김경종
526 lines
23 KiB
Markdown
526 lines
23 KiB
Markdown
<!-- source-page: 551 -->
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<details>
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<summary>text_image</summary>
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a
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a/2
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B
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x
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b
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A
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y
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</details>
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Figure P12–9
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<details>
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<summary>text_image</summary>
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p
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r
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r
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z
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</details>
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Figure P12–10
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12.11 A square steel plate 2 m by 2 m and 10 mm thick at the bottom of a tank must support salt water at a height of 3 m, as shown in Figure P12–11. Assume the plate to be built in (fixed all around). The plate allowable stress is 100 MPa. Let $E = 2 0 0 \mathrm { G P a } , \nu = 0 . 3$ for the steel properties. The weight density of salt water is $1 0 . 0 5 4 ~ \mathrm { k N } / \mathrm { m } ^ { 3 }$ . Determine the maximum principal stress in the plate and compare to the yield strength.
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12.12 A stockroom floor carries a uniform load of $p = 8 0 ~ \mathrm { { l b / f t ^ { 2 } } }$ over half the floor as shown in Figure P12–12. The floor has opposite edges clamped and remaining edges and midspan simply supported. The dimensions are 10 ft by 20 ft. The floor thickness is 6 in. The floor is made of reinforced concrete with $E = 3 \times 1 0 ^ { 6 }$ psi and $\nu = 0 . 2 5$ . Determine the maximum deflection and maximum principal stress in the floor.
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<details>
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<summary>text_image</summary>
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▽
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3 m
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2 m
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2 m
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</details>
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Figure P12–11
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<details>
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<summary>text_image</summary>
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p
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10 ft
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10 ft
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x
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z
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</details>
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<details>
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<summary>text_image</summary>
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10 ft
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y
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x
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</details>
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Figure P12–12
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<!-- source-page: 552 -->
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# Heat Transfer and Mass Transport
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# Introduction
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In this chapter, we present the first use in this text of the finite element method for solution of nonstructural problems. We first consider the heat-transfer problem, although many similar problems, such as seepage through porous media, torsion of shafts, and magnetostatics [3], can also be treated by the same form of equations (but with different physical characteristics) as that for heat transfer.
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Familiarity with the heat-transfer problem makes possible determination of the temperature distribution within a body. We can then determine the amount of heat moving into or out of the body and the thermal stresses.
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We begin with a derivation of the basic differential equation for heat conduction in one dimension and then extend this derivation to the two-dimensional case. We will then review the units used for the physical quantities involved in heat transfer.
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In preceding chapters dealing with stress analysis, we used the principle of minimum potential energy to derive the element equations, where an assumed displacement function within each element was used as a starting point in the derivation. We will now use a similar procedure for the nonstructural heat-transfer problem. We define an assumed temperature function within each element. Instead of minimizing a potential energy functional, we minimize a similar functional to obtain the element equations. Matrices analogous to the stiffness and force matrices of the structural problem result.
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We will consider one-, two-, and three-dimensional finite element formulations of the heat-transfer problem and provide illustrative examples of the determination of the temperature distribution along the length of a rod and within a twodimensional body and show some three-dimensional heat transfer examples as well.
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Next, we will consider the contribution of fluid mass transport. The onedimensional mass-transport phenomenon is included in the basic heat-transfer differential equation. Because it is not readily apparent that a variational formulation is possible for this problem, we will apply Galerkin’s residual method directly to the
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<!-- source-page: 553 -->
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differential equation to obtain the finite element equations. (You should note that the mass transport stiffness matrix is asymmetric.) We will compare an analytical solution to the finite element solution for a heat exchanger design/analysis problem to show the excellent agreement.
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Finally, we will present some computer program results for two-dimensional heat transfer.
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# 13.1 Derivation of the Basic Differential Equation
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# One-Dimensional Heat Conduction (without Convection)
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We now consider the derivation of the basic differential equation for the onedimensional problem of heat conduction without convection. The purpose of this derivation is to present a physical insight into the heat-transfer phenomena, which must be understood so that the finite element formulation of the problem can be fully understood. (For additional information on heat transfer, consult texts such as References [1] and [2].) We begin with the control volume shown in Figure 13–1. By conservation of energy, we have
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$$
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E _ {\text { in }} + E _ {\text { generated }} = \Delta U + E _ {\text { out }} \tag {13.1.1}
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$$
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or $q _ { x } A d t + Q A d x d t = \Delta U + q _ { x + d x } A d t$ ð13:1:2Þ
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where
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$E _ { \mathrm { i n } }$ is the energy entering the control volume, in units of joules (J) or kW h or Btu.
