Question

Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 160 kPa. Determine the page 253temperature drop during this process and the final specific volume of the refrigerant.

Answer 1

The correct solution will be "0.0344 m³/kg". The further explanation is given below.

Step-by-step explanation:

The given values are:

`P_1=700 \ Kpa`

`=0.7 \ Kpa`

From table,

`T_1=26.69^(\circ) C`

`h_f=88.82 \ kJ/kg`

`P_2=160 \ Kpa`

`h_f=31.21 \ kJ/kg`

`T_(sat) = -15.6^(\circ)C`

Now,

`\Delta T=T_(in)-T{out}`

`=-15.6-26.69`

`=-42.29^(\circ)C`

`V=V_f+x(V_g-V_f)`

`=V_f+(h-h_f)/(h_g-h_f)(V_g-V_f)`

On putting the value, we get

`=(0.0007437)+((88.82-31.21)/(241.11-31.21))(0.12348-0.0007437)`

`=0.0344 \ m^3/Kg`