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A 1-m3 tank holds a two-phase liquidvapor mixture of
Chapter 3, Problem 3.16(choose chapter or problem)
A 1-m3 tank holds a two-phase liquid–vapor mixture of carbon dioxide at \(-17^{\circ} \mathrm{C}\). The quality of the mixture is 70%. For saturated carbon dioxide at \(-17^{\circ} \mathrm{C}, v_{\mathrm{f}}=0.9827 \times 10^{-3} \mathrm{m}^{3} / \mathrm{kg} \text { and } v_{\mathrm{g}}=1.756 \times 10^{-2} \mathrm{~m}^{3} / \mathrm{kg}\). Determine the masses of saturated liquid and saturated vapor, each in kg. What is the percent of the total volume occupied by saturated liquid?
Questions & Answers
QUESTION:
A 1-m3 tank holds a two-phase liquid–vapor mixture of carbon dioxide at \(-17^{\circ} \mathrm{C}\). The quality of the mixture is 70%. For saturated carbon dioxide at \(-17^{\circ} \mathrm{C}, v_{\mathrm{f}}=0.9827 \times 10^{-3} \mathrm{m}^{3} / \mathrm{kg} \text { and } v_{\mathrm{g}}=1.756 \times 10^{-2} \mathrm{~m}^{3} / \mathrm{kg}\). 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:Step 1 of 3
Consider the given data:
The volume of the liquid is \({v_f} = 0.9827 \times {10^{ - 3}}\;\frac{{{{\rm{m}}^3}}}{{{\rm{kg}}}}\)
The volume of the vapor is \({v_g} = 1.756 \times {10^{ - 2}}\;\frac{{{{\rm{m}}^3}}}{{{\rm{kg}}}}\).
The dryness fraction is \(x = 70\% = 0.7\).
Calculate the specific volume of the two-phase liquid-vapor mixture.
\({v_T} = {v_f} + x\left( {{v_g} - {v_f}} \right)\)
Plug the provided values.
\({v_T} = 0.9827 \times {10^{ - 3}} + 0.7\left( {1.756 \times {{10}^{ - 2}} - 0.9827 \times {{10}^{ - 3}}} \right)\)
\( = 0.9827 \times {10^{ - 3}} + 0.7 \times \left( {11.60411 \times {{10}^{ - 3}}} \right)\)
\( = 0.9827 \times {10^{ - 3}} + 11.60411 \times {10^{ - 3}}\)
\( = 0.01258\;\frac{{{{\rm{m}}^3}}}{{{\rm{kg}}}}\)
Calculate the total mass of the mixture.
\(m = \frac{V}{{{v_T}}}\)
Here, \(V\) is the total volume.
Substitute \(1\;{{\rm{m}}^3}\) for \(V\)and \(0.01258\;\frac{{{{\rm{m}}^3}}}{{{\rm{kg}}}}\) for \({v_T}\).
\(m = \frac{1}{{0.01258}}\)
\( = 79.46\;{\rm{kg}}\)