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$\Delta U$ is the change in stored energy, in units of kW h (kWh) or Btu.
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$q _ { x }$ is the heat conducted (heat flux) into the control volume at surface edge x, in units of $\mathrm { k W } / \mathrm { m } ^ { 2 }$ or $\mathrm { B t u } / ( \mathrm { h } \mathrm { - } \mathrm { f t } ^ { 2 } )$ .
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$q _ { x + d x }$ is the heat conducted out of the control volume at the surface edge $x + d x$ .
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t is time, in h or s (in U. S. customary units) or s (in SI units).
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<details>
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<summary>text_image</summary>
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Insulated boundary
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qₓ → Q → qₓ + dx
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Insulated boundary
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dx
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</details>
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Figure 13–1 Control volume for one-dimensional heat conduction
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<!-- source-page: 554 -->
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Q is the internal heat source (heat generated per unit time per unit volume is positive), in $\mathrm { k W } / \mathrm { m } ^ { 3 }$ or $\mathrm { B t u } / ( \mathrm { h } \mathrm { - } \mathrm { f t } ^ { 3 } )$ (a heat sink, heat drawn out of the volume, is negative).
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A is the cross-sectional area perpendicular to heat flow q, in $\mathrm { m } ^ { 2 }$ or $\mathrm { f t } ^ { 2 }$ .
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By Fourier’s law of heat conduction,
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$$
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q _ {x} = - K _ {x x} \frac {d T}{d x} \tag {13.1.3}
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$$
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where
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$K _ { x x }$ is the thermal conductivity in the x direction, in $\mathrm { k W } / ( \mathrm { m } \cdot { } ^ { \circ } \mathrm { C } )$ or Btu/(h-ft-F).
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T is the temperature, in C or F.
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dT=dx is the temperature gradient, in C/m or F/ft.
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Equation (13.1.3) states that the heat flux in the x direction is proportional to the gradient of temperature in the x direction. The minus sign in Eq. (13.1.3) implies that, by convention, heat flow is positive in the direction opposite the direction of temperature increase. Equation (13.1.3) is analogous to the one-dimensional stress/strain law for the stress analysis problem—that is, to $\sigma _ { x } = E ( d u / d x )$ . Similarly,
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$$
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q _ {x + d x} = - K _ {x x} \frac {d T}{d x} \bigg | _ {x + d x} \tag {13.1.4}
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$$
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where the gradient in Eq. (13.1.4) is evaluated at x þ dx. By Taylor series expansion, for any general function f ðxÞ, we have
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$$
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f _ {x + d x} = f _ {x} + \frac {d f}{d x} d x + \frac {d ^ {2} f}{d x ^ {2}} \frac {d x ^ {2}}{2} + \dots
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$$
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Therefore, using a two-term Taylor series, Eq. (13.1.4) becomes
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$$
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q _ {x + d x} = - \left[ K _ {x x} \frac {d T}{d x} + \frac {d}{d x} \left(K _ {x x} \frac {d T}{d x}\right) d x \right] \tag {13.1.5}
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$$
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The change in stored energy can be expressed by
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$$
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\Delta U = \text { specific heat } \times \text { mass } \times \text { change in temperature }
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$$
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$$
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= c (\rho A d x) d T \tag {13.1.6}
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$$
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where c is the specific heat in $\mathbf { k W } \cdot \mathbf { h } / ( \mathbf { k g } \cdot { ^ { \circ } C } )$ or $\mathrm { B t u / ( s l u g – ^ { \circ } F ) }$ , and $\rho$ is the mass density in $\mathrm { k g } / \mathrm { m } ^ { 3 }$ or slug/ft3. On substituting Eqs. (13.1.3), (13.1.5), and (13.1.6) into Eq. (13.1.2), dividing Eq. (13.1.2) by A dx dt, and simplifying, we have the onedimensional heat conduction equation as
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$$
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\frac {\partial}{\partial x} \left(K _ {x x} \frac {\partial T}{\partial x}\right) + Q = \rho c \frac {\partial T}{\partial t} \tag {13.1.7}
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$$
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For steady state, any differentiation with respect to time is equal to zero, so Eq. (13.1.7) becomes
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$$
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\frac {d}{d x} \left(K _ {x x} \frac {d T}{d x}\right) + Q = 0 \tag {13.1.8}
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$$
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<!-- source-page: 555 -->
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<details>
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<summary>text_image</summary>
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T = T_B
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S_1
