Answer to Question #272777 in Thermal Power Engineering for Gioh

Question

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 = 30^oC

      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.

Answer 1

Mass flow rate of air (M_a) =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 =. $\frac{326.4}{3600}=0.09067kg/s$

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

Where $N_e=$ Engine speed

$N_r=$ 2 for a four stroke engine

$N_e=\frac{2400}{60}=40rps$

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

PV =MRT

Where P=1 bar =$1×10^5Pa$

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

T= 30°c $=273.15+30=303.15$

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

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

V_d = $\frac{0.004123}{0.78}$

$=0.005286$

$V_d=\frac{ALN_{e}n}{n_r}$

Where A= cross section of cylinder

L= Stroke of cylinder

n= Number of cylinders

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

Let D= bore

D =0.8L

$AL=\frac{\pi×0.64L^2.L}{4}=6.608×10^{-5}$

$L^3={1.315×10^{-4}}$

$L=0.051$