Hydraulics 2
Hydraulics 2 - Fluid Flow Measurement, Measuring Devices, and Unsteady Flow
Fluid Flow Measuring Devices
Venturi Meter
A device composed of a U-manometer used for measuring flow.
Invented by Clemens Herschel in 1881, named after Venturi.
Venturi principle: Increase in kinetic energy per unit weight equals the decrease in potential energy per unit weight.
Qa = CA2\sqrt{2g}, where:
1 - inlet values
2 - throat values
Q = AV
Steps in Solving Venturi Meter Problems:
Set up Bernoulli's Energy Equation involving the inlet and the throat.
Use the continuity equation to relate the velocity of the inlet to the velocity of the throat or vice versa.
For the pressure head difference, perform summation of pressure head from the inlet to the throat.
Sample Problem:
A venturi meter is used to measure flowrate in a water treatment plant. Inlet diameter is 200 mm, throat diameter is 50 mm. The gage fluid in the manometer is mercury, and it deflects by h = 150 mm. Determine the discharge through the venturi meter.
Pitot Tube
Tube with circular cross-section bent in L-shape, used for measuring velocity of liquids in open flow.
First used by Henri Pitot in 1732.
V1 = \sqrt{2g(\frac{Sg}{S_f} - 1)h}, where:
h = deflection of gage liquid
S_g = spec. gr. of gage fluid
S_f = spec. gr. of flowing fluid
V1 = \sqrt{2g(\frac{\gammag}{\gamma} - 1)h}
Sample Problems:
A pitot tube is inserted in a 200-mm diameter pipe. Determine the discharge in cubic meters per second given the water column heights in the piezometer and pitot tube.
If the flowing fluid is water, the gage liquid is mercury, and the pitot tube is centered inside the pipe, determine the centerline velocity if the mercury deflection was recorded at 30 centimeters.
Nozzle
A converging tube installed at the end of a pipe or hose to increase the issuing jet's velocity.
Bernoulli's Energy Equation is applied between the base and the tip.
Energy of the Jet: E = \frac{(V_{jet})^2}{2g}
Power: P = Q\gamma E
Problem:
For water shooting out of a pipe and nozzle, neglecting losses:
Find the velocity of the jet.
Find the power exerted by the jet in HP.
Find the maximum height from the nozzle that the water can reach.
Orifice
Any opening having a closed perimeter, made in a wall or partition used for measuring flow of fluids.
Shapes: circular, square, or rectangular.
Circular sharp-crested orifice is most widely used because of its simplicity.
Note: if the tank's area is more than 16 times the orifice area, neglect the velocity of water in the tank.
Device Coefficients
Coefficient of Discharge (C or Cd): C = \frac{Actual\ Discharge (Qa)}{Theoretical\ Discharge (Q_t)}
Coefficient of Velocity (Cv): Cv = \frac{Actual\ Velocity (Va)}{Theoretical\ Velocity (Vt)} = \sqrt{2gH}
Coefficient of Contraction (Cc): Cc = \frac{Area\ of\ vena\ contracta (a)}{Area\ of\ opening (A)}
Va = Cv\sqrt{2gH}
Qa = (CA)Cv\sqrt{2gH} or Q_a = CA\sqrt{2gH}
Sample Problems:
Calculate the discharge in liters per second through a 100 mm diameter orifice under a head of 5.5 m of water. Assume C = 0.61 and C_v = 0.98.
A 75-mm diameter orifice discharges 0.02114 m³/s of water under a head of 3 m. The diameter of the jet at vena contracta is 60 mm. Compute the coefficient of velocity.
An open cylindrical tank, 2.4 m in diameter and 6-meter tall has 1 m Glycerin (s = 1.5); 2.5 m of water; and 1.5 m of oil (s = 0.82). Determine the discharge through the 125 mm diameter located at the bottom of the tank. Assume C = 0.65.
Calculate the discharge in liters per second through a 100 mm diameter orifice under a head of 5.5 m of water. The reservoir is enclosed, and the existing pressure at the freeboard is 20 kPa. Assume Cc = 0.61 and Cv = 0.98.
Weirs
Overflow structures built across open channels for measuring or controlling the flow of liquids.
Nappe: the overflowing stream in a weir.
Crest: the edge or top surface of a weir which the flowing liquid contacts.
Dropdown Curve: the downward curvature of the liquid surface before the weir.
