Basic Mechanical Engineering (3110006)

Given data
$\mathrm{m} = 1 \mathrm{kg}$
$\mathrm{p} = 15 \mathrm{bar} = 1500 \mathrm{kPa}$
${\mathrm{T}}_{\sup } = 400° \mathrm{C} = 673 \mathrm{K}$
$\mathrm{x} = 0.9$
${\mathrm{C}}_{\mathrm{ps}} = 2.1 \mathrm{kJ} / \mathrm{kg} \mathrm{K}$

To Find: ${\mathrm{u}}_{\sup } = ? , {\mathrm{u}}_{\mathrm{wet}} = ?$

From steam table at 15 bar, We have,  
${\mathrm{t}}_{\mathrm{s}} = 198.3°\mathrm{C} = 471.3 \mathrm{K}$
${\mathrm{h}}_{\mathrm{f}} = 844.7 \mathrm{kJ}/\mathrm{kg}$
${\mathrm{h}}_{\mathrm{fg}} = 1945.2 \mathrm{kJ}/\mathrm{kg}$
${\mathrm{h}}_{\mathrm{g}} = 2789.9 \mathrm{kJ}/\mathrm{kg}$
${\mathrm{v}}_g = 0.1317 {\mathrm{m}}^3/\mathrm{kg}$

Now,

$\frac{{\mathrm{v}}_{\sup }}{{\mathrm{T}}_{\sup }} = \frac{{\mathrm{v}}_{\mathrm{g}}}{{\mathrm{T}}_{\mathrm{s}}}$

${\mathrm{v}}_{sup} = {\mathrm{v}}_{\mathrm{g}} \times \frac{{\mathrm{T}}_{\sup }}{{\mathrm{T}}_{\mathrm{s}}}$

${\mathrm{v}}_{sup}= 0.1317 \times \frac{673}{ 471.3 }$

${\mathrm{v}}_{\sup } = 0.188 {\mathrm{m}}^3/\mathrm{kg}$

Now,

${\mathrm{h}}_{\sup } = {\mathrm{h}}_{\mathrm{g}} + {\mathrm{C}}_{\mathrm{ps}} ( {\mathrm{T}}_{\sup } - {\mathrm{T}}_{\mathrm{s} })$

${\mathrm{h}}_{sup} = 2789.9 + 2.1 ( 673 - 471.3 )$

${\mathrm{h}}_{sup} = 3213.47 \mathrm{kJ}/\mathrm{kg}$

Now,

${\mathrm{u}}_{\sup } = {\mathrm{h}}_{\sup } - {\mathrm{pv}}_{\sup }$

${\mathrm{u}}_{sup} = 3213.47 - 1500 \times 0.188$

${\mathrm{u}}_{sup} = 2913.47 \mathrm{kJ}/\mathrm{kg}$

Now,

${\mathrm{u}}_{\mathrm{wet}} = {\mathrm{h}}_{\mathrm{wet}} - {\mathrm{pv}}_{\mathrm{wet}}$

${\mathrm{u}}_{wet} = ( {\mathrm{h}}_{\mathrm{f}} + \mathrm{x} {\mathrm{h}}_{\mathrm{fg}} ) - \mathrm{p} ( \mathrm{x} {\mathrm{v}}_{\mathrm{g}} )$

${\mathrm{u}}_{wet} = ( 844 + 0.9 \times 1945.2 ) - 1500 ( 0.9 \times 0.1317 )$

${\mathrm{u}}_{\mathrm{wet}} = 2417.59 \mathrm{kJ} / \mathrm{kg}$