Question #272777

A four-stroke petrol engine has the following data:

      Number of cylinders=4

           Fuel used = 19.2 kg/hour

           Fuel to air ratio = 1:16 by weight

          Suction pressure = 1bar

          Suction temperature = 30oC

      Stroke to bore ratio = 1.25

      Engine speed = 2400rpm

       Volumetric Efficiency = 78%

Find the stroke length of engine[m]. Write the answer with 3 decimal places and do not add the unit.


1
Expert's answer
2021-11-30T14:16:01-0500

Mass flow rate of air (Ma) =19.2×16=307.2 kg/hr


Total rate of injection of air and fuel =307.2+19.2=326.4kg/hr


In kg/sec =. 326.43600=0.09067kg/s\frac{326.4}{3600}=0.09067kg/s


Mass flow rate =Mass.NeNr=\frac{Mass.N_e}{N_r}


Where Ne=N_e= Engine speed

Nr=N_r= 2 for a four stroke engine


Ne=240060=40rpsN_e=\frac{2400}{60}=40rps


Mass M =0.09067×240=0.004534kg=\frac{0.09067×2}{40}=0.004534kg



PV =MRT


Where P=1 bar =1×105Pa1×10^5Pa


R=R= Gas constant =300J/kg/k

T= 30°c =273.15+30=303.15=273.15+30=303.15




V=0.004534×300×303.15100000=0.004123m3V=\frac{0.004534×300×303.15}{100000}\\=0.004123m^3



Swept volume = volumeefficiency\frac{volume}{efficiency}

Vd = 0.0041230.78\frac{0.004123}{0.78}


=0.005286=0.005286


Vd=ALNennrV_d=\frac{ALN_{e}n}{n_r}


Where A= cross section of cylinder

L= Stroke of cylinder

n= Number of cylinders



AL=0.005286×240×4=6.608×105AL=\frac{0.005286×2}{40×4}=6.608×10^{-5}


Let D= bore

D =0.8L


AL=π×0.64L2.L4=6.608×105AL=\frac{\pi×0.64L^2.L}{4}=6.608×10^{-5}



L3=1.315×104L^3={1.315×10^{-4}}


L=0.051L=0.051


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