Answer to Question #215966 in Molecular Physics | Thermodynamics for Wavie

Question

A quality governed four stroke, single cylinder gas engine has a bore of 146 mm and a stroke of 280 mm. At 475 rev/min and full load the net load on the friction brake is 433 N, and the torque arm is 0.45 m. The indicator diagram gives a net area of 578 mm2 and a length of 70 mm with a spring rating of 0.815 bar per mm. Calculate the indicated power, the brake power and the mechanical efficiency.

Answer 1

Four stroke single cylinder gas engine

n = no. of cylinders = 1

Dia. of cylinder bore = 146 mm =0.146m

Length of stroke = L = 280 mm

= 0.28 m

N = 475 revolution/min

= 475/ 60 rev/sec. = 7.5416 rev/sec.

Full load, break load = W = B = 433 N Torque arm = R = 0.45 m

Net area of indicator diagram

= 578 mm² = 578 * 10^-6 m²

Length of indicating diagram = 70 mm = 70 * 10 ^-3 m

Spring rating constt. = 0.815 bar/mm

= 815 bar/m

Indicator power

$i_p=\frac{P_iALnN}{2}$

This formula is four stroke engine but

$P_i=$ indicated pressure=?

$P_i=\frac{578\times10^{-6}\times815}{70\times10^{-3}}=6.73bar$

Area

$A=\frac{\pi d^2}{4}=0.01674m^2$

$$i_p=\frac{673\times105\times0.01674\times0.28\times7.916}{2}=12485.4watt=12.4854Kwatt$$

$T=w\times R=194.85Nm$

$b_p=2\pi TN=9.691Kwatt$

Now

$\eta_{mech}=\frac{b_p}{i_p}=\frac{9.691}{12.485}=0.776$

$\eta_{mech}=77.6$ %