Question

An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20°C to 1000 kPa and 300°C. Determine (a) the work required by the compressor, in kJ/kg, and (b) the power required to drive the air compressor, in kW. Air = 0.287 kPa m3 /kgK. cp,avg = 1.02 kJ/kgC

Answer 1

Answers:

a) $290.875 kJ/kg$

b) $4.07 kW$

Explanation:

a) Since air is considered an ideal gas, its gas constant $R$ is established in a Thermodynamic table as $R=0.287 \frac{kPa m^{3}}{kg K}$.

On the other hand, we have to convert the temperatures from Celsius to Kelvin:

$T_{1}=20\°C + 273.15 K=293.15 K \approx 295 K$

$T_{2}=300\°C + 273.15 K=573.15 K \approx 580 K$

Then, we can find the enthalpy $h$ from another Thermodynamic table (A-17) with this given temperatures in Kelvin:

For $T_{1}\approx 295 K$, $h_{1}=295.17 kJ/kg$ (1)

For $T_{2}\approx 580 K$, $h_{2}=586.04 kJ/kg$ (2)

Now, the work required by the compressor (the work input) $W_{in}$ is given by the following equation for $1 kg$ of air (assuming there is only one inlet and one exit in this system, where the volume is controlled):

$W_{in}=\Delta h=h_{2}-h_{1}$ (3)

$W_{in}=586.04 kJ/kg - 295.17 kJ/kg$

$W_{in}=290.875 kJ/kg$  (4) This is the work required by the air compressor

b) The power input $\dot{W}$ is given by:

$\dot{W}=\dot{m}W_{in}$ (5)

Where $\dot{m}$ is the mass flow rate at the inlet:

$\dot{m}=\frac{\dot{V_{1}}}{V_{1}}$ (6)

We already know the volume flow rate $\dot{V_{1}}$:

$\dot{V_{1}}=10\frac{L}{s}. \frac{1000 mL}{1 L} . \frac{10^{-6} m^{3}}{1 mL}=0.01 m^{3}$ (7)

The volume at the inlet $V_{1}$ is given by the Ideal gas law equation:

$V_{1}=\frac{R T_{1}}{P_{1}}$ (8)

$V_{1}=\frac{(0.287 \frac{kPa m^{3}}{kg K})(295 K)}{120 kPa}$ (9)

$V_{1}=0.705 m^{3}/kg$ (10)

Substituting (7) and (10) in (6):

$\dot{m}=\frac{0.01 m^{3}}{0.705 m^{3}/kg}$ (11)

$\dot{m}=0.014 kg/s$ (12)

Substituting (12) in (5):

$\dot{W}=(0.014 kg/s)(290.875 kJ/kg)$ (5)

Finally:

$\dot{W}=4.07 kK/s = 4.07 kW$