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 heat tranfer question shibashankar Member Posts: 65 Threads: 27 Joined: Aug 2009 Reputation: 0 08-16-2017, 08:38 PM Net radiation heat transfer between any two surfaces can be expressed as Qi j = Ai Ji Fi j Aj Jj Fi j (W) ( engel, Heat and Mass Transfer, p 729 (13-28)) It is common to assume the surfaces of an enclosure to be opaque, diffuse and gray. That is surfaces are nontransparent, they are diffuse emitters and diffuse reflector, and their radiation properties are independent of wavelength. For thermal equilibrium net heat transfer becomes zero. Qi j =0 and rearranging the equation Ai Ji Fi j Aj Jj Fi j =0 Ai Ji Fi j = Aj Jj Fi j Ji Aj Fj i __ = __ Jj Ai Fi j Right side of the equation represents the geometric effects and left side of the equation represents surface emission and reflection effects. Applying reciprocity relation, right side is equal one for diffuse assumption and for thermal equilibrium Ji and Jj must be equal then it results equal temperature. Real surfaces are not emits diffusely. So right side of the equation is not equal 1. Then for thermal equilibrium two surfaces must be different temperature. This contradicts second law of thermodynamics. How it can be explain? It can be said that radiocity assumption is not useful for this situation but why? Let us have concentric cylinders. Fig. 1 fig. 2 Inside cylinder i has Fi j = 1 because all radiation leaving surface i reaches to the surface j. With the diffuse assumption for j surface some part of radiation reaches to i surface and some part turns itself. If surfaces i and j has nondiffuse character and as it is seen on fig 2, point p do not emits along two arcs of ab and cd. Than reciprocity does not work. For particular equation becomes Ji ( ) Aj Fj i __ = __ Jj ( ) Ai Fi j Experimental results show that with nondiffuse emittance right side of equation is not equal one (because emittance is changing with the angle) and during thermal equilibrium Ji( ) and Jj ( ) must not be equal. As a result for thermal equilibrium real surfaces can have different temperature. This means that heat can flow lower temperature to higher temperature which is violate the second law of thermodynamic. Is it this reasoning true? Or which condition is not taken into the consideration. In general if it is used some assumption and write equality then any real case it is observed changes the equality but in thermodynamics it is assumed diffusive emittanced and write above equation, in considering real case it is not changed. Another example, most efficient cycle is Carnot cycle. Any case differ Carnot cycle condition lowers the efficiency. Radiative heat transfer calculation used without any change for assumed case and real case. Halil brahim SOYLU « Next Oldest | Next Newest »