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kenny6666 [7]
3 years ago
8

If the local atmospheric pressure is 14.6 psia, find the absolute pressure (in psia) in a column of glycerin (rho = 74.9 lbm/ft^

3) at depth of 27in.
Engineering
1 answer:
Novay_Z [31]3 years ago
8 0

Answer:

52.2538 psia

Explanation:

The absolute pressure at depth of 27 inches can be calculated by:

Pressure = Local pressure + Gauge pressure

Also,

P_{gauge}=\rho\times g\times h

Where,

\rho is the density of glycerin (\rho=74.9\ lbm/ft^3)

g is the gravitational acceleration = 32.1741 ft/s²

h = 27 in

Also, 1 in = 1/12 ft

So,

h = 27 / 12 ft = 2.25 ft

So,

P_{gauge}=74.9\times 32.1741\times 2.25\ lbf/ft^2=5422.1402\ lbf/ft^2

Also,

1 ft = 12 inch

1 ft² = 144 in²

So,

P_{gauge}=5422.1402\ lbf/ft^2=\frac {5422.1402\ lbf}{144\ in^2}=37.6538\ lbf/in^2=37.6538\ psia

Local pressure = 14.6 psia

So,

<u>Absolute pressure = 14.6 psia + 37.6538 psia=52.2538 psia</u>

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A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
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Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

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solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

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here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

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R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

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3 years ago
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Answer:

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Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

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Also final diameter d(f) = 19.9837 mm

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Poisson said that V = Е(elasticity)/Е(long)

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Also G = Е/2. (1+V)

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