AppliedThermodynamics 5th ed. by Eastop and Mcconkey Problem 12.14 A single acti

Question

AppliedThermodynamics 5th ed. by Eastop and Mcconkey Problem

12.14

A single acting air motor has a bore of 380mm and stroke of 610mm. The supply pressure and temperature is 6.2 bar and 150 oC respectively. The air motor has a speed of 200 rpm, back pressure of 1.03 bar and cut-off ratio of 0.46. The clearance volume is 20% of the swept volume and mechanical efficiency is 95%. The law of compression and expansion can be expressed by pV1.35 = constant. Considering the temperature and pressure of the air in the clearance space at the beginning of admission are 6.2 bar and 150 oC respectively, calculate;

(a) Air consumption

(b) The air temperature after blow down

(c) The fraction of stroke travelled by the piston before recompression begins

(d) The shaft power developed.

Explanation / Answer

Swept volume = pi/4*Bore^2*Stroke = 3.14/4*0.38^2*0.61 = 0.069146 m^3

Clearance volume, V5 = V6 = 20% of swept volume = 0.2*0.069146 = 0.01383 m^3

Cut-off volume V1 = 46% of swept volume + clearance volume = 0.46*0.069146 + 0.01383 = 0.04564 m^3

Volume at end of expansion V2 = Swept Volume + clearance volume = 0.069146 + 0.01383 = 0.083 m^3

Pressure at end of expansion P2 = P1(V1/V2)^1.35

P2 = 6.2*(0.04564 / 0.083)^1.35

P2 = 2.76 bar

Temperature at end of expansion T2 = T1(V1/V2)^(n-1)

T2 = (150+273)*(0.04564/0.083)^(1.35-1)

T2 = 343 K = 70 deg C

a) Air consumption = [P1*V1 / (R*T1)]*N

Air consumption = [6.2*10^5*0.04564 / (287*(150+273)]*200 = 46.62 kg/min

b)

Temperature after blowdown T3 = T2*(P3/P2) = 343*(1.03 / 2.76) = 128 K = -145 deg C

c)

P5 = P4(V4/V5)^n

6.2 = 1.03*(V4 / 0.01383)^1.35

V4 = 0.0522 m^3

V4 - V5 = (1-x)*swept volume

0.0522 - 0.01383 = (1-x)*0.069146

Fraction of stroke travelled by piston before recompression begines x = 0.44 or 44 %

d)

Indicated power = area of cycle = P1(V1 - V6) + (P1V1-P2V2)/ (n-1) - P3(V3 - V4) - (P5V5 - P4V4)/(n-1)

Ind.work = [6.2*(0.04564 - 0.01383) + (6.2*0.04564 - 2.76*0.083)/(1.35-1) - 1.03*(0.083-0.0522) - (6.2*0.01383 - 1.03*0.0522)/(1.35-1)] *10^2 kJ

Ind.work = [0.197 + 0.154 - 0.0317 - 0.0914]*10^2 kJ

Ind.work = 22.8 kJ

Shaft power = Mech efficiency * Indiacated power = 0.95*22.8 = 21.65 kJ = 21.65*200/120 kW = 36.1 kW

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