Question #228985
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.
1
Expert's answer
2021-08-25T12:25:48-0400

Quantity of air delivered by the compressor

m=1260=0.2  kg/sm= \frac{12}{60}= 0.2 \;kg/s

Conditions of air at the inlet 1:

Pressure p1=1  bar=105  N/m2p_1=1 \;bar = 10^5 \;N/m^2

Specific volume v1=0.5  m3/kgv_1=0.5 \;m^3/kg

Velocity C1=12  m/sC_1=12 \;m/s

Temperature at inlet:

T1=p1v1R=105×0.5287=174.55  KT_1= \frac{p_1v_1}{R} = \frac{10^5 \times 0.5}{287}=174.55 \;K

Conditions of air at the outlet 2:

Pressure p2=8  bar=8×105  N/m2p_2= 8 \;bar = 8 \times 10^5 \;N/m^2

Specific volume v2=0.14  m3/kgv_2=0.14 \;m^3/kg

Velocity C2=90  m/sC_2= 90 \;m/s



Increase in enthalpy of air passing through the compressor,

(h2h1)=150  kJ/kg(h_2-h_1)=150 \;kJ/kg

Heat lost to the surroundings,

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

(i) Motor power required to drive the compressor:

Applying energy equation to the system,

m(h1+C122+Z1g)+Q=m(h2+C222+Z2g)+WZ1=Z2m(h1+C122)+Q=m(h2+C222)+WW=m[(h1h2)+C12C222]+Q=0.2[150+1229022×1000]+(11.67)=42.46  kJ/s=42.46  kWm(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, d1d2:\frac{d_1}{d_2}:

The mass of air passing through the compressor is given by

m=A1C1v1=A2C2v2A1A2=C2C1×v1v2=9012×0.50.14=26.78(d1d2)2=26.78d1d2=5.175m = \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.


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