Heat Transfer Transient Conduction

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

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 = θi) @ t=0) t < 0 T = Ti Ti T∞ < Ti t ≥ 0 T = T(t) T(t) Introduce and or

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

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:

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?

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

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

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)

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:

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?

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?

Problem 5.10