A pipe of 10 cm inner diameter is used to send crude oil over distance of 400 meters. The entire pipe was laid horizontal. The viscosity of the oil is 10 cp. The density of oil is 800 kg/m3 . a. (20 pts) The desired volumetric flow rate is 0.1 m3 /min. What is the Reynolds number of this flow? Is the flow laminar or turbulent? What is the pressure difference needed to generate this flow rate? b. (15 pts) Three month later, the operator found that they had to triple the pressure difference to maintain the flow rate at 0.1 m3 /min. The operator thought that wax deposition from the oil had reduced the inner diameter of the tube. Based on this assumption, can you estimate the reduced inner diameter of the tube? What is the Reynolds number of the flow?

Answers

Answer 1

Answer:

See explaination

Explanation:

Looking at Reynolds number, we can go ahead and describe the Reynolds number as a dimensionless value that is used to determine whether the fluid is exhibiting laminar flow (R less than 2300) or turbulent flow (R greater than 4000). Laminar flow is when a fluid moves smoothly and is predictable.

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A Pipe Of 10 Cm Inner Diameter Is Used To Send Crude Oil Over Distance Of 400 Meters. The Entire Pipe
A Pipe Of 10 Cm Inner Diameter Is Used To Send Crude Oil Over Distance Of 400 Meters. The Entire Pipe

Related Questions

A museum has three rooms, each with a motion sensor (m0, m1, and m2) that outputs 1 when motion is detected. At night, the only person in the museum is one security guard who walks from room to room. Using the combinational design process, create a circuit that sounds an alarm (by setting an output A to 1) if motion is ever detected in more than one room at a time (i.e., in two or three rooms), meaning there must be an intruder or intruders in the museum.

a. Describe the behavior of this problem with a truth table.

b. Find the minimal logic expression based on the truth table. You can use either K-map or Boolean algebra.

c. Construct the gate-level circuit schematic using the logic expression derived in problem b.

Answers

Answer:

a) see attachment

b) A= m0m1+ m1m2+ m0m2

see attachment for K-map

c) see attachment

Explanation:

a) see attachment for truth table

b) see attachment for k-map

A= m0m1+ m1m2+ m0m2

c) see attachment for gate level circuit

Air enters the compressor of a cold air-standard Brayton cycle with regeneration at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rate of entropy production in the regenerator, in kW/K

Answers

(a) The thermal efficiency of the cycle.

The thermal efficiency of the cycle is given by
NThermal= net work/ heat supplied
NThermal= WNet/QH
=2026.08/4547.82= 44.55%

(b) The backwork ratio:
compressor work/ turbine work
=1760.76/3786.84= 0.465

(c) The net power developed:
WNet= WT-WC
WNet= 3786.84- 1760.76= 2026.08 kW

A large plate is at rest in water at 15?C. The plate is suddenly translated parallel to itself, at 1.5 m/s. The resulting fluid movement is not exactly like that in a b.l. because the velocity profile builds up uniformly, all over, instead of from an edge. The governing transient momentum equation, Du/Dt = ?(?2u/?y2), takes the form 1 ? ?u ?t = ?2u ?y2


`Determine u at 0.015 m from the plate for t = 1, 10, and 1000 s. Do this by first posing the problem fully and then comparing it with the solution in Section 5.6. [u 0.003 m/s after 10 s.]

Answers

Answer:

Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s

   

Explanation:

