Question

A 4 cylinder 4S, SI engine develops a max brake torque of 160 Nm at 3000 rpm. Calculate the engine displacement, bore and stroke. The BMEP at the max engine torque point is 960 Kpa. Assume square engine​

Answer 1

Given:-

No. of cylinders, k=4

Brake torque, $T_{b}=160N-m$

Brake mean effective pressure, bmep=960kPa

To Find:-

Engine displacement,bore and stroke

Explanation:

Let engine displacement be $V_{s}$ and bore,stroke be D & L respectively.

Brake torque, $T_{b}=160N-m$ = bmep$\times V_{s}$

$160=960\times10^{3}\times V_{s}\Rightarrow V_{s}=0.1667\times10^{-3}m^{3}\\\\V_{s}=\frac{\pi}{4}\times D^{2}\times L\times\frac{4}{2}\\\\0.1667\times10^{-3}=\frac{\pi}{4}\times D^{3}\times2\\\\Stroke, L=Bore, D=0.04734m=47.34mm$

Answer 2

No. of cylinders, k=4

Brake torque, T_{b}=160N-mT

b

=160N−m

Brake mean effective pressure, bmep=960kPa

To Find:-

Engine displacement,bore and stroke

Explanation:

Let engine displacement be V_{s}V

s

and bore,stroke be D & L respectively.

Brake torque, T_{b}=160N-mT

b

=160N−m = bmep\times V_{s}×V

s

\begin{gathered}160=960\times10^{3}\times V_{s}\Rightarrow V_{s}=0.1667\times10^{-3}m^{3}\\\\V_{s}=\frac{\pi}{4}\times D^{2}\times L\times\frac{4}{2}\\\\0.1667\times10^{-3}=\frac{\pi}{4}\times D^{3}\times2\\\\Stroke, L=Bore, D=0.04734m=47.34mm\end{gathered}

160=960×10

3

×V

s

⇒V

s

=0.1667×10

−3

m

3

V

s

=

4

π

×D

2

×L×

2

4

0.1667×10

−3

=

4

π

×D

3

×2

Stroke,L=Bore,D=0.04734m=47.34mm