Question

A piston-cylinder assembly contains ammonia, initially at a temperature of-20°C and a quality of 70%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 180°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg.

Answer 1

w = -28.8 kJ/kg

q = 723.13 kJ/kg

Step-by-step explanation:

Given :

Initial properties of piston cylinder assemblies

Temperature,
`T_(1)` = -20°C

Quality, x = 70%

= 0.7

Final properties of piston cylinder assemblies

Temperature,
`T_(2)` = 180°C

Pressure,
`P_(2)` = 6 bar

From saturated ammonia tables at
`T_(1)` = -20°C we get

`P_(1)` =
`P_(sat)` = 1.9019 bar

`v_(f)` = 0.001504
`m^(3)` / kg

`v_(g)` = 0.62334
`m^(3)` / kg

`u_(f)` = 88.76 kJ/kg

`u_(g)` = 1299.5 kJ/kg

Therefore, for initial state 1 we can find

`v_(1)` =
`v_(f)`+x (
`v_(g)`-
`v_(f)`

= 0.001504+0.7(0.62334-0.001504)

= 0.43678
`m^(3)` / kg

`u_(1)` =
`u_(f)`+x (
`u_(g)`-
`u_(f)`

= 88.76+0.7(1299.5-88.76)

=936.27 kJ/kg

Now, from super heated ammonia at 180°C, we get,

`v_(2)` = 0.3639
`m^(3)` / kg

`u_(2)` = 1688.22 kJ/kg

Therefore, work done, W = area under the curve

`w = \left ((P_(1)+P_(2))/(2)  \right )\left ( v_(2)-v_(1) \right )`

`w = \left ((1.9019+6* 10^(5))/(2) \right )\left ( 0.3639-0.43678\right )`

`w = -28794.52` J/kg

= -28.8 kJ/kg

Now for heat transfer

`q = (u_(2)-u_(1))+w`

`q = (1688.2-936.27)-28.8`

= 723.13 kJ/kg