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Solution Thermodynamics: Applications
Chapter 12-H-x diagrams
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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.
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H = x1H1+x2H2+DH H2 = 0 for pure liquid water At 32F) H1 =0 for pure H2SO4 at 77F Pure H2SO4 Pure water
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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
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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
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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
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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 (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
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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 = 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
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