Heat Conduction (Fourier’s Law, analogous to Fick’s Law)
qx” = -k | {dT/dx} , x is the direction of heat transfer, qx” is the flux,
energy/(area.time)
Note the analogy to NA = -D | {dCA/dZ}
Convection – energy is transferred by the bulk motion of the fluid
Newton’s Law of Cooling: q” = h(Ts - T∞)
Where the heat transfer coefficient “h” is analogous to the mass transfer coefficient. Ts
and T∞ are the temperature at the surface and bulk (outer flow) region respectively.
Diagram
Forced convection – external means (fan, pump) is used
Natural convection – flow is induced by buoyancy forces
Radiation – energy emitted by matter. For the frequently encountered case of radiation
exchange between a small surface at Ts and a much larger surface at Tsurr completely
surrounding the small surface
qrad” = Єσ(Ts4 – Tsurr4) ; T(inK), σ = 5.67 x 10-8 w/(m?K4)
Є values are listed in Appendix A
The above equation also assumes Є is equal to absorptivity (α)
Radiation (Example Problem)
A small oxidized horizontal metal tube with an OD of 0.0254 m and 0.61 m long with a
surface temp. of 588K is in a very large furnace enclosure with fire-brick walls and the
surrounding air at 1088K. The emissivity of the metal tube is 0.60 at 1088K and 0.46 at
588K. Calculate the heat transfer rate to the tube (in W(J/s)) by radiation.
qrad = A,Єσ(Ts4 – Tsurr4)
This equation is approximate where the emissivity is taken to be equal to absorptivity and
that a single emissivity value at Tsurr is used.
π(0.0254)(0.61)(0.6)(5.67 x 10-8)((588)4 – (1088)4)
Note: π(0.0254)(0.61) is πDL
(0.6) is Є
(5.67 x 10-8) is σ(w/(m2K4))
qrad = -2130w
Heat Flux at a Surface
-k | {dT/dx}|x=0 = qs” ||| DIAGRAM
Insulated or adiabatic surface: dT/dx}|x=0 = 0 ||| DIAGRAM
Convection surface condition
-k | {dT/dx}|x=0 = h(T∞ – T(0))
Temperature Distribution (One Dimensional Steady-State Heat Conduction)
{d{k | dT/dx}/dx} = 0 (general derivation, p. 63)
Alternate derivation
Consider a surface perpendicular to x direction of thickness δx ||| DIAGRAM
energy in – energy out
-k | {dT/dx}|x – (-k | {dT/dx}|x+δx) + energy generation or??? source = energy,
accumulation or storage
For no energy generation and accumulation (steady-state)
-k | {dT/dx}|x + (-k | {dT/dx}|x+δx) = 0
When δx -> 0, the above equation becomes
{d({k | dT/dx})/dx} = 0
Or??? {k | dT/dx} = constant => linear temperature distribution
Heat transfer rate = qx = -kA | dT/dx
Or??? qx = (kA/L)(Ts,1 – Ts,2)
Also from convection surface condition
qx = h1A(T∞,1 – Ts,1)
h2A(Ts,2 - T∞,1)
DIAGRAM
Define “thermal resistance”
Conduction; Rt,cond = (Ts,1 – Ts,2)/qx = L/(kA)
Convection; Rt,conv = (Ts - T∞)/q = 1/(hA)
Since qx is constant (why?)
qx = (T∞,1 – Ts,1) / (1/(h1A)) = (Ts,1 – Ts,2) / (L/(kA)) = (Ts,2 - T∞,2) / (1/(h2A))
also qx = (T∞,1 – T∞,2) / (Rtot) , (overall driving force) / (overall resistance)
where Rtot = 1/( h1A) + L/(kA) + 1/( h2A)
other variations: [Also note qx” = qx/A]
qx = (T∞,1 – Ts,1) / (1/(h1A) + L/(kA))
qx = (Ts,1 - T∞,2) / (L/(kA) + 1/(h2A))
:. qx” = (T∞,1 – Ts,2) / (1/(h1) + L/k)
qx” = (Ts,1 - T∞,2) / (L/k + 1/h2)
The Energy(Heat Diffusion) Equation (p.61-65), “Fundamentals of Heat & Mass
Transfers”
Rectangular coordinates
(∂/(∂x))(k(((∂T))/(∂x)))+(∂/(∂y))(k(((∂T))/(∂y)))+(∂/(∂z))( k(((∂T))/(∂x)))+q{with a little
dot over it} = PCp | {(∂T)/(∂T)}
Cylindrical coordinates (see Fig. 2.9 for coordinate description)
(1/r)(∂/(∂r))(kr (∂T)/(∂r)) + (1/rsquared)(2/∂Φ)(k((∂T)/∂Φ)) + (2/∂z)(k((∂T)/∂z)) + q{with
a little dot over it} = PCp | {(∂T)/(∂T)}
Spherical coordinates (see Fig.2.10)
(1/r2)(∂/(∂r))((kr2)((∂T)/(∂r))) + 1/(r2sin2θ)(∂/∂Φ)(k((∂T)/∂Φ)) +
(1/r2sinθ)(∂/(∂θ)(ksinθ((∂T)/(∂θ)) + q{with a little dot over it} = PCp | {(∂T)/(∂T)}
When the differential equation is 2nd order in the spatial coordinate (consider x in
rectangular coordinates) you will need two boundary (or surface)conditions to solve the
problem, i.e. T(o,t) = To ; -k(((∂T)) / (∂x))|x=L = he(T(l,t)-T∞)
In Addition
he differential equation is first order with respect to time hence you will need one
condition (initial condition) for time in order to obtain the solution, i.;e. T(x,o) = To.
