Question

How do you determine the properties of steam or refrigerant 134a from the properties if you know only specific internal energy and specific volume (or specific enthalpy and specific entropy) at the thermodynamic state?

Answer 1

• If the specific internal energy is given then we have to compare the given specific value of internal energy and the actual specific value of internal energy. If the given value is less than the original value then this is the case of compressed liquid state and if the value is greater than the original value then this is the case for superheated vapour state.
• The same will occurs for specific volume case.
• If the specific entropy is given then similarly again we have to compare with the actual value of specific entropy, if the given specific entropy is less then the actual specific entropy then this is the for compressed liquid state and if the given specific entropy is greater than the actual specific entropy then this is the case for superheated vapour state.
• The same will occurs for specific enthalpy.

In your example, the value of specific volume is missing so I give you another example

for water at 200kPa specific volume is 0.5 m³ /kg

the value of the specific volume of saturated liquid and vapour is v_f = 0.001061 m³ /kg

and v_g = 08857m³ /kg respectively.

So now, the quality x= (v-v_f.)(v-v_g)

x= 0.564 m³ /kg

From the quality, we can find all the other properties, let we will find the enthalpy

h=h_f + x h_fg

or, h= 504.70 + (0.564)*(2201.9)

or, h= 1746.6 kJ/kg