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Elena L [17]
2 years ago
5

Liquefied Natural Gas (LNG) is a natural gas in its liquid form that is clear, colorless, odorless, non-corrosive, and non-toxic

. A) TrueB) False
Engineering
1 answer:
ZanzabumX [31]2 years ago
6 0

Liquefied Natural Gas (LNG) can be defined as a natural gas which in liquid form appear clear and colorless. It is odorless, non-toxic, and non-corrosive. Therefore, the given statement is A) True.

  • LNG or Liquified Natural Gas is a fossil fuel that is produced after the compression of organic matter in the form of algae and phytoplankton.
  • LNG consists of 95% methane gas.
  • The combustion of LNG produces carbon dioxide and water vapors.
  • It burns with a least pollution thus called as cleanest fossil fuel.
  • The liquefaction of the natural gas takes place at -160 degree Celsius. The liquefaction of the gas causes it to transport easily in gas tanks.
  • LNG is colorless, and clear.
  • LNG does not possess any smell and it is non-corrosive to metallic tanks.
  • LNG is also non-toxic.

Learn more about natural gas:

brainly.com/question/12200462

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What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5×10-4
krok68 [10]

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}

<u>Where:</u>

<em>ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)</em>

<em>σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi) </em>

<em>2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in) </em>

Hence, the  maximum stress (σ) is:

\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa    

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!

5 0
3 years ago
Which type of blade is used with a demolition saw?
Scrat [10]
The correct answer is An abrasive wheel.
5 0
3 years ago
Una frase de: ama la vida quien___________________________________
TEA [102]

Answer:

A phrase from: who loves life

Explanation:

5 0
2 years ago
Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.
saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 \mathbf{m^{3}}

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; v_1 = vg_1 = 0.0996 \mathbf{m^{3}} / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \mathbf{m^{3}} / kg

From temperature T₂ = 200⁰ C

v_f_2 = 0.0016 \ m^3/kg  

vg_2 = 0.127 \ m^3/kg  

Since  vf_2 < v_2 , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}

x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}

x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}

\mathsf{x_2 =0.78}

At temperature T₂, the specific internal energy u_f_2 = 850.6 \ kJ/kg , also ug_2 = 2594.3 \ kJ/kg

Thus,

u_2 = uf_2 + x_2 (ug_2 -uf_2)

u_2 =850.6  +0.78 (2594.3 -850.6)

u_2 =850.6  +1360.086

u_2 =2210.686 \ kJ/kg

<u>At state 3:</u>

Temperature T_3=T_2 = 200 ^0 C ,

V_3 = 2V_1 = 0.06 \ m^3

Specific volume v_3 = 0.2  \ m^3/kg

Thus; vg_3 =vg_2 = 0.127 \ m^3/kg ,

SInce v_3 > vg_3, therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at v_3 = 0.2  \ m^3/kg and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \ m^3/kg

The specific internal energy u_3 at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

u_2-u_1

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

u_3-u_2

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0
3 years ago
An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

3 0
2 years ago
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