Question

A 1-m3 tank holds a two-phase liquid-vapor mixture of carbon dioxide at – 17 °C. The quality of the mixture is 70%. For saturated carbon dioxide at – 17 °C, vf = 0.9827×10-3 m3/kg and vg =1.756×10-2 m3/kg. Determine the masses of saturated liquid and saturated vapor, each in kg. What is the percent of the total volume occupied by saturated liquid?

Answer 1

Answer:

a)  $m_v= 56.16 Kg$

    $m_l= 14.04 Kg$

b)  $\mu=1.37\%$

Explanation:

From the question we are told that:

Volume of tank $V_t=1m^3$

Temperature of $CO_2=-17^oC$

Quality of the mixture $Q= 70%$

Specific Volume constants at $-17^oC:$

 $v_f = 0.9827*10^{-3} m3/kg$

 $v_g =1.756*10^{-2} m3/kg.$

Generally the equation for Specific Volume is mathematically given by

 $v = v_f + x (v_g -v_f)$

 $v= (0.9827 * 10^{-3} ) + 0.8 * (17.56 * 10^{-3} -0.9827 * 10^{-3})$

 $v= 0.014244 m3/Kg$

Generally the equation for Mass is mathematically given by

 $m=\frac{v'}{v}$

 $m=\frac{1}{0.014244}$

 $m=70.20 Kg$

Generally the Mass of saturated Vapor is mathematically given by

 $m_v=0.8 * (70.202)$

 $m_v= 56.16 Kg$

Generally the Mass of saturated Liquid is mathematically given by

 $m_l = (70.20 Kg)-(56.16 Kg)$

 $m_l= 14.04 Kg$

b)

Generally the equation for Volume is mathematically given by

 $v_l = m_l x v_f$

 $v_l= (14.04 Kg) (0.9827 x 10{-3} m3/kg)$

 $v_l= 0.01379 m^3$

Therefore Percentage of liquid

 $\mu = \frac{v_l}{v} * 100 \%$

 $\mu= [(0.01379 m^3)/(1 m^3)] *100 \%$

 $\mu=1.37\%$