Answer to Question #228985 in Molecular Physics | Thermodynamics for Unknown346307

Question

12 kg of air per minute is delivered by a centrifugal air compressor. The
inlet and outlet conditions of air are C1 = 12 m/s, p1 = 1 bar, v1 = 0.5 m3/kg and C2 = 90 m/s,
p2 = 8 bar, v2 = 0.14 m3/kg. The increase in enthalpy of air passing through the compressor is
150 kJ/kg and heat loss to the surroundings is 700 kJ/min.
Find : (i) Motor power required to drive the compressor ;
(ii) Ratio of inlet to outlet pipe diameter.
Assume that inlet and discharge lines are at the same level.

Answer 1

Quantity of air delivered by the compressor

$m= \frac{12}{60}= 0.2 \;kg/s$

Conditions of air at the inlet 1:

Pressure $p_1=1 \;bar = 10^5 \;N/m^2$

Specific volume $v_1=0.5 \;m^3/kg$

Velocity $C_1=12 \;m/s$

Temperature at inlet:

$T_1= \frac{p_1v_1}{R} = \frac{10^5 \times 0.5}{287}=174.55 \;K$

Conditions of air at the outlet 2:

Pressure $p_2= 8 \;bar = 8 \times 10^5 \;N/m^2$

Specific volume $v_2=0.14 \;m^3/kg$

Velocity $C_2= 90 \;m/s$

Increase in enthalpy of air passing through the compressor,

$(h_2-h_1)=150 \;kJ/kg$

Heat lost to the surroundings,

$Q= -700 \;kJ/min = -11.67 \;kJ/s$

(i) Motor power required to drive the compressor:

Applying energy equation to the system,

$m(h_1+ \frac{C_1^2}{2} + Z_1g) + Q = m(h_2 + \frac{C_2^2}{2} + Z_2g)+ W \\ Z_1=Z_2 \\ m(h_1 + \frac{C_1^2}{2}) + Q = m(h_2 + \frac{C_2^2}{2}) + W \\ W = m[(h_1-h_2) + \frac{C^2_1-C^2_2}{2}] +Q \\ = 0.2[-150 + \frac{12^2 -90^2}{2 \times 1000}] + (-11.67) \\ = -42.46 \;kJ/s = -42.46 \;kW$

Motor power required (or work done on the air) = 42.46 kW

(ii) Ratio of inlet to outlet pipe diameter, $\frac{d_1}{d_2}:$

The mass of air passing through the compressor is given by

$m = \frac{A_1C_1}{v_1}= \frac{A_2C_2}{v_2} \\ \frac{A_1}{A_2}= \frac{C_2}{C_1} \times \frac{v_1}{v_2} = \frac{90}{12} \times \frac{0.5}{0.14}=26.78 \\ (\frac{d_1}{d_2})^2 = 26.78 \\ \frac{d_1}{d_2}= 5.175$

Hence ratio of inlet to outlet pipe diameter = 5.175.