conduction & convection

Conduction & Convection

Quiz 9 – 2014.01.27

TIME IS UP!!!

A flat furnace wall is constructed with a 4.5-inch layer of refractory brick (k = 0.080 Btu/ft·h·F) backed by a 9-inch layer of common brick (k = 0.800 Btu/ft·h·F) and a 2-inch layer of silica foam (k = 0.032 Btu/ft·h·F). The temperature of the inner face of the wall is 1200F, and that of the outer face is 170F. 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?

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

Geometric Considerations

Heat Conduction Through Concentric Cylinders

Geometric Considerations

Heat Conduction Through Concentric Cylinders

Geometric Considerations

Heat Conduction Through Concentric Cylinders

Geometric Considerations

Heat Conduction Through Hollow Spheres

𝐴=4𝜋𝑟 2

𝑄𝐴=−𝑘 𝑑𝑇

𝑑𝑟

∫ 𝑄4𝜋𝑟 2

𝑑𝑟=−𝑘∫𝑑𝑇

Integrating both sides: ( 1𝑟1−

1𝑟2 )𝑄=− 4𝜋𝑘 (𝑇2−𝑇 1)

Geometric Considerations

Heat Conduction Through Hollow Spheres

𝐴=4𝜋𝑟 2

𝑄𝐴=−𝑘 𝑑𝑇

𝑑𝑟

∫ 𝑄4𝜋𝑟 2

𝑑𝑟=−𝑘∫𝑑𝑇

Rearranging: 𝑄=−4𝜋𝑟1𝑟2

𝑘 (𝑇2−𝑇 1

𝑟2−𝑟1 )

Geometric Considerations

Heat Conduction Through Hollow Spheres

Define a geometric mean area:

𝐴𝐺𝑀=4𝜋𝑟1𝑟2

…and a geometric mean radius:

𝑟 𝐺𝑀=√𝑟 1𝑟2 𝑄=−𝑘 𝐴𝐺𝑀(𝑇 2−𝑇1

𝑟2−𝑟1 )𝐴𝐺𝑀=4𝜋 𝑟𝐺𝑀

2

*Final form

𝑄=−4𝜋𝑟1𝑟2

𝑘 (𝑇2−𝑇 1

𝑟2−𝑟1 )

Shell Balance

Plane Wall/Slab

Shell Balance

Plane Wall/Slab

Shell Balance

Plane Wall/Slab

Shell Balance

Plane Wall/Slab

Shell Balance

Cylinder

Shell Balance

Sphere

Heat Transfer Coefficient

Where:Q = heat flow rateA = heat transfer areah = heat transfer coefficientTw = temperature at solid wallTf = temperature at bulk fluid 𝑄=h 𝐴 (𝑇𝑤−𝑇 𝑓 )

Convection Heat Transfer

Useful Conversion:

1 𝐵𝑡𝑢h𝑟 𝑓 𝑡 2° 𝐹

=5.6783 𝑊𝑚2𝐾

Heat Transfer Coefficient

Where:Q = heat flow rateA = heat transfer areah = heat transfer coefficientTw = temperature at solid wallTf = temperature at bulk fluid

𝑄=h 𝐴 (𝑇𝑤−𝑇 𝑓 )Convection Heat Transfer

𝑄=𝑇𝑤−𝑇 𝑓

1h𝐴

Driving force

Thermal Resistance

Heat Transfer Coefficient

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.

2D pDt

v g ve.g. Navier-

Stokes Eq’n

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.

2D pDt

v g v 𝒗=𝒗 (𝑥 , 𝑦 , 𝑧 , 𝑡 )??

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.

2D pDt

v g vInertial

Forces

Pressure ForcesViscous Forces

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+¿¿

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

Dimensionless Groups

Inertial Forces

Pressure Forces

Viscous Forces

+¿Inertial Forces

Reynolds Number, Re – the ratio of inertial to viscous forces.

𝑅𝑒=𝐷𝑣 𝜌𝜇 =

𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙𝑣𝑖𝑠𝑐𝑜𝑢𝑠

Euler Number, Eu – the ratio of pressure to inertial forces.

𝐸𝑢=𝑃𝜌𝑣2

=𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙

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

𝑅𝑒=𝐷𝑣 𝜌𝜇

𝐸𝑢=𝑃𝜌𝑣2

Dimensionless Groups

Useful dimensionless groups for Heat Transfer:

Dim. Group Ratio EquationPrandtl, Pr molecular diffusivity of momentum /

molecular diffusivity of heat

Nusselt, Nu heat convection / heat conduction

cP = 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/m2-K)

Dimensionless Groups

Correlations for Heat Transfer Coefficients:

Dittus-Boelter Equation

Sieder-Tate Equation

𝑁𝑢=0.023𝑅𝑒0.8𝑃𝑟𝑛

𝑁𝑢=0.023𝑅𝑒0.8𝑃𝑟 1 /3𝜙𝑣

𝜙𝑣=( 𝜇𝜇𝑤 )

0.14

n = 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)

Dimensionless Groups

Exercise!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.