CHME 1150 Assignment 4

pdf

School

Northern Alberta Institute of Technology *

*We aren’t endorsed by this school

Course

1150

Subject

Mechanical Engineering

Date

Feb 20, 2024

Type

pdf

Pages

12

Uploaded by ammarrajeh74

Report
Dashboard / My courses / CHME1150.1221.A01.REG.A02.REG.X01.REG.X02.REG.X03.REG / Online Assignments / Assignment 4, Chapter 5 (Buoyancy) Started on Monday, 17 October 2022, 11:09 AM State Finished Completed on Wednesday, 19 October 2022, 6:46 PM Time taken 2 days 7 hours Grade 36.00 out of 36.00 ( 100 %) Feedback
Question 1 Correct Mark 4.00 out of 4.00 70678412: A03Q11 4.1 Buoyancy of instrument package Category: Assignment 04 The instrument package shown in the following figure weighs 175 N. Calculate the tension in the cable in N if the package is completely submerged in seawater having a specific weight of 10.05 kN/m . (provide your answer ( numerical value and units ) in N with a precision of two decimal places) Answer: 639.05 N 3 The correct answer is: 639.05 N
Question 2 Correct Mark 4.00 out of 4.00 70678410: A03Q13 4.3 Buoyancy of steel pipe Category: Assignment 04 A certain standard steel pipe has an outside diameter of 168 mm, and a 0.98 m length of the pipe weighs 277 N. Would the pipe float or sink in glycerin (sg = 1.26) if its ends are closed? (Answers: " 1 " = It will float , " -1 " = It will sink ) -1 1 Your answer is correct. The correct answer is: -1
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Question 3 Correct Mark 4.00 out of 4.00 70678413: A04Q01 4.4 Cylindrical float Category: Assignment 04 A cylindrical float has a 10-in diameter and is 12 in long. What should be the specific weight of the float material if it is to have 0.91 of its volume below the surface of a fluid with a specific gravity of 0.78? (provide your answer ( numerical value and units ) in lb/ft with a precision of two decimal places ) Answer: 44.29 lb/ft3 3 The correct answer is: 44.29 lb/ft3
Question 4 Correct Mark 4.00 out of 4.00 70678423: A04Q02 4.6 Level indicator specific weight Category: Assignment 04 A float to be used as a level indicator is being designed to float in oil, which has a specific gravity of 0.83. It is to be a cube 100 mm on a side, and is to have 87 mm submerged in the oil. Calculate the required specific weight of the float material. (provide your answer ( numerical value and units ) in kN/m with a precision of two decimal places ) Answer: 7.08 kN/m 3 The correct answer is: 7.08 kN/m3
Question 5 Correct Mark 4.00 out of 4.00 70678431: A04Q03 4.8 Partially submerged pump Category: Assignment 04 Answer: 12.75 lb
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
The correct answer is: 12.75 lb
Question 6 Correct Mark 4.00 out of 4.00 70678433: A04Q04 4.10 Cylindrical drum Category: Assignment 04 A cylindrical drum is 2 ft in diameter, 2.15 ft long, and weighs 31.3 lb when empty. Aluminum weights are to be placed inside the drum to make it neutrally buoyant in fresh water. What volume of aluminum will be required if it weighs 0.100 lb/in ? (provide your answer ( numerical value and units ) in ft with a precision of two decimal places ) Answer: 2.26 ft3 3 3 The correct answer is: 2.26 ft3
Question 7 Correct Mark 4.00 out of 4.00 70678436: A04Q05 4.14 Specific gravity for hydrometer's top mark Category: Assignment 04 Answer: 0.9297 Equation of static equilibrium in the vertical direction : F v = 0 F b on hydrometer W hydrometer = 0 F b on hydrometer = γ fluid × V displaced V displaced = V bulb + V stem / fluid V cylinder = π D 2 cylinder 4 L cylinder sg fluid = γ fluid γ ref ( H 2 O , 4 C ) γ fluid = sg fluid × γ ref ( H 2 O , 4 C ) The equilibrium equation gives : sg fluid × γ ref ( H 2 O , 4 C ) V bulb + V stem / fluid = W hydrometer Or sg fluid × γ ref ( H 2 O , 4 C ) π D 2 bulb 4 L bulb + π D 2 stem 4 L stem / fluid = W hydrometer ( ) ( )
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Solving for sg fluid : sg fluid = W hydrometer γ ref ( H 2 O , 4 C ) π D 2 bulb 4 L bulb + π D 2 stem 4 L stem / fluid = 4 × 0.042682 lb π 62.4 lb ft 3 (1 in ) 2 × 1.5 in + (0.25 in ) 2 × (1.30 in + 1 in − 0.40 in ) 1 ft 12 in 3 = 0.9297 The correct answer is: 0.9297 ( ) ( ) ( )
Question 8 Correct Mark 4.00 out of 4.00 70678447: A04Q06 4.15 Specific gravity for hydrometer's bottom mark Category: Assignment 04 Answer: 0.9776 The correct answer is: 0.9776
Question 9 Correct Mark 4.00 out of 4.00 70678435: A04Q07 4.22 Specific weight of cylinder Category: Assignment 04 Answer: 6.63 kN/m The correct answer is: 6.63 kN/m3 ◄ Assignment 3, Chapter 3 (Pressure Measurement) Jump to... Assignment 5, Chapter 5 (Buoyancy)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help