How does the clearance volume affect the efficiency of the Otto cycle?

Engineering

1 answer:

eduard2 years ago

7 0

Answer:

Explanation:

A smaller clearance volume means a higher compression. A higher compression means better thermal efficiency. However a compression ratio too high might be troublesome, as it can cause accidental ignition of the fuel-air mix. This is the reason why Otto cycle engines have lower compressions that Diesel engines. In a Diesel engine the mix ignites by compression instead of a spark.

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In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had

kipiarov [429]

Answer:

Moisture/ water content w = 26%

Void ratio , e = 0.73

Explanation:

Initial mass of saturated soil w1 = mass of soil - weight of container

= 113.27 g - 49.31 g = 63.96 g

Final mass of soil after oven w2 = mass of soil - weight of container

= 100.06 g - 49.31 g = 50.75

Moisture /water content, w = $\frac{w1-w2}{w2}$ = $\frac{63.96-50.75}{50.75}$ = 0.26 = 26%

Void ratio = water content X specific gravity of solid

= 0.26 X 2.80 =0.728

5 0

2 years ago

At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

$F=ma$ for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that $W = mg$

For simplicity we work with $g =9.807$ $\frac{m}{s^{2}}$ despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula $g = a-bz$ the "normal" or "standard" weight of an object is given by $W = mg = ma$ when $z = 0$ , so we need to find the value of $z$ that makes $W = m(a-bz) = 0.997ma$ meaning that the original weight decrease by a 0.30%, so now we operate...

$m(a-bz) = 0.997ma$ now we group like terms on the same sides $ma(1-0.997) = mbz$ we cancel equal tems on both sides and obtain that $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$