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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 heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei
miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

4 0
4 years ago
Why is it important to cut all the way through an electrical wire on the first try?
lbvjy [14]

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I always thought it was so that the older wire could not have a problem and have another electrician must come back and fix it.

Explanation:

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svp [43]

Answer:

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Tensile stress is referred as a deforming force, in which force acts perpendicular to the surface and pull an object apart, attempting to elongate it.

The tensile stress is a type of normal stress, in which a perpendicular force creates the stress to an object’s surface.

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