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Process of Liquefying of air composed of 79% Nitrogen & 21% Oxygen from Gaseous to Liquid Phases as Follows:

1) When Temperature is higher than T>83 k they both are gas.

2) at T=83 k and less slightly the phase transformation take place to right O2 and to Left N2. as expected N2 Liquefy faster than O2, Where mixture are presents of both substances.

3)Finally at T=73 k both Substances are liquid with slight percentage of O2 gas.

Solved! How do i plot last part of the code my solution on same plot.

Here's the question to calcify things

enter image description here

Here is my full code!

t1[x_] := 77.4 + 25.27*x - 33.66*x^2 + 51.52*x^3 - 42.65*x^4 + 12.32*x^5; 

t2[x_] := 77.4 + 8.372*x - 6.162*x^2 + 14.62*x^3 + 2.201*x^4 - 6.235*x^5; 


    p1 = Plot[{t1[x], t2[x]}, {x, 0, 1}, Frame -> True, PlotRange -> {{0, 1}, {76, 92}}, 
   PlotLegends -> "Expressions", ImageSize -> Large]

no = 40*0.21; 
nn = 40*0.79; 

x1 = FindRoot[t1[x] == 78, {x, 1}]
x2 = FindRoot[t2[x] == 83, {x, 1}]
xg = x /. x1
xl = x /. x2


sol = Flatten[Thread[Solve[{xg == ngo/(ngn + ngo), xl == nlo/(nln + nlo), 
       no == ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}, Reals]]] /. Rule -> Equal

Plot must be similar to Figure

enter image description here

enter image description here

enter image description here

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  • $\begingroup$ You have not yet received a solution to the problem. Use xg = x /. FindRoot[t1[x] == 78, {x, 1}]; xl = x /. FindRoot[t2[x] == 83, {x, 1}]; sol = NSolve[{xg == ngo/(ngn + ngo), xl == nlo/(nln + nlo), no == ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}, Reals] $\endgroup$ Mar 19, 2019 at 12:24
  • $\begingroup$ i did it please check right now. Thanks for passing brother :) $\endgroup$
    – nufaie
    Mar 19, 2019 at 12:28
  • $\begingroup$ {ngo, nlo, ngn, nln} - is it the number of moles of oxygen and nitrogen in the gas and liquid phase? $\endgroup$ Mar 19, 2019 at 12:34
  • $\begingroup$ yes sir exactly $\endgroup$
    – nufaie
    Mar 19, 2019 at 12:35
  • $\begingroup$ it's ok if you eliminate after solution two point of nitrogen {ngn, nln} $\endgroup$
    – nufaie
    Mar 19, 2019 at 12:43

1 Answer 1

1
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t1[x_] := 77.4 + 25.27*x - 33.66*x^2 + 51.52*x^3 - 42.65*x^4 + 12.32*x^5; 
t2[x_] := 77.4 + 8.372*x - 6.162*x^2 + 14.62*x^3 + 2.201*x^4 - 6.235*x^5; 

pp1 = Plot[{t1[x], t2[x]}, {x, 0, 1}, Frame -> True, PlotRange -> {{0, 1}, {76, 92}}, 
    PlotLegends -> "Expressions", ImageSize -> Large];


"Number of moles 21% nOxygen 79% nNitrogen"
no = 40*0.21; 
nn = 40*0.79; 



"Part Liquefying of Oxygen"


####PURE*GAS####
xg1 = x /. FindRoot[t1[x] == 83, {x, 1}]

sol1 = Flatten[Thread[Solve[{xg1 == ngo/(ngn + ngo), xl1 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g1 = Graphics[{Arrow[{{xg1, 90}, {xg1, 83}}]}]; 

p1 = Graphics[{Blue, PointSize[0.01], Point[{xg1, 83}]}];



####MIX####

xm1 = x /. FindRoot[t2[x] == 83, {x, 0.2, 0.6}]

sol2 = Flatten[Thread[Solve[{xg1 == ngo/(ngn + ngo), xl1 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g2 = Graphics[{Arrow[{{xg1, 83}, {xm1, 83}}]}];

p2 = Graphics[{Green, PointSize[0.01], Point[{xm1, 83}]}]; 



####PURE*LIQUID####

xl1 = x /. FindRoot[t2[x] == 78, {x, 0.6, 0}]

sol3 = Flatten[Thread[Solve[{xg1 == ngo/(ngn + ngo), xl1 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g3 = Graphics[{Arrow[{{xm1, 83}, {xl1, 78}}]}];

p3 = Graphics[{Red, PointSize[0.01], Point[{xl1, 78}]}];




"Part Liquefying of Nitrogen"


####PURE*GAS####
xg2 = x /. FindRoot[t1[x] == 83, {x, 1}]

sol4 = Flatten[Thread[Solve[{xg2 == ngo/(ngn + ngo), xl2 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g4 = Graphics[{Arrow[{{xg2, 90}, {xg2, 83}}]}];

p4 = Graphics[{Blue, PointSize[0.01], Point[{xg2, 83}]}]; 


####MIX####

xm2 = x /. FindRoot[t1[x] == 78, {x, 0.2, 0.6}]

sol5 = Flatten[Thread[Solve[{xg2 == ngo/(ngn + ngo), xl2 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g5 = Graphics[{Arrow[{{xg2, 83}, {xm2, 78}}]}]; 

p5 = Graphics[{Green, PointSize[0.01], Point[{xm2, 78}]}];


####PURE*LIQUID####
xl2 = x /. FindRoot[t2[x] == 78, {x, 0.6, 0}]
sol6 = Flatten[Thread[Solve[{xg2 == ngo/(ngn + ngo), xl2 == (nlo/(nln + nlo))*no == 
        ngo + nlo, nn == ngn + nln}, {ngo, nlo, ngn, nln}]]] /. Rule -> Equal

g6 = Graphics[{Arrow[{{xm2, 78}, {xl2, 78}}]}]; 

p6 = Graphics[{Red, PointSize[0.01], Point[{xl2, 78}]}]; 



"Graph"
    Show[{pp1, g1, g2, g3, p1, p2, p3, g4, g5, g6, p4, p5, p6}, 
      {FrameLabel -> {"X", "T(k)"}}]

enter image description here

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