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Two kg of a two phase liquid vapor mixture of carbon dioxide(co2) exists at a - 40C in a 0.05 tank

User Origds
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The question is incomplete. Here is the complete question.

Two kg of a two phase liquid vapor mixture of carbon dioxide (CO₂) exists at -40°C in a 0.05m³ tank. Determine the quality of the mixture, if the values of specific volume for saturated liquid and saturated vapor CO₂ at -40°C are
v_(f) = 0.896 x 10⁻³m³/kg and
v_(g)= 3.824 x 10⁻²m³/kg, respectively.

Answer: x = 1

Step-by-step explanation: In a phase change of a pure substance, at determined pressure and temperature, the substance exists in two different phases: saturated liquid and saturated vapor.

Quality (x) is the ratio of saturated vapor in the mixture and can be written as:


x=(m_(vapor))/(m_(liquid)+m_(vapor))

It has value between 0 and 1: x = 0 for saturated liquid and x = 1 for saturated vapor.

When related with volumes, quality is rearranged as:


x=(v-v_(f))/(v_(g)-v_(f))

Solving for x:


x=(0.05-0.896.10^(-3))/(3.824.10^(-2)-0.896.10^(-3))


x=(0.049104)/(0.037344)

x = 1.3

Quality of mixture of carbon dioxide is x = 1, which means it's for saturated vapor.

User Mukama
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