Answer 7_4_6

(a)  The energy supplied = calorific value x rate of flow

= 45 x 10⁶ J/kg x (2 x 10^-2 kg/ min ÷ 60 s) =  15 800 J/s

(b) Engine goes through the power cycle once every two revolutions.  There are 15 cycles per second:

Indicated power = 380 J x 15 s^-1 = 5700 J/s

(c) Thermal efficiency = indicated power ÷ input power = 5700 ÷ 15 000 = 0.38 (38 %)

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