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Diffstat (limited to '3012/CH11/EX11.4')
| -rwxr-xr-x | 3012/CH11/EX11.4/Ex11_4.sce | 35 |
1 files changed, 35 insertions, 0 deletions
diff --git a/3012/CH11/EX11.4/Ex11_4.sce b/3012/CH11/EX11.4/Ex11_4.sce new file mode 100755 index 000000000..0d233d37f --- /dev/null +++ b/3012/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,35 @@ +// Given:- +// Analysis +// For comparison, Table A-2 gives at 100C, +hgf =2257.00 // in kj/kg +ugf = 2087.6 // in kj/kg +sgf = 6.048 // in kj/kg.K +// Values +printf( ' From table, hg-hf = %.2f',hgf); +printf( ' From table, ug-uf = %.2f',ugf); +printf( ' From table, sg-sf = %.2f',sgf); + +// Part(a) +T = 373.15 // in kelvin +// If we plot a graph between temperature and saturation pressure using saturation pressure–temperature data from the steam tables, the desired slope is: +delpbydelT = 3570.00 // in N/(m^2.K) +vg = 1.673 // in m^3/kg +vf = 1.0435e-3 // in m^3/kg +// Calculations +// From the Clapeyron equation +hgf = T*(vg-vf)*delpbydelT*10**-3 // in kj/kg +// Result +printf( '\n Part(a)using Clapeyron equation, hg-hf = %.2f KJ/kg', hgf); + +// Part(b) +psat = 1.014e5 // in N/m^2 +hgf = 2256.00 // can be obtained using IT software in kj/kg +// Calculations +ugf = hgf - psat*(vg-vf)/10**3 // in kj/kg +// Result +printf( '\n Part(b)ug-uf = %.2f KJ/kg',ugf) +// Part(c) +// Calculation +sgf =hgf/T // in kj/kg.K +// Result +printf( '\n Part(c)sg-sf = %.2f KJ/kg. k',sgf) |