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Transient Conduction - CHAPTER 5 TRANSIENT CONDUCTION In...

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C HAPTER 5: T RANSIENT C ONDUCTION In practical terms, very few processes operate at a true steady-state. Variations in operation may arise from “noise” caused by controller accuracy and/or mechanical variation inherent to the different pieces of equipment that make up the process barb4right Processes such as this may be considered to be at steady state For other processes, disturbances consistently arise, causing the controller to continuously adjust to search for the specified set-point barb4right Processes such as these virtually always operate in transient mode. dT/dt 0
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D EALING W ITH N ON -S TEADY S TATE C ONDUCTION Transient conduction may be simple or extremely difficult to consider, depending on the assumptions applied. Consider the case of suddenly changing the temperature of one surface of an object. If temperature gradients within the solid may be neglected (small, high k ) barb2right lumped capacitance method If the only significant temperature gradient is 1D barb2right approximations of heat diffusion equation solutions If significant temperature gradients exist in 2-3 dimensions barb2right finite element/finite difference method
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L UMPED C APACITANCE M ETHOD The lumped capacitance method can be used to find dT/dt if it is assumed that the temperature T ( x,y,z ) is identical throughout the entire object at any instantaneous time point . With no spatial variation in T , d 2 T/dx 2 , d 2 T/dy 2 , and d 2 T/dz 2 are all zero . This assumption is reasonable when R (conduction through body) ( Δ x / kA ) << R (heat loss at surface) (1/ hA ) barb2right if the body is very small barb2right if the thermal conductivity (k) is very large ( k barb2right ) barb2right if h, h r are small = + + + q z T y T x T k & 2 2 2 2 2 2 t T c p ρ
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Common situation: an object initially at temperature T 1 is immersed in a cooling fluid at temperature T 2 barb2right the object loses heat by convection from the surface to the surroundings If no internal heat generation is taking place ( q = 0), the energy balance is: st conv out E E E & & & = = (heat loss by convection = internal energy change) ( ) dt dT V c T t T hA p s ρ = ) ( c p = heat capacity (per unit mass) V = volume of solid body Q: Why can’t we directly use the general heat diffusion equation to do this balance? T i T(t) T < T i t < 0 t ≥ 0 .
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We can solve this equation in the same manner as we did for fins. Let = T T θ , such that ( ) V c T T hA dt dT P s ρ = barb2right θ ρ θ V c hA dt d P s = Separating variables, = t p s dt V c hA d i 0 ρ θ θ θ θ No x, y, or z dependence on temperature barb2right no boundary conditions required to solve problem Time dependence on temperature ( first derivative) barb2right we need one initial condition to solve this problem: @ t= 0, T=T i , or T i -T = θ i Integrating (LHS = ln( θ )-ln( θ o ) = ln( θ/ θ o ) and taking exponentials: = t V c
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