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

Answer:

A) Flow rate = 2.783 kg/s

B) Power input = 1088.65 Kw

Explanation:

A) P1 = 100KPa; v = 30m/s; T1 = 20°C = 273 + 20k = 293K; A = 0.078 m²; R= 0.287 kPa.m³/kg

The mass flow rate is given as;

m' = ρ1V1' = (P1A1v1)/RT1

Thus, m' = (100 x 0.078 x 30)/(0.287 x 293) = 234/84.091 = 2.783 kg/s

B) The power input is going to be derived from the energy balance equation which is;

m'(h1 + v1²/2) + W' = m'h2

Factorizing out, we have;

W' = m'(h2 - h1 - v1²/2)

Now we have to find the specific enthalpies at temperature of 293K and 400 + 273 = 673K

Thus, looking at the table i attached for ideal gas properties of air,

At T =293K,when we interpolate, we have h1 approximately = 293. 166 Kj/kg and at T=673K,when we interpolate, we have h2 approximately = 684.344 Kj/Kg

Thus, since W' = m'(h2 - h1 - v1²/2)

W' = 2.783(684.344 - 293. 166) = 1088.65 Kw