Question 2 (Multiple Choice Worth 3 points)

Pipelines are a useful means of transporting oil because they: Multiple select question. are fast never fail to deliver are chea

Rufina [12.5K]

Pipelines are a useful means of transporting oil because they offer low maintenance and dependable transportation for a narrow but important range of products.

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

A pipeline is a system of connected pipelines that can be either underground or out in the environment. These pipelines are used to transport or distribute water, gas, and oil.

The options are attached

a. Pipelines provide jobs for consumers because of the resurgence of exploration and drilling in North America.

b. Pipelines are versatile, carrying more ton-miles than any other mode of transport over more than 2 million miles of pipeline.

c. Pipelines have more locations than water carriers.

d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.

Thus, the correct option is d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.

Learn more about Pipelines

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7 0

2 years ago

Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c

Nitella [24]

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress acting on the bottom plate = 157.5 N/m²

(c) The velocity along the centerline of the channel = 0.93 m/s

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

= 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

= (1.8225*10⁻⁷/1.14) * (35000)

= 1.60*10⁻⁷ * 35000

= 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ = h*(Δp/L)

= 0.0045* 35000

= 157.5 N/m²

(c) Calculating the velocity along the centre of the channel using the formula;

u(max) = h²/2μ)* (Δp/L)

= (0.0045²/2*0.38) * 35000

=2.664*10⁻⁵ *35000

= 0.93 m/s

7 0

3 years ago

Coal fire burning at 1100 k delivers heat energy to a reservoir at 500 k. Find maximum efficiency.

Marizza181 [45]

Answer:

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

<em>This is a Carnot system</em>

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - $\frac{Tc}{Th}$

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - $\frac{500}{1100}$

Eff = 1 - 0.45 = 0.55 or<em> 55%</em>

7 0

3 years ago

A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle

Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

$X_{s} =1.23(\frac{V_{line}^{2} }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms$

the network voltage phase is

$V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV$

the power transmitted is equal to:

$|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV$

the line induced voltage is

$|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV$

7 0

3 years ago

An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

P= 800 x 108.3

P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

$a=18.66\ m/s^2$

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0

3 years ago