Question

A four-cylinder, four-stroke spark-ignition engine operates at 2800 RPM. The processes within each cylinder can be modeled as an air-standard Otto cycle with a temperature of 280 K and a pressure of 70 kPa at the beginning of the compression process. The compression ratio is 10, and the total engine volume is 2555.6 cm3 . Suppose heat generated from the combustion process (i.e. heat addition per unit mass) is 1080 kJ/kg and air is an ideal gas having variable specific heats. Determine: (a) Total mass of air in the engine (in kg) (6 Points) (b) Determine the power developed by the engine (in hp). (24 Points)

Answer 1

Answer:

a) Total mass of air in the engine (in kg) = 0.00223 kg

b) power developed by the engine (in hp) = 180.61 hp

Explanation:

Initial temperature, $T_{1} = 280 K$

Initial pressure, $P_{1} = 70 kPa$

Compression ratio, r = 10

The initial total engine volume, $V_{1} = 2555.6 cm^{3} = 2555.6 * 10^{-6} m^{3}$

a) Total mass of air in the engine

Using the gas equation, $P_{1} V_{1} = mRT_{1}$

Where R = 0.287 kJ/kg-K

70 * 2555.6 * 10⁻⁶ = m * 0.287 * 280

$m = \frac{70 * 2555.5 * 10^{-6} }{0.287 * 280}$

m = 0.00223 kg

b) Power developed by the engine

Heat generated due to combustion, $Q_{in} = 1080 kJ/kg$

$\frac{T_{2} }{T_{1} }= (\frac{V_{1} }{V_{2} })^{\gamma -1}$

Compression ratio, $r = \frac{V_{2} }{V_{1} } = 10$

$\frac{T_{2} }{T_{1} }=10^{0.4}\\\frac{T_{2} }{280 }=10^{0.4}\\T_{2} = 280 * 10^{0.4}\\T_{2} = 703.328 K$

$Q_{in} = c_{v} (T_{3} -T_{2} )$

Where Specific capacity of air,  $c_{v} = 0.718 kJ/kg-K$

$1080 = 0.718 (T_{3} -703.328 )\\1504.18 + 703.328 = T_{3}\\ T_{3} = 2207.51 K$

$\frac{T_{4} }{T_{3} } = (\frac{V_{2} }{V_{1} } )^{\gamma -1} \\\frac{T_{4} }{T_{3} } = (1/10 )^{0.4} \\\frac{T_{4} }{2207.51 } = (1/10 )^{0.4} \\T_{4} = 2207.51 * (1/10 )^{0.4} \\T_{4} = 878.82 K$

$Q_{out} = c_{v} (T_{4} - T_{1} )\\Q_{out} = 0.718 (878.82 - 280 )\\Q_{out} = 430 kJ/kg$

$w_{net} = Q_{in} - Q_{out}\\w_{net} = 1080 - 430\\w_{net} = 650 kJ/kg$

There are 4 cylinders, k = 4

N = 2800/2

N = 1400

Power developed by the engine,

$P =\frac{ mw_{net} Nk}{60} \\P =\frac{ 0.00223*650* 1400*4}{60} \\P = 134.68 kW$

1 kW = 1.34102 hp

P = 134.68 * 1.34102

P = 180.61 hp