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L
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q_x = +q_x^*
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x
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S_2
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</details>
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<details>
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<summary>text_image</summary>
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qₓ* = 0
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(insulated)
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L
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T = T_B
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S₁
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</details>
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Figure 13–2 Examples of boundary conditions in one-dimensional heat conduction
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For constant thermal conductivity and steady state, Eq. (13.1.7) becomes
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$$
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K _ {x x} \frac {d ^ {2} T}{d x ^ {2}} + Q = 0 \tag {13.1.9}
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$$
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The boundary conditions are of the form
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$$
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T = T _ {B} \quad \text { on } S _ {1} \tag {13.1.10}
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$$
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where $T _ { B }$ represents a known boundary temperature and $S _ { 1 }$ is a surface where the temperature is known, and
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$$
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q _ {x} ^ {*} = - K _ {x x} \frac {d T}{d x} = \text { constant } \quad \text { on } S _ {2} \tag {13.1.11}
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$$
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where $S _ { 2 }$ is a surface where the prescribed heat flux $q _ { x } ^ { * }$ or temperature gradient is known. On an insulated boundary, $q _ { x } ^ { * } = 0$ . These different boundary conditions are shown in Figure 13–2, where by sign convention, positive $q _ { x } ^ { * }$ occurs when heat is flowing into the body, and negative $q _ { x } ^ { * }$ when heat is flowing out of the body.
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# Two-Dimensional Heat Conduction (Without Convection)
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Consider the two-dimensional heat conduction problem in Figure 13–3. In a manner similar to the one-dimensional case, for steady-state conditions, we can show that for material properties coinciding with the global x and y directions,
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$$
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\frac {\partial}{\partial x} \left(K _ {x x} \frac {\partial T}{\partial x}\right) + \frac {\partial}{\partial y} \left(K _ {y y} \frac {\partial T}{\partial y}\right) + Q = 0 \tag {13.1.12}
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$$
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with boundary conditions
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$$
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T = T _ {B} \quad \text { on } S _ {1} \tag {13.1.13}
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$$
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$$
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q _ {n} = q _ {n} ^ {*} = K _ {x x} \frac {\partial T}{\partial x} C _ {x} + K _ {y y} \frac {\partial T}{\partial y} C _ {y} = \text { constant } \quad \text { on } S _ {2} \tag {13.1.14}
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$$
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<details>
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<summary>flowchart</summary>
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```mermaid
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graph TD
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A["q_x"] --> B["Q"]
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B --> C["q_x+dx"]
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B --> D["q_y+dy"]
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B --> E["q_y"]
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```
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</details>
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Figure 13–3 Control volume for two-dimensional heat conduction
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<!-- source-page: 556 -->
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<details>
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<summary>text_image</summary>
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q_n = +q_n^*
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y
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n
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θ
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x
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s_2
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</details>
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Figure 13–4 Unit vector normal to surface $S _ { 2 }$
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where $C _ { x }$ and $C _ { y }$ are the direction cosines of the unit vector n normal to the surface $S _ { 2 }$ shown in Figure 13–4. Again, $q _ { n } ^ { * }$ is by sign convention, positive if heat is flowing into the edge of the body.
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# 13.2 Heat Transfer with Convection
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For a conducting solid in contact with a fluid, there will be a heat transfer taking place between the fluid and solid surface when a temperature difference occurs.
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The fluid will be in motion either through external pumping action (forced convection) or through the buoyancy forces created within the fluid by the temperature differences within it (natural or free convection).