Head: the distance between the liquid surface and the crest of the weir, measured before the dropdown curve.
Suppressed Weir: full width weir having weir length being equal to the width of the channel.
Contracted Weir: weir having sides sharp edged.
Coefficient of Weir:
Qa = \frac{2}{3}C\sqrt{2g}L[H + \frac{(Va)^2}{2g}]^{\frac{3}{2}}When the velocity of approach is equal to zero:
Q_a = \frac{2}{3}C\sqrt{2g}L(H)^{\frac{3}{2}}C_w = \frac{2}{3}C\sqrt{2g}
Francis Formula
Q_a = 1.84LH^{\frac{3}{2}}
Rectangular Contracted Weir
Effective Length (L') of weir if contraction is present:
L' = L - 0.10H (Single contraction)
L' = L - 0.20H (Double Contraction)
Triangular Weir (V-Notch Weir)
For a Vertex angle = 90°,
Qa = \frac{8}{15}C\sqrt{2g} tan(\frac{\theta}{2}) H^{\frac{5}{2}} Qa = 1.40H^{\frac{5}{2}}
Trapezoidal Weir
Q_a = \frac{2}{3}C\sqrt{2g}L(H)^{\frac{3}{2}} + \frac{8}{15}C\sqrt{2g} tan(\frac{\theta}{2}) H^{\frac{5}{2}}
Cipoletti Weir (Trapezoidal weir where tan(\frac{\theta}{2}) = \frac{1}{4})
Q_a = 1.859LH^{\frac{3}{2}}
Sample Problems:
Determine the discharge through a rectangular weir having a crest length of 3 m and the head acting over the weir is 0.5 m. Use Francis formula.
A flow of 10.9 m³/s passes over a suppressed weir that is 4.88 m long. The total depth upstream must not exceed 2.44 meters. Using a coefficient of discharge equal to 1.85.
Determine the head on a 45° V-notch weir for a discharge of 200 L/s. Use C = 0.57.
Unsteady Flow: Weirs and Orifice
Unsteady flow occurs when the flow rate (Q) at the cross-section varies with time.
The flow through orifices, weirs, or tubes is said to be steady only if the total head producing flow, H, is constant.
Unsteady Flow: Orifice
For tanks with constant cross-sectional area and the outflow is through an orifice or tube, with no inflow, the time for the head to change from h1 to h2 is:
t = \frac{2As}{CAo\sqrt{2g}}(\sqrt{h1} - \sqrt{h2}), where:
A_s = surface area of tank
A_o = area of orifice
h_1 = initial head
h_2 = final head
C = meter coefficient
If the liquid flows through a submerged orifice or tube connecting two tanks or containers, the time for the head to change from h1 to h2 is:
t = \frac{2A{s1}A{s2}}{CAo\sqrt{2g}(A{s1} + A{s2})}(\sqrt{h1} - \sqrt{h_2})
Unsteady Flow: Weir
For weirs with channels/reservoirs with constant cross-sectional area and the outflow is through a crest length of L, with no inflow, the time for the head to change from h1 to h2 is:
t = \frac{2As}{CWL} (H2^{-\frac{1}{2}}-H1^{-\frac{1}{2}})
where:A_s = surface area of tank;
C_w - weir coefficient;
h1 initial head; h2 - final head;
L - crest length
Problems
A cylindrical tank 6 m in diameter and 4 m in height is filled with water. If an orifice 200 mm in diameter with a meter coefficient equal to 0.86 is placed 0.5 m from the bottom of the tank, (a) how long will it take for the water surface to drop by 1 m? (b) How long will it take to empty the tank?
A spillway with crest at elevation 125 m controls a reservoir 5 hectares in area. If the length of spillway is 20 m, find the time it takes for the water to be drawn from elevation 127.8 m to elevation 125.5 m. Use Francis Formula.
Two vertical cylindrical tanks having 2 meters and 3 meters respectively are connected with a 200-mm-diameter tube at its lower portion and having meter coefficient of 0.60. When the tube is closed, the water surface in tank 1 is 5 meters above the second tank. How long will it take after opening the tube, for the water surface in the second tank to rise by one meter?
Practice Problems
A series of practice problems with multiple-choice answers are provided, covering Venturi meters, orifices, nozzles, weirs, and unsteady flow scenarios. These problems are intended for review and exam preparation.