Question

A four stroke gasoline engine has a compression ratio of 10:1 with 4 cylinders of total displacement 2.3 L. the inlet state is 280 K, 70 kPa and the engine is running at 2100 RPM with the fuel adding 1400 kJ/kg in the combustion process. What is the network in the cycle and how much power is produced?

Answer 1

Answer:

Net work, W = 842.52 kJ/kg

Power, P = 32.82 kW

Given:

compression ratio, r = 10:1

Total volume displaced, $V_{d} = 2.3 l = 2.3\times 10^{- 3} m^{3}$

$T_{inlet}$ = 280 K

$P_{inlet}$ = 70 kPa

Speed, N = 2100 rpm = 35 rps

Heat energy added, Q = 1400 kJ/kg

Solution:

Now, calculation of the overall efficiency:

$\eta = 1- r^(1 - k)$

where

k = 1.4

$\eta = 1- 10^(0.4) = 0.6018$

Also,

Net work done, W = $\eta Q = 0.6018\times 1400 = 842.52 kJ/kg$

Now, specific volume is given by:

$P_{inlet}\vartheta = RT_{inlet}$

$70\times 10^{3}\vartheta = 0.287\times 280$

$\vartheta = 1.148 m^{3}/kg$

Now, mean effective pressure, P' can be calculated as:

$W = P'(\vartheta - \vartheta')$

$\frac{W}{(\vartheta - \vartheta')} = P'$

$\frac{842.52}{\vartheta(1 - \frac{\vartheta'}{\vartheta})} = P'$

(Since, r = $\frac{\vartheta}{\vartheta'}$)

$\frac{842.52}{\vartheta(1 - \frac{1}{r})} = P' = \frac{842.52}{1.148(1 - \frac{1}{10})}$

P' = 815.45 kPa

Now, for power produced, P:

P = $\frac{1}{2}P'V_{d} N$

P = $\frac{1}{2}\times 815.45\times 2.3\time 10^{- 3}\times 35$

P = 32.82 kW