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Unformatted text preview: C HAPTER 5: T RANSIENT C ONDUCTION In practical terms, very few processes operate at a true steadystate. 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 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 setpoint Processes such as these virtually always operate in transient mode. dT/dt 0 D EALING W ITH N ONS 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 ) lumped capacitance method If the only significant temperature gradient is 1D approximations of heat diffusion equation solutions If significant temperature gradients exist in 23 dimensions finite element/finite difference method 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) ( x / kA ) << R (heat loss at surface) (1/ hA ) if the body is very small if the thermal conductivity (k) is very large ( k ) if h, h r are small q z T y T x T k 2 2 2 2 2 2 t T c p Common situation: an object initially at temperature T 1 is immersed in a cooling fluid at temperature T 2 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 ) ( c p = heat capacity (per unit mass) V = volume of solid body Q: Why cant we directly use the general heat diffusion equation to do this balance? T i T(t) T < T i t < t . We can solve this equation in the same manner as we did for fins. Let T T , such that V c T T hA dt dT P s V c hA dt d P s Separating variables, t p s dt V c hA d i No x, y, or z dependence on temperature no boundary conditions required to solve problem Time dependence on temperature ( first derivative) we need one initial condition to solve this problem: @ t= 0, T=T i , or T iT = i Integrating (LHS = ln(...
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This note was uploaded on 01/09/2012 for the course CHEM ENG 2A04 taught by Professor Toddhoare during the Spring '10 term at McMaster University.
 Spring '10
 TODDHOARE

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