Lab Manual

HYDRAULIC ENGINEERING LABORATORY EXPERIMENTS FOR CIVE 330, HYDRAULICS I

Department of Civil and Architectural Engineering Drexel University Revised, June 2000

LABORATORY MANUAL FOR CIVE 330: HYDRAULICS I

Hydraulic Engineering Laboratory Department of Civil and Architectural Engineering Drexel University Revised: June 2000

Experiment 1 Center of Pressure on Partially and Fully Submerged Plates n Purpose: To determine the center of pressure on a partially submerged and fully submerged plane surface.

Figure 1-1 Hydrostatic Pressure Apparatus n Procedure: 1. Place the quadrant on the two dowel pins and, using clamping screw, fasten it to the balance arm. 2. Level the Plexiglas tank by adjusting the screwed feet. The level is indicated on the circular spirit level. 3. Hang the balance pan and make the balance arm horizontal by moving the counter balance weight. 4. Measure a, L, d, b as shown in Figure 1-1. 5. Close the drain cock and fill the tank with water until the water level reaches the bottom edge of the quadrant. Level the arm by moving the counterbalance weight. 6. Place 50 grams on the balance pan and slowly add water to the tank until the balance arm is again horizontal. Record the water level (y) on the quadrant and the weight on the balance pan (W = mg). 7. Repeat Step 6 for several increments placing about 50 grams on the balance pan for each step until the water level reaches the top of the quadrant end face. Repeat Step 6 one more time so that the quadrant end face is totally submerged for this last run. 8. Remove each increment of weight and allow the water to drain until the balance arm is level. Note the weights and water levels for each increment as the weights are removed.

n Interpretation of Results: You want to find the center of pressure on the plate for each reading taken during filling and draining the tank. To do this, take moments about the pivot. Thus, − mg ( L) + F (a + d − z ) = 0 (1-1) in which z = the height of the center of pressure above the bottom of the plate. The force on the submerged plate is given by, 1 F = ρgyA with y = y and A = by (1-2) 2 Therefore, y2 (1-3) F = ρgb 2 Substituting,  ρgy 2b  (a + d − z ) = 0 (1-4) − mg ( L) +   2  Solving for z we get, 2mL z = a+d − 2 (1-5) ρy b Note that ρ = 1 gm/cm3 or 1000 kg/m3.

For each of the readings obtained during filling and draining the tank calculate the height above the bottom of the plate of the center of pressure (z) and plot the calculated values of z against y. Fit a straight line to the data. n Questions: 1. What is the slope of the straight line? 2. How far above the bottom of the plate should the center of pressure be? 3. Theoretically, what should the value of the slope be? Did you get this value? If not, why not? 4. If the plate had been a isosceles triangle with its base at the bottom, what would the theoretical slope of the line be? n Data: Water temperature= a= 10.2 cm; L=27.5 cm; d= 10.0 cm; b=7.5 cm Tank Filling m (gm) y (cm)

Tank Draining m (gm)

y (cm)

Experiment 2 Transition from Laminar to Turbulent Flow - Reynolds Number n Purpose: To observe the transition of laminar flow through a tube into transitional and turbulent flow.

Figure 2-1 Laminar, transitional and turbulent conditions in tube.

Figure 2-2 Dye Injection System n Procedure:

Close the dye flow control valve and fill the reservoir with dye. Check whether the dye injector is just above the bellmouth entry. Close the flow control valve. Open the bench inlet valve and fill the head tank to the over flow level; close the inlet valve. 5. Measure the temperature of the water. 6. Open the inlet valve and flow control valve slightly until water just trickles from the outlet pipe. Open the control valve a little more and adjust the dye valve until slow flow with a line of dye down the tube is achieved. Measure and record the flow rate with a stopwatch and graduated measuring cylinder. 1. 2. 3. 4.

