Question

An insulated 8m* rigid tank contains air, modeled as an ideal gas, at 600kPa and 400K. A valve connected to the tank is opened and air is allowed to escape until the pressure drops to 200kPa. The air temperature is held constant during the process by an electric heater placed in the tank. Determine the electrical energy supplied to the air during this procesS. Internal energy of the initial air is U = 286.1 6hĺ kg ; The enthalpy of the exiting air is Hvit400.98HJ/ kg

Answer 1

Step-by-step explanation:

We assumes that all the electrical energy produced will be converted into the heat energy.

Electrical energy input = Heat energy absorbed by object = Heat energy required = Q

As, Q = U - W

= U - RT
`ln (P_(2))/(P_(1))`

Since, it is given that U = 286.1 kJ/kg,
`P_(1)` is 600 kPa,
`P_(2)` is 200 kPa, and T is 400 K.

Therefore, putting these given values into the above formula is as follows.

Q = U - RT
`ln (P_(2))/(P_(1))`

=
`286.1 kJ/kg - 8.314 kJ/kmol K * 400 K * ln (200)/(600)`

=
`286.1 kJ/kg - (8.314)/(28.97) kJ/kg K * 400 K * ln (200)/(600)`

= 412.27 kJ/kg

Thus, we can conclude that the electrical energy supplied to air is 412.27 kJ/kg.