A Fourcylinder sparkignition engine has a compression ratio

A Four-cylinder spark-ignition engine has a compression ratio of 8, and each cylinder has a max. volume of 0.6 L. At the beginning of the compression process, the air is at 98 kPa and 17 oC, and the max. temperature in the cycle is 1800 K. Assuming the engine to operate on the ideal Otto cycle, determine (a) the amount of heat supplied per cylinder, (b) the thermal efficiency and the number of revolutions per minute required for a net power output of 60 kW. Assume variable specific heats for air.

Solution

y= 1.4

cp of air at 1600 K is taken as 1.11 kJ /kg K

cv = .718 kJ /kg K

compression ratio, r = 8

v1 / v2 = 8

Also, v1 - v2 = 0.6 L = 0.0006m^3

v2 = 8.57 x 10^-5 m^3

v1 = 6.85 x 10^-4 m^3

p1 = 98kPa

t1 = 17 C = 290 K

m =p1* v1 / R * t1 = 8.065 x 10^-4 kg

t2 = t1 x r^(y-1) = 666.24 K

p2 = m R t2 / v2 = 1799.57 kPa

vx = v2 = 8.57 x 10^-5 m^3

px = 2 * p2 = 3599.13  kPa

tx = px * vx /mR = 1332.57 K

Q2x =m cv (tx - t2) = 8.065 x 10^-4 x .718 (1332.57 - 666.24) = 0.386 kJ

p3 = px = 3599.13 kPa

t3 = t max = 1800 C = 2073 K

Qx3 = m cp(t3 - tx) = 0.663 kJ

v3 = m R t3 / p3 = 1.333 x 10^-4 m^3

cut off ratio, rc= v3 / vx = 1.55

t4 = t3 x (rc / r)^(y-1) = 1075.2

Q added = Q2x + Qx3 = 0.386 kJ + 0.663 kJ = 1.049 kJ

Q rejected = m cv (t4 -t1) = 0.455 kJ

Efficiency = 1 - (Q rej / Q added) = 1 - (.455 / 1.049) = 0.5663 = 56.63 %

W net (cyl) = Q in - Q out = .594 kJ

MEP = W net (cyl) / (v1 - v2) = 0.99 kPa

no. of revolutions / min= 2 Wnet x 60 / no of cyl * W net(cyl) = 2 x 60 x 60 / (4 x .594) = 3030 rpm