# Question When a three-cylinder, four-stroke cycle, SI engine, operating at $$4000 \mathrm{RPM}$$ is connected to an eddy current dynamometer. $$70.4 \mathrm{~kW}$$ of power is dissipated by the dynamometer. The engine has a total displacement vohlme of $$2.4$$ liters and a mechanical efficiency of $$82 \%$$ at $$4000 \mathrm{RPM}$$. Example Problem 2-3 Because of heat and mechanical losses, the dynamomeref (actual power engine). 1. power lost to friction in engine (power recorde 2. brake mean effective pressure 3. engine torque at $$4000 \mathrm{RPM}$$ 4. engine specific volume

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Transcribed Image Text: When a three-cylinder, four-stroke cycle, SI engine, operating at $$4000 \mathrm{RPM}$$ is connected to an eddy current dynamometer. $$70.4 \mathrm{~kW}$$ of power is dissipated by the dynamometer. The engine has a total displacement vohlme of $$2.4$$ liters and a mechanical efficiency of $$82 \%$$ at $$4000 \mathrm{RPM}$$. Example Problem 2-3 Because of heat and mechanical losses, the dynamomeref (actual power engine). 1. power lost to friction in engine (power recorde 2. brake mean effective pressure 3. engine torque at $$4000 \mathrm{RPM}$$ 4. engine specific volume
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Transcribed Image Text: When a three-cylinder, four-stroke cycle, SI engine, operating at $$4000 \mathrm{RPM}$$ is connected to an eddy current dynamometer. $$70.4 \mathrm{~kW}$$ of power is dissipated by the dynamometer. The engine has a total displacement vohlme of $$2.4$$ liters and a mechanical efficiency of $$82 \%$$ at $$4000 \mathrm{RPM}$$. Example Problem 2-3 Because of heat and mechanical losses, the dynamomeref (actual power engine). 1. power lost to friction in engine (power recorde 2. brake mean effective pressure 3. engine torque at $$4000 \mathrm{RPM}$$ 4. engine specific volume
&#12304;General guidance&#12305;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