
	10	128	128	256		10	512	128	256
0.5	0	128	128	512	0.5	0	128	128	512
	5	128	128	512		5	256	128	512
	10	128	128	512		10	256	128	512

The governing Navier-Stokes equations with the incompressible condition in the cylindrical coordinate system are expressed as

(4) $\frac{\partial \mathit{u}}{\partial \mathit{t}}=-\mathit{u}\frac{\partial \mathit{u}}{\partial \mathit{x}}-\mathit{v}\frac{\partial \mathit{u}}{\partial \mathit{r}}-\frac{1}{\mathit{r}}\mathit{w}\frac{\partial \mathit{u}}{\partial \mathit{\theta }}-\frac{\partial \mathit{p}}{\partial \mathit{x}}+\mathit{\nu }\frac{{\partial }^2\mathit{u}}{\partial {\mathit{x}}^2}+\frac{\mathit{\nu }}{\mathit{r}}\frac{\partial }{\partial \mathit{r}}\left(\mathit{r}\frac{\partial \mathit{u}}{\partial \mathit{r}}\right)+\frac{\mathit{\nu }}{{\mathit{r}}^2}\frac{{\partial }^2\mathit{u}}{\partial {\mathit{\theta }}^2},$

(5) $\frac{\partial \mathit{v}}{\partial \mathit{t}}=-\mathit{u}\frac{\partial \mathit{v}}{\partial \mathit{x}}-\mathit{v}\frac{\partial \mathit{v}}{\partial \mathit{r}}-\frac{1}{\mathit{r}}\mathit{w}\frac{\partial \mathit{v}}{\partial \mathit{\theta }}+\frac{{\mathit{w}}^2}{\mathit{r}}-\frac{\partial \mathit{p}}{\partial \mathit{r}}+\mathit{\nu }\frac{{\partial }^2\mathit{v}}{\partial {\mathit{x}}^2}+\frac{\mathit{\nu }}{\mathit{r}}\frac{\partial }{\partial \mathit{r}}\left(\mathit{r}\frac{\partial \mathit{v}}{\partial \mathit{r}}\right)+\frac{\mathit{\nu }}{{\mathit{r}}^2}\frac{{\partial }^2\mathit{v}}{\partial {\mathit{\theta }}^2}-\frac{\mathit{\nu v}}{{\mathit{r}}^2}-\frac{2\mathit{\nu }}{{\mathit{r}}^2}\frac{\partial \mathit{w}}{\partial \mathit{\theta }},$

(6) $\frac{\partial \mathit{w}}{\partial \mathit{t}}=-\mathit{u}\frac{\partial \mathit{w}}{\partial \mathit{x}}-\mathit{v}\frac{\partial \mathit{w}}{\partial \mathit{r}}-\frac{1}{\mathit{r}}\mathit{w}\frac{\partial \mathit{w}}{\partial \mathit{\theta }}-\frac{\mathit{vw}}{\mathit{r}}-\frac{1}{\mathit{r}}\frac{\partial \mathit{p}}{\partial \mathit{\theta }}+\mathit{\nu }\frac{{\partial }^2\mathit{w}}{\partial {\mathit{x}}^2}+\frac{\mathit{\nu }}{\mathit{r}}\frac{\partial }{\partial \mathit{r}}\left(\mathit{r}\frac{\partial \mathit{w}}{\partial \mathit{r}}\right)+\frac{\mathit{\nu }}{{\mathit{r}}^2}\frac{{\partial }^2\mathit{w}}{\partial {\mathit{\theta }}^2}-\frac{\mathit{\nu w}}{{\mathit{r}}^2}+\frac{2\mathit{\nu }}{{\mathit{r}}^2}\frac{\partial \mathit{v}}{\partial \mathit{\theta }},$

(7) $\frac{\partial \mathit{u}}{\partial \mathit{x}}+\frac{1}{\mathit{r}}\frac{\partial \mathit{rv}}{\partial \mathit{r}}+\frac{1}{\mathit{r}}\frac{\partial \mathit{w}}{\partial \mathit{\theta }}=0.$

The velocity components u, v, w are the axial, radial and azimuthal velocities, respectively. In the fully-developed turbulent flow, the mean quantities are stationary and homogeneous in the axial and azimuthal directions. The equations of the mean velocity fields are written by

(8) $0=-\frac{\mathit{dP}}{\mathit{dx}}+\frac{1}{\mathit{r}}\frac{\partial }{\partial \mathit{r}}\left\{\mathit{r}\left(\mathit{\nu }\frac{\partial \mathit{U}}{\partial \mathit{r}}-\overline{\mathit{u}\prime \mathit{v}\prime }\right)\right\},$

(9) $0=\frac{1}{{\mathit{r}}^2}\frac{\partial }{\partial \mathit{r}}\left\{{\mathit{r}}^2\left(\mathit{\nu r}\frac{\partial }{\partial \mathit{r}}\left(\frac{\mathit{W}}{\mathit{r}}\right)-\overline{\mathit{v}\prime \mathit{w}\prime }\right)\right\}.$

Here, $\overline{\mathit{u}\prime \mathit{v}\prime }$ and $\overline{\mathit{v}\prime \mathit{w}\prime }$ are the shear components of the Reynolds stress. In the case of the laminar flow vanishing the Reynolds stress in (8) and (9), the solution of U is

(10) $\mathit{U}=\frac{{\mathit{R}}_{\mathit{out}}^2}{4\mathit{\nu }}\frac{\mathit{dP}}{\mathit{dx}}\left\{{\left(\frac{\mathit{r}}{{\mathit{R}}_{\mathit{out}}}\right)}^2-1+\frac{1-{\mathit{\alpha }}^2}{log\mathit{\alpha }}log\frac{\mathit{r}}{{\mathit{R}}_{\mathit{out}}}\right\},$

and those of W are

(11) $\mathit{W}=\frac{{\mathit{W}}_0}{1-{\mathit{\alpha }}^2}\left(\frac{\mathit{r}}{{\mathit{R}}_{\mathit{out}}}-{\mathit{\alpha }}^2\frac{{\mathit{R}}_{\mathit{out}}}{\mathit{r}}\right),$

in the outer-wall rotation and

(12) $\mathit{W}=\frac{\mathit{\alpha }{\mathit{W}}_0}{1-{\mathit{\alpha }}^2}\left(\frac{{\mathit{R}}_{\mathit{out}}}{\mathit{r}}-\frac{\mathit{r}}{{\mathit{R}}_{\mathit{out}}}\right),$

in the inner-wall rotation. The velocities U and W in the laminar rotating turbulent flow through concentric annuli are independent of each other. When α is small, the first linear term in Eqn.11 is dominant and the outer-wall rotation in the laminar flow produces an approximately forced vortex. On the other hand, the laminar solution in the inner-wall rotation (12) almost represents a free vortex near the inner wall.

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