he 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

Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Braja M. Das, Nagaratnam Sivakugan
Chapter6: Hydraulic Conductivity
Section: Chapter Questions
Problem 6.14P
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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
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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