Question

# Water at 100$^{\circ} \mathrm{C}$, quality 50% in a rigid box is heated to 110$^{\circ} \mathrm{C}$. How do the properties (P, v, x, u, and s) change (increase, stay about the same, or decrease)?

Solution

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We are given following data for Water in rigid tank:

$T_1=100\text{ C}$

$T_2=110\text{ C}$

$x_1=0.5$

From Saturated Water table B.1.1. corresponding to $T_1=100 C$ we can find stage 1 pressure saturation and evaporation specific volume, internal energy and entropy:

$P_1=101.3\text{ kPa}$

$u_f=418.91\frac{\text{ kJ}}{\text{ kg}}$

$u_{fg}=2087.58\frac{\text{ kJ}}{\text{ kg}}$

$\nu_f=0.001044\frac{\text{ m}^3}{\text{ kg}}$

$\nu_{fg}=1.67185\frac{\text{ m}^3}{\text{ kg}}$

$s_f=1.3068\frac{\text{ kJ}}{\text{ kg K}}$

$s_{fg}=6.0480\frac{\text{ kJ}}{\text{ kg K}}$

Calculating initial specific internal energy:

$u_1=u_f+x_1\cdot u_{fg}=418.91+0.5\cdot 2087.58=1462.70\frac{\text{ kJ}}{\text{ kg}}$

Calculating initial specific volume:

$\nu_1=\nu_f+x_1\cdot \nu_{fg}=0.001044+0.5\cdot 1.67185=0.837\frac{\text{ m}^3}{\text{ kg}}$

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