MIT2_094S11_lec15

MIT2_094S11_lec15 - 2.094 — Finite Element Analysis of...

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Unformatted text preview: 2.094 — Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 15 - Field problems Prof. K.J. Bathe MIT OpenCourseWare Heat transfer, incompressible/inviscid/irrotational flow, seepage flow, etc. Reading: Sec. 7.2-7.3 • Differential formulation • Variational formulation • Incremental formulation • F.E. discretization 15.1 Heat transfer Assume V constant for now: S = Sθ ∪ Sq θ(x, y, z, t) is unkown except θ|Sθ = θpr . In addition, q s |Sq is also prescribed. 15.1.1 Differential formulation I. Heat flow equilibrium in V and on Sq . ∂θ II. Constitutive laws qx = −k ∂x . ∂θ qy = −k ∂ y ∂θ qz = −k ∂z (15.1) (15.2) III. Compatibility: temperatures need to be continuous and satisfy the boundary conditions. 61 MIT 2.094 15. Field problems Heat flow equilibrium gives � � � � � � ∂ ∂θ ∂ ∂θ ∂ ∂θ k + k + k = −q B ∂x ∂x ∂y ∂y ∂z ∂z (15.3) where q B is the heat generated per unit volume. Recall 1D case: unit cross-section dV = dx · (1) (15.4) q |x − q |x+dx + q B dx = 0 � ∂ qx q |x − q |x + dx + q B dx = 0 ∂x (15.5) � − � ∂ ∂x ∂ ∂x −k � k ∂θ ∂x ∂θ ∂x � � (15.6) B� � dx dx � +q � =0 (15.7) = −q B (15.8) We also need to satisfy ∂θ = qS ∂n k (15.9) on Sq . 15.1.2 � θ Principle of virtual temperatures ∂ ∂x � ∂θ k ∂x � + ··· + q B � =0 (15.10) � ( θ�S = 0 and θ to be continuous.) θ � � θ V ∂ ∂x � � � ∂θ k + · · · + q B dV = 0 ∂x (15.11) 62 MIT 2.094 15. Field problems Transform using divergence theorem (see Ex 4.2, 7.1) � � � �T Sq � B θ kθ θq dV + θ q S dSq ���� dV = V V heat flow ⎛ ∂θ ∂x ⎞ ⎜ ⎜ θ =⎜ ⎜ ⎝ ∂θ ∂y (15.12) Sq ⎟ ⎟ ⎟ ⎟ ⎠ � (15.13) ∂θ ∂z ⎡ k k=⎣ 0 0 0 k 0 ⎤ 0 0⎦ k (15.14) Convection b oundary condition � � qS = h θe − θS (15.15) where θe is the given environmental temperature. Radiation � � �4 � 4 q S = κ∗ (θr ) − θS � � �2 � � r �� � 2 = κ∗ (θr ) + θS θ + θS θr − θS � � = κ θr − θS (15.16) (15.17) (15.18) where κ = κ(θS ) and θr is given temperature of source. At time t + Δt � � � � T t+Δt t+Δt � S t+Δt B θ k θ dV = θ q dV + θ t+Δtq S dSq V V t+Δt θ = tθ + θ Let t+Δt (i) or θ t+Δt (0) with θ = t+Δt (i−1) θ (15.19) Sq (15.20) + Δθ (i) (15.21) t =θ (15.22) From (15.19) � �T (i) θ t+Δtk(i−1) Δθ � dV V � � �T t+Δt � (i−1) = θ t+Δtq B dV − θ t+Δtk(i−1) θ dV V V � � �� � (i−1) (i) S + θ t+Δth(i−1) t+Δtθ e − t+Δtθ S + ΔθS dSq (15.23) Sq where the ΔθS (i) term would be moved to the left-hand side. We considered the convection conditions � � � S θ t+Δth t+Δtθ e − t+Δtθ S dSq (15.24) Sq The radiation conditions would be included similarly. 63 MIT 2.094 15. Field problems F.E. discretization ˆ θ = H1x4 · t+Δtθ4x1 t+Δt � ˆ θ2x1 = B2x4 · t+Δtθ4x1 t+Δt S ˆ θ = H S · t+Δtθ t+Δt for 4-node 2D planar element (15.25) (15.26) (15.27) For (15.23) � ⎛ ⎞ � gives (i) θ t+Δtk(i−1) Δθ � dV =⇒ ⎝ ���� t+Δtk(i−1) �B dV ⎠ Δ �ˆ(i) BT θ ��� � �� � ��� V �T V � 4x2 θ t+Δtq B dV ⇒ V � 2x4 2x2 (15.28) 4x1 H T t+Δtq B dV (15.29) V � θ � T t+Δt (i−1) t+Δt � (i−1) k θ �� dV ⇒ V B T t+Δtk(i−1) B dV � V � θ S T t+Δt (i−1) h � t+Δt e θ− � t+Δt S (i−1) θ t+Δt ˆ(i−1) � + ΔθS (i) �� dSq θ �� known (15.30) � =⇒ Sq ⎛ � Sq 15.2 ⎛ ⎞⎞ (15.31) T ˆ ˆ HS H S � t+Δth(i−1) ���� ⎝ t+Δtθ e − ⎝ t+Δtθ (i−1) +Δ �ˆ(i) ⎠⎠ dSq θ � �� ��� � �� � � �� � 4x1 1x4 4x1 4x1 4x1 Inviscid, incompressible, irrotational flow 2D case: vx , vy are velocities in x and y directions. �·v =0 ∂ vx ∂ vy + =0 ∂x ∂y ∂ vx ∂ vy − =0 ∂y ∂x or Reading: Sec. 7.3.2 (15.32) (incompressible) (15.33) (irrotational) (15.34) Use the potential φ(x, y ), vx = ⇒ ∂φ ∂x vy = ∂φ ∂y (15.35) ∂2φ ∂2φ + 2 = 0 in V ∂ x2 ∂y (15.36) (Same as the heat transfer equation with k = 1, q B = 0) 64 MIT OpenCourseWare http://ocw.mit.edu 2.094 Finite Element Analysis of Solids and Fluids II Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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This note was uploaded on 12/29/2011 for the course ENGINEERIN 2.094 taught by Professor Prof.klaus-jürgenbathe during the Spring '11 term at MIT.

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