du/u² = dt = ∫du/2.3183 = ∫dt

0.4313 u = t + c

(a) t = 0, u= 15m/s, c = 0.647

u = t+c/0.4313 = t + 0.647/0.4313

(a) when t= 1   u = 1+ 0.647/0.4313 = 3.8187m/s

(b) when t= 10   u = 10 + 0.647/0.4313 = 24.0858m/s

(c)when t= 1000  u = 1000 + 0.647/0.4313 = 3220.071m/s

public class Point { private int x; private int y; public void setx(int x) this.x = x; public void setY(int y) this.y=y; public void setXY(int x, int y) X=y; ysy; public boolean IsEqual(Point p) return ((p.x == x) && (p.y == y)); public void DisplayPoint() System.out.println("(" + x + " +X+ + y + ")"); public void movePoint(int deltax, int deltay) {!! 'x' is increased by deltax and y is increased by deltay public void SetLocation(Point P) {//make point to have the specified location P. public static void main(String[] args) { // TODO Auto-generated method stub Point P1 = new Point(); Point P2 = new Point(); P1.setx(5); P1.setY(6); P2.setx(5); P2.setY(6); System.out.println(P1.IsEqual(P2)); P2.setXY(3,4); System.out.println(P1.IsEqual(P2)); Given the class 'Point' which has two member variables x, and y. What are x and y for Point object P1 after the end of the main method ? (Pay close attention to the setXY method) x=6, y=5 Ox=6, y= 6 x=5, y= 6 x=5, y= 5

Answers

Answer:

x =5

y =6

Explanation:

The reason behind that is that while using the reference P2 we have setted the values of X and Y both as 5 and 6 respectively in the main function.

After setting we are calling the setXY function which sets the value to 3 and 4 for X and Y respectively but does not implement in the value to the instance.

Apart from which setX and setY function are setting the values with respect to the instance variable using "this" keyword.

Hence, the values 5 and 6 are been displayed at the end of the main method.

Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa, which is supercritical, so no condensation occurs. The inlet is at 220°C and the exit is at 50°C. Find the specific heat transfer.

Answers

Answer:

The answer which is a calculation can be found as an attached document

Explanation:

Answer:

Answer is -286.78 kJ/kg

Refer below.

Explanation:

Refer to the pictures for brief explanation.

For some material, the heat capacity at constant volume Cv at 34 K is 0.81 J/mol-K, and the Debye temperature is 306 K. Estimate the heat capacity (in J/mol-K) (a) at 44 K, and (b) at 477 K.

Answers

Answer:

a. Heat Capacity = 1.756J/mol-K

b. Heat Capacity = 24.942J/mol-k

Explanation:

Given

Constant volume Cv = 0.81J/mol-k

T1 = 34K

Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K

First, The value of the temperature-independent constant.

Using Cv = AT³

Make A the subject of formula

A = Cv/T³

Substitute each values

A = 0.81/34³

A = 0.000020608589456543

A = 2.061 * 10^-5J/mol-k

The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature

Cv = AT³

So, The heat capacity when T = 44k is then calculated as

Cv = 2.061 * 10^-5 * 44³

Cv = 1.755522084266232

Cv = 1.756J/mol-K

(b) at 477 K.

Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:

Cv = 3R

Where R = universal gas constant

R = 8.314J/mol-k

Cv = 3 * 8.314

Cv = 24.942J/mol-k

Answer:

a) 1.75 J/mol-k

b) 24.94 J/mol-k

Explanation:

We are given:

Cv = 0.81J/mol-k

Debye temperature = 306k

Heat capacity varies with temperature and the relationship between them is given as:

Cv = AT³

To find temperature independent constant A, we have:

[tex] A = \frac {C_v}{T^3} [/tex]

[tex] A = \frac{0.81J/mol-k}{34^3} [/tex]

[tex] A = 2.06*10^-^5 [/tex]

a) for temperature at 44k

Cv = AT³

[tex] C_v = 2.06*10^-^5 * 44^3 [/tex]

Cv = 1.75 J/mol-k

b) for T at 477k

Since the temperature 447k is higher than Debye's temperature, the specific heat will be given as:

Cv = 3R

Where R = universal gas constant = 8.314 J/mol-k

Therefore,

Cv = 3 * 8.314 J/mol-k

Cv = 24.94 J/mol-k

A cylindrical rod of 1040 Steel originally 10.5 mm in diameter is to be cold worked by drawing to a final diameter of 8.5 mm. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 750 MPa and a ductility of at least 12 %EL are desired. Explain, with details and calculations, how this may be accomplished.

Answers

Answer:

Check the explanation

Explanation:

The desired Cold working (which is any process or procedure of metalworking in which the metal body is formed at a lower temperature beyond its recrystallization, typically at the ambient temperature.) tensile strength in excess of 750 MPa and a ductility of at least 12 %EL can be seen in the attached image below.

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