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Pressure and Temperature

An air water ...

Question

An air water mixture is contained in a rigid, closed vessel with a volume of $35 m^{3}$ at 1.985 bar, 120°C and $ϕ = 10$ %. The mixture is cooled at constant volume until its temperature is reduced to 22°C. What is the specific volume of vapour when condensation just begins to occur?

7.981 m^{3} / k g

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9.145 m^{3} / k g

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12.202 m^{3} / k g

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10.38 m^{3} / k g

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Solution

The correct option is B $9.145 m^{3} / k g$

$ϕ = \frac{P_{v 1}}{P_{s 1}}$

$\Rightarrow P_{v 1} = ϕ P_{s 1}$

= 0.10 × 1.985 = 0.1985 bar

The specific volume of the vapour at state 1 equals the specific volume of the vapour at state 1 since this is a constant volume process.

$v_{1}^{'} = \frac{(¯ R / M_{o}) T_{1}}{P_{v 1}}$

$= (\frac{8314}{18}) \times \frac{393}{0.1985 \times 10^{5}} = 9.145 m^{3} / k g$

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An air water mixture is contained in a rigid, closed vessel with a volume of 35 m3 at 1.985 bar, 120°C and ϕ=10%. The mixture is cooled at constant volume until its temperature is reduced to 22°C. What is the specific volume of vapour when condensation just begins to occur?

The correct option is B 9.145 m3/kg ϕ=Pv1Ps1 ⇒ Pv1=ϕPs1 = 0.10 × 1.985 = 0.1985 bar The specific volume of the vapour at state 1 equals the specific volume of the vapour at state 1 since this is a constant volume process. v′1=(¯R/Mo)T1Pv1 =(831418)×3930.1985×105=9.145 m3/kg

An air water mixture is contained in a rigid, closed vessel with a volume of $35 m^{3}$ at 1.985 bar, 120°C and $ϕ = 10$ %. The mixture is cooled at constant volume until its temperature is reduced to 22°C. What is the specific volume of vapour when condensation just begins to occur?