106efficiency (max) = 1-T1/T2 = 1-(350+273)/(1400+273) = 0.628
Now heat supplied (1hr) = 0.44*41870 = 18422.8 kJ
work done (1hr) = 3.4*3600 kJ = 12240 kJ
efficiency = 12240/18422.8 > 0.628
so it is NOT genuine
edit: missed out the time..
An engineer claims to have developed an engine which develops 3.4 kW while consuming 0.44 Kg of fuel of calorific value of calorific value of 41870 kJ / kg in one hour. The maximum and minimum temperatures recorded in the cycle are 1400 °C & 350 °C respectively. is the claim of the engineer genuine ?
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6 Answers
106
1@ Ashish Your answer is not correct .
Read it carefully and explain you answer !!!
11Efficiency = 1400-350/1673= 1050/1673 = 0.628 =E
Heat supplied = 0.44*41870kJ in 1 hour
Work done = 3.4K*3600 in 1Hour
Effi.=e=3.4K*3600/0.44*41870K = 0.664
e>E Hence it is not genuine
1I'm getting "the claim of the engineer is genuine"
Kuch mistake kari hogi maine...[2]