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Published byEfren Damon Modified about 1 year ago

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Conduction & Convection

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Quiz 9 – TIME IS UP!!! A flat furnace wall is constructed with a 4.5-inch layer of refractory brick (k = Btu/ft·h· F) backed by a 9-inch layer of common brick (k = Btu/ft·h· F) and a 2-inch layer of silica foam (k = Btu/ft·h· F). The temperature of the inner face of the wall is 1200 F, and that of the outer face is 170 F. a.What is the temperature of the interface between the refractory brick and the common brick? b.What would be the temperature of the outer face if the silica foam is placed between the two brick layers?

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Outline 2.Conduction Heat Transfer 2.1. Series/Parallel Resistances 2.2. Geometric Considerations 3.Convection Heat Transfer 3.1. Heat Transfer Coefficient 3.2. Dimensionless Groups for HTC Estimation

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Geometric Considerations Heat Conduction Through Concentric Cylinders

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Geometric Considerations Heat Conduction Through Concentric Cylinders

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Geometric Considerations Heat Conduction Through Concentric Cylinders

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Geometric Considerations Heat Conduction Through Hollow Spheres Integrating both sides:

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Geometric Considerations Heat Conduction Through Hollow Spheres Rearranging:

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Geometric Considerations Heat Conduction Through Hollow Spheres Define a geometric mean area: …and a geometric mean radius: *Final form

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Shell Balance Plane Wall/Slab

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Shell Balance Plane Wall/Slab

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Shell Balance Plane Wall/Slab

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Shell Balance Plane Wall/Slab

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Shell Balance Cylinder

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Shell Balance Sphere

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Heat Transfer Coefficient Where: Q = heat flow rate A = heat transfer area h = heat transfer coefficient T w = temperature at solid wall T f = temperature at bulk fluid Convection Heat Transfer Useful Conversion:

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Heat Transfer Coefficient Where: Q = heat flow rate A = heat transfer area h = heat transfer coefficient T w = temperature at solid wall T f = temperature at bulk fluid Convection Heat Transfer Driving force Thermal Resistance

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Heat Transfer Coefficient

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Dimensionless Groups Thermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms. Mechanism Ratio Analysis 1. Heat, mass, and momentum transport are described by differential equations of change. e.g. Navier- Stokes Eq’n

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Dimensionless Groups Thermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms. Mechanism Ratio Analysis 2. However, these equations are complex and most of the time difficult to solve/integrate.

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Dimensionless Groups Thermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms. Mechanism Ratio Analysis 3. But still, valuable information is described in these equations, relating the different forces. Inertial Forces Pressure Forces Viscous Forces

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Dimensionless Groups Thermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms. Mechanism Ratio Analysis 4. In the Mechanism Ratio Analysis, solving the equations of change is replaced by empiricism. Inertial Forces Pressure Forces Viscous Forces

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Dimensionless Groups Thermal conductivities are easy to determine by calorimetric experiments, but heat transfer coefficients require the analysis of transfer mechanisms. Mechanism Ratio Analysis 5. This is done by just taking the ratio of the mechanisms and making them into dimensionless groups. Inertial Forces Pressure Forces Viscous Forces Inertial Forces

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Dimensionless Groups Inertial Forces Pressure Forces Viscous Forces Inertial Forces Reynolds Number, Re – the ratio of inertial to viscous forces. Euler Number, Eu – the ratio of pressure to inertial forces.

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Dimensionless Groups Reynolds 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, 1980

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Dimensionless Groups Useful dimensionless groups for Heat Transfer: Dim. GroupRatioEquation Prandtl, Pr molecular diffusivity of momentum / molecular diffusivity of heat Nusselt, Nu heat convection / heat conduction c P = specific heat (J/kg-K) μ = viscosity (Pa-s) D = characteristic length (diameter) (m) k = thermal conductivity (W/m-K) h = heat transfer coefficient (W/m 2 -K)

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Dimensionless Groups Correlations for Heat Transfer Coefficients: Dittus-Boelter Equation Sieder-Tate Equation n = 0.4 when fluid is heated n = 0.3 when fluid is cooled (for forced convection/ turbulent, horizontal tubes) (for forced convection/ turbulent, Re > & 0.5 < Pr < 100)

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Dimensionless Groups Exercise! An organic liquid enters a 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 Btu/lb-°F, 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.

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