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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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Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

3 0
3 years ago
In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
taurus [48]

Answer:

<em>No, the velocity profile does not change in the flow direction.</em>

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>

3 0
3 years ago
Do you know anything about Android graphics?
Mashutka [201]

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Android has got visually appealing graphics and mind blowing animations.

The Android framework provides a rich set of powerful APIS for applying animation to UI elements and graphics as well as drawing custom 2D and 3D graphics.

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7 0
3 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
3 years ago
Explain by Research how a basic generator works ? using diagram<br>​
natulia [17]
Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry
8 0
3 years ago
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