Question #128455

Freon 12 is contained in a sealed glass container at 50°C. As it is cooled, vapor droplets arc noted condensing on the sidewalls at 20°C. Find the original pressure in the container. Arts. 650 kPa

Expert's answer

At state 1

Temperature = T_{1} = 50^{0}C

Pressure = P_{1} = ?kpa

Specific volume = V_{1} = ? m^{3}/kg

At state 2

Temperature = T_{2} = 20^{0}C

Pressure = P_{2} = ?kpa

Specific volume = V_{2} = ? m^{3}/kg

Solution

Control volume = fixed volume (v) and fixed mass (m) at 50^{0}C

Process- cooling 20^{0}C at constant volume we assume saturated process

At state 2

v_{2} at 20^{0}C

So, v_{2} = 0.0308m^{3}/kg

At state 1

v_{1} = v_{2} = 0.0308m^{3}/kg

From super heated table of R-12

at pressure of 625 kpa volume is 0.0305 m^{3}/kg

at pressure of 600kpa volume is 0.0489 m^{3}/kg

so interpolating between those two points we will get pressure at state 1

i.e P_{1}

p_{1} = 625 + {0.0489/0.0305 -0.0308/0.0305} (625-600)

P_{1} = 649.789Kpa = 650kpa

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