A tank of volume 0.05 m³ contains a mixture of saturated water and saturated steam at 200°C. The mass of the liquid present is 8 kg. The entropy (in kJ/kg K) of the mixture is ___________ (correct to two decimal places).

Property data for saturated steam and water are:

At 200°C, P_sat = 1.5538 MPa

v_f = 0.001157 m³/kg, v_g = 0.12736 m³/kg

S_fg=4.1014 kJ/kg K, s_f =2.3309 kJ/kg K

The correct answer is 2.49.

Step-by-Step Explanation:

1. Identify the given data:
Total volume of the tank, $V=0.05m^3$
Temperature of the mixture, $T=200°C$
Mass of the liquid phase, $m_L=8\text{kg}$
Specific volume of saturated liquid at 200°C, $v_f=0.001157m^3/\text{kg}$
Specific volume of saturated vapor at 200°C, $v_g=0.12736m^3/\text{kg}$
Specific entropy of saturated liquid, $s_f=2.3309\text{kJ/kg K}$
Specific entropy of vaporization, $s_{\mathrm{fg}}=4.1014\text{kJ/kg K}$

2. Calculate the volume occupied by the liquid phase ($V_L$):
$V_L=m_L\times v_f$
Substituting the given values:
$V_L=8\times 0.001157=0.009256m^3$

3. Calculate the volume occupied by the vapor phase ($V_v$):
The remaining volume in the tank is occupied by the steam (vapor):
$V_v=V-V_L$
Substituting the values:
$V_v=0.05-0.009256=0.040744m^3$

4. Calculate the mass of the vapor phase ($m_v$):
$m_v=\frac{V_v}{v_g}$
Substituting the values:
$m_v=\frac{0.040744}{0.12736}\approx 0.3199\text{kg}$

5. Calculate the dryness fraction ($x$):
The dryness fraction of the mixture is the ratio of the mass of vapor to the total mass of the mixture:
$x=\frac{m_v}{m_v+m_L}$
Substituting the values:
$x=\frac{0.3199}{0.3199+8}\approx 0.0384$

6. Calculate the specific entropy of the mixture ($s$):
$s=s_f+x\times s_{\mathrm{fg}}$
Substituting the values:
$s=2.3309+0.0384\times 4.1014$
$s=2.3309+0.1575=2.4884\text{kJ/kg K}$

Rounding to two decimal places, the entropy of the mixture is 2.49 kJ/kg K.