Question

g Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 132.82 kPa. Determine the temperature drop during this process and the final specific volume of the refrigerant. The inlet enthalpy of R-134a is 88.82 kJ/kg and saturation temperature is 26.69°C.

Answer 1

Answer:

$\Delta T=-46.73^0C$

$vmix=0.04481\frac{m^3}{kg}$

Explanation:

Hello,

In this case, since the process is isoenthalpic the given enthalpy remains constant, besides, the temperature at the 132.82 kPa is computed by interpolating the following data:

$\left[\begin{array}{ccc}P(kPa)&T(^0C)\\120&-22.32\\140&-18.77\\132.82&x=-20.044\end{array}\right]$

Thus, the temperature drop is:

$\Delta T=T_2-T_1=-20.044^0C-26.69^0C=-46.73^0C$

Moreover, the volume is obtained by, at first, computing the thermodynamic data at -20.044 °C by interpolating:

$\left[\begin{array}{ccccc}T(^0C)&hf(kJ/kg)&hfg(kJ/kg)&vf(m^3/kg)&vg(m^3/kg)\\-22.32&22.49&214.48&0.0007324&0.16212\\-18.77&27.08&212.08&0.0007383&0.14014\\-20.044&x=25.43&y=212.94&z=0.0007362&w=0.14802\end{array}\right]$

Then, by knowing the mixture's quality based on the mixture's constant enthalpy:

$x=\frac{88.82-25.43}{212.91}=0.30$

And finally, the mixtures volume:

$vmix=0.0007362+0.3*0.14802=0.04481\frac{m^3}{kg}$

Best regards.