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3 years ago

CNG is a readily available alternative to

Engineering

1 answer:

Veseljchak [2.6K]3 years ago

3 0

Answer:

Basics. CNG is a readily available alternative to gasoline that's made by compressing natural gas to less than 1% of its volume at standard atmospheric pressure. ... Expanding the numbers of CNG fueling stations would allow for the increase of CNG vehicles on U.S. roads.

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A fluid flows steadily through a pipe with a uniform cross sectional area. The density of the fluid decreases to half its initia

Vikentia [17]

Answer:

c. V2 equals V1

Explanation:

We can answer this question by using the continuity equation, which states that:

$A_1 v_1 = A_2 v_2$ (1)

where

A1 is the cross-sectional area in the first section of the pipe

A2 is the cross-sectional area in the second section of the pipe

v1 is the velocity of the fluid in the first section of the pipe

v2 is the velocity of the fluid in the second section of the pipe

In this problem, we are told that the pipe has a uniform cross sectional area, so:

A1 = A2

As a consequence, according to eq.(1), this means that

v1 = v2

so, the velocity of the fluid in the pipe does not change.

5 0

3 years ago

A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 90mm OD, a 1.65

Andru [333]

Given:

Outer Diameter, OD = 90 mm

thickness, t = 0.1656 mm

Poisson's ratio, $\mu$ = 0.334

Strength = 320 MPa

Pressure, P =3.5 MPa

Formula Used:

1). $Axial Stress_{max}$ = $P[\frac{r_{o}^{2} + r_{i}^{2}}{r_{o}^{2} - r_{i}^{2}}]$

2). factor of safety, m = $\frac{strength}{stress_{max}}$

Explanation:

Now, for Inner Diameter of cylinder, ID = OD - 2t

ID = 90 - 2(1.65) = 86.7 mm

Outer radius, $r_{o}$ = 45mm

Inner radius, $r_{i}$ = 43.35 mm

Now by using the given formula (1)

$Axial Stress_{max}$ = $3.5[\frac{45^{2} + 43.35^{2}}{45^{2} - 43.35^{2}}]$

$Axial Stress_{max}$ = $3.5\times 26.78$ =93.74 MPa

Now, Using formula (2)

factor of safety, m = \frac{320}{93.74} = 3.414

7 0

3 years ago

A 3-m wide rectangular channel has a flow velocity of 1.8 m/s when the depth of flow is 1.2 m. what will be the flow velocity wh

valina [46]

Answer:

The flow velocity reduces to 0.72 m/s

Explanation:

According to the equation of continuity discharge in the channel should remain same

Thus we have

$A_{1}V_{1}=A_{2}V_{2}$

For a rectangular channel we have $Area=Depth\times Width$

Applying values in the continuity equation and since the width of the channel remains constant 3.0 m we have

$3\times 1.8\times 1.2=3\times 3\times V_{2}\\\\\therefore V_{2}=\frac{6.48}{9}=0.72m/s$

6 0

3 years ago

Consider a two-dimensional incompressible velocity potential phi = ???????? cos theta + ????????theta, where B and L are constan

scoundrel [369]

solution:

Increasing the magnification and decreasing the field view

we are given:

a(t)=4*t^2-)

where t= time in seconds, and a(t) = acceleration as a function of time.

and

x(0)=-2 )

x(2) = -20 )

where x(t) = distance travelled as a function of time.

need to find x(4).

from (1), we express x(t) by integrating, twice.

velocity = v(t) = integral of (1) with respect to t

v(t) = 4t^3/3 - 2t + k1 )

where k1 is a constant, to be determined.

integrate (4) to find the displacement x(t) = integral of (4).

x(t) = integral of v(t) with respect to t

= (t^4)/3 - t^2 + (k1)t + k2 ) where k2 is another constant to be determined.

from (2) and (3)

we set up a system of two equations, with k1 and k2 as unknowns.

x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 )

substitute (6) in (3)

x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20

16/3 -4 + 2k1 -2 = -20

2k1 = -20-16/3 +4 +2 = -58/3

k1 = -29/3 )

thus substituting (6) and (7) in (5), we get

x(t) = (t^4)/3 - t^2 - 29t/3 + 2 )

which, by putting t=4 in (8)

x(4) = (4^4)/3 - (4^2 - 29*4/3 +2

= 86/3, or

= 28 2/3, or

= 28.67 (to two places of decimal)

7 0

3 years ago

A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M

Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of $300\times 100 mm^2$

Force(F)=8 MN

Therefore stress due to this Force( $\sigma$ )

$\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}$

$\sigma =266.66 \times 10^{6} Pa$

$\sigma =266.66 MPa$

Since induced stress is greater than Yield stress therefore Plastic deformation occurs

8 0

3 years ago