CHME 1150 Assignment 6

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Northern Alberta Institute of Technology *

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1150

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Mechanical Engineering

Date

Feb 20, 2024

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pdf

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Uploaded by ammarrajeh74

Dashboard / My courses / CHME1150.1221.A01.REG.A02.REG.X01.REG.X02.REG.X03.REG / Online Assignments / Assignment 6, Chapter 6 (Flow of Fluids & Bernoulli's Energy Equation) Question 1 Correct Mark 4.00 out of 4.00 Started on Wednesday, 9 November 2022, 3:38 PM State Finished Completed on Friday, 11 November 2022, 1:02 PM Time taken 1 day 21 hours Grade 46.00 out of 46.00 ( 100 %) Feedback 70679101: A05Q09 5.53 Smaller tubing size with maximum velocity 2022 Category: A06Q09 5.53 Smallest tubing size From the list of standard hydraulic steel tubing in the appendix "Dimensions of Steel and Copper Hydraulic Tubing - Inch-based sizes" (Table G.1, online textbook p 496; Data Book p 15) and "Dimensions of Steel and Copper Hydraulic Tubing - Metric-based sizes" (Table G.2, online textbook p 496; Data Book p 16) select the smallest size that would carry 3.26 L/min of oil with a maximum velocity of 0.41 m/s from either Table G.1 or Table G.2. Select the correct calculated cross flow area (A), the Flow Area, the outside diameter (OD) and the inside diameter (ID). A = 1.325e-4 m2, Flow area = 1.407e-4 m2, OD = 15.88 mm & ID = 15.75 mm A = 1.325e-4 m2, Flow area = 1.247e-4 m2, OD = 15.00 mm & ID = 12.60 mm A = 1.325e-4 m2, Flow area = 1.407e-4 m2, OD = 15.88 mm & ID = 13.39 mm  A = 1.131e-4 m2, Flow area = 1.242e-4 m2, OD = 15.88 mm & ID = 12.57 mm A = 1.948e-4 m2, Flow area = 1.948e-4 m2, OD = 15.88 mm & ID = 15.75 mm Your answer is correct. The correct answer is: A = 1.325e-4 m2, Flow area = 1.407e-4 m2, OD = 15.88 mm & ID = 13.39 mm

Question 2 Correct Mark 6.00 out of 6.00 70678505: A05Q10A 5.54 Velocity in Schedule 40 pipe Category: Assignment 06 A standard 6-in Schedule 40 steel pipe is carrying 115 gal/min of water. The pipe then branches into two standard 3-in pipes. If the flow divides evenly between the branches, calculate the velocity of flow in the 6-in pipe in ft/s . (round the final answer to three significant figures ) Answer: 1.28 ft/s  The correct answer is: 1.28 ft/s

Question 3 Correct Mark 4.00 out of 4.00 70678520: A06Q02 5.58 Throat section in venturi Category: Assignment 06 Q1 = 0.06725 m3/s; Calc. Throat Flow Area = 7.269e-4 m2; OD = 32.00 mm; Wall thickness = 2.000 mm Q1 = 0.01970 m3/s; Calc. Throat Flow Area = 7.269e-4 m2; OD = 32.00 mm; Wall thickness = 2.000 mm  Q1 = 0.05366 m3/s; Calc. Throat Flow Area = 0.001980 m2; OD = 32.00 mm; Wall thickness = 2.000 mm Q1 = 0.01970 m3/s; Calc. Throat Flow Area = 8.066e-4 m2; OD = 32.00 mm; Wall thickness = 2.000 mm Q1 = 0.06725 m3/s; Calc. Throat Flow Area = 0.002482 m2; OD = 60.00 mm; Wall thickness = 2.000 mm Q1 = 0.05366 m3/s; Calc. Throat Flow Area = 0.001980 m2; OD = 50.00 mm; Wall thickness = 2.000 mm Your answer is correct. The correct answer is: Q1 = 0.01970 m3/s; Calc. Throat Flow Area = 7.269e-4 m2; OD = 32.00 mm; Wall thickness = 2.000 mm

