Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2ExplanationWe have four-stroke cycle, SI engine, operating at 4000rpm. \( \begin{align*} \mathrm{{P}_{{{\left.{d}{y}\right.}{n}{m}{o}{m}{e}{t}{e}{r}}}} &= \mathrm{{70.4}{k}{W}} \end{align*} \)engine has a total displacement volume , 2.4 l\( \begin{align*} \mathrm{{V}} &= \mathrm{{2.4}\times{10}^{{-{{3}}}}{m}^{{3}}} \end{align*} \)(1) -We need to determine power lost due to friction. Explanation:Please refer to solution in this step.Step2/2\( \begin{align*} \mathrm{\eta} &= \mathrm{\frac{{P}_{{{\left.{d}{y}\right.}{n}}}}{{{B}.{P}.}}} \end{align*} \)Break power, \( \begin{align*} \mathrm{{B}{P}} &= \mathrm{\frac{{{70.4}{k}{W}}}{{0.93}}={75.69}{k}{W}} \end{align*} \)Now calculate Indicated power, \( \begin{align*} \mathrm{\eta_{{{m}{e}{c}{h}}}} &= \mathrm{\frac{{{B}{P}}}{{{I}{P}}}} \end{align*} \)IP = Indicated Power\( \begin{align*} \mathrm{{0.82}} &= \mathrm{\frac{{75.69}}{{{I}{P}}}}\\[3pt]\mathrm{{I}{P}} &= \mathrm{{92.30}{k}{W}} \end{align*} \)HenceFriction loss can be determined, \( \begin{align*} \mathrm{{I}{P}} &= \mathrm{{B}{P}+{F}{P}} \end{align*} \)FP = Friction Power\( \begin{align*} \mathrm{{F}{P}} &= \mathrm{{92.30}-{75.69}}\\[3pt]\mathrm{{F}{P}} &= \ma ... See the full answer