Engineering, 08.10.2020 01:01 k11kiki

An engine manufacturer wants to develop a gasoline engine that produces 300 hP at 6000 rpm. They can achieve 10 bar BMEP at peak power with a naturally aspirated engine, and 20bar BMEP with a turbocharged engine. Most modern light-duty engines have displacements of ~0.5L per cylinder. Determine the displacement and minimum number of cylinders (rounded to nearest whole number) required to meet these specs for: a. A naturally aspirated engine. b. A turbocharged engine.

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Answer from: laylay1734

a). 5 cylinders

b). 3 cylinders

Explanation:

To develop a gasoline engine that produces

Power, P = 300 hp

= 300 x 746 = 223710 watt

at speed, N = 600 rpm = 100 rps

and stroke volume $V_s=0.5$ L

= $0.5 \times 10^{-3}\ m^3$

a). A naturally aspirated engine

BMEP at peak power = 10 bar

= $10 \times 10^5\ N/m^2$

Suction volume, V = $V_s \times N$

= $0.5 \times 10^{-3} \times 100 = 0.05\ m^3/s$

Power produced in one engine cylinder , p = BMEP x V

= $10 \times 10^5 \times 0.05$

= 50000 watts

No. of cylinders required, n = $\frac{P}{p}$

= $\frac{223710}{50000}$

= 4.47

So number of cylinders ≈ 5 nos.

b). A turbo charged engine

BMEP = 20 bar = $20 \times 10^5\ N/m^2$

and Volume V = $0.05\ m^3 /s$

Power produced in one engine cylinder, p = BMEP x V

= $20\times 10^5 \times 0.05$

= 100000 watts

Therefore, number of cylinders, n = $\frac{P}{p}$

= $\frac{223710}{100000}$

= 2.23

So number of cylinders ≈ 3 nos.

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An engine manufacturer wants to develop a gasoline engine that produces 300 hP at 6000 rpm. They can...

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