Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Basic Terminologies
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Common Units Of Pressure
Pressure is something that we experience in daily life but fail to recognize it as our body is so accustomed to it. For example, the atmospheric air applies pressure on our body, but we don’t feel it as we are exposed to it since birth and our body is ha…
Question

The diameter of a pipe carrying water changes
gradually from 6 inches at a lower point to 18 inches
at a higher point by 15 ft. What will be the difference
in pressure, in psi, between these points if 6.2 cfs is
flowing considering no loss in energy.

 

please use the photo for ung format solutions and i request that the solution is encoded so that i can understand it. Please input figures. Thank you

Sample problems: 1. A liquid (sp gr 2.0) is flowing in a 50 mm diameter pipe. The total energy at a given point is found to be 75 J/kg. The elevation of the pipe above the datum line is 3 m, and the pressure in the pipe is 65.5 kPa. Compute the velocity of flow and the horsepower in the stream at that point. Given: EGL = energy gradient line V2 2g HGL hydraulic gradient line E 50 mm ø pipe Q • point Datum line E = 75 J/kg = 75 N.m/kg (1kg/9.81N) = 7.645 m Z = 3 m p = 65.5 KPa = 65.5 KN/m² D = 50 mm = 0.050 m Sp. Gr. = 2.0 w = 2 (9810) = 19620 N/m3 = 19.62 KN/m³ %3D Required: V and WHP Solution: V2 + E = 2g + Z V2 65.5 + 19.62 7.645 = +3 2g V2 = 7.645 – 6.338 2g V = V(2g)(1.307) = 5.064m/s wQH WHP = 746 WAVE 19620 (4) D?(5.064)(7.645) WHP = 746 746 2 50 19620 (4) To00) (5.064)(7.645) 1000. WHP = 1.999hp %D 746
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Transcribed Image Text:Sample problems: 1. A liquid (sp gr 2.0) is flowing in a 50 mm diameter pipe. The total energy at a given point is found to be 75 J/kg. The elevation of the pipe above the datum line is 3 m, and the pressure in the pipe is 65.5 kPa. Compute the velocity of flow and the horsepower in the stream at that point. Given: EGL = energy gradient line V2 2g HGL hydraulic gradient line E 50 mm ø pipe Q • point Datum line E = 75 J/kg = 75 N.m/kg (1kg/9.81N) = 7.645 m Z = 3 m p = 65.5 KPa = 65.5 KN/m² D = 50 mm = 0.050 m Sp. Gr. = 2.0 w = 2 (9810) = 19620 N/m3 = 19.62 KN/m³ %3D Required: V and WHP Solution: V2 + E = 2g + Z V2 65.5 + 19.62 7.645 = +3 2g V2 = 7.645 – 6.338 2g V = V(2g)(1.307) = 5.064m/s wQH WHP = 746 WAVE 19620 (4) D?(5.064)(7.645) WHP = 746 746 2 50 19620 (4) To00) (5.064)(7.645) 1000. WHP = 1.999hp %D 746
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