Question

During the power stroke in a four-stroke automobile engine, the pis-ton is forced down as the mixture of combustion products and air undergoes an adiabatic expansion. Assume (1) the engine is running at 2500 cycles/min; (2) the pressure immediately before the expansion is 20.0 atm; (3) the volumes of the mixture immediately before and after the expansion are 50.0 cm3and 400 cm3, respectively ; (4) the time interval for 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.

Answer 1

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This question is very wordy. You just have to cut through the words and get to the important information here.

Adiabatic expansion...   two things you need to know:

   p_i V_i^1.4 = p_f V_f^1.4    and     Work = (5/2) (p_iV_i - p_f V_f )

Also, they ask you for the power generated. This is just the work divided by the time.

So you have to do three things here:

1.   find the time

2. find the final pressure (they give you initial pressure and both volumes)

3. calculate the work

At that point,   power = work /time    and you're done

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time for one cycle = 1 min /2500 = 60 sec / 2500 = 0.024 seconds

You are told the expansion is 1/4 of this time, so

time for expansion   = (1/4) *0.024 =   0.006 seconds

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Now, find the final pressure...

Be careful! The initial gauge pressure is  20.0 atm, so the initial ABSOLUTE pressure is 21.0 atm

And now...

    p_f = p_i ( V_i / V_f)^1.4 =   21.0 * ( 50.0 / 400)^1.4

          =   1.1426 atm

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Now... you can calculate the work done. First, you have toconvert some units:

   p_i V_i = 21.0atm * 50.0 cc (101300 Pa / 1 atm ) * ( 1 m³ /1000000 cc )

           =    106.365 Joules

p_f V_f =  1.1426atm * 400 cc (101300 Pa / 1 atm ) * ( 1m³ / 1000000 cc )

           =    46.298 Joules

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And then

work = (5/2) ( 106.365 - 46.298 ) =    150.17Joules

And finally

   avg power = work / time = 150.17 J / 0.006 s

                 =   25028.33 Watts =   25.028 kW

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