Shell Energy Balances and Temperature Distribution in Solids and Laminar Flows

Upload
imranunar 
Category
Documents

view
363 
download
54
Embed Size (px)
Transcript of Shell Energy Balances and Temperature Distribution in Solids and Laminar Flows
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 firstorder 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 firstorder 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 solidfluid 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 (ohm1 cm1). 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 firstorder 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 firstorder 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 r0)
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 10.21 and 10.22 (page # 295) and try to solve those by yourself.
o Self study the sections 10.3, 10.4, 10.5, 10.6