ME310_HW02_Hydrostatics_solution

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ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 1/8 INDIANA UNIVERSITY-PURDUE UNIVERSITY INDIANAPOLIS Purdue School of Engineering & Technology Department of Mechanical and Energy Engineering Student’s Name: ME 31002/EEN 31000: Fluid Mechanics Homework #2: Hydrostatics Due: Sat, Feb 3, 2024 Use the given units in the problem statement. Do not convert to a different unit system. Show your work: state the governing equations and values utilized in your solution. Box your answers. Score Problem Grade 1 of 10 2 of 10 3 of 10 4 of 10 5 of 10 Total of 50

ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 2/8 Student’s name: Problem 1 (10 points): An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown liquid is 1.5 m and the depth of the oil (specific weight = 8.5 kN/m3 ) floating on top is 5.0 m. A pressure gage connected to the bottom of the tank reads 65 kPa. What is the specific gravity of the unknown liquid? Ans. : ?? = 1.53 . Solution The gage pressure at the bottom is 𝑝 ?𝑜𝑡 = ? 1 ℎ 1 + ? 2 ℎ 2 where ? 1 = 8.5 kN/m3, ℎ 1 = 5 m, ℎ 2 = 1.5 m, and ? 2 is unknown. It is also known that 𝑝 ?𝑜𝑡 = 65 kPa. Then, ? 2 = 1 ℎ 2 (𝑝 ?𝑜𝑡 − ? 1 ℎ 1 ) = 1 (1.5 m) (65 kN m 2 − (8.5 kN m 3 ) (5 m)) ∴ ? 2 = 15 kN m 3 The specific gravity is ?? = ? 2 ? 𝐻 2𝑂 = 15 kN m 3 9.8 kN m 3 ∴ ?? = 1.53

ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 3/8 Student’s name: Problem 2 (10 points): Knowing that the air pressure at sea level is 29.92 inHg, determine the height at with the air pressure drops to 25.69 inHg as follows: (a) assume that the air is an incompressible fluid, (b) assume that air is compressible and there is no change in temperature, and (c) assume that air is compressible and the temperature with altitude. Ans. : (a) ℎ = 3904 ft , (b) ℎ = 4213 ft , (c) ℎ = 4153 ft . Solution (a) Assuming that the air is an incompressible fluid: 𝑝 = 𝑝 ? − ? ? ℎ or ℎ = 𝑝 ? − 𝑝 ? ? where the pressures are 𝑝 = (25.69 inHg) ? 𝐻𝑔 = (25.69 in) (847 lb ft 3 ) ( 1 ft 12 in ) ∴ 𝑝 = 1813 lb ft 2 𝑝 ? = (29.92 inHg) ? 𝐻𝑔 = (29.92 in) (847 lb ft 3 ) ( 1 ft 12 in ) ∴ 𝑝 ? = 2112 lb ft 2 and the specific weight of air is (Table 2.1) ? ? = 0.07647 lb ft 3 Replacing, ℎ = 𝑝 ? − 𝑝 ? ? = 2112 lb ft 2 − 1813 lb ft 2 0.07647 lb ft 3 ∴ ℎ = 3904 ft (b) Assuming that air is compressible and there is no change in temperature: 𝑝 = 𝑝 ? exp [− ?ℎ ?? ? ] or ℎ = − ?? ? ? ln 𝑝 𝑝 ?

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ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 4/8 Student’s name: where ? ? = 518.67 °R , ? = 32.2 ft/s 2 , and ? = 1716 ft ⋅ lb slug ⋅ °R ⁄ . Replacing, ℎ = − (1716 ft ⋅ lb slug ⋅ °R ) (518.67 °R) 32.2 ft s 2 ln 1813 2112 ∴ ℎ = 4213 ft (c) Assume that air is compressible and temperature changes with altitude. 𝑝 = 𝑝 ? (1 − ?ℎ ? ? ) 𝑔 𝑅𝛽 or ℎ = ? ? ? [1 − ( 𝑝 𝑝 ? ) 𝑅𝛽 𝑔 ] where the lapse rate is ? = 0.00357 °R/ft . Replacing ?? ? = (1716 ft ⋅ lb slug ⋅ °R ) (0.00357 °R ft ) 32.2 ft s 2 = 0.1903 and ℎ = (518.67 °R) (0.00357 °R/ft) [1 − ( 1813 2112 ) 0.1903 ] ∴ ℎ = 4153 ft

ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 5/8 Student’s name: Problem 3 (10 points): A U-tube manometer is connected to a closed tank containing air and water as shown in the figure below. At the closed end of the manometer, the air pressure is 16 psia. Determine the reading on the pressure gage for a differential reading of 4 ft on the manometer. Express your answer in psi (gage). Assume standard atmospheric pressure and neglect the weight of the air columns in the manometer. Ans .: 4.66 psig. Solution : The air gage pressure is 𝑝 air = 16 psia − 14.7 psi = 1.3 psig and 𝑝 air = 1.3 lb in 2 ( 12 in 1 ft ) 2 = 187.2 lb ft 2 The water gage pressure is 𝑝 w = 𝑝 air + (4 ?𝑡)? + (2 ?𝑡)? 𝐻 2 𝑜 𝑝 w = 187.2 lb ft 2 + (4 ft) (90 lb ft 3 ) + (2 ft) (62.4 lb ft 3 ) ∴ 𝑝 w = 672 lb ft 2 = 4.66 psig

ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 6/8 Student’s name: Problem 4 (10 points): Two pipes are connected by a manometer as shown in the figure below. Determine the pressure difference, 𝑝 ? − 𝑝 ? , between the pipes. Ans. : −3.332 kPa. Solution : 𝑝 ? = 𝑝 ? + ? 𝐻 2 𝑂 (0.5 + 0.6) − ?(0.6) + ? 𝐻 2 𝑂 (1.3 − 0.5) and ? = ?? ? 𝐻 2 𝑂 where ?? = 2.6 and ? 𝐻 2 𝑂 = 9.8 kN/m3. 𝑝 ? = 𝑝 ? + ? 𝐻 2 𝑂 (1.1) − ? 𝐻 2 𝑂 (2.6)(0.6) + ? 𝐻 2 𝑂 (0.8) Then, 𝑝 ? − 𝑝 ? = ? 𝐻 2 𝑂 (−1.1 + 1.56 − 0.8) ∴ 𝑝 ? − 𝑝 ? = −3.332 kPa

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ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 7/8 Student’s name: Problem 5 (10 points): Determine the magnitude and direction of the force that must be applied to the bottom of the gate shown in the figure to keep the gate closed. Ans. : ? = 14.37 kN. Solution Using the pressure prism (see Figure), where the resulting force is ? 𝑖 is the is the volume of the pressure prism, and ? = 9.8 kN/m 3 , one obtains, ? 1 = (9.8 kN m 3 ) (1.3 m)(0.8 m)(2 m) = 20.38 kN ? 2 = 1 2 (9.8 kN m 3 ) (0.8 m)(0.8 m)(2 m) = 6.27 kN

ME 31002/EEN 31000: Fluid Mechanics, Homework #2, Due: Sat, Feb 3, 2024. Page 8/8 Student’s name: Sum of moments with respect to the hinge, yields, ?𝑑 = ? 1 𝑑 1 + ? 2 𝑑 2 where 𝑑 = 0.8 m, 𝑑 1 = 0.8 2 ⁄ = 0.4 m and 𝑑 2 = 0.8 (2 3 ⁄ ) = 5.33 m. Replacing, ? = (20.38 kN)(0.4 m) + (6.27 kN)(5.33 m) (0.8 m) ? = 14.37 kN