Question

Nitrogen gas is compressed adiabatically in a steady-flow, steady-state process from 0.1 MPa and 25^oC at a rate of 0.2 kg/s. The air exits the compression at a temperature of 125^oC. Assuming cold-air constant properties, use the values found in Table A-2(a) and determine the exit pressure [kPa] by using the polytropic relationship using n=k.   

Answer 1

Answer:

The solution is given below in the fig.

Here the mass flow rate is put in the place of volume because kg/s or kg/m^3 remains constant during the process and hence no effect on the output.

Answer 2

To determine the exit pressure (P_exit) of nitrogen gas using the polytropic relationship with n = k (cold-air constant), we can use the following formula for an adiabatic process:

P_exit = P_initial * (T_exit / T_initial)^k

where: P_exit is the exit pressure of nitrogen gas (in kPa) P_initial is the initial pressure (0.1 MPa) T_exit is the exit temperature (125°C) T_initial is the initial temperature (25°C) k is the cold-air constant, which can be found in Table A-2(a) (assume constant for nitrogen gas)

First, let's convert the temperatures to Kelvin:

T_exit = 125 + 273.15 = 398.15 K T_initial = 25 + 273.15 = 298.15 K

Next, let's find the value of the cold-air constant (k) for nitrogen gas from Table A-2(a). I don't have access to Table A-2(a) as it is not included in the response. However, for nitrogen gas, the specific heat ratio (γ) is approximately 1.4, which is equal to k.

Now, we can calculate the exit pressure (P_exit):

P_exit = 0.1 MPa * (398.15 K / 298.15 K)^1.4

P_exit ≈ 0.1 MPa * (1.333)^1.4

P_exit ≈ 0.1 MPa * 1.737

P_exit ≈ 0.1737 MPa

Finally, let's convert the exit pressure to kPa:

P_exit ≈ 0.1737 MPa * 1000 kPa/MPa

P_exit ≈ 173.7 kPa

The exit pressure of nitrogen gas is approximately 173.7 kPa.