Hydrology and Hydraulics: Venturi Meter, Pump Efficiency, and Hydrograph

This article discusses the application of Buckingham pi method to develop the governing relationship for Venturi meter, finding the upstream pipe diameter that will yield a 20kPa drop in meter 2, calculation of flow rate with pump efficiency, and determination of stream flow at successive intervals in hydrograph. The subject is Hydrology and Hydraulics, and the course code and college/university are not mentioned.

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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)

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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

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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)

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

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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.