Open the flow control valve a little more and observe the dye streak. Again measure the flow rate and record the condition of the dye streak. Continue opening the flow control valve in small increments, record the condition of the dye streak and measure the flow rate at each step. Continue this process until the dye streak breaks up indicating fully turbulent flow. 8. Reverse the process by decreasing the flow in small increments, recording the dye streak condition and measuring the flow rate at each step. 7.

n Interpretation of Results:

Calculate Reynolds number for each flow condition investigated above. Notes:

ℜ=

VD

ν

(2-1)

in which ℜ = the Reynolds number, V = the flow velocity, D = the tube diameter and ν = the kinematic viscosity of water at the measured temperature. Note also that the V can be found from the continuity equation, Q = VA where A = the cross-sectional area of the tube. For the experimental apparatus, D = 1 cm. Notes:

Laminar flow denotes a steady condition where all streamlines follow parallel paths. Under this condition the dye will remain easily identifiable as a solid core. Turbulent flow denotes an unsteady condition where streamlines interact causing shear plane collapse and mixing of the fluid. Under this condition the dye will become dispersed as mixing occurs. Transitional flow conditions occur when the flow is neither laminar nor turbulent flow but is in the “transition” stage of going from laminar to turbulent. Under this condition the dye will appear as a wandering dye stream prior to its dispersion across the flow tube at the onset of turbulence. n Questions:

1. At what Reynolds number does the flow just start to become turbulent as you increase the flow rate (the onset of turbulence)? What is the Reynolds number when the flow becomes fully turbulent? 2. At what Reynolds number does the flow again become laminar as you decrease the flow rate? 3. Do you get the same Reynolds number for the transition from laminar to turbulent flow as you get for the transition back from turbulent to laminar flow? Explain your answer.

4. Compare the Reynolds numbers calculated above with the values given in your textbook for the transitions. n Data: Water temperature= ν= D= 1 cm

Vol. collected (mL)

Increasing Q Time of collection (sec)

Flow condition

Vol. collected (mL)

Decreasing Q Time of collection (sec)

Flow condition

Experiment 3 Verification of Bernoulli's Theorem n Purpose: The purpose of this experiment is to illustrate Bernoulli’s Theorem by demonstrating the relationship between pressure head and kinetic energy head for a conduit of varying cross-section. n Pre-Lab Setup 1. Set up the Bernoulli apparatus on the working surface and level it. 2. Connect the supply hose to the inlet stub and tighten the hose. 3. If not already open, open the drain cock on the outlet tank. n Procedure: 1. Close the main control valve and start the pump. 2. Regulate the pump flow to fill the header tank and maintain it at a steady level. The flow through the channel will be quite rapid and the pressure at the throat may be too low to show on the piezometer tube. 3. Increase the back pressure in the channel and the outlet tank by slowly closing the drain cock. This will tend to raise the level in the outlet tank so the pump flow control valve should also be carefully regulated. 4. Adjust both pump flow and drain cock until there is the widest possible difference in pressure between the inlet and throat of the channel, with the water level visible in every piezometer tube. 5. Measure the volumetric flow rate with a graduated cylinder and stop watch. 6. Measure the height of the water level in each piezometer tube and record on the data sheet together with the corresponding distance from the channel entrance. 7. Measure the height of the water level in both the inlet and outlet tank. 8. Switch off the pump and close the main valve. n Questions: (interpretation of results) 1. Using your measured discharge rate, calculate the velocity at the throat of the flow conduit if the cross-sectional area of the throat is 40.32 mm2 (0.4032 cm2) 2. Calculate the total head, H, at the throat. (The total head is the sum of the measured pressure head and the velocity head at the throat.) 3. Plot the total head, H, as a function of distance, x, where x = 0 at the inlet, x = 2.5 cm at the first tube, etc., x = 15.0 cm at the throat and x = 30.0 cm at the outlet. 4. What is the head loss between the inlet and the throat? 5. What is the head loss between the throat and the outlet? 6. Assume that the total head varies linearly between x = 0 and x = 15.0 cm and also from x = 15.0 cm to x = 30.0 cm, and determine the total head at each piezometer tube. 7. Determine the velocity head at each piezometer tube. 8. Determine the velocity at each piezometer tube.

