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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Engineering Thermodynamics by Onkar Singh Chapter 6 Example 19")
+m=3;//mass of wet steam in kg
+p=1.4;//pressure of wet steam in bar
+V1=2.25;//initial volume in m^3
+V2=4.65;//final volume in m^3
+T=400;//temperature of steam in degreee celcius
+disp("from steam table,vg=1.2455 m^3/kg,hf=457.99 KJ/kg,hfg=2232.3 KJ/kg")
+vg=1.2455;
+hf=457.99;
+hfg=2232.3;
+disp("specific volume of wet steam in cylinder,v1=V1/m in m^3/kg")
+v1=V1/m
+disp("dryness fraction of initial steam(x1)=v1/vg")
+x1=v1/vg
+x1=0.602;//approx.
+disp("initial enthalpy of wet steam,h1=hf+x1*hfg in KJ/kg")
+h1=hf+x1*hfg
+disp("at 400 degree celcius specific volume of steam,v2=V2/m in m^3/kg")
+v2=V2/m
+disp("for specific volume of 1.55 m^3/kg at 400 degree celcius the pressure can be seen from the steam table.From superheated steam tables the specific volume of 1.55 m^3/kg lies between the pressure of 0.10 Mpa (specific volume 3.103 m^3/kg at400 degree celcius)and 0.20 Mpa(specific volume 1.5493 m^3/kg at 400 degree celcius)")
+disp("actual pressure can be obtained by interpolation")
+p2=.1+((0.20-0.10)/(1.5493-3.103))*(1.55-3.103)
+disp("p2=0.20 MPa(approx.)")
+p2=0.20;
+disp("saturation temperature at 0.20 Mpa(t)=120.23 degree celcius from steam table")
+t=120.23;
+disp("finally the degree of superheat(T_sup)in K")
+disp("T_sup=T-t")
+T_sup=T-t
+disp("final enthalpy of steam at 0.20 Mpa and 400 degree celcius,h2=3276.6 KJ/kg from steam table")
+h2=3276.6;
+disp("heat added during process(deltaQ)in KJ")
+disp("deltaQ=m*(h2-h1)")
+deltaQ=m*(h2-h1)
+disp("internal energy of initial wet steam,u1=uf+x1*ufg in KJ/kg")
+disp("here at 1.4 bar,from steam table,uf=457.84 KJ/kg,ufg=2059.34 KJ/kg")
+uf=457.84;
+ufg=2059.34;
+u1=uf+x1*ufg
+disp("internal energy of final state,u2=u at 0.2 Mpa,400 degree celcius")
+disp("u2=2966.7 KJ/kg")
+u2=2966.7;
+disp("change in internal energy(deltaU)in KJ")
+disp("deltaU=m*(u2-u1)")
+deltaU=m*(u2-u1)
+disp("form first law of thermodynamics,work done(deltaW)in KJ")
+disp("deltaW=deltaQ-deltaU")
+deltaW=deltaQ-deltaU
+disp("so heat transfer(deltaQ)in KJ")
+deltaQ
+disp("and work transfer(deltaW)in KJ")
+deltaW
+disp("NOTE=>In book value of u1=1707.86 KJ/kg is calculated wrong taking x1=0.607,hence correct value of u1 using x1=0.602 is 1697.5627 KJ/kg")
+disp("and corresponding values of heat transfer= 4424.2962 KJ and work transfer=616.88424 KJ.")
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