Answer:
v_water=1,0432e-3 m3/kg
v_r22=0,801412 ft3/lb
v_ammonia=1,9167 ft3/lb
Explanation:
To calculate the specific volumes of pure substances like water, refrigerant 22 and Ammonia, the respective properties tables are needed. In my case, I took the values from the following reference: Fundamentals of Engineering Thermodynamics 7th Edition, Moran, Shapiro et al.
In the case of the water, the specific volume v_f can be taken from the saturated water (liquid-vapor), pressure table (Table A-3). Within the table we look for the pressure value and then the v_f specific volume. The values are the following ones:
[tex]P_{w}=100 kPa \\T_{w}=20C\\v_{w}=v_{f}=1,0432*10^{-3}m^{3}/kg[/tex]
Speaking of the refrigerant 22 is a little bit more complicated because of the quality value. First we have identify the following values for the 40 lbf/in2 pressure value (Table A-8E):
[tex]v_{f}=0,01198 ft^{3}/lb\\v_{g}=1,3277 ft^{3}/lb[/tex]
Then the specific volume can be calculated as follows:
[tex]v_{r22} =v_{f}+x(v_{g} -v_{f})=0,01198+0,6(1,3277-0,01198)=0,801412 ft^{3}/lb[/tex]
Regarding the Ammonia, the saturation temperature of this substance is 96,31 °F which means that at 195 °F, we would have superheated ammonia vapor. As the values on the table for the 20 lbf/in2 pressure, are not available at 195 °F we have to look for the two values that the suggested temperature is between (In this case 180 and 220 °F). The following values, for the 20 lbf/in2 pressure value, are needed (Table A-15E):
[tex]T_{1}=180F\\v_{1}=1,8599 ft^{3}/lb\\T_{2}=220F\\v_{2}=2,0114 ft^{3}/lb[/tex]
Then the specific volume can be calculated as follows:
[tex]m=(v_{2}-v_{1})/(T_{2}-T_{1})=(2,0114-1,8599)/(220-180)=3,7875*10^{-3}\\ v_{ammonia}=m(T-T_{1})+v_{1}=3,7875*10^{-3}(195-180)+1,8599=1,9167ft^{3}/lb[/tex]
