Question

Un fierbator cu benzina ( putere calorica q=45 MJ/kg) avand randamentul 50% este folosit pentru a fierbe apa aflata la 100 grade.Stiind ca fierbatorul consuma m=120g de benzina pe ora intr-un minut se obtine o masa de vapori (lamba v=2,25 MJ/kg) de :

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Answer 1

[tex]\displaystyle Se~da:\\ \\ q=45MJ=45\times 10^6J\\ \\ \eta=50\%=0,5\\ \\ m=120\frac gh=0,002\frac {kg}{min}\\ \\ t=1min\\ \\ \lambda_v=2,25MJ=2,25\times 10^6J\\ \\ m_v=?J\\ \\ \\ Formule:\\ \\ \eta=\frac{Q_u}{Q_t}\\ \\ \eta=\frac{m_v\times\lambda_v}{q\times m\times t}\\ \\ m_v=\frac{\eta\times q\times m\times t}{\lambda_v}\\ \\ \\ Calcule:\\ \\ m_v=\frac{0,5\times 45\times10^6\times 0,002\times 1}{2,25\times 10^6}=0,02kg[/tex]