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Conduction & Convection. Quiz 9 – 2014.01.27. - PowerPoint PPT Presentation

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Conduction & ConvectionQuiz 9 2014.01.27TIME IS UP!!!A flat furnace wall is constructed with a 4.5-inch layer of refractory brick (k = 0.080 Btu/fthF) backed by a 9-inch layer of common brick (k = 0.800 Btu/fthF) and a 2-inch layer of silica foam (k = 0.032 Btu/fthF). The temperature of the inner face of the wall is 1200F, and that of the outer face is 170F. What is the temperature of the interface between the refractory brick and the common brick?What would be the temperature of the outer face if the silica foam is placed between the two brick layers?2OutlineConduction Heat Transfer2.1. Series/Parallel Resistances2.2. Geometric ConsiderationsConvection Heat Transfer3.1. Heat Transfer Coefficient3.2. Dimensionless Groups for HTC EstimationGeometric ConsiderationsHeat Conduction Through Concentric Cylinders

We arranged it like this so that we can introduce a new concept called the logarithmic mean area (next slide).4Geometric ConsiderationsHeat Conduction Through Concentric Cylinders

Emphasize why dx became dr in cylindrical coordinates. The direction of heat flow is now radial.Make the students guess the equation for area (A = 2*pi*r*L) before flashing the equation. This should bethe area perpendicular to the flow. You may also opt the students to integrate before flashing the final equation.5Geometric ConsiderationsHeat Conduction Through Concentric Cylinders

Emphasize why dx became dr in cylindrical coordinates. The direction of heat flow is now radial.Make the students guess the equation for area (A = 2*pi*r*L) before flashing the equation. This should bethe area perpendicular to the flow. You may also opt the students to integrate before flashing the final equation.6Geometric ConsiderationsHeat Conduction Through Hollow Spheres

Integrating both sides:7Geometric ConsiderationsHeat Conduction Through Hollow Spheres

Rearranging:8Geometric ConsiderationsHeat Conduction Through Hollow SpheresDefine a geometric mean area:and a geometric mean radius:*Final formThe final form is made analogous to the form of Fouriers Law for a flat slab. The thermal resistance is now deltaR/kA(GM), analogous to deltaX/kA in flat slabs.9Shell BalancePlane Wall/Slab

10Shell BalancePlane Wall/Slab

11Shell BalancePlane Wall/Slab

12Shell BalancePlane Wall/Slab

13Shell BalanceCylinder

14Shell BalanceSphere

15Heat Transfer CoefficientWhere:Q = heat flow rateA = heat transfer areah = heat transfer coefficientTw = temperature at solid wallTf = temperature at bulk fluidConvection Heat TransferUseful Conversion:

In the figure, 2 fluids surround the wall with different HTCs.Usually, the fluid closest to the wall forms a film, which is why HTC is sometimes referred as a film coefficient.16Heat Transfer CoefficientWhere:Q = heat flow rateA = heat transfer areah = heat transfer coefficientTw = temperature at solid wallTf = temperature at bulk fluidConvection Heat TransferDriving forceThermal ResistanceEmphasize that 1/hA is analogous to deltaX/kA and that they also have the same units.17Heat Transfer Coefficient

From last meeting, the thermal conductivities of these materials were discussed. It is evident now that convection dominates over conduction in the case of fluids, as seen by the magnitude of the HTCs.18Dimensionless GroupsThermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms.Mechanism Ratio Analysis1. Heat, mass, and momentum transport are described by differential equations of change.

e.g.Navier-Stokes EqnSource: Foust Chapter 13, Fundamentals of Transfer MechanismsThis is a prelude to dimensionless groups, and their use in estimating transfer coefficients (not only heat).These numbered points are pre-requisites to why we use dimensionless ratios.19Dimensionless GroupsThermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms.Mechanism Ratio Analysis2. However, these equations are complex and most of the time difficult to solve/integrate.

