Shell Energy Balances andTemperature Distributions in

Solids and Laminar Flow

Introduction Shell Energy balance

Temperature Distribution Solids and laminar Flow

Energy In Energy Out

Temperature distribution perpendicular to flow

Approach

(i) An energy balance is made over a thin slab or shell perpendicular to the direction of the heat flow, and this balance leads to a first-order differential equation from which the heat flux distribution is obtained;

(ii) Then into this expression for the heat flux, we substitute Fourier's law of heat conduction, which gives a first-order differential equation for the temperature as a function of position. The integration constants are then determined by use of boundary conditions for the temperature or heat flux at the bounding surfaces.

Heat conduction problems are solved by following approach

SHELL ENERGY BALANCES; BOUNDARY CONDITIONS

(1)

SHELL ENERGY BALANCES; BOUNDARY CONDITIONSConvective Transport of energy+ Molecular Transport (heat

conduction)+ Molecular Work = Combined Energy Flux (e)

SHELL ENERGY BALANCES; BOUNDARY CONDITIONS The energy production term in Eq. (1) includes:

1. The degradation of electrical energy into heat,2. The heat produced by slowing down of neutrons and nuclear

fragments liberated in the fission process,3. The heat produced by viscous dissipation, and4. The heat produced in chemical reactions.

SHELL ENERGY BALANCES; BOUNDARY CONDITIONS After Eq. (1) has been written for a thin slab or shell of material,

the thickness of the slab or shell is allowed to approach zero. This procedure leads ultimately to an expression for the

temperature distribution containing constants of integration, which we evaluate by use of boundary conditions.

The commonest types of boundary conditions are: The temperature may be specified at a surface. The heat flux normal to a surface may be given (this is equivalent to

specifying the normal component of the temperature gradient). At interfaces the continuity of temperature and of the heat flux normal to the

interface are required. At a solid-fluid interface, the normal heat flux component may be related to

the difference between the solid surface temperature To and the "bulk" fluid temperature Tb (Newton’s law of cooling)

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCE The system: An electric wire of circular cross section with

radius R and electrical conductivity ke (ohm-1 cm-1). Through this wire there is an electric current with current

density I (amp/cm2). The transmission of an electric current is an irreversible

process, and some electrical energy is converted into heat (thermal energy).

The expression for rate of heat production per unit volume is given by:

The quantity Se is the heat source resulting from electrical dissipation.

ee K

IS

2

(2)

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCE

o The temperature rise in the wire is not so large that the temperature dependence of either the thermal or electrical conductivity need be considered.

o The surface of the wire is maintained at temperature To

Assumptions

Aim Derivation of expression for radial temperature distribution

within the wire.

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCEShell considerations

oA cylindrical shell of thickness Δr and length L

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCESince v = 0 the only contributions to the energy balance are

)2())2( rrrr rLqqrL Rate of heat in across cylindrical surface at r

(3)

Rate of heat out across cylindrical surface at r + Δr

)2())()(2( rrrrrr rLqqLrr (4)

Rate of thermal energy production by electrical dissipation

eSLrr )2( (5)

[The notation qr means "heat flux in the r direction," and (…)|r+Δr means "evaluated at r + Δr.“]

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCEWe now substitute these quantities into the energy balance of Eq.(1). Division by 2πLΔr and taking the limit as Δr goes to zero gives:

rSr

rqrqe

rrrrr

r

))()(lim

0(6)

The expression on the left side is the first derivative of rqr w.r.t. r, so that Eq. (6) becomes

rSrqdr

der )( (7)

This is a first-order differential equation for the energy flux, and it may be integrated to give

r

CrSq e

r1

2 (8)

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCEApplying BC1: if r = 0 => C1=0, so Eq. (8) becomes:

(9)

This states that the heat flux increases linearly with r

We now substitute Fourier's law in the form q = – k (dT/dr) into Eq. (9) to obtain

2

rSq e

r

2

rS

dr

dTk e

When k is assumed to be constant, this first-order differential equation can be integrated to give

(10)

2

2

4C

k

rST e (11)

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCEApplying BC2: if r = R => T=TO, so C2 becomes:

22

0 14 R

r

k

RSTT e

02

2 )4/( TkRSC e Hence Eq. (11) becomes

(12)

Eq.(12) gives the temperature rise as a parabolic function of the distance r from the wire axis.

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCEo Once the temperature and heat flux distributions are known,

various information about the system may be obtained. Maximum Temperature Rise (at r-0)

Average Temperature Rise

Heat flow at the surface (for a length L of wire)

(13)

(14)

(15)

HEAT CONDUCTION WITH ANELECTRICAL HEAT SOURCE

o Home Work:

o See the solved examples 10.2-1 and 10.2-2 (page # 295) and try to solve those by yourself.

o Self study the sections 10.3, 10.4, 10.5, 10.6