An internal combustion engine follows the air-standard Otto cycle, Isotropic com
ID: 1852078 • Letter: A
Question
An internal combustion engine follows the air-standard Otto cycle, Isotropic compression, constant-volume heat exchange, isotropic expansion, and constant-volume heat transfer. At the beginning of the compression process (state 1) the air is @ 80 degrees F and 1 atm. At the end of the heat transfer to the system (state 3) the air temperature is 2,790 degrees F. The compression ratio is 9.. A) Calculate the pressure (atm) and temperature (R) of the air at each of the 4 states of the cycle B) Calculate the thermal efficiency of the engineExplanation / Answer
GIVEN:
Otto cycle (air, ideal gas)
T1 = 80 [F] = 540 [R]
P1 = 1 [atm]
T3 = 2700 [F] = 3160 [R]
r = 9
PV = RT
V = RT/P
R = Universal Gas Constant / Molar Weight of air
For universal gas constants, go here: http://en.wikipedia.org/wiki/Gas_constant
R = 0.73024 [ft3atm°R1lb-mol1] * (2.20462 lbmol/gmol) * 28.97 [g/mol]
R = 0.055571 [ft3atm°R1]
V1 = 0.055571 [ft3atm°R1] * (540[R]) / 1[atm]
V1 = V3 = 30 [ft3]
V2 = V1 / rp
V2 = 30 / 9
V2 = V4 = 3.3333 [ft3]
n = 1 - ((1/r)^(k-1))
n = 1 - ((1/9)^(1.4-1))
n = 0.5847
n = 58.5%
T2 = T1 / (1-n)
T2 = 540[R] / (1-0.5847)
T2 = 1300[R]
P2 = P1 * [(T2/T1)^((k-1)/k)]
P2 = 1[atm] * [(1300/540)^((1.4-1)/1.4)]
P2 = 1.28 [atm]
T3 = 3160 [R]
P3 = P2 * (T3/T2)
P3 = 1.28 [atm] * (3160/1300)
P3 = 3.11 [atm]
T4 = T3 * (T1 / T2)
T4 = 3160 * (540 / 1300)
T4 = 1312 [R]
P4 = P1 * (T4 / T1)
P4 = 1 [atm] * (1312 / 540)
P4 = 2.42 [atm]
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