Question

Thermodynamics

A piston-cylinder system contains 0.005 m3 of liquid water and 0.8 m3 of water vapor in equilibrium at 600 kPa. Heat is transferred at constant pressure until the temperature reaches 200ºC. a- Determine the initial thermodynamic state of the system. b- Find the initial temperature of water. c- Calculate the total mass of water in the cylinder. d- Find the final volume of the system. e- Sketch the above thermodynamic process on a P-V phase diagram.

Answer 1

The process is isobaric and that the cylinder is a closed system.

Since the process is isobaric, state 2 is a saturated vapor at 300 kPa and its properties can be found in the saturation pressure table.

V1 (liquid) = 0.005 m3

V1 (vapour) = 0.8 m3

P1 = P2 = 600 kPa

T2 = 200 oC

(b). Liquid water and water vapor exists in equilibrium at 600 kPa.

So initial temperature = T(saturation) at 600 kPa

T(initial) = T(saturation) at 600 kPa

T(initial) = 158.83 oC

(c). At 600 kPa pressure and 158.83 oC saturation temperature:

V (saturated liquid) = 0.001101 m3/kg

V (saturated vapor) =  0.31560 m3/kg

Mass of saturated liquid = 0.005 / 0.001101

= 4.54 kg

Mass of saturated vapor = 0.8 / 0.31560

= 2.53 kg

Mass fraction of vapor = 2.53 / (4.54 + 2.53)

= 0.358 kg vapor / kg

Total mass of water in the cylinder = 4.54 + 2.53

= 7.07 kg

(d). T2 = 200 oC

P2 = 600 kPa

Since, T2 > T(saturation)

From superheated stem table:

specific volume, $\hat{V}$2 = 0.35212 m3/kg

Final volume, V2 = mass * $\hat{V}$2

= 7.07 * 0.35212

V2 = 2.49 m3

(e). P-V phase diagram:

(kPa) T1 158.83 oC Critical point vapor Subcooled 2 iquid Two-Phase Envelope Saturated vapor Saturated liquid V (m3/kg)