In the hydrodynamic entrance region of a pipe with a steady flow of an incompressible liquid A. The average velocity increases with distance from the entrance.
B. The average velocity stays the same with distance from the entrance.
C. The maximum velocity increases with distance from the entrance.
D. The maximum velocity decreases with distance from the entrance.
E. B and C
F. B and D

Answers

Answer 1

Answer:

D. The maximum velocity decreases with distance from the entrance.

Explanation:

This is because over time, the pressure with with the incompressible liquid enters decreases with distance from the entrance

Answer 2

Answer:

C. The maximum velocity increases with distance from the entrance

Explanation:

As the fluid particles moves into the pipe, the layer of fluid in contact with the surface of the pipe come to a complete stop. This layer also causes the fluid

particles in the adjacent layers to gradually slow down as a result of friction between fluid molecules, leaving the fluid at the center of the pipe with the maximum velocity.

Since the fluid is incompressible, to make up for this velocity reduction, the velocity of the fluid at the mid-

section of the pipe has to increase to keep the mass flow rate through the

pipe constant. As a result, a velocity gradient develops along the pipe and the maximum velocity which is at the center of the pipe increases with distance from entrance.

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8. 15 A manual arc welding cell uses a welder and a fitter. The cell operates 2,000 hriyr. The welder is paid $30/hr and the fitter is paid $25/hr. Both rates include applicable overheads. The cycle time to complete one welded assembly is 15. 4 min. Of this time, the arc-on time is 25%, and the fitter's participation in the cycle is 30% of the cycle time. A robotic arc welding cell is being considered to replace this manual cell. The new cell would have one robot, one fitter, and two workstations, so that while the robot is working at the first sta tion, the fitter is unloading the other station and loading it with new components. The fitter's rate would remain at $25/hr. For the new cell, the production rate would be eight welded assemblies per hour. The arc-on time would increase to almost 52%, and the fitter's participation in the cycle would be about 62%. The installed cost of the robot and worksta tions is $158,000. Power and other utilities to operate the robot and arc welding equipment will be $3. 80/hr, and annual maintenance costs are $3,500. Given a 3-year service life, 15% rate of return, and no salvage value, (a) determine the annual quantity of welded assem blies that would have to be produced to reach the break-even point for the two methods. (b) What is the annual quantity of welded assemblies produced by the two methods work. Ing 2,000 hryr?

Answers

The annual quantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

To determine the break-even point between the manual arc welding cell and the robotic cell, we need to calculate the total costs for each method and then equate them.

For the manual arc welding cell:

Labor cost per hour = (welder's hourly rate x arc-on time) + (fitter's hourly rate x fitter's participation in the cycle) = ($30 x 0.25) + ($25 x 0.3) = $11.25

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $11.25 x 15.4 / 60 = $2.89

Overhead cost per welded assembly = (labor cost per hour x (1 - arc-on time - fitter's participation in the cycle)) x cycle time per assembly / 60 = ($30 x 0.45) x 15.4 / 60 = $4.68

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly = $2.89 + $4.68 = $7.57

Total cost per hour = total cost per welded assembly x production rate = $7.57 x 8 = $60.56

Total cost per year = total cost per hour x hours of operation per year = $60.56 x 2,000 = $121,120

For the robotic arc welding cell:

Labor cost per hour = fitter's hourly rate x fitter's participation in the cycle = $25 x 0.62 = $15.50

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $15.50 x 15.4 / 60 = $3.97

Overhead cost per welded assembly = power and utility cost per hour + annual maintenance cost / production rate = $3.80 + $3,500 / (8 x 2,000) = $3.80 + $0.22 = $4.02

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly + (installed cost / (production rate x service life)) = $3.97 + $4.02 + ($158,000 / (8 x 3)) = $3.97 + $4.02 + $6,208.33 = $14.19

Total cost per hour = total cost per welded assembly x production rate = $14.19 x 8 = $113.52

Total cost per year = total cost per hour x hours of operation per year = $113.52 x 2,000 = $227,040

To find the break-even point, we set the total cost of the manual arc welding cell equal to the total cost of the robotic arc welding cell and solve for the annualquantity of welded assemblies:

$121,120 + x($7.57) = $227,040 + x($14.19)

$7.57x - $14.19x = $227,040 - $121,120

$-6.62x = $105,920

x = $105,920 / $6.62

x = 15,982.7

Therefore, the annualquantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

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If new research showed that the standard heights for crest vertical curve design were H1=3.0 ft and H2=1.5 ft, respectively, by what percent would the minimum length of curvature increase or decrease for a crest curve with a stopping sight distance of 500 ft and grades of +3% and -2%, respectively.

Answers

Answer:

A = 5

S<L, L = 714.89ft

S>L, L = 650.29ft

L = 115.85ft

Percentage min. Length of curvature = 6.2 %

Explanation: see explanation at the attached file

_________ items are similar to the free issue items, but their access is limited. (CLO5) a)-Bin stock items free issue b)-Bin stock controlled issue c)-Critical or insurance spares d)-Rebuildable spares e)-consumables

Answers

Answer:

a)-Bin stock items free issue

Explanation:

Bin stock items free issue items are similar to the free issue items, but their access is limited.

Bin stock items free issue items are similar to the free issue items, but their access is limited.

A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa

Answers

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto (1)/(V)$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 150 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v L

$V_2$ = final volume of gas = $(v)/(3)L$

$150* v=P_2* (v)/(3)$

$P_2=450kPa$

Therefore, the new pressure of the gas will be 450 kPa.

An electric current of 237.0 mA flows for 8.0 minutes. Calculate the amount of electric charge transported. Be sure your answer has the correct unit symbol and the correct number of significant digits x10

Answers

Answer:

amount of electric charge transported = 1.13 × $10^(-2)$ C

Explanation:

given data

electric current = 237.0 mA = 0.237 A

time = 8 min = 8 × 60 sec = 480 sec

solution

we get here amount of electric charge transported that is express as

amount of electric charge transported = electric current × time ...........1

put here value and we get

amount of electric charge transported = 0.237 × 480

amount of electric charge transported = 113.76 C

amount of electric charge transported = 1.13 × $10^(-2)$ C

Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model.Technician B says that hybrid vehicles have 12-volt and high voltage batteries.

Who is right?

Answers

Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.

What are hybrid vehicle?

Hybrid vehicle are defined as a powered by a combustion engine and/or a number of electric motors that draw power from batteries. A gas-powered car simply has a traditional gas engine, but a hybrid car also features an electric motor.

One important advantage of hybrid cars is their capacity to reduce the size of the main engine, which improves fuel efficiency. Many hybrid vehicles employ electric motors to accelerate slowly at first until they reach higher speeds. They then use gasoline-powered engines to increase fuel efficiency.

Thus, technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.

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