R11.3. Diffusion and Reaction Facilitated Heat Transfer

Transcription

1 Chapter 11 Professional Reference Shelf R11.3. Diffusion and Reaction Facilitated Heat Transfer When diffusion is coupled with a reversible reaction contained between two surfaces, there is an increase in the rate of energy exchange between the two surfaces. R Fundamentals of The Energy Flux Consider the flux of energy, e z, across the plane of area c shown below. Ignoring any radiative heat transfer, the energy crossing the plane is the sum of the conductive energy transport, Q, and convective transport resulting from the flow of all the components, F j H j. Energy crossing the plane = Q + F j H j Where: F j = molar flow rate of component j, (mol/s) H j = enthalpy of component j, (kj/mol) ÝQ = conductive heat flow, (kj/s) If one now divides by the area, c, of the plane, the energy flux, e, is e = q + W j H j (R11.3-1) Where: q = conductive transport = Q/ c q = k T, The molar flux of crossing the plane is kj s m (R11.3-) For species W j = F j c = molar flux of component j, (mol/m s) dy W = D B + y ( W + W B ) (R11.3-3) In writing shell energy balances, one now only needs to consider the energy flux e. 1

2 R11.3.B. Conductive Heat Transfer One means of accelerating conductive energy transport in a gas phase is by enthalpy transport. L q T1 T 1 > T 0 T0 Consider two horizontal parallel plates separated by a small distance, L. n inert non-reacting gas is contained between the two plates of which the top plate is heated and maintained at temperature T 1 and the bottom plate cooled and maintained at temperature T 0. Energy is transported from the upper plate to the lower plate solely by conduction. For constant thermal conductivity, the heat flux is We are neglecting any convective that might occur. R11.3.C. Facilitated Heat Transfer q = k ( L T 1 T 0 ) (R11.3-4) Consider what happens when the inert gas is replaced by a dissociating gas, e.g., N O 4. exo NO N O 4 (R11.3-5) endo Near the upper, high-temperature plate, the equilibrium dictates that N O 4 will dissociate to form NO while near the plate the lower temperature shifts the equilibrium so that the NO reacts to form N O 4. s a result the concentration of NO will be higher at the upper plate than at the lower plate, while the concentration of N O 4 will be greater at the lower plate and less at the upper plate. These concentration gradients cause N O 4 to diffuse toward the upper plate and NO molecules to diffuse toward the lower plate. Consider the path of two molecules that have just been formed by the dissociation of one N O 4 molecule at the top plate. These two molecules diffuse toward the bottom plate where they combine to form one molecule of N O 4. Because the reaction of NO going to N O 4 is exothermic, the heat of reaction is released near the cooler plate. The N O 4 molecule then diffuses toward the upper plate where it dissociates to form NO. This dissociation reaction is endothermic. Thus, energy is absorbed by the dissociation reaction at the top plate by the resulting molecules which diffuse to the bottom plate to release the energy upon recombination reaction at the

3 lower plate. The heat energy flux from the top to the bottom plate is substantially greater when the inert gas is replaced by a dissociating gas. R11.3.D. Example of Diffusion and Reaction Facilitated Heat Transfer Consider the one.-dimensional, steady-state transfer of energy through a gaseous binary mixture of NO and N O 4, a rapidly reacting system which has been studied experimentally for this enthalpy transport effect. Suppose that the mixture is confined between two horizontal, parallel, and inert solid walls ( 0 y L), the upper wall (y = 0) being at a higher temperature than the lower wall (y = 0), such that there is no tendency for natural convection to occur. NO = N O 4 B B z = 0 T = T1 T 1 > T 0 B z = L T = T 0 Figure RE Schematic of facilitated heat transfer B Taking as our basis of calculating B Problem Statement The following reaction takes place in between the two parallel plates shown in Figure R NO N O 4 ( H Rx < 0) ssuming that the reaction is at equilibrium everywhere, with P N O 4 = K P P NO then P i = y i P T0 ( ) (RE11.3-1) K P P T0 = K = y N O y NO = y y B 3

4 where y is mole fraction, and assuming that the gases behave ideally, (a) derive an expression for the effective thermal conductivity k eff at any point y, where e z = energy flux = k eff (RE11.3-) such that (k eff k), where k is the ordinary (or frozen ) thermal conductivity of the (unreacting) gas mixture, is independent of composition ( y NO ). Neglect sensible heat effects ( C P = 0) and take H R constant. Solution (b) Make a rough sketch of (k eff k) vs. T indicating any peculiarities. Vant Hoff s Law dlnk eq = dlnk = H Rx RT (RE11.3-3) Energy Balance The energy flux is K = y N O 4 ( y NO ) = y B y dlnk = H Rx RT e z = q z + W jz H j (RE11.3-4) (RE11.3-5) (RE11.3-6) z=0 e e z z+ z z-l Balancing a segment between z and z + z e z e z+ z = 0 (RE11.3-7) Divide by y and take the limit Integrating de z = 0 4 (RE11.3-8)

5 e z = constant = k T z + W zh + W Bz H B (RE11.3-9) For every mole of B that diffuses upward, moles of diffuse downward. W z = W Bz e z = k T W H B H = k z W H Rx where W W z dy W z = cd B W = cd B 1 y dy The energy flux is W z = cd B 1 y e = k T z + cd B 1 y y z H Rx y z (RE ) (RE ) We now need to evaluate y at equilibrium y y = 1 y B K = 1 y (RE ) Solving for y 1+ 4K 1 y = K Differentiating the mole fraction of wrt z. dy = dy dk dk = K Substituting (RE ) into (RE11-.15) ( ) 1+ 4K 1+ K 1+ 4K dk (RE ) (RE ) 5

6 e = k cd B 1 y H Rx ( ) K 1+ 4K 1+ 4K 1+ K Using Van t Hoffs Equation and the chain rule dlnk = dlnk = H Rx RT e z = k + 1 cd B 1+ 4K 1 4K H Rx RT e z = k eff ( k eff k)= cd B H Rx 1 RT 1+ 4K 1+ 4K (1) s T () s T 0 (1) Then K 0 () Then K dlnk ( ) K 1+ 4K 1+ 4K 1+ K 1+ K 1+ 4K 1+ 4K and ( k eff k) 0 and ( k eff k) 0 Note: (1) K goes almost to zero, albeit a very small constant 1 () Note K 0 faster than T so that (K/T ) goes to zero. (RE ) (RE11.3-0) (RE11.3-1) We see the enhancement in the thermal conductivity goes through a maximum Figure RE11.3- Enhanced conductive heat transfer. t high temperatures N O 4 is completely dissociated, or at low temperatures NO is completely dimerized so that in these instances 6

7 k eff k since the enthalpy exchange mechanism no longer occurs in a non-dissociating gas. 7