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We will now consider the derivation of the basic differential equation for onedimensional heat conduction with convection. Again we assume the temperature change is much greater in the x direction than in the y and z directions. Figure 13–5 shows the control volume used in the derivation. Again, by Eq. (13.1.1) for conservation of energy, we have
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$$
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q _ {x} A d t + Q A d x d t = c (\rho A d x) d T + q _ {x + d x} A d t + q _ {h} P d x d t \tag {13.2.1}
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$$
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In Eq. (13.2.1), all terms have the same meaning as in Section 13.1, except the heat flow by convective heat transfer is given by Newton’s law of cooling
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$$
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q _ {h} = h (T - T _ {\infty}) \tag {13.2.2}
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$$
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where
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h is the heat-transfer or convection coefficient, in $\mathbf { k W } / ( \mathbf { m } ^ { 2 } \cdot { } ^ { \circ } \mathbf { C } )$ or $\mathrm { B t u } / ( \mathrm { h - f t ^ { 2 } - ^ { \circ } F } )$ .
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T is the temperature of the solid surface at the solid/fluid interface.
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$T _ { \infty }$ is the temperature of the fluid (here the free-stream fluid temperature).
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<details>
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<summary>text_image</summary>
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q_h
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h, T_∞
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q_x
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Q
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A
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q_x+dx
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q_h
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T_∞
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dx
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</details>
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Figure 13–5 Control volume for one-dimensional heat conduction with convection
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<!-- source-page: 557 -->
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<details>
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<summary>text_image</summary>
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S₃
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h
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T
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Insulated boundary
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T∞
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(Streamlines)
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</details>
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Figure 13–6 Model illustrating convective heat transfer (arrows on surface $S _ { 3 }$ indicate heat transfer by convection)
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P in Eq. (13.2.1) denotes the perimeter around the constant cross-sectional area A.
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Again, using Eqs. (13.1.3)–(13.1.6) and (13.2.2) in Eq. (13.2.1), dividing by A dx dt, and simplifying, we obtain the equation for one-dimensional heat conduction with convection as
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$$
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\frac {\partial}{\partial x} \left(K _ {x x} \frac {\partial T}{\partial x}\right) + Q = \rho c \frac {\partial T}{\partial t} + \frac {h P}{A} (T - T _ {\infty}) \tag {13.2.3}
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$$
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with possible boundary conditions on (1) temperature, given by Eq. (13.1.10), and/or (2) temperature gradient, given by Eq. (13.1.11), and/or (3) loss of heat by convection from the ends of the one-dimensional body, as shown in Figure 13–6. Equating the heat flow in the solid wall to the heat flow in the fluid at the solid/fluid interface, we have
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$$
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- K _ {x x} \frac {d T}{d x} = h (T - T _ {\infty}) \quad \text { on } S _ {3} \tag {13.2.4}
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$$
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as a boundary condition for the problem of heat conduction with convection.
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# 13.3 Typical Units; Thermal Conductivities, K; and Heat-Transfer Coefficients, h
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Table 13–1 lists some typical units used for the heat-transfer problem.
