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//(12.8) An air–water vapor mixture is contained in a rigid, closed vessel with a volume of 35 m3 at 1.5 bar, 120C, and psi = 10%. The mixture is cooled at constant volume until its temperature is reduced to 22C. Determine (a) the dew point temperature corresponding to the initial state, in C, (b) the temperature at which condensation actually begins, in C, and (c) the amount of water condensed, in kg.
//solution
//variable initialization
V = 35 //volume of the vessel in m^3
p1 = 1.5 //in bar
T1 = 120 //in degree celcius
psi1 = .1
T2 = 22 //in degree celcius
//part(a)
//The dew point temperature at the initial state is the saturation temperature corresponding to the partial pressure pv1. With the given relative humidity and the saturation pressure at 120C from Table A-2
pg1 = 1.985
pv1 = psi1*pg1 //partial pressure in bar
//Interpolating in Table A-2 gives the dew point temperature as
T = 60 //in degree celcius
printf('the dew point temperature corresponding to the initial state, in degee celcius is: %f',T)
//part(b)
Rbar = 8314 //universal gas constant
Mv = 18 //molar mass of vapor in kj/kmol
vv1 =((Rbar/Mv)*(T1+273))/(pv1*10^5) //the specific volume of the vapor at state 1 in m^3/kg
//Interpolation in Table A-2
Tdash = 56 //in degrees
printf('\n\nthe temperature at which condensation actually begins in degree celcius is: %f',Tdash)
//part(c)
mv1 = V/vv1 //initial amount of water vapor present in kg
//from table
vf2 = 1.0022e-3
vg2 = 51.447
vv2 = vv1 //specific volume at final state
x2 = (vv2-vf2)/(vg2-vf2) //quality
mv2 = x2*mv1 //the mass of the water vapor contained in the system at the final state
mw2 = mv1-mv2
printf('\n\n the amount of water condense in kg is: %f',mw2)
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