B40DM FORMULAE AND CHARTS Momentum Transport: Equation of Continuity: ππ ππ ππ ππ ππ£π₯ ππ£π¦ ππ£π§ + π£π₯ + π£π¦ + π£π§ +π( + + )=0 ππ‘ ππ₯ ππ¦ ππ§ ππ₯ ππ¦ ππ§ Navier-Stokes Equation: • Rectangular coordinates (x, y, z): x-component: π π·π£π₯ ππ π 2 π£π₯ π 2 π£π₯ π 2 π£π₯ = πππ₯ − +π( 2 + + ) π·π‘ ππ₯ ππ₯ ππ¦ 2 ππ§ 2 π π·π£π¦ π 2 π£π¦ π 2 π£π¦ π 2 π£π¦ ππ = πππ¦ − +π( 2 + + ) π·π‘ ππ¦ ππ₯ ππ¦ 2 ππ§ 2 y-component: z-component: π π·π£π§ ππ π 2 π£π§ π 2 π£π§ π 2 π£π§ = πππ§ − +π( 2 + + ) π·π‘ ππ§ ππ₯ ππ¦ 2 ππ§ 2 Heat Transport: Convective-Diffusion Heat Equation: • Rectangular coordinates (x, y, z): ππ ππ ππ ππ π2π π2 π π2 π ππΆπ ( ππ‘ + vx πx + vy πy + vz πz ) = k [ππ₯ 2 + ππ¦ 2 + ππ§ 2 ] + π + πΜ • Cylindrical coordinates (r, π, π³): ππΆπ ( ππ ππ vθ ππ ππ π 2 π 1 ∂T 1 π 2 π π 2 π + vr + + vz ) = k [ 2 + + + ] + π + πΜ ππ‘ πr r πθ πz πr r πr r 2 πθ2 ππ§ 2 • Spherical coordinates (r, π, ∅): ππΆπ ( ππ ππ vθ ππ v∅ ππ ) + vr + + ππ‘ πr r πθ rsinπ π∅ 1 π 2 ∂π 1 π ∂π 1 π 2π )+ 2 (sinθ ) + 2 2 ( 2 )] + π + πΜ = k [ 2 (r π ππ ππ π sinπ ππ ππ π sin π π∅ Others: βπ΄π‘ π − π∞ − ππππ =π ππ − π∞ βπ π΅π = π΄ π πΌπ‘ πΌπ‘ = 2 (π/π΄) (π₯1 )2 πΉπ = π= π∞ − π π∞ − ππ π= πΌπ‘ π₯1 2 π= π₯ π₯1 π= π βπ₯1 ππ − π(π₯, π‘) π₯ = erfβ‘( ) ππ − ππ 2√πΌπ‘ π£= π π πΌ= π πππ ππ = π£ πππ = πΌ π ππ’ = βπΏ π Mass Transport: • Rectangular coordinates (x, y, z): π2π ππΆπ΄ π2π π2 π = π·π΄π΅ [ ππ₯ 2π΄ + ππ¦ 2π΄ + ππ§ 2π΄] ππ‘ • Cylindrical coordinates (r, π, π³): ππΆπ΄ π 2 ππ΄ 1 ∂ππ΄ 1 π 2 ππ΄ π 2 ππ΄ = π·π΄π΅ [ 2 + + + ] ππ‘ πr r πr r 2 πθ2 ππ§ 2 • Spherical coordinates (r, π, ∅): ππΆπ΄ 1 π ∂ππ΄ 1 π ∂ππ΄ 1 π 2 ππ΄ )+ 2 (sinθ ) + 2 2 ( 2 )] = π·π΄π΅ [ 2 (r 2 ππ‘ π ππ ππ π sinπ ππ ππ π sin π π∅ Others: ππ΄,π§ = −π·π΄π΅ πππ΄ ππ΄ + (ππ΄,π§ + ππ΅,π§ ) ππ§ π π= πΆπ΄,∞ − πΆπ΄ πΆπ΄,∞ − πΆπ΄,0 π= π·π΄π΅ π‘ π₯1 2 π= π₯ π₯1 π= π·π΄π΅ ππ π₯1 πΆπ΄,π − πΆπ΄ (π₯, π‘) π₯ = erfβ‘( ) πΆπ΄,π − πΆπ΄,0 2√π·π΄π΅ π‘ ππ = π π = π·π΄π΅ ππ·π΄π΅ πΏπ = πΌ π = π·π΄π΅ πππ π·π΄π΅ πβ = ππ πΏ π·π΄π΅ ππ΄ = ππ (ππ΄,π − ππ΄,∞ ) ππ π£∞ ππ π£∞ πΆ β = 2π = ππ£ π Sc=Pr=1 ∞ π = πΆπ /2 π£π₯ 1+( )(ππ−1) π£∞ Pr=1 πΆπ /2 πβ ππ = = π πππ π£∞ 1 + 5 × πΆ /2 {ππ − 1 + ln [1 + 5 (ππ − 1)]} √ π 6 π πΆ ππ· = π£ π ππ 2/3 = 2π β‘ 0.6<Sc<2500 ∞ 1 πβπ₯ = 0.332π ππ₯2 ππ 2/3 Figure A.1 Central temperature history of various solids with initial uniform temperature, To and constant surface temperature, Ts. π₯ Table A.1 Error Function erf (2 πΌπ‘). √ Figure A.2 Unsteady state transport in large flat slab. Figure A.3 Unsteady state transport in a long cylinder. Figure A.4 Unsteady state transport in a sphere. Figure A.5 Center temperature history for an infinite plate. Figure A.6 Center temperature history for an infinite cylinder. Figure A.7 Center temperature history for a sphere. Figure A.8 Charts for solution of unsteady transport problem: flat plate. Figure A.9 Charts for solution of unsteady transport problem: cylinder. Figure A.9 Charts for solution of unsteady transport problem: sphere.