Question

un motor termic dezvolta o putere de 60kW .randamentul lui este de 25% .citi litri de benzina a consumat moptorul in 10 ore?Puterea calorica a benzinei este de 4.6*10la a 7 j/kg.

Answer 1

ro=m/V⇒V=m/ro
ro benzina=700 kg/m³
q=4.6*10^7J/kg=Qtot/m
P =60kW=60000W=L/t⇒L=P*t=60000W*10 h=60000W*36000s=216*10^7J
L=Qutil=216*10^7J
randamentul=Qutil*100%/Qtot⇒
Qtot=Qutil*100%/randanent=216*10^7J*100%/25%
Qtot=864*10^7J
q=Qtot/m⇒m=Qtotal/q=864*10^7J/4.6*10^7J/kg=187.82kg≈188kg
V=m/ro=188kg/700 kg/m³=0.268m³
1m³=10³dm³
1dm³=1l
0.268m³=268l