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Table 13–2 lists some typical thermal conductivities of various solids and liquids. The thermal conductivity K, in Btu/(h-ft-F) or $\mathbf { W } / ( \mathbf { m } \cdot { } ^ { \circ } \mathbf { C } )$ , measures the
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Table 13–1 Typical units for heat transfer
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<table><tr><td>Variable</td><td>SI</td><td>U.S. Customary</td></tr><tr><td>Thermal conductivity, K</td><td>kW/(m·°C)</td><td>Btu/(h-ft-°F)</td></tr><tr><td>Temperature, T</td><td>°C or K</td><td>°F or °R</td></tr><tr><td>Internal heat source, Q</td><td>kW/m3</td><td>Btu/(h-ft3)</td></tr><tr><td>Heat flux, q</td><td>kW/m2</td><td>Btu/(h-ft2)</td></tr><tr><td>Convection coefficient, h</td><td>kW/(m2·°C)</td><td>Btu/(h-ft2-°F)</td></tr><tr><td>Energy, E</td><td>kW·h</td><td>Btu</td></tr><tr><td>Specific heat, c</td><td>(kW·h)/(kg·°C)</td><td>Btu/(slug-°F)</td></tr><tr><td>Mass density, ρ</td><td>kg/m3</td><td>slug/ft3</td></tr></table>
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<!-- source-page: 558 -->
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Table 13–2 Typical thermal conductivities of some solids and fluids
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<table><tr><td>Material</td><td>K [Btu/(h-ft-°F)]</td><td>K [W/(m·°C)]</td></tr><tr><td colspan="3">Solids</td></tr><tr><td>Aluminum, 0°C (32°F)</td><td>117</td><td>202</td></tr><tr><td>Steel (1% carbon), 0°C</td><td>20</td><td>35</td></tr><tr><td>Fiberglass, 20°C (68°F)</td><td>0.020</td><td>0.035</td></tr><tr><td>Concrete, 0°C</td><td>0.468–0.81</td><td>0.81–1.40</td></tr><tr><td>Earth, coarse gravelly, 20°C</td><td>0.300</td><td>0.520</td></tr><tr><td>Wood, oak, radial direction, 20°C</td><td>0.098</td><td>0.17</td></tr><tr><td colspan="3">Fluids</td></tr><tr><td>Engine oil, 20°C</td><td>0.084</td><td>0.145</td></tr><tr><td>Dry air, atmospheric pressure, 20°C</td><td>0.014</td><td>0.0243</td></tr></table>
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Table 13–3 Approximate values of convection heat-transfer coefficients (from Reference [1])
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<table><tr><td>Mode</td><td>h [Btu/(h-ft2-°F)]</td><td>h [W/(m2·°C)]</td></tr><tr><td>Free convection, air</td><td>1–5</td><td>5–25</td></tr><tr><td>Forced convection, air</td><td>2–100</td><td>10–500</td></tr><tr><td>Forced convection, water</td><td>20–3,000</td><td>100–15,000</td></tr><tr><td>Boiling water</td><td>500–5,000</td><td>2,500–25,000</td></tr><tr><td>Condensation of water vapor</td><td>1,000–20,000</td><td>5,000–100,000</td></tr></table>
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amount of heat energy (Btu or W h) that will flow through a unit length (ft or m) of a given substance in a unit time (h) to raise the temperature one degree $( ^ { \circ } \mathrm { F } \ \mathrm { o r } \ ^ { \circ } \mathrm { C } )$ .
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Table 13–3 lists approximate ranges of values of convection coefficients for various conditions of convection. The heat transfer coefficient h, in $\mathrm { B t u } / ( \mathrm { h - f t ^ { 2 } - ^ { \circ } F } )$ or W/ $( \mathbf { m } ^ { 2 } \cdot ^ { \circ } \mathbf { C } )$ , measures the amount of heat energy (Btu or W h) that will flow across a unit area $\left( \mathrm { f t } ^ { 2 } \ \mathrm { o r } \ \mathrm { m } ^ { 2 } \right)$ of a given substance in a unit time (h) to raise the temperature one degree $( ^ { \circ } \mathrm { F } \ \mathrm { o r } \ ^ { \circ } \mathrm { C } )$ .
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Natural or free convection occurs when, for instance, a heated plate is exposed to ambient room air without an external source of motion. This movement of the air, experienced as a result of the density gradients near the plate, is called natural or free convection. Forced convection is experienced, for instance, in the case of a fan blowing air over a plate.
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# 13.4 One-Dimensional Finite Element Formulation Using a Variational Method
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The temperature distribution influences the amount of heat moving into or out of a body and also influences the stresses in a body. Thermal stresses occur in all bodies that experience a temperature gradient from some equilibrium state but are not free
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<!-- source-page: 559 -->
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to expand in all directions. To evaluate thermal stresses, we need to know the temperature distribution in the body. The finite element method is a realistic method for predicting quantities such as temperature distribution and thermal stresses in a body. In this section, we formulate the one-dimensional heat-transfer equations using a variational method. Examples are included to illustrate the solution of this type of problem.
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# Step 1 Select Element Type
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The basic element with nodes 1 and 2 is shown in Figure 13–7(a).
