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Chap08prac - This one way to make “dry ice.” You...

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P. 8.1) CO2 at 15MPa and 25 ° C is throttled to 0.1Mpa. Determine the temperature and fraction vapor. (NOTE: there is a typo. Part (c ) makes no sense at 1.5MPa.) Solution: Ebal for a valve, 0 = H (a) assume ideal gas, ( ) K T T Cp H 298 0 2 = = = (b) by PREOS.XLS with Ref=224.1K,0.1MPa, “Liquid”. H2 = H1 = 9695 J/mol, then “T2sat” = 184.1K. q=(9695-0)/(17066-0)= 57% Note: This is just an estimate, because CO2 is not really liquid at 0.1MPa Note2: If you solved this using 1.5MPa, you get 281K. ( c ) Note: for the problem statement at 1.5MPa, there is no difference from (b). At 5.27bar, 216.6K, Hsolid=HsatVap-Hsub, Hsub=Hvap+Hfus=17829-2420+43.2*44*4.184=23362 Hsolid=17829-23362= -5533. C-Ceq ln(Psat/Pref)= -(Hsub/R)*(1/T-1/Tref) 1/T=1/216.6-ln(1/5.27)/(23362/8.314)=0.005208 T=192 At 1bar, 192K=HsatVap=17322 Hsolid=17322-23362= -6040 q = (9695+6040)/(23362) = 67% FYI: This one way to make “dry ice.”
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Unformatted text preview: This one way to make “dry ice.” You throttle/spray it into “bag” and compress it into a block. See also CO2 chart in Perry’s Handbook, Chapter 3. P8.2) CO is liquefied from 300bar,150K to 1bar. Compute q and Sgen. Soln: Ref=82.0K,0.1MPa,liq. H(150,30)=4139. Ebal: ∆ H=0; q=(4139-0)/(6048-0)=68% Sf=0.68*73.83+0 = 50.2; S(150,30)=27.365 ⇒ Sgen=50.2-27.4 = 22.8 J/mol-K. P8.3) Same CO using 90% efficient turbine. ∆ Srev=0 ⇒ q=27.635/73.83=0.374; Wrev=0.374*6048-4139= -1877J/mol ⇒ Wact=-1689 Hout=4139-1689=2450 ⇒ q = 2450/6048=40.5% ⇒ 1-q=59.5% P8.4) Methane at 300K, 250bar is liquefied (~Linde style) with outlet pressure = 30bar. Compute fraction liquefied. Soln: Ref=111.8K,1bar; qH8+(1-q)H6=H3=10654; Tsat(30bar)=176.7K HsatVap=8289;HsatLiq=4104; H8=H(300,3MPa)=13455; q=(10654-4104)/(13455-4104)=70% ⇒ 1-q=30%...
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