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Lesechka [4]
3 years ago
12

In the well-insulated trans-Alaska pipeline, the high viscosity of the oil and long distances cause significant pressure drops,

and it is reasonable to question whether flow work would be significant. Consider an 150 km length of pipe of diameter 1.2 m, with oil flow rate 500 kg/s. The oil properties are 900 kg/m3, 2000 J/kg·K, 0.765 N·s/m2. Calculate the pressure drop, the flow work, and the temperature rise caused by the flow work. Recall from chapter 1, for steady flow, one inlet, one outlet, no heat transfer, no work, and negligible kinetic and protential energy changes, the enthalpy of the fluid remains constant. Enthalpy is the sum of internal energy and flow work.
Engineering
1 answer:
s2008m [1.1K]3 years ago
5 0

Our values are defined by,

L=150*10^3m

\Phi = 1.2m

\dot{m}=700kg/s

\rho_o=900km/m^3

Sp=2000J/KgK

\mu_k = 0.765Ns/m^2

With this data we can easily calculate first the mean velocity of the flow,

The velocity is given by,

\nu=\frac{\dot{m}}{\rho_o A_c}

\nu = \frac{\dot{m}}{\rho_o \pi/4\Phi^2}

\nu = \frac{4*700}{900\pi(1.2)^4}

\nu=0.688m/s

With the mean velocity is now possible determine if the flow is Laminar or turbulent, for that we need Number Reynolds.

Remember that a Laminar Flow is given when Reynolds number is minor to 2300, then

Re_x = \frac{4\dot{m}}{\pi D\mu_k}

Re_x = \frac{4*700}{\pi*1.2*0.765}

Re_x = 970.8798

Re_x < 2300, then the flow is laminar.

For this section we will consider the Pressure drop to calculate the flow Work and the Total Energy conservation of this flow.

To calculate the Pressure drop we need the Friction factor, which is given by,

f= \frac{64}{Re}=\frac{64}{970.8789}

f= 0.06591

Then the Pressure drop is given by,

\Delta P = \frac{f\rho_o \nu^2 L}{2D}

\Delta P = \frac{0.06591*900*(0.688)^2*183*10^3}{2*1.2}

\Delta P = 2.141*10^6Pa

We can now calculate the Flow Work given by,

W_f = \frac{\Delta P\dot{m}}{\rho_o}

W_f = \frac{2.141*10^6*700}{900}

W_f = 1.6651*10^6J

From energy conservation we have

W_f=q

W_f = \dot{m}c_p\Delta T

Re-arrange for \Delta T

\Delta T = \frac{W_f}{\dot{m}c_p}

\Delta T = \frac{1.56*10^6}{700*2000}

\Delta T = 1.1894\°c

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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
IrinaVladis [17]

Answer:

The original length of the specimen l_{o} = 104.7 mm

Explanation:

Original diameter d_{o} = 30 mm

Final diameter d_{1} = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length l_{1} = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × 10^{3} M pa

Shear modulus G = 25.4 G pa = 25.4 × 10^{3} M pa

We know that the relation between the shear modulus & elastic modulus is given by

G = \frac{E}{2(1 + \mu)}

25.5 = \frac{65.5}{2 (1 + \mu)}

\mu = 0.28

This is the value of possion's ratio.

We know that the possion's ratio is given by

\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }

{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}

{\frac{change \ in \ length}{l_{o} } = 0.00476

\frac{l_{1} - l_{o}  }{l_{o}  } = 0.00476

\frac{l_{1} }{l_{o} } = 1.00476

Final length l_{o} = 105.2 m

Original length

l_{o} = \frac{105.2}{1.00476}

l_{o} = 104.7 mm

This is the original length of the specimen.

5 0
3 years ago
In a creep test, increasing the temperature will (choose the best answer) A. increase the instantaneous initial deformation B. i
Hitman42 [59]

Answer:

All of the above

Explanation:

firstly, a creep can be explained as the gradual deformation of a material over a time period. This occurs at a fixed load with the temperature the same or more than the recrystallization temperature.

Once the material gets loaded, the instantaneous creep would start off and it is close to electric strain. in the primary creep area, the rate of the strain falls as the material hardens. in the secondary area, a balance between the hardening and recrystallization occurs. The material would get to be fractured hen recrstallization happens.  As temperature is raised the recrystallization gets to be more.

8 0
2 years ago
What does it mean to say that PEER is a data-driven, consumer-centric, and comprehensive system?
Reika [66]

Answer:

have you heard of gnoogle?

Explanation:have you heard of goongle?

3 0
3 years ago
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X+3=2<br>x=??<br><br><br><br>No spamming​
PtichkaEL [24]

Answer:

x+3=2

x=2-3꧁

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4 0
3 years ago
An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of
uranmaximum [27]

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

\Delta L=L\alpha\Delta T

Now by putting the values

\Delta L=750\times \25.5\times 10^{-6}\times 200

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

3 0
3 years ago
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