Source: Foust Chapter 13, Fundamentals of Transfer MechanismsThis is a prelude to dimensionless groups, and their use in estimating transfer coefficients (not only heat).These numbered points are pre-requisites to why we use dimensionless ratios.20Dimensionless GroupsThermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms.Mechanism Ratio Analysis3. But still, valuable information is described in these equations, relating the different forces.

Inertial ForcesPressure ForcesViscous ForcesSource: Foust Chapter 13, Fundamentals of Transfer MechanismsThis is a prelude to dimensionless groups, and their use in estimating transfer coefficients (not only heat).These numbered points are pre-requisites to why we use dimensionless ratios.21Dimensionless GroupsThermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms.Mechanism Ratio Analysis4. In the Mechanism Ratio Analysis, solving the equations of change is replaced by empiricism.Inertial ForcesPressure ForcesViscous ForcesSource: Foust Chapter 13, Fundamentals of Transfer MechanismsThis is a prelude to dimensionless groups, and their use in estimating transfer coefficients (not only heat).These numbered points are pre-requisites to why we use dimensionless ratios.22Dimensionless GroupsThermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms.Mechanism Ratio Analysis5. This is done by just taking the ratio of the mechanisms and making them into dimensionless groups.Inertial ForcesPressure ForcesViscous ForcesInertial ForcesSource: Foust Chapter 13, Fundamentals of Transfer MechanismsThis is a prelude to dimensionless groups, and their use in estimating transfer coefficients (not only heat).These numbered points are pre-requisites to why we use dimensionless ratios.23Dimensionless GroupsInertial ForcesPressure ForcesViscous ForcesInertial ForcesReynolds Number, Re the ratio of inertial to viscous forces.Euler Number, Eu the ratio of pressure to inertial forces.By the mechanism ratio analysis, it is assumed that the ratio between forces and mechanisms are enough to describe heat, mass, and momentum transport. When we say describe, it means to provide Transfer Coefficients.24Dimensionless GroupsReynolds Number, Re the ratio of inertial to viscous forces.Euler Number, Eu the ratio of pressure to inertial forces.If the phenomenon is so complex that negligible knowledge can be gained from the investigation of the differential equations, then empirical processes are available for evolving dimensionless groupings of the involved variables. Foust, 1980The ratio terms in dimensionless groups help interpret the contribution of each mechanism/force to the phenomenon.25Dimensionless GroupsUseful dimensionless groups for Heat Transfer:cP = specific heat (J/kg-K) = viscosity (Pa-s)D = characteristic length (diameter) (m)k = thermal conductivity (W/m-K)h = heat transfer coefficient (W/m2-K)These dimensionless groups will be needed for the correlations in estimating heat transfer coefficients.Also, convince the students that the equations agree with the ratios and that they are dimensionless.26Dimensionless GroupsCorrelations for Heat Transfer Coefficients:Dittus-Boelter EquationSieder-Tate Equationn = 0.4 when fluid is heatedn = 0.3 when fluid is cooled(for forced convection/ turbulent, horizontal tubes)(for forced convection/ turbulent, Re > 10000 & 0.5 < Pr < 100)Note that h can be computed from the Nusselt number.The phi here is the viscosity correction factor and mu-w is viscosity at the wall.The viscosity at the wall and the bulk viscosity differs because of temperature differences.27Dimensionless GroupsExercise!An organic liquid enters a 0.834-in. ID horizontal steel tube, 3.5 ft long, at a rate of 5000 lb/hr. You are given that the specific heat, thermal conductivity, and viscosity of the liquid is 0.565 Btu/lb-F, 0.0647 Btu/hr-ft-F, and 0.59 lb/ft-hr, respectively. All these properties are assumed constant. If the liquid is being cooled, determine the inside-tube heat transfer coefficient using the Dittus-Boelter Equation.Answer: 497.8 Btu/hr-ft2-degFSource: Di ko na maalala eh. Baka imbento ko lang din nung 133.28

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