Question

The exergy of 0.4 kg of wet steam at 0. 1 bar and 0.85 quality is ____ kJ .
(Assume dead state of 1 bar and 300 K)

Data :
At 1 bar, 300 K
$h_0=113.2kJ/kg;s_0=0.395kJ/kgK$

At 0.1 bar, 0.85 quality

$h_1=192kJ/kg;h_g=2582kJ/kg$

$s_f=0.649kJ/kgK;s_g=8.148kJ/kgK$

• A. 48.8 kJ
• B. 57.9 kJ
• C. 11 2.6 kJ
• D. 93.8 kJ

Answer 1

The correct option is A 48.8 kJ

At dead state: $T_0=300K,h_0=113.2kJ/kg,s_0=0.395kJ/kgK,m=0.4kg$

Exergy of system, $\varphi =m\left[\right(h-h_0)-T_0(s-s_0\left)\right]$

Now, $h_0-T_0{s}_0=113.2-(300\times 0.395)$

$=-5.3kJ/kg$

At 0.1 bar, 0.85 steam quality,

$h=h_f+x[h_g-h_f]$

$=192+0.85[2582-192)$

$\Rightarrow$ $h=2223.5kJ/kg$

$s=s_f+0.85[s_g-s_f]$

$=0.649+0.85[8.148-0.649]$

$=7.023kJ/kgK$

Now, $h-T_{0}s=2223.5-300\times 7.023$

$=116.6kJ/kg$

Now, Exergy, $\varphi =m[116.6-(-5.3\left)\right]$

$=0.4\times 121.9=48.76kJ$