1. Heat Transfer
Transient Conduction

2. General Considerations
Transient or unsteady conduction is time-dependent conduction as a result of changes with time in the boundary conditions or heat generation of a system.
During a transient process, the temperature distribution is time-dependent in a system or medium (dT/dx  0)
Applications of transient conduction include material processing (quenching, solidification, purification of alloys, etc).
For 1-D conduction with no heat generation:
Exact solutions are often complicated and difficult to obtain. We therefore use simpler approaches, which are in general preferable.
Solutions to transient conduction problems include:
I. Lumped Capacitance Method (LCM)
small temperature gradients within a solid
II. Exact or Graphical Methods
large temperature gradients within the solid

3. Lumped Capacitance Method for 1-D Transient Conduction
Consider a hot metal at Ti at t = 0 that is quenched by immersing it in a liquid at T at t =0
Assume T in solid is spatially uniform (Tf(x))
Energy balance on the solid gives
The differential equation can be solved by separating variables and integrating from the initial conditions (T=Ti (θ = t=0)
t < 0
T = Ti Ti
T∞ < Ti
t ≥ 0
T = T(t)
T(t)
Introduce
and or

4. 1 θ θi t = thermal time constant increasing t (t1 < t2 < t3)
Thermal resistance due to convection heat transfer
Lumped thermal capacitance of a solid
increasing t
(t1 < t2 < t3)
e-1 = 0.37
e-4 = 0.02
e-5 = 0.01 1
As Ct and Rt increase, t increases and the solid responds more slowly to a change in its thermal environment (time to reach thermal equilibrium will increase) θ θi t
This characteristic is analogous to a voltage decay when a capacitor with capacitance Ct is discharged through a resistor with resistance Rt in electrical RC circuits.
At t < 0, the valve is closed and the solid is charged to θi
At t ≥ 0, the valve is opened and discharge occurs through the thermal resistance
Thermal Circuit
Analogy

5. Total heat transfer (Q)
Conservation of energy:
This is often written as:
Q  Positive for cooling, Est (internal energy) decreases
Q  Negative for heating, Est (internal energy) increases
The total energy transfer (Q) is:

6. Validity for the lumped Capacitance Method
The magnitude of (Ts,1-Ts,2) compared to (Ts,2-T) can be estimated by performing a S.S. surface energy balance at x = L:
x=0
T∞ , h T
T s,2
T s,1 L
qcond
qconv
With which temperature profile would the LCM be most applicable?   
= Biot Number (Bi)
When is LCM applicable?

7. t T(x,0) = Ti Bi >> 1 T = T(x,t) Rcond >> Rconv T ∞
T ≈ T(t)
Rcond << Rconv
Bi ≈ 1
Rcond ≈ Rconv
Definitions:
Biot Number (Bi) = Resistance due to conduction
Resistance due to convection
Fourier Number (Fo) = dimensionless time
where Lc = characteristic length, α = thermal diffusivity, t = time, h = convective heat transfer coefficient, k = thermal conductivity

8. Lc = characteristic length = Volume of Solid = V Total surface Area As
When Using L.C.M
Time constant
Conclusion:
LCM solutions for transient temperature response of a solid (1-D):
For a plane wall of thickness = 2L
For a Long Cylinder
For a sphere or
To be used only when Bi  0.1

9. General Lumped Capacitance Method for 1-D Transient Conduction
Transient conduction may occur in a solid material due to simultaneous or combined effects of convection and radiation heat transfer, applied heat flux to a surface and heat generation
Applying conservation of energy
The above equation is nonlinear 1st order, non-homogeneous ode, and exact solution cannot be obtained  numerical solution
Surroundings
ρ, c, V, T(0) = Ti
qrad’’
Eg, Est
qs’’
qconv’’
Tsur
As,h
A s(c,r)

10. Exact solutions for simplified cases
No applied heat flux or heat generation, and negligible or nonexistent convection, the governing equation reduces to:
Which can be separated as:
Integrating and arranging yields:
T cannot be obtained explicitly from the above equation, however for Tsur = 0 (radiation to space), we have:

11. Case 2 Negligible radiation with h independent of time
Introduce θ = T - T∞ , so that dθ/dt = dT/dt, the ge reduces to:
where and
Introduce
Separating variables and integrating yields:
Substituting θ’ and θ, we get:
What if b=0?
When is this equation (and all LCM equations) applicable?

12. Problem 5.9
Carbon steel (AISI 1010) shafts of 0.1 m diameter are heat treated in a gas-fired furnace whose gases are at 1200 K and provide a convection coefficient of 100 W/(m2·K). If the shafts enter the furnace at 300K, how long must they remain in the furnace to achieve a centerline temperature of 800K?

13. Problem 5.10