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<details>
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<summary>text_image</summary>
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x̂
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L
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t₁ t₂
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1 2
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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T = N₁t₁ + N₂t₂
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t₁
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1
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x̂
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L
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t₂
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2
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(b)
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</details>
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Figure 13–7 (a) Basic one-dimensional temperature element and (b) temperature variation along length of element
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# Step 2 Choose a Temperature Function
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We choose the temperature function T [Figure 13–7(b)] within each element similar to the displacement function of Chapter 3, as
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$$
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T (x) = N _ {1} t _ {1} + N _ {2} t _ {2} \tag {13.4.1}
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$$
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where $t _ { 1 }$ and $t _ { 2 }$ are the nodal temperatures to be determined, and
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$$
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N _ {1} = 1 - \frac {\hat {x}}{L} \quad N _ {2} = \frac {\hat {x}}{L} \tag {13.4.2}
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$$
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are again the same shape functions as used for the bar element. The ½N
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matrix is then given by
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$$
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[ N ] = \left[ 1 - \frac {\hat {x}}{L} \quad \frac {\hat {x}}{L} \right] \tag {13.4.3}
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$$
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and the nodal temperature matrix is
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$$
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\{t \} = \left\{ \begin{array}{l} t _ {1} \\ t _ {2} \end{array} \right\} \tag {13.4.4}
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$$
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In matrix form, we express Eq. (13.4.1) as
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$$
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\{T \} = [ N ] \{t \} \tag {13.4.5}
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$$
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<!-- source-page: 560 -->
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# Step 3 Define the Temperature Gradient=Temperature and Heat Flux=Temperature Gradient Relationships
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The temperature gradient matrix fgg, analogous to the strain matrix feg, is given by
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$$
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\{g \} = \left\{\frac {d T}{d \hat {x}} \right\} = [ B ] \{t \} \tag {13.4.6}
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$$
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where ½B
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is obtained by substituting Eq. (13.4.1) for $T ( \hat { x } )$ into Eq. (13.4.6) and differentiating with respect to x^, that is,
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$$
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[ B ] = \left[ \frac {d N _ {1}}{d \hat {x}} \quad \frac {d N _ {2}}{d \hat {x}} \right]
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$$
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Using Eqs. (13.4.2) in the definition for $[ B ]$ , we have
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$$
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[ B ] = \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \tag {13.4.7}
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$$
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The heat flux/temperature gradient relationship is given by
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$$
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q _ {x} = - [ D ] \{g \} \tag {13.4.8}
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$$
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where the material property matrix is now given by
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$$
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[ D ] = \left[ K _ {x x} \right] \tag {13.4.9}
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$$
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# Step 4 Derive the Element Conduction Matrix and Equations
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Equations (13.1.9)–(13.1.11) and (13.2.3) can be shown to be derivable (as shown, for instance, in References [4–6]) by the minimization of the following functional (analogous to the potential energy functional $\pi _ { p } )$ :
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$$
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\pi_ {h} = U + \Omega_ {Q} + \Omega_ {q} + \Omega_ {h} \tag {13.4.10}
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$$
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where $U = \frac { 1 } { 2 } \int \int \limits _ { V } \left[ K _ { x x } \left( \frac { d T } { d x } \right) ^ { 2 } \right] d V$
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$$
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\Omega_ {Q} = - \iiint_ {V} Q T d V \quad \Omega_ {q} = - \iint_ {S _ {2}} q ^ {*} T d S \quad \Omega_ {h} = \frac {1}{2} \iint_ {S _ {3}} h (T - T _ {\infty}) ^ {2} d S \tag {13.4.11}
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$$
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and where $S _ { 2 }$ and $S _ { 3 }$ are separate surface areas over which heat flow (flux) $q ^ { * } \left( q ^ { * } \right.$ is positive into the surface) and convection loss $h ( T - T _ { \infty } )$ are specified. We cannot specify $q ^ { * }$ and h on the same surface because they cannot occur simultaneously on the same surface, as indicated by Eqs. (13.4.11).
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Using Eqs. (13.4.5), (13.4.6), and (13.4.9) in Eq. (13.4.11) and then using Eq. (13.4.10), we can write $\pi _ { h }$ in matrix form as
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$$
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\begin{array}{l} \pi_ {h} = \frac {1}{2} \iiint_ {V} \left[ \{g \} ^ {T} [ D ] \{g \} \right] d V - \iiint_ {V} \left\{t \right\} ^ {T} [ N ] ^ {T} Q d V \\ - \iint_ {S _ {2}} \{t \} ^ {T} [ N ] ^ {T} q ^ {*} d S + \frac {1}{2} \iint_ {S _ {3}} h [ (\{t \} ^ {T} [ N ] ^ {T} - T _ {\infty}) ^ {2} ] d S \tag {13.4.12} \\ \end{array}
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$$
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