Respuesta :
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, [tex]T_{1} = 280 K\\[/tex]
Initial pressure, [tex]P_{1} = 70 kPa[/tex]
Compression ratio, r = 10
The initial total engine volume, [tex]V_{1} = 2555.6 cm^{3} = 2555.6 * 10^{-6} m^{3}[/tex]
a) Total mass of air in the engine
Using the gas equation, [tex]P_{1} V_{1} = mRT_{1}[/tex]
Where R = 0.287 kJ/kg-K
70 * 2555.6 * 10⁻⁶ = m * 0.287 * 280
[tex]m = \frac{70 * 2555.5 * 10^{-6} }{0.287 * 280}[/tex]
m = 0.00223 kg
b) Power developed by the engine
Heat generated due to combustion, [tex]Q_{in} = 1080 kJ/kg[/tex]
[tex]\frac{T_{2} }{T_{1} }= (\frac{V_{1} }{V_{2} })^{\gamma -1}[/tex]
Compression ratio, [tex]r = \frac{V_{2} }{V_{1} } = 10[/tex]
[tex]\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[/tex]
[tex]Q_{in} = c_{v} (T_{3} -T_{2} )[/tex]
Where Specific capacity of air, [tex]c_{v} = 0.718 kJ/kg-K[/tex]
[tex]1080 = 0.718 (T_{3} -703.328 )\\1504.18 + 703.328 = T_{3}\\ T_{3} = 2207.51 K[/tex]
[tex]\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[/tex]
[tex]Q_{out} = c_{v} (T_{4} - T_{1} )\\Q_{out} = 0.718 (878.82 - 280 )\\Q_{out} = 430 kJ/kg[/tex]
[tex]w_{net} = Q_{in} - Q_{out}\\w_{net} = 1080 - 430\\w_{net} = 650 kJ/kg[/tex]
There are 4 cylinders, k = 4
N = 2800/2
N = 1400
Power developed by the engine,
[tex]P =\frac{ mw_{net} Nk}{60} \\P =\frac{ 0.00223*650* 1400*4}{60} \\P = 134.68 kW[/tex]
1 kW = 1.34102 hp
P = 134.68 * 1.34102
P = 180.61 hp