9. Determine the cross-sectional area of the flow at each piezometer tube and plot that area as a function of x. 10. Calculate the degree of pressure recovery. See Appendix 2. What does this indicate about the energy of the fluid as it passes through contractions and expansions? Notes:

1. Bernoulli’s equation is derived by integrating the equations of fluid motion. Assumptions used to obtain the simplified version of the equation are that the fluid is inviscid and incompressible and that the flow is steady. Bernoulli’s equation is a mathematical statement of the work-energy principle which directly corresponds to the equations of motion. This principle states that the work done on a particle is equal to the change in kinetic energy of the particle. Along a streamline, v2 + + z = const . γ 2g p

(3-1)

2. Conservation of Mass For a given cross-sectional area the product of the velocity and density is proportional to the mass flow rate.

M = ρQ = ρvAn v=

(mass / time) (vol / time) Q M = = = An An ρAn (mass / vol ) An

(3-2) (3-3)

Q = vAn (continuity equation for incompressible fluid) where, M = mass flow rate, Q = volumetric flow rate, v = average velocity, An = area normal to the direction of flow, and ρ = mass density. Between any two points in the flow, Inflow = Outflow. Therefore,

M in = M out

(3-4)

ρ 1v1 A1 = ρ 2 v 2 A2

(3-5)

which for an incompressible fluid becomes,

v1 A1 = v 2 A2 = Q

(3-6)

1

1

2

3

4

Figure 3-1

5

6

7

8

9

10

11

Bernoulli Apparatus

If the cross-sectional area decreases, the velocity must increase to satisfy continuity. Applying Bernoulli’s equation to a flow where there is no change in elevation (z = constant), a decrease in velocity must be accompanied by an increase in pressure and vice versa. Bernoulli’s equation expresses the conservation of energy and that the work done on the fluid shows up as a change in kinetic and/or potential energy. n Data: Water temperature= γ= Q= mL/sec Width of channel= 6.72 mm (for cross-check ONLY) Ht. of top of channel from given datum= 50.5 mm Ht. of water in inlet tank= Ht. of water in outlet tank=

Tube no. 1 2 3 4 5 6 7 8 9 10 11

Ht. of channel x-section, d (mm) 15.0 13.0 12.0 10.0 8.0 6.0 8.0 10.5 12.0 13.5 15.5

X-section area (mm2) (for cross-check ONLY)

Ht. of water in tube from datum (cm)

Ht. of water in tube from mid-ht. of channel (cm)

Experiment 4 Measuring Velocity with a Pitot Tube n Purpose: The purpose of this experiment is to show how a pitot tube can be used to measure flow velocity and to compare the measured velocity profile in a closed conduit with the profile assumed for an ideal fluid. n Procedure:

1. Using the crank, elevate the base of the channel to a height of 4 inches. Insure that the sluice gate is all the way in its upward position.

2. Start the pump. Adjust the inlet and outlet valves so that the inlet reservoir is at a height of 9.5 inches and the level of the outlet reservoir is at about 9 inches. Do not readjust heights once you have started to take measurements. 3. Measure the flow rate once the system is in equilibrium. 4. Measure the water level in the piezometer and the pitot tubes. 5. Keeping the flow rate constant, repeat Step 3 for the following bed levels: a) 3 inches, b) 2 inches, c) 1 inch Make sure that the horizontal portion of the pitot tube is midway between the top and the bottom of the channel for each bed level. (As the base of the channel is adjusted, the level in the inlet and outlet reservoirs will decrease since the cross-section of the flow area is increasing at the throat.) 6. Reset the bed to 4 inches. Repeat Step 3 with the horizontal portion of the pitot tube at the following levels: a) 4.1 inches, b) 4.3 inches, c) 4.7 inches, d) 5.0 inches, e) 5.2 inches f) 5.5 inches, g) 5.7 inches, h) 6.0 inches, i) 6.2 inches 7. Change the flow rate by adjusting the inlet and outlet valves. Make sure the difference between the levels of the inlet and outlet reservoirs is about 0.5 inch. Repeat Steps 5 and 6 for one additional flow rate. (Be sure to measure the flow rate in each case.) n Questions: (interpretation of results)

