Solution Thermodynamics: Applications Chapter 12-H-x diagrams

Enthalpy-concentration diagrams H-x diagram; one curve at each temperature Basis of the diagram: H1 = 0 for some state of species 1; H2=0 for some state of species 2.

H = x1H1+x2H2+DH H2 = 0 for pure liquid water At 32F) H1 =0 for pure H2SO4 at 77F Pure H2SO4 Pure water

H = x1H1+x2H2+DH For any isotherm, you can find DH at a given concentration H1x1+H2X2 DH <0 exothermic solution DH >0 endothermic solution

Adiabatic mixture of solutions A and B Mixing is at constant P or steady flow with no shaft work or changes in PE or KE Solution A Solution B H1 H2 xC DHt =Q =0 With an enthalpy balance and a mass balance, you can show that : Final solution And it can be shown that the three points: (HA, xA); (HB,xB), and (HC, xC) are in straight line

Evaporator, concentrates NaOH solution from 10% (by weight) to 50%, removes vapor. The evaporator operates at constant P. Find the heat transfer rate in the evaporator. Get enthalpies from H-x diagram and from steam tables and solve for Q. Read examples 12.7, 12.8, 12.9

Get H1 from Fig.12.17 (pure H2SO4 at 100F) 150 lbm of H2SO4 are mixed with 350 lbm of an aqueous solution containing 25 wt% H2SO4 in an isothermal process at 100oF. What is the heat effect? Get H1 from Fig.12.17 (pure H2SO4 at 100F) Get H2 for the 25% solution at 100oF Get the wt% of the final solution: (100x m1+25x m2)/(m1+m2) = 47.5% Get m3 and H3 (47.5% at 100F) Solve for Q=m3H3-(m1H1+m2H2)=38150 BTU H1 =8, H2 =-23

Find enthalpies H1 and H2 from Figure 12.19 400 lbm of 37.5% aqueous solution NaOH at 130F is mixed with 175 lbm of 10 wt% solution at 200F. (a) what is the heat effect if the final temperature is 80F. Find enthalpies H1 and H2 from Figure 12.19 Get wt % final solution; 27.32% Get final mass, m3=575 lbm Get final enthalpy, H3 (27.3% and 80 F) Calculate Q =-43025 BTU (b) if the mixing is adiabatic, what is the final T Get H3 from enthalpy balance, and then read T from graph 12.19 H1 =100, H2 = 152 Q=0, H = 116, T= 165 F