If we burn 1 gr of ethyl alcohol (C2H5OH) this will translate into a release of about 50 W of thermal energy of the flame that raise the temperature of the hot and cold reservoirs of a Sterling engine at 55 °C and 95 °C respectively. If at steady state the engine turns at about 240 rpm, its gears are shaped as rings of radius ~2 cm and weigh 50 gr, how efficient (in %) is the transfer of energy via flame?

0.116%

Explanation:

efficiency of sterling engine = 1 - ( 55 + 273 )k / ( 95 + 273 ) k

                                               = 1 - ( 328 / 368 ) = 0.10869

Determine how efficient the transfer of energy via flame is

apply this relation ; energy of steady state / energy output   ( 1 )

where;  energy output = efficiency of sterling engine * energy

                                     = 0.10869 * 50 = 5.4345 J/sec

energy of steady state = 1/2 * I * w^2   ( 2 )

I = moment of inertia of engine = 2 * 10^-5 kgm^2

back to equation 2 :  

energy of steady state = 1/2 * ( 2 * 10^-5 ) * ( 240 * 2π / 60 )^2

                                      = 630.88 * 10^-5  J /sec

Input values into equation 1

= [ (630.88 * 10^-5 )  / 5.4345  ] * 100

= 0.116%

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml