During the power stroke in a four-stroke automobile engine, the piston is forced
ID: 1455547 • Letter: D
Question
During the power stroke in a four-stroke automobile engine, the piston is forced down as the mixture of combustion products and air undergoes an adiabatic expansion. Assume (1) the engine is running at 2 500 cycles/m in; (2) the gauge pressure immediately before the expansion is 20.0 atm; (3) the volumes of the mixture immediately before and after the expansion are 50.0 cm3 and 400 cm3, respectively (Fig. P21.31); (4) the time interval for the expansion is one-fourth that of the total cycle; and (5) the mixture behaves like an ideal gas with specific heat ratio 1.40. Find the average power generated during the power stroke.Explanation / Answer
Assuming reversible operation the work done by the expanding gas is given by the integral
..... V
W = P dV
..... V
(V initial volume, V final volume)
To solve it you need to relate pressure to volume throughout the whole expansion.
Pressure and volume for an ideal gas undergoing an adiabatic, reversible process satisfies the condition:
PV^k = constant
(with heat capacity ratio k)
Hence:
PV^k = PV^k
So you can replace pressure in the work integral by
PV^k = PV^kV^-k
=>
..... V
W = PV^kV^-k dV = PV^k(V^(1-k) - V^(1 - k))/(1 - k)
..... V
= P V ( (V/V)^(1-k) - 1)/(1 - k)
= P V (1 - (V/V)^(k-1) / (k - 1)
for k=1.4
W = 2.5 P V (1 - (V/V)^0.4)
Initial absolute pressure, i.e. gauge pressure plus atmospheric pressure is
P = 21 atm = 21 101325 Pa = 2,127,825 Pa
Initial volume
V = 50cm³ = 5×10 m3
The work done by the piston in each power stroke is:
W = 2.5 2,127,825 Pa V (1 - (50/400)^0.4) = 60 Pam³ = 60 J
The average power generated
P = W/t_stroke
Power stroke time is one fourth of the cycle time
t_stroke = (1/4)t_cycle = (1/4)(1/2500 min¹)
=>
P = 60 J 4 2500 min¹
= 600,000 Jmin¹
= 10000 Js¹
= 10 kW
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