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Question 4 Correct Mark 6.00 out of 6.00 70678508: A06Q03 5.60 Gasoline flow through contraction Category: Assignment 06 Answer: 242 kPa  Bernoulli ′ s equation : z 1 + v 2 1 2 g + P 1 γ = z 2 + v 2 2 2 g + P 2 γ z 1 = z 2 Q 1 = A 1 × v 1 v 1 = Q 1 A 1 Q 1 = A 2 × v 2 v 2 = Q 1 A 2 sg fluid = γ fluid γ ref ( H 2 O , 4 ∘ C ) γ fluid = sg fluid × γ ref ( H 2 O , 4 ∘ C ) From page 359: A 1 = 1.744 × 10 − 2 m 2 and A 2 = 4.347 × 10 − 3 m 2 Therefore from the Bernoulli ′ s equation : Q 1 A 1 2 2 g + P 1 sg fluid × γ ref ( H 2 O , 4 ∘ C ) = Q 1 A 2 2 2 g + P 2 sg fluid × γ ref ( H 2 O , 4 ∘ C ) Solving for P 2 : P 2 = sg fluid × γ ref ( H 2 O , 4 ∘ C ) × Q 2 1 2 × g 1 A 2 1 − 1 A 2 2 + P 1 = ( ) ( ) ( )

0.67 × 9.81 kN m 3 × 0.11 m 3 s 2 2 × 9.81 m s 2 1 1.744 × 10 − 2 m 2 2 − 1 4.347 × 10 − 3 m 2 2 + 443 kPa = 242 kPa The correct answer is: 242 kPa ( ) ( ( ) ( ) )

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Question 5 Correct Mark 4.00 out of 4.00 70678509: A06Q04 5.61 Pressure after an expansion Category: Assignment 06 Answer: 33.0 kPa  The correct answer is: 33.0 kPa

Question 6 Correct Mark 8.00 out of 8.00 70678518: A06Q05 5.66 Volume flow rate from nozzle Category: Assignment 06 Q = 0.0084160 m3/s; PA = 40.623 kPa; PB = 32.285 kPa Q = 0.0084160 m3/s; PA = 40.559 kPa; PB = 32.221 kPa  Q = 0.0072573 m3/s; PA = 32.298 kPa; PB = 23.959 kPa Q = 0.0035637 m3/s; PA = 14.116 kPa; PB = 5.7772 kPa Q = 0.0084160 m3/s; PA = 40.371 kPa; PB = 32.032 kPa Your answer is correct. **************************************************************************************

The correct answer is: Q = 0.0084160 m3/s; PA = 40.559 kPa; PB = 32.221 kPa

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Question 7 Correct Mark 4.00 out of 4.00 70678507: A06Q06 5.70 Siphon volume flow rate (copy) Category: Assignment 06 Answer: 3.22 m  The correct answer is: 3.22 m

Question 8 Correct Mark 6.00 out of 6.00 70678524: A06Q07 5.76 Venturi's manometer deflection Category: Assignment 06 PA - PB = 62.500 kPa; h = 508.06 mm PA - PB = 73.242 kPa; h = 508.06 mm PA - PB = 73.242 kPa; h = 595.38 mm  PA - PB = 73.242 kPa; h = 635.07 mm PA - PB = 78.125 kPa; h = 635.07 mm Your answer is correct.

Question 9 Correct Mark 4.00 out of 4.00 70678526: A06Q08 5.93 Application of Torricelli's Theorem Category: Assignment 06 The correct answer is: PA - PB = 73.242 kPa; h = 595.38 mm Answer: 3.50 psig  The correct answer is: 3.50 psig

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◄ Assignment 6A, Chapter 6 (Flow of Fluids) Jump to... Assignment 7, Chapter 6 (Flow of Fluids & Bernoulli's Energy Equation) & 7 (General Energy Equation) ►