The energy equation for the cylindrical geometry:
1/r | {dkr | {dT/dr}/dr} = 0
Diagram
{dkr | {dT/dr}/dr} = 0
kr | {dT/dr} = constant
r | {dT/dr} = C1 (C1 = constant/k, i.e another constant)
{TS1∫TS2} dT = C1 {r1∫r2} dr/r ; TS2 – TS1 = C1ln(r2/r1)
C1 = (Ts2 – Ts1)/ln (r2/r1)
Or dT/dr = C1/r = (TS2 – TS1) / (rln(r2 / r1))
Fourier’s Law; qr = -kA | {dT/dr} = -k2πrL | {dT/dr}
qr = 2πLk((TS2 – TS1) / (ln(r2 / r1))) (Note the heat transfer rate (W, or J/s is constant!)
Or qr = (TS1 – TS2) / Rt, cond {cond is thermal resistance for radial conduction}
Rt, cond = (ln(r2 / r1)) / (2πLk)
If the fluid temperatures inside and outside the pipe are t∞,1 and t∞,2 , in analogy to the
plane wall, one can write:
qr = (T∞,1 – T∞,2) / (1/(2πr1Lh1) + (ln(r2 / r1))/(2πkL) + 1/(2πr2Lh2))
Or qr = (T∞,1 – T∞,2) / Rtot
Where Rtot = 1/(2πr1Lh1) + (ln(r2 / r1))/(2πkL) + 1/(2πr2Lh2)
Often an overall heat transfer coefficient “U” is defined, either based on inner surface
area (A1) or outer surface area (A2). Let’s base “U” on inner surface area => Rtot = 1 /
(U1A1)
1 / (U1A1) = 1/(U12πr1L) = 1/(2πr1Lh1) + (ln(r2 / r1))/(2πkL) + 1/(2πr2Lh2)
1/U1 = 1/h1 + (r1ln(r2/r1)) / k + r1 / (r2h2)
Or U1 = 1 / (1/h1 + (r1/k)ln(r2/r1) + r1/(r2h2))
For the more general case of a composite system (such as one or more layers of insulation
can be placed around the pipe) see Equation 3.31 in the text
U1 = 1/(1/h1 + (r1/kA)ln(r2/r1) + (r1/kB)ln(r3/r2) + (r1/kC)ln(r4/r3) + (r1/r4)(1/h4))
The next recitation will involve solving a transient heat conduction problem. In transient
heat conduction problems, the temperature at any particular location in the solid varies
with respect to time. Typically, the temperature of the solid will not be spatially uniform
but in many applications, it may be assumed to be so. This assumption simplifies the
treatment of the transient heat conduction problem since the problem can now be
formulated as an ordinary differential equation where the solid temperature is a function
of time rather than the more complicated case of a partial differential equation where the
solid temperature is a function of time as well as distance. Here we will first illustrate the
criterion for assuming uniform solid temperature, i.e. the lumped capacitance method or
the lumped parameter method.
Criterion for the Validity of the Lumped Capacitance method for Transient Conduction
The criterion that will be developed here is for the steady-state conduction process but is
also applicable for the transient process.
DIAGRAM
Steady-state conduction through a plane wall of thickness L, consider the surface
(boundary) condition
-k | {dT/dx} |x=L = h(Ts,2 – T∞)
For steady-state conduction dT/dx is linear
(k(Ts,1 – Ts,2))/L = h(Ts,2 – T∞)
(Ts,1 – Ts,2) / (Ts,2 - T∞) = (hL) / k = Bi
For Bi C 0.1, “lumped” method may be used. This criterion can be generalized to any
shape by using a characteristic length “Lc” defined as Lc = (volume of the object / surface
area of the object) h(Lc/k) < 0.1, for a sphere Lc = (sphere diameter) / 6
Heat Conduction (Transient)
Consider the problem of a long cylindrical object initially at T0 being immersed in a
medium (such as H2O or air) at constant temperature of T∞. It was established that if the
condition (hLc) / k < 0.1 is satisfied, lumped parameter approach may be used. Otherwise
it is necessary to solve the following differential equation
PCp | {(∂T)/(∂T)} = (1/r)(∂/(∂r))(kr(∂T)/(∂r))
Or for constant “k”
PCp | {(∂T)/(∂T)} = k(1/r)(∂/(∂r))(kr(∂T)/(∂r))
With the following initial & boundary conditions
At t=0, T = T0 for all r
At r=R, -k((∂T)/(∂r)) = h(T – T∞) for t>0
At r=0, (∂T)/(∂r) = 0 for all t
Or // T is finite for all t