A tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains 0.03 m3 of refrigerant-134a that is a saturated liquid at 0.9 MPa, while the other side is evacuated. The partition is now removed, and the refrigerant fills the entire tank. If the final state of the refrigerant is 208C and 280 kPa, determine the volume of the tank

V = 4.859[m^3]

Explanation:

In order to solve this problem, we must find the mass of the refrigerant, as there are no leaks or leaks of refrigerant, we can conclude that the mass is conserved throughout the process.

For saturation conditions, we can find the specific volume, in the attached table of saturation values for pressure of 0.9 [MPa] we can see the specific volume.

with the volume of 0.03 [m^3], we can find the mass.

Now using the superheat tables for the 134a we can find the specific volume

Since it is difficult to find the specific volume value for a pressure of 280 [kPa] & T = 208[C], with over-heated refrigerant. We go to the diagram of this refrigerant to see if the condition is over heating exists to these conditions.

In the second image attached we can see that, there is the superheating condition, highlighted at the end of the red line. And that the specific volume is approximately equal to 0.15[ m^3/kg].

Therefore we can take a normal table of coolant overheating 134a to P= 0.28[ MPa] and make an extrapolation.

the extrapolation will be with the following values:

130 = 0.11512

140 = 0.11818

208 = 0.139 [m^3/kg]

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid