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2 years ago

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A liquid with a specific gravity of 2.6 and a viscosity of 2.0 cP flows through a smooth pipe of unknown diameter, resulting in

a pressure drop of 0.183 lb/in? for 1.73 mi. What is a pipe diameter in inches if the mass rate of flow is 7000 lb/h?

1 answer:

sasho [114]2 years ago

6 0

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A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,

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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 $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$

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3 years ago

What is the area enclosed by the cycle area of the Carnot cycle illustrating on a P-V diagram?

Inga [223]

Answer:

The work of the cycle.

Explanation:

The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.

The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.

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3 years ago