1
$\begingroup$
isoth = ContourPlot[ {v p } , {v, .12, 2}, {p, .12, 2}]
adiab = ContourPlot[ {v^1.4  p} , {v, .12, 2}, {p, .12, 2}]
Show[{isoth, adiab}, PlotRange -> All]

In an attempt to see heat/work thermodynamics in a Carnot cycle, could not plot hyperbolae type curves together. Please help to Show how the first plot can be made transparent when superimposed on the second and also how to refine level differences say to one fourth the present given default value.

EDIT1:

RegionPlot per Jose Antonio's suggestion for chosen constants

enter image description here

$\endgroup$
  • $\begingroup$ Add the option , ContourShading -> False for the contour plots. $\endgroup$ – Jason B. Nov 14 '17 at 18:31
  • $\begingroup$ Not sure if it's what you want, you may use e.g. Contours -> 20, to refine level differences $\endgroup$ – egwene sedai Nov 14 '17 at 18:43
  • $\begingroup$ Thanks Both fixes have been helpful. Any given answer I accept $\endgroup$ – Narasimham Nov 14 '17 at 18:52
1
$\begingroup$

If you know the equations of your adiabatics and isothermals, you can use:

isoth1 = 1; 
adiab1 = 1; 
isoth2 = 1.5; 
adiab2 = 1.5; 
RegionPlot[v p > isoth1 && v^(1/4) p > adiab1 && v p < isoth2 && v^(1/4) p < adiab2,
{v, .12, 2.5}, {p, .12, 2.5}, PlotPoints -> 100,
FrameLabel -> (Style[#, 14, Bold] & /@ {"V", "P"}), RotateLabel -> False]

enter image description here

$\endgroup$
  • $\begingroup$ RegionPlot .Yes looking for that exactly, before going in for ContourPlot ! $\endgroup$ – Narasimham Nov 14 '17 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.