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3241004551 [841]
4 years ago
8

Consider the 3.8-L engine used in the engine laboratory. At 2500 RPM the engine has a volumetric efficiency of 85%, air/fuel rat

io is 14, and the measured torque is 200 lb-ft and the friction torque is 38 lb-ft. Assuming the combustion efficiency is 89%, calculate: (a) Indicated thermal efficiency (b) Mechanical efficiency (c) isfc and bmep
Engineering
1 answer:
AnnZ [28]4 years ago
3 0

Answer:

A number of engine geometry properties and performance parameters obtained from dynamometer testing are given in Table below:

Please calculate:

i)                 Cylinder bore, bore to stroke ratio, connecting rod to crank radius ratio, and clearance volume in each engine cylinder.

ii)                Brake power output at 3800 rpm and the torque output at 6000 rpm.

iii)              Brake, indicated and friction mean effective pressures at both 3800 rpm and 6000 rpm.

iv)              Brake specific fuel consumption in kg/(kW h) and brake thermal efficiency at 3800 rpm.

v)                Volumetric efficiency at 3800 rpm.

vi)              Air to fuel ratio and the relative air to fuel ratio at 3800 rpm. Is this a lean or rich mixture?

Engine type Displacement/ size Number of cylinders Stroke length Connecting rod length Compression ratio Maximum brake power6000 rpm Maximum [email protected] 3800 rpm Mass flow rate of fuel 3800 rpm Volumetric flow rate of air 3800 rpm Mechanical efficiency @ 3800 rpm Mechanical efficiency @ 6000 rpm Calorific value of fuel Ambient air temperature Ambient air pressure Specific gas constant for air Gasoline, naturally aspirated, four-stroke 1.329 L 4 (Straight) 85 mm 141 mm 11.5 73 kW 128 N m 3.2 g/s 37.5 L/s 0.88 0.72 46 MJ/kg 10 °C 101.3 kPa 287.1 J/(kg K)

Explanation:

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A coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is placed in a magnetic field of 0.56 T.
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Explanation:

it is given that diameter = 8.6 cm

radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m

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area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}

magnetic field =0.56 T

maximum torque= BINASINΘ  for maximum torque sinΘ=1

so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm

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