To determine The resultant force due to oil in the tank and the location of the resultant force from the top of the tank. Explanation Given: Base of triangular portion of trapezoidal section ( b ) is 0.75 m. Height of the tank ( h ) is 4 m. Length of the side AD of bottom trapezoid ( L A D ) is 2.5 m. Length of the side CB of bottom trapezoid ( L C B ) is 1 m. Density of oil ( ρ 0 ) is 900 kg / m 3 . Acceleration due to gravity ( g ) is 9.81 m / s 2 . Variation of height where the pressure acts is y . Calculation: Determine the slope of the bottom trapezoidal section. m = V H Here, m is the slope, H is the horizontal distance, and V is the vertical distance. Assume the slope of trapezoidal plate as 2:1. m = 2 1 For slope 2:1, the vertical height is 2 times the horizontal distance. Determine the height of bottom trapezoidal section ( h A B ). h A B = 2 × b h A B = 2 × 0.75 = 1.5 m Draw the cross section of the bottom trapezoidal plate and establish the x and y axis as shown in Figure (1). Determine the height of point A from the top of the tank ( h A ) . h A = h − h A B h A = 4 m − 1.5 m = 2.5 m Determine the equation for the side AB of the plate. y − h A x − ( L A D 2 ) = h − h A ( L C B 2 ) − ( L A D 2 ) y − 2.5 m x − ( 2.5 m 2 ) = 4 m − 2.5 m ( 1 m 2 ) − ( 2.5 m 2 ) 2 y − 5 2 x − 2.5 = 1.5 ( − 1.5 2 ) ( − 1.5 ) ( 2 y − 5 ) = 1.5 × 2 ( 2 x − 2.5 ) − 3 y + 7.5 = 6 x − 7.5 6 x + 3 y = 15 2 x + y = 5 2 x = 5 − y The differential element is shaded in Figure (1). Determine the area of the differential element ( d A ) . d A = ( x + x ) d y d A = 2 x d y d A = ( 5 − y ) d y Determine the pressure ( p ) acting on the differential element

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Pressure Vessels

The term pressure vessel refers to a closed container that contains gas or liquid at a different pressure than environment pressure. The pressure vessels have a wide scale of applications in our day-to-day lives and other industrial places. Some common e…

Videos

Question

Chapter 2, Problem 69P

To determine

The resultant force due to oil in the tank and the location of the resultant force from the top of the tank.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2, Problem 68P

Chapter 2, Problem 70P

Students have asked these similar questions

Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/s

Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4

Problem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…

Chapter 2 Solutions

Fluid Mechanics (2nd Edition)

Ch. 2 - Prob. 1FP Ch. 2 - The container is partially filled with oil, water,...Ch. 2 - The U-tube manometer is filled with mercury,...Ch. 2 - The tube is filled with mercury from A to B, and...Ch. 2 - The air pressure in the pipe at A is 300 kPa....Ch. 2 - Determine the absolute pressure of the water in...Ch. 2 - The bin is 1.5 m wide and is filled with water to...Ch. 2 - The bin is 2 m wide and is filled with oil to the...Ch. 2 - The 2-m-wide container is filled with water to the...Ch. 2 - Determine the resultant force of the water acting...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.