Explanation Given: Distances from left bank, corresponding depth and velocities are given below, 0 , 1.0 , 0 : 2.3 , 1.30 : 20 , 3.0 , 1.54 ; 30 , 27 , 1.90 ; 40 , 2.4 , 1.95 ; 50 , 3.0 , 1.60 ; 60 , 3.1 , 1.70 ; 70 ; 80 , 2.8 , 1.54 ; 90 , 1.24 ; 100 , 2.0 , 0.58 ; 108 , 2.2 , 0.28 ; 11 , 1.5 , 0. Concept Used: Volume computations are required to determine the capacities of bins, tanks, reservoirs and buildings and to check stockpiles of coal, gravel, and other materials. Calculation: Calculate the water discharge of a river, W a t e r d i s c h a r g e = V a v g A ..... ( 1 ) A = D e p t h o f o n e e n d + D e p t h o f another e n d D i s tan c e between e n d 2 .... ( 2 ) V a v g = v e l o c i t y o f o n e e n d + v e l o c i t y o f another e n d 2 ........ ( 3 ) Area is calculating in equation ( 2 ) , A = 1 + 2.3 2 10 A = 16.5 f t 2 V a v g = 0 + 1.3 2 V a v g = 0.65 f t / sec Now calculate water discharge using equation ( 1 ) , W a t e r d i s c h a r g e = V a v g A W a t e r d i s c h a r g e = 0.65 × 16.5 W a t e r d i s c h a r g e = 10.725 f t 3 / sec Volume discharge for second section is given by, Area is calculating in equation ( 2 ) , A = 2.3 + 3 2 10 A = 26.5 f t 2 V a v g = 1.3 + 1.54 2 V a v g = 1.42 f t / sec Now calculate water discharge using equation ( 1 ) , W a t e r d i s c h a r g e = V a v g A W a t e r d i s c h a r g e = 1.42 × 26.5 W a t e r d i s c h a r g e = 37.63 f t 3 / sec Volume discharge for third section is given by, Area is calculating in equation ( 2 ) , A = 3 + 2.7 2 10 A = 28.5 f t 2 V a v g = 1.9 + 1.54 2 V a v g = 1.72 f t / sec Now calculate water discharge using equation ( 1 ) , W a t e r d i s c h a r g e = V a v g A W a t e r d i s c h a r g e = 1.72 × 28.5 W a t e r d i s c h a r g e = 49.02 f t 3 / sec Volume discharge for forth section is given by, Area is calculate in equation ( 2 ) , A = 2.7 + 2.4 2 10 A = 25.5 f t 2 V a v g = 1.9 + 1.95 2 V a v g = 1.92 f t / sec Now calculate water discharge using equation ( 1 ) , W a t e r d i s c h a r g e = V a v g A W a t e r d i s c h a r g e = 1.92 × 25.5 W a t e r d i s c h a r g e = 48.96 f t 3 / sec Volume discharge for fifth section is given by, Area is calculate in equation ( 2 ) , A = 2.4 + 3 2 10 A = 27 f t 2 V a v g = 1.6 + 1.95 2 V a v g = 1.78 f t / sec Now calculate water discharge using equation ( 1 ) , W a t e r d i s c h a r g e = V a v g A W a t e r d i s c h a r g e = 1.78 × 27 W a t e r d i s c h a r g e = 48.06 f t 3 / sec Volume discharge for sixth section is given by, Area is calculate in equation ( 2 ) , A = 3.1 + 3 2 10 A = 30.5 f t 2 V a v g = 1.6 + 1.7 2 V a v g = 1

Volume in f t 3 / sec .

Elementary Surveying: An Introduction To Geomatics (15th Edition)

15th Edition

ISBN: 9780134604657

Author: Charles D. Ghilani

Publisher: PEARSON

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Highway Drainage

Highway drainage is the process of extracting and regulating the excess surface water on the pavement by giving the desired slope. The construction of a highway drain helps to divert the water present on the surface, and sub-grade by the means of slope. …

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Chapter 26, Problem 26.28P

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Volume inft3/sec.

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Chapter 26, Problem 26.27P

Chapter 27, Problem 27.1P

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This is an old exam practice question. The answer key says the answer is Pmax = 52.8kN but I am confused how they got that.

F12-45. Car A is traveling with a constant speed of 80 km/h due north, while car B is traveling with a constant speed of 100 km/h due east. Determine the velocity of car B relative to car A. pload Choose a File Question 5 VA - WB VBA V100 111413 + *12-164. The car travels along the circular curve of radius r = 100 ft with a constant speed of v = 30 ft/s. Determine the angular rate of rotation è of the radial liner and the magnitude of the car's acceleration. Probs. 12-163/164 pload Choose a File r = 400 ft 20 pts

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