Report

Overall Shell Energy Balance Forms of Energy Generation: (Se) 1. Degradation of electrical energy to heat (Sn) 2. Heat from nuclear source (by fission) (Sv) 3. Heat from viscous dissipation Let S = rate of heat production per unit volume (W/m3) Energy Generation Electrical Heat Source Consider an electrical wire (solid cylinder): Shell Heat Balance: 2 − 2 +∆ + 2∆ = 0 Rate of Heat IN: 2 Rate of Heat OUT: 2 Generation: 2∆ +∆ Electrical Heat Source Rate of Heat IN 2 Area perpendicular to qr at r = r = (2) ∙ The Shell: Rate of Heat IN: 2 Rate of Heat OUT: 2 Generation: 2∆ +∆ Electrical Heat Source Rate of Heat OUT 2 +∆ Area perpendicular to qr at r = r + dr = (2( + ∆)) ∙ +∆ The Shell: Rate of Heat IN: 2 Rate of Heat OUT: 2 Generation: 2∆ +∆ Electrical Heat Source Generation = Volume X Se = + ∆ 2 − 2 ∴ = 2∆ Too small = 2 + 2∆ + ∆ 2 − 2 ∴ = 2∆ ∙ The Shell: Rate of Heat IN: 2 Rate of Heat OUT: 2 Generation: 2∆ +∆ Electrical Heat Source Consider an electrical wire (solid cylinder): Shell Heat Balance: 2 2 − 2 +∆ − 2 + 2∆ = 0 +∆ = − 2∆ Dividing by 2∆: − ∆ +∆ = − Q: Why did we divide by 2∆ and not by 2∆? Electrical Heat Source Consider an electrical wire (solid cylinder): We now have: − ∆ +∆ = − Taking the limit as ∆ → 0: = − Q: Is this correct? NO! Electrical Heat Source Consider an electrical wire (solid cylinder): We now have: − ∆ +∆ = − We must adhere to the definition of the derivative: lim ∆→0 − ∆ +∆ − ∆ +∆ = + = = Electrical Heat Source Consider an electrical wire (solid cylinder): = We now have: Integrating: 1 = + 2 Boundary conditions: = 0, = , = = 0 Note: The problem statement will tell you hints about what boundary conditions to use. Electrical Heat Source Consider an electrical wire (solid cylinder): We now have: 1 = + 2 Applying B.C. 1: = 0, = Because q has to be finite at r = 0, all the terms with radius, r, below the denominator must vanish. Therefore: 1 = 0 = 2 Electrical Heat Source Consider an electrical wire (solid cylinder): We now have: = 2 Substituting Fourier’s Law: − = 2 − = 2 − 2 = + 2 4 Electrical Heat Source Consider an electrical wire (solid cylinder): We now have: − 2 = + 2 4 Applying B.C. 2: = , − 2 0 = + 2 4 2 2 = 0 + 4 This is it! But, we rewrite it into a nicer − 2 2 form… = + + 0 4 4 = 0 Electrical Heat Source Consider an electrical wire (solid cylinder): Temperature Profile: 2 − 0 = 1− 4 2 Important assumptions: 1. Temperature rise is not large so that k and Se are constant & uniform. 2. The surface of the wire is maintained at T0. 3. Heat flux is finite at the center. Electrical Heat Source Other important notes… Let: = electrical conductivity = current density 2 1 Ω ∙ = voltage drop over a length These imply the following : 2 = = 2 − 0 = 1− 4 2 2 2 2 = 4 42 Electrical Heat Source Heat flux profile: = 2 The stress profile versus the temperature profile: Temperature Profile: 2 − 0 = 1− 4 2 Electrical Heat Source Quantities that might be asked for: 1. Maximum Temperature Substituting r = 0 to the profile T(r): 2 = 0 + 4 2. Average Temperature Rise − 0 = 2 − 0 0 0 2 0 0 2 1 − 0 = = 8 2 3. Heat Outflow Rate at the Surface = = = 2 2 = = 2 ∙ Electrical Heat Source Examples for Review: Example 10.2-1 and Example 10.2-2 Bird, Stewart, and Lightfoot, Transport Phenomena, 2nd Ed., p. 295 Nuclear Heat Source Consider a spherical nuclear fuel assembly (solid sphere): Before doing a balance, let: = volumetric heat rate of production within the fissionable material only 0 = volumetric heat rate of production at r = 0 Sn depends on radius parabolically: = a dimensionless positive constant Nuclear Heat Source Consider a spherical nuclear fuel assembly (solid sphere): Before doing a balance, let: = temperature profile in the fissionable sphere () = temperature profile in the Alcladding () = heat flux in the fissionable sphere () = heat flux in the Al cladding Nuclear Heat Source Consider a spherical nuclear fuel assembly (solid sphere): For the fissionable material: () 4 2 () − 4 2 +∆ + 4 2 ∆ = 0 Rate of Heat IN: () 2 4 Rate of Heat OUT: 4 2 Generation: 4 2 ∆ () +∆ Electrical Heat Source Generation = Volume X Sn 4 = + ∆ 3 3 − 3 ∴ = 4 2 ∆ Too small 4 = 3 + 3 2 ∆ + 3 ∆ 3 2 + ∆ 3 − 3 ∴ = 4 2 ∆ Rate of Heat IN: () 2 4 Rate of Heat OUT: 4 2 Generation: 4 2 ∆ () +∆ Nuclear Heat Source For the fissionable material: () 4 2 () − 4 2 +∆ No generation here! + 4 2 ∆ = 0 For the Al cladding: () 4 2 Dividing by 4∆: () 2 2 +∆ − ∆ () − 4 2 +∆ =0 Dividing by 4∆: () = 2 2 2 − +∆ ∆ =0 Nuclear Heat Source For the fissionable material: () 4 2 () − 4 2 +∆ No generation here! + 4 2 ∆ = 0 For the Al cladding: () 4 2 Taking ∆ → 0: 2 () = 2 () − 4 2 Taking ∆ → 0: 2 () =0 +∆ =0 Nuclear Heat Source For the fissionable material: () 4 2 () − 4 2 +∆ No generation here! + 4 2 ∆ = 0 For the Al cladding: () 4 2 Taking ∆ → 0: 2 () = 0 1 + () − 4 2 Taking ∆ → 0: 2 2 2 () =0 +∆ =0 Nuclear Heat Source For the fissionable material: () 4 2 () − 4 2 +∆ No generation here! + 4 2 ∆ = 0 For the Al cladding: () 4 2 Integrating: () − 4 2 Integrating: +∆ =0 Nuclear Heat Source Boundary Conditions: = 0, () 1 =0 Integrating: For the fissionable material Boundary Conditions: () = () , () 1 = 0 1 + 3 5 Integrating: For the Al cladding () = 3 Nuclear Heat Source For the fissionable material Inserting Fourier’s Law: For the Al cladding Inserting Fourier’s Law: Nuclear Heat Source For the fissionable material For the Al cladding Boundary Conditions: Boundary Conditions: R(C) At r = R(F), T(F) = T(C) R(F) At r = R(C), T(C) = T0 Nuclear Heat Source For the fissionable material For the Al cladding Overall Shell Energy Balance Recall the Overall Shell Energy Balance: Q by Convective Transport W by Molecular Transport Q by Molecular Transport W by External Forces Energy Generation SteadyState! Overall Shell Energy Balance We need all these terms for viscous dissipation: Q by Convective Transport Q by Molecular Transport How can we account for all these terms at once? W by Molecular Transport Combined Energy Flux Vector We introduce something new to replace q: Combined Energy Flux Vector: Heat Rate from Molecular Motion Convective Energy Flux 1 2 = + + ∙ + 2 Work Rate from Molecular Motion Combined Energy Flux Vector We introduce something new to replace q: Combined Energy Flux Vector: Recall the molecular stress tensor: When dotted with v: = + ∙ = + [ ∙ ] Substituting into e: 1 2 = + + + [ ∙ ] + 2 Combined Energy Flux Vector We introduce something new to replace q: Combined Energy Flux Vector: 1 2 = + + + [ ∙ ] + 2 Simplifying the boxed expression: + = + = + = Finally: 1 2 = + + [ ∙ ] + 2 Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: = Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: We now make a shell balance shown in red on the left. Rate of Energy IN: Rate of Energy OUT: When the combined energy flux vector is used, the generation term will automatically appear from e. +∆ Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: +∆ − =0 = 1 =0 We now make a shell balance shown in red on the left. Rate of Energy IN: Rate of Energy OUT: When the combined energy flux vector is used, the generation term will automatically appear from e. +∆ Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: 1 2 = + + [ ∙ ] + 2 = 1 Fourier’s Law: Newton’s Law: = − = − When the combined energy flux vector is used, the generation term will automatically appear from e. Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: = 1 Substituting the velocity profile: Integrating: − − = 1 − − =− 2 2 = 1 2 1 − + 2 2 When the combined energy flux vector is used, the generation term will automatically appear from e. Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: Boundary Conditions: =− 2 2 1 − + 2 2 After applying the B.C.: − 0 1 2 = − 0 2 − 0 1− + When the combined energy flux vector is used, the generation term will automatically appear from e. Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: =− 2 2 1 − + 2 2 Q: So where is Sv? = After applying the B.C.: − 0 1 2 = − 0 2 − 0 1− + When the combined energy flux vector is used, the generation term will automatically appear from e. 2 Viscous Dissipation Source Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders: Temperature Profile: − 0 1 2 = − 0 2 − 0 1− + New Dimensionless Number: Dim. Group Brinkman, Br Ratio viscous heat dissipation/ molecular heat transport Equation 2 − 0 Viscous Dissipation Source Scenarios when viscous heating is significant: 1. Flow of lubricant between rapidly moving parts. 2. Flow of molten polymers through dies in highspeed extrusion. 3. Flow of highly viscous fluids in high-speed viscometers. 4. Flow of air in the boundary layer near an earth satellite or rocket during reentry into the earth’s atmosphere.