Question

Ammonia is contained in a closed, rigid 5-m3 tank initially a two-phase liquid-vapor mixture with a pressure of 2 bar. The tank contents are heated and at the final state the pressure is 4 bar and the specific volume is 0.39550 m3/kg. The effects of kinetic and potential energy are negligible. For the ammonia as the closed system, determine the amount of energy transfer by heat, in kJ per kg of ammonia.

Answer 1

`619.917 kJ/kg`

Step-by-step explanation:

As the tank is closed and rigid, so, the mass inside the tank is fixed and the volume,
`V`, remains constant which is

`V=5 m^3`

For the pressure, P=2 bar, the thermodynamic state of ammonia on the boiling curve are

The specific volume of the liquid ammonia,
`v_f=0.0015 m^3/kg`

The specific volume of the vapor ammonia,
`v_g=0.595 m^3/kg`

The specific enthalpy of the liquid ammonia,
`h_f=113.464 kJ/kg`

The specific enthalpy of the vapor ammonia,
`h_g=1439.291 kJ/kg`

At pressure, P=4 bar

The given specific volume,
`v= 0.39550 m^3/kg`.

Let
`v_1` and
`v_2` be the specific volumes of Ammonia in the tank. As the tank is closed, so there is no flow of Ammonia to or from the tank. Hence the mass remains constant. Also, the volume of the tank is constant, so the specific volume of Ammonia in the tank remains constant.

Hence,
`v_1=v_2=0.39550 m^3/kg`

Let
`x_1` be the vapor quality for initial condition, so

`x_1=(v_1-v_f)/(v_g-v_f)`

`\Rightarrow x_1= (0.39550-0.0015)/(0.595-0.0015)=0.664`

So, the initial specific energy, the liquid-vapor mixture of Ammonia have, is

`e_1=h_f+x_1(h_g-h_f)`

`\Rightarrow e_1=113.464+0.664(1439.291-113.464)=993.813 kJ/kg`

Now, at pressure, P=4 bar and
`v_2=0.39550 m^3/kg`, total amount exist in the vapor phase having the specific enthalpy
`1613.73 kJ/kg`

So, the final energy, Ammonia have, is

`e_2=1613.73 kJ/kg`

So, change in energy per kilogram of Ammonia is

`\Delta e=e_2-e_1=1613.73-993.813=619.917 kJ/kg`.

As the change of kinetic and potential energy are negligible and work done by the system (Ammonia in the tank) is zero due to the rigid tank, so the energy added to the tank can cause only change in energy of the system (Ammonia in the tank).

Hence, the amount of energy transfer by heat

`= \Delta e= 619.917 kJ/kg`.