Hydrology and Hydraulics: Venturi Meter, Pumping, & Rainfall Analysis

This assignment solution delves into the principles of hydrology and hydraulics, focusing on practical applications and problem-solving. It begins with an analysis of Venturi meters, employing the Buckingham Pi method to establish governing relationships and determine upstream pipe diameters for specific pressure drop scenarios. Subsequently, the solution addresses a pumping system problem, calculating the flow rate required to pump water from a lower reservoir to an elevated tank, considering pump efficiency and head loss. Finally, the assignment tackles a rainfall analysis problem, determining the total depth of storm rainfall and stream flow at successive intervals using hydrograph ordinates and accounting for base flow and storm losses. Desklib offers a wealth of similar solved assignments and study resources for students.

Surname 1
Student Name
Instructor
Course Details
Date
Hydrology and Hydraulics
Question 1
The pressure drop in a Venturi meter varies only with the fluid density, velocity of approach and
the diameter ratio of the meter. Venturi meter 1 in water at 250C shows a 10kPa drop when the
approach velocity is 6m/s. Meter 2, geometrically similar to meter 1, is used in a 0.15M3/s flow
of kerosene (density = 800kg/m3).
i. Use the Buckingham pi method to develop the governing relationship
ii. Find the upstream pipe diameter that will yield 20kPa drop in meter 2.
Solution
Given that
n=5 and independent quantities are 3 that is (D, V and Density)
But using the buckingham pi theorem
3.14=3.14 (f ( D , V , density , P))
This equals D a V b c (ML−1 T −2)
( L)a(¿−1) b( ML−3) c( ML−1 T −2)=3.14
But it is established that a+ b−3 c=1
Solving for a, b and c we obtain (Thompson p.20)
-b-2=0
c=-1
b=-2
a=0
But the Function is given as = /( V2)
(b)

Surname 2
Given that At T= 25C, V=6m/s and P1=10 kPa
Additionally, Q=0.15m3/sec and P2=20 Kpa
But the function is given as
P m/(mV m 2)=P p /( p V p 2)
10000/(997 X 6 x 6)=20000 /(800 xV 2)
V=9.47m/sec
But also flow rate discharge Q= AxV
0.15=0.785 xDxDx 9.47
D=0.14m
Question 2
Water is being pumped from a lower reservoir to an elevated tank as shown below. The pump is
80% efficient and rated at 200kW. Determine the flow rate if the total head loss from point 1 to 2
is 15m.
Solution
Given that Z1= 50m, Z2 = 100m and h2 = 15m

Surname 3
But P = Q r−¿ E P¿
Thus, 0.80 x 200=∅ x 0.87 x 9.79 x EP
EP= 160
8.51 ∅ = 18.80
∅
Thus, it follows that
Using the relationship, ¿ ¿ ¿
We obtain EP=126.12 ∅ 2
Hence, 50+ 18.80
∅ =126.12∅ 2 +100+15
18.80
∅ =126.12∅ 2 +65
126.12∅2 −18.80
∅ +65=0
Using substitution in the quadratic equation to solve we obtain (Houghtalen p.12)
At , Q= 0.210 =1.96
Q= 0.22 = 7.29
Q= 0.24 =18.7
Q = 0.2 = 0
Hence, q = 0.200m3/s
Question 3:
The successive 3-h ordinates of a 6-unit hydrograph for a particular catchment are 0. 10, 30, 50,
40, 30, 20, 10, and 0m3/s, respectively. The peak observed due to a 6-h storm was 200m3/s.
Assume a constant base flow of 7m3/s and an average storm loss of 5mm/h.
Determine the total depth of storm rainfall (mm)
Determine the total depth of the storm rainfall (mm)

Surname 4
Determine the stream flow at successive 3-h intervals
Solution
Effective rainfall depth = 30mm
Thus, the Storm flow = 156m3/ Sec

Surname 5
Work Cited
Houghtalen, Robert J., Ned HC Hwang, and A. Osman Akan. Fundamentals of hydraulic
engineering systems. Prentice Hall, 2016.
Thompson, Stephen A. Hydrology for water management. CRC Press, 2017.