1. Plot the average velocity vs. the centerline velocity determined by the pitot tube (see Note). This graph should contain 8 data points, 4 for each flow rate. The data points for each flow rate should be distinguished by using different symbols. Compute the flow rate from the measured velocity distribution. Discuss the difference between the centerline velocity measured by the pitot tube and the average velocity calculated from the flow rate. The graph should have a line with a slope of 45° representing perfect agreement so that the comparison can be made. 2. Plot the velocity profile for each flow. Comment on the shape of the profiles. Remember that the conduit is rectangular in shape and that the flow is probably turbulent so that the profile will generally not be parabolic. Note: Figure 1 shows a closed conduit completely filled by a flow of water whose velocity is to be measured. The piezometer, a vertical tube connected to a hole in the smooth roof of the conduit, indicates that the water is under pressure. If the hole does not alter the flow pattern, the column of water, p/ρg, in the piezometer indicates the static head of the streamtube adjacent to the wall. Moreover, if all the stream tubes in the conduit flow parallel to the walls, they will exert no centrifugal force and the pressure distribution throughout will be hydrostatic. Thus the static pressure head, p/ρg, along a streamtube a distance y' above the bed will be related to the piezometer column by, p'/(ρg) + y’ = p/(ρg) + y thus, p'/(ρg) = p/(ρg) + (y - y') (4-1) The total energy is the sum of the pressure, potential and kinetic energies of the fluid within the streamtube. One form of Bernoulli's equation expresses the sum of the energy per unit weight of fluid, H, as, H = p'/(ρg) + (y' + z) + v2/(2g) (4-2) Each term is considered to be a form of head since it has dimensions of length. Thus, H is known as the total head, p'/ρg is the pressure head, (y' + z) is the potential head, and v2/(2g) is the velocity head. the sum of the pressure head and the potential head is known as the piezometric head, h.

Near the tip of the horizontal portion of the pitot tube the flow along the stream tube is reduced to zero velocity at what is the stagnation point. The local static pressure, pS , and the total head measured by the pitot tube is represented by the terms, pS /(ρg) + (y' + z) = H (4-3) Assuming that no loss of total head occurs along the streamtube, Bernoulli's equation states, p'/(ρg) + (y' + z) + v2/(2g) = pS /(ρg) + (y' + z) (4-4) or, h + v2/(2g) = H (4-5) Thus, v = 2 g ( H − h) (4-6)

Generally, for real fluid flows, the velocity and the total head are not constant across a section of conduit (from one streamtube to another). However, the pitot tube can still be used satisfactorily to measure velocity where the pressure distribution is nearly hydrostatic (where the streamtubes are nearly parallel). The average velocity is calculated as, V = Q/A = Q/(W y)

(4-7)

Figure 4-1 Velocity Measurement with a Pitot Tube in which V = the average velocity in the conduit cross-section, Q = the discharge, W = the width of the conduit (1.5 inches), and y = the depth of the conduit (the vertical height of the channel opening) which varies with your experimental setup.

Experiment 5 Discharge Through an Orifice n Purpose: This experiment will relate the flow through a circular orifice to the total head. The experimental flow rates will be compared to the theoretical flow rates and the flow coefficient determined. n Procedure: 1. Fit the 5 mm orifice into the threaded hole in the side of the tank. 2. Start the pump and regulate the flow rate to maintain the water level in the tank steady at 50 cm. 3. Record the volume of water discharged through the orifice with a graduated cylinder for a time of five seconds. 4. Repeat Step 3 for water levels of 45, 40, 35, 30, 25, 20, and 10 cm. 5. Raise the water level to 50 cm. 6. Stop the pump and record the time required for the tank to empty. Be careful in specifying the time when the tank is said to be empty. (When the jet stops.) 7. Repeat Step 6 for heights of 40, 30, 20, and 10 cm. 8. Repeat Steps 1 through 7 for the 8 mm orifice. 9. Close the feed valve and switch off the pump. n Data: Diameter of the tank = 9 cm. Diameter of the orifices = 5 mm and 8 mm. n Data Analysis: 1. Calculate the theoretical flow rate Qt. Plot the theoretical curve of the flow rate Qt vs. the total head, H, for both the 5 mm and 8 mm diameter orifices. H should range from 0 to 50 cm and Qt should be in cubic cm per second.

