Question

A piston-cylinder device contains 2.8 kg of water initially at 400 °C and 1.2 MPa. The water is allowed to cool at constant pressure until 28% ofits mass condenses into liquid. a) Evaluate the final temperature. b) Calculate the initial and final volumes (m3) c) Calculate the enthalpy at the initial and final states (kJ)

Answer 1

a).Final temperature,
`T_(2)` = 180°C

b).Initial Volume,
`V_(1)` = 0.713412
`m^(3)`

Final Volume,
`V_(2)` = 0.33012
`m^(3)`

c). Initial enthalpy,
`H_(1)` =9129.68 kJ

Final enthalpy,
`H_(1)` =6234.76 kJ

Step-by-step explanation:

Given :

Total mass, m= 2.8 kg

Initial temperature,
`t_(i)` = 400°C

Initial pressure,
`p_(i)` = 1.2 MPa

Therefore from steam table at 400°C, we can find--

`h_(1)` = 3260.6 kJ/kg

`v_(1)` = 0.25479
`m^(3)` / kg

Now it is mentioned that 28% of the mass is condensed into liquid.

So, mass of liquid,
`m_(l)` = 0.28 of m

= 0.28 m

mass of vapour,
`m_(v)` = 0.72 m

∴ Dryness fraction, x =
`(m_(v))/(m_(l)+m_(v))`

=
`(0.72 m)/(0.28 m+0.72 m)`

= 0.72

a). The final temperature can be evaluated from the steam table at 1.2 MPa,

`h_(2)` = 2226.7 kJ/kg

`v_(2)` = 0.1179
`m^(3)` / kg

Final temperature,
`T_(2)` = 180°C

b). We know
`v_(1)` = 0.25479
`m^(3)` / kg

∴ Initial Volume,
`V_(1)` =
`v_(1)` x m

`V_(1)` = 0.25479 x 2.8

`V_(1)` = 0.713412
`m^(3)`

We know,
`v_(2)` = 0.1179
`m^(3)` / kg

∴ Final Volume,
`V_(2)` =
`v_(2)` x m

`V_(2)` = 0.1179 x 2.8

`V_(1)` = 0.33012
`m^(3)`

c). We know,

`h_(1)` = 3260.6 kJ/kg

∴ Initial enthalpy,
`H_(1)` =
`h_(1)` x m

= 3260.6 x 2.8

= 9129.68 kJ

`h_(2)` = 2226.7 kJ/kg

∴ Final enthalpy,
`H_(1)` =
`h_(2)` x m

= 2226.7 x 2.8

= 6234.76 kJ