Question

Air is compressed by an adiabatic compressor from 100 kPa and 20°C to 1.8 MPa and 400°C. Air enters the compressor through a 0.15-m2 opening with a velocity of 30 m/s. It exits through a 0.078-m2 opening. Calculate the mass flow rate of air and the required power input. The constant pressure s

Answer 1

Answer:

(a) the mass flow rate of air is 5.351 kg/s

(b) the input power required is 2090.786 kW

Explanation:

Given;

initial pressure, P₁ = 100 kPa

initial temperature, T₁ = 20 °C

Final pressure, P₂ = 1.8 MPa

Final temperature, T₂ = 400 °C

Inlet area of the compressor = 0.15 m²

outlet area of compressor = 0.078 m²

velocity of air = 30 m/s

Part (a) mass flow rate of air through the inlet

Mass flow rate = Area x velocity = density x volumetric rate

m = Av = ρV

from ideal gas law, PV = nRT and ρ = m/V

substitute these values in the above equations, we will have;

$m = \frac{PAv}{RT}$

$m = \frac{100000*0.15*30}{287*(20+273)}= 5.351 \ kg/s$

Part (b) the required power input

$W + m(h_1+\frac{v_1{^2}}{2}) = mh_2$

where;

W is the input power

m is the mass flow rate

h₁ is the initial enthalpy

h₂ is the final enthalpy

initial and final enthalpy are obtained from steam table using interpolation;

h₁ = 293.166 kJ

h₂ = 684.344 kJ

$W + m(h_1+\frac{v_1{^2}}{2}) = mh_2\\\\W = mh_2 - m(h_1+\frac{v_1{^2}}{2})\\\\W = 5.351 ( 684.344) - 5.351 (293.166 + \frac{30^2}{2000}) \\\\W = 3661.925 \ kW -1571.139 \ kW\\\\W = 2090.786 \ kW$

Answer 2

The mass flow rate of air and the required power input are 5.35 kg/s and 2,090.40 kW respectively.

Given the following data:

• Initial pressure = 100 kPa.

• Initial temperature = 20°C to K = 273 + 20 = 293 K.

• Final pressure = 1.8 MPa.

• Final temperature = 400°C to K = 273 + 400 = 673 K.

• Inlet area = 0.15 $m^2$.

• Velocity = 30 m/s.

• Outlet area = 0.078 $m^2$.

Scientific data:

• Initial enthalpy = 293.166 kJ.

• Final enthalpy = 684.344 kJ.

• Ideal gas constant = 287 J/kg.K.

How to calculate the mass flow rate of air.

For an ideal gas, the mass flow rate of air is given by this formula:

$M=Av = \rho V\\\\M=\frac{PAv}{RT}$

Substituting the given parameters into the formula, we have;

$M=\frac{100\times 10^3 \times 0.15 \times 30}{287 \times 293} \\\\M=\frac{450000}{84091}$

M = 5.35 kg/s.

How to calculate the power input.

Mathematically, the input power for an adiabatic process is given by this formula:

$P=Mh_2-M(h_1+\frac{v^2}{2000} )\\\\P=5.35(684.344)-5.35(293.166+\frac{30^2}{2000} )\\\\P=3661.2404-5.35(293.166+0.450)\\\\P=3661.2404-1570.8456$

P = 2,090.40 kW.

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