2. On the same graphs plot the experimental values of the flow rate Qe vs. H. Comment on the differences between the theoretical and experimental values of Q. 3. For each orifice, plot a graph of Qe vs. H on log-log paper or plot the logarithms of Qe and H on arithmetic graph paper. From the plot determine the value of k (y intercept) and n (slope). Compare these values to the theoretical values. See Appendix 1 and Note 1. 4. For each orifice determine the value of the discharge coefficient, Cd, for each measured value of H. Plot Cd vs. H. Describe how Cd varies with respect to H. See Notes 2 and 3.

5. When the column is draining, the flow rate equals the rate of change of volume with respect to time. Use the equation for the discharge coefficient and by integration show, H1 − H 2 t 2 − t1 = 2 AC (5-1) Cd Ao 2 g 6. Construct a table with the following information:

(

Initial Height (cm) 50 40 30 20 10

Final Height (cm) 40 30 20 10 0

Measured time to drop (sec)

)

Theoretical time to drop (sec)

Discuss how the measured times compare with the theoretical times as the water level in the column drops. Notes:

1. The flow through an orifice can be described by an equation of the form, Q = k Hn. Given the values of Q and H, the values of k and n can be determined by plotting Q vs. H on log-log paper. Alternatively, on linear scale graph paper, the logarithm of Q can be plotted against the logarithm of H to find k and n. Either common or natural logarithms can be used but common logarithms (to base 10) are preferred. A best-fit straight line is drawn through the points; n is then the slope of the line and the logarithm of k is the y intercept at H = 1 (log H = 0). 2. If an ideal fluid is assumed with no energy losses, flow through an orifice is described by, Qt = Ao 2 gH (5-2) This equation is derived from Bernoulli's equation. Assuming no losses, Bernoulli’s equation is, p A v 2A p v2 + + z A = B + B + zB (5-3). γ 2g γ 2g Only the third, fifth and sixth terms of eq. (3-3) are non-zero terms, therefore, v2 (z A − zB ) = B (5-4). 2g Letting H = (zA - zB), v2 H = B or v B = 2 gH (5-5). 2g

Note that the velocity at the orifice depends only on the head and not on the orifice area. From continuity, v = Q/Ao, therefore, Q = Ao 2 gH (which is equation 5-2) (5-2) Therefore, theoretically, in the logarithmic plot of the data, n = 0.5 (5-6) and k = Ao 2 g (5-7) 3. The experimental flow rate through an orifice for a real fluid will be less than the theoretical rate. To determine the experimental flow rate a discharge coefficient is used so that, Qe = Cd Qt (5-8) Therefore, Qe Cd = (5-9) Ao 2 gH Notation:

t = time, Ac = area of column, Ao = area of orifice, H = available head, Cd = discharge coefficient, g = acceleration of gravity, Qt = theoretical discharge rate given by eq. (5-2), Qe = experimentally measured flow rate, n = exponent, and k = coefficient.

Figure 5-1 Orifice Flow

Experiment 6 Measuring Flow Rates with a Venturi Meter n Purpose: The Venturi meter is widely used as a flow measurement device. It can easily be inserted in a pipeline with only a small energy loss. In this experiment we will measure the flow rates through the Venturi and compare them with those calculated from the pressure readings at the inlet and the throat. n Procedure: 1. Close the pump flow control valve and start the pump. 2. Close the outlet valve about 3 1/4 turns from the full open position. 3. Regulate the pump flow to produce the maximum possible differential water level on piezometers 2 and 3 at the Venturi inlet and throat respectively. Take care to avoid water flowing up the piezometers into the manifold. 4. Allow a steady flow rate to be established through the entire circuit. 5. Measure the flow rate through the outlet and the level of water in each piezometer. 6. Carefully regulate the flow rate so that the differential between the inlet and the throat is reduced in about 5 equal steps. 7. Measure all piezometric heights and the flow rate for each step. n Questions: (interpretation of results) 1. By plotting a graph of the logarithm of the measured flow rate vs. the logarithm of the measured pressure drop, verify the relationship between Qmeas and Hmeas derived in Note 1 below. 2. Calculate the meter constant (Cm) and the meter coefficient (k). Show that the intercept with the y-axis on the graph plotted in question 1 above is equal to log (kCm). See Note 2 below. 3. Calculate the theoretical flow rate and compare the values with the measured flow rates. Comment on any discrepancies. 4. Calculate the pressure recovery. See Appendix (A-2.) 5. Discuss the performance of the Venturi meter. See any text on fluid mechanics. Notes:

1. The expression for the theoretical flow rate can be derived from Bernoulli's equation. p2 /(ρg) + v22 /(2g) + z2 = p3 /(ρg) + v32 /(2g) + z3 (6-1) Qtheor = v2 A2 = v3 A3 (6-2) For a horizontal channel, z 2 = z3 (6-3). 2 2 p2 − p 3 v 3 − v 2 (6-4) = ρg 2g Substituting for the velocity into Eq. 6-1, and multiplying top and bottom by A22,

( p2 − p3 ) Q 2 (r 2 − 1) (6-5) = ρg A22 (2 g ) in which r = the ratio of the inlet area to the throat area, r = A2 /A3 (6-6) and theoretically, (p2 - p3)/(ρg) = htheor = the difference in piezometric pressure head and A2 = the inlet area. Rearranging gives, 2 ghtheor (6-7) Qtheor = A2 (r 2 − 1) 2. Since A2 , g and r are constant for a given Venturi meter, the expression (Eq. 6-7) becomes, Qtheor = Cm htheor (6-8) in which, 2g Cm = A2 (6-9) 2 (r − 1) where Cm is known as the meter constant. (Note that Cm is not dimensionless!) Due to friction and the contraction losses, the actual pressure drop, H, will be greater than the theoretical pressure drop, h. The actual flow rate is given by the equation, Qmeas = kCm H meas (6-10) in which, htheor k= (6-11). H meas k is called the meter coefficient which is always less than one.

datum 15 mm

Figure 6-1 Diagram of Venturi Meter

n Data: Water temperature= ρ= Set Flow rate Vol. Time Q (mL) (sec) (mL/sec) 1 2 3 4 5

#1

Piezometer height (cm) #2 #3 #4

Average= Note: Convert all units consistently.

k

Qtheor (mL/sec)

Experiment 7 Energy Loss in a Hydraulic Jump n Purpose: The purpose of this experiment is to examine the transition from supercritical (rapid) flow to subcritical (slow) flow in an open channel and to analyze the resulting energy changes. n Procedure: 1. Adjust the tailgate of the channel so that it causes no obstruction at the outlet end. This is done by placing the tailgate into its lowest position. 2. Open the sluice gate to a height of approximately 3/4 inch. Keep the flow rate at about 12 gal/min. Adjust the flow so that the inlet reservoir has sufficient head to cause supercritical flow to exist along the entire length of the channel. Insure that no overflow occurs from the rear of the inlet reservoir. 3. By adjusting the tailgate, create a hydraulic jump in the central portion of the channel. 4. Measure: a) the depth of flow in front of (y1) and behind (y2) the jump, b) the levels in the pitot tubes on each side of the jump, by placing them 20, 40, 60 and 80 percent of the flow depth from the water surface, c) the depth of flow at the vena contracta, and d) the opening of the sluice gate. 5. Keeping the flow rate constant and changing the sluice gate opening to 1 or 1 1/2 inch, repeat Steps 1 through 4 at least 3 times until you obtain a situation where you cannot obtain a hydraulic jump by tailgate adjustment. 6. For 1 additional flow rate, create a hydraulic jump in the central portion of the channel and measure the depth of flow in front of (y1) and behind (y2) the jump. n Questions: (interpretation of results)

1. Draw the velocity distribution profiles and compute the average velocities at the cross sections on either side of the hydraulic jump. Compare the profile values with the average velocity. 2. Calculate the energy loss in the hydraulic jump using Bernoulli’s Theorem (Eq. 7-1) and the average velocity. Compare this to the difference in heights in the pitot tubes. 3. Plot the relationships between the ratio of conjugate depths and the corresponding Froude number. See Eqs. 7-11 and 7-12. 4. Calculate the critical depths. Do they fall within the expected ranges? To answer this question, a) Draw the specific force curve for one value of Q. b) Draw the specific energy curve for the same value of Q. (F = γd2/2) 5. Calculate the coefficient of contraction for the sluice gate by comparing the vena contracta with ygate. (C = yvc/ygate) 6. What percent of the upstream energy is dissipated? Is the hydraulic jump an efficient energy dissipator?

Notes:

1. In open channel flow, the free surface coincides with the hydraulic grade line provided that the pressure distribution is hydrostatic. The pressure distribution will be hydrostatic if the accelerations and curvature of the streamlines in a vertical plane are negligible and if the slope of the bed is small (< 10%). Under these conditions the expression for the total energy may be written as, (7-1) H = p/γ + z + v2/(2g) = y + z + v2/(2g) in which y = the vertical distance from the bed to the water surface (depth of flow) and z is defined as the height of the bed above the datum. If we consider the channel bed as the datum (z = 0), the Eq. 7-1 reduces to, (7-2) H = y + v2/(2g) = E where E is termed the specific energy. For this experiment, since z = 0, the total energy and the specific energy are the same. Furthermore, for this experiment with a rectangular channel cross-section with constant width b, (area of cross-section), A = by (7-3), (flow rate or discharge), Q = v A = v b y (7-4), (discharge per unit width), q = Q/b = v b y/b = v y (7-5). Therefore,

(7-6). E = y + q2/(2gy2) The relationship expressed by Eq. 7-6 is cubic in terms of y and there must be three solutions for a given set of values for E and q. However, only two of the solutions are real. Thus, there are two possible depths of flow for a given specific energy level, E, and discharge, q. The two depths are referred to as alternate depths. This provides for two regimes of flow, either slow and deep, or fast and shallow, referred to as subcritical flow and supercritical flow, respectively. If there is a transition from one regime to the other for a given E and q, then the flow must go through an intermediate condition known as critical flow. This critical condition describes the state of flow at which the specific energy is a minimum for a given flow rate per unit width, q. Conversely, at this critical condition for a given E, the flow rate, q, must reach a maximum value. The depth at which critical discharge occurs is called the critical depth, yc , and the velocity is the critical velocity, vc . From Eq. 7-6 assuming critical conditions, (7-7) vc2 = g yc 3 2 (7-8) q = g yc and, (7-9). yc = 2/3 E From Eq. 7-7, at critical flow conditions, (7-10) Fr2 = vc2 /gyc = 1 in which Fr = the Froude number. At the critical flow condition, Fr = 1. When Fr < 1 the flow is subcritical and when Fr > 1 the flow is supercritical. The hydraulic jump allows a reasonably abrupt transition from supercritical to subcritical flow. It can be accompanied by considerable energy loss and turbulence. The energy loss

varies with the Froude number of the jump and is generally an unknown. The energy loss can be found by computing the flow energy upstream of the jump and the reduced flow energy downstream of the jump. The energy loss is the difference between the two energy levels. Conservation of momentum can be used to derive the equations for the upstream and downstream depths which are needed to calculate the energies and the energy loss. The relationships between the upstream and downstream depths are given by, y2 1 = 1 + 8 Fr21 − 1 (7-11) y1 2 y1 1 = 1 + 8 Fr22 − 1 (7-12) y2 2 where the subscripts 1 and 2 apply to conditions upstream and downstream of the jump respectively. y1 and y2 are termed the conjugate or sequent depths. The upstream Froude number is given by, v Fr1 = 1 (7-13) gy1 and the downstream Froude number is given by, v Fr 2 = 2 (7-14) gy 2 (Note that the conjugate depths are not the same as the alternate depths discussed above since there is a loss of energy from the upstream side to the downstream side of the hydraulic jump.)

[ [

] ]

Figure 7-1 Hydraulic Jump

n Data: Water temperature=

Q Sluice (gpm) Opening (in) 11 0.75 1.00 1.10 12 0.75 1.00 1.10

Depth of ven.contr. , yvc (in)

y1 (in )

y2 (in )

Pitot tube levels (in) Pitot tube 1 Pitot tube 2 0.2y 0.4y 0.6y 0.8y 0.2y 0.4y 0.6y 0.8y

Experiment 8 Flow Over a Weir n Purpose: The purpose of this experiment is to verify the discharge equation and to determine the discharge coefficient for a sharp crested weir. The flow of water over a weir depends on the shape of the weir and the height of the water level above the sill or notch of the weir. The data from the tests described below are used to verify the theoretical relