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1 year ago

Types of lubricants on the market include:

Engineering

1 answer:

zhenek [66]1 year ago

8 0

Answer:

1. Oil

2. Grease

3. Penetrating Lubricants

4. Dry Lubricants

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A ___ is a type of purlin used as a horizontal stiffener between columns around the perimeter of a building.

BabaBlast [244]

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girt can be the answer u know

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2 years ago

A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to __________ the border.

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A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to demarcate the border.

<h3>What is a border?</h3>

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A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to demarcate the border.

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2 years ago

Ninety-five percent of the acetone vapor in an 85 vol.% air stream is to be absorbed by countercurrent contact with pure water i

Sati [7]

Answer:

Explanation:

.......................................................................................................................

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3 years ago

Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th

Nikitich [7]

Answer:

critical stress = 595 MPa

Explanation:

given data

fracture toughness = 74.6 MPa- $\sqrt{m}$

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress = $\frac{k}{f\sqrt{\pi a}}$ ......................1

put here value and we get

critical stress = $\frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}$

critical stress = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

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3 years ago

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f

dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, $head\ loss = \frac{f L V^2}{( 2 g D)}$

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

$V = \frac{Q}{A}$

$= \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s$

obtained Darcy friction factor

calculate Reynold number (Re) ,

$Re = \frac{\rho V D}{\mu}$

where, $\rho$ = density of water

$\mu$ = Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2

so reynold number is

$Re = \frac{1000\times 2.015\times 0.318}{0.001}$

= 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

$HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}$

HL = 1.64 m

7 0

3 years ago