Question

For each case, determine the specific volume at the indicated state. a. Water at p = 1 bar, T = 20°C. Find , in m3/kg. b. Refrigerant 22 at p = 40 lbf/in.2, x = 0.6. Find , in ft3/lb. c. Ammonia at p = 200 lbf/in.2, T = 195°F. Find , in ft3/lb.

Answer 1

v_water=1,0432e-3 m3/kg

v_r22=0,801412 ft3/lb

v_ammonia=1,9167 ft3/lb

Step-by-step 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:

`P_(w)=100 kPa \\T_(w)=20C\\v_(w)=v_(f)=1,0432*10^(-3)m^(3)/kg`

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):

`v_(f)=0,01198 ft^(3)/lb\\v_(g)=1,3277 ft^(3)/lb`

Then the specific volume can be calculated as follows:

`v_(r22) =v_(f)+x(v_(g) -v_(f))=0,01198+0,6(1,3277-0,01198)=0,801412 ft^(3)/lb`

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):

`T_(1)=180F\\v_(1)=1,8599 ft^(3)/lb\\T_(2)=220F\\v_(2)=2,0114 ft^(3)/lb`

Then the specific volume can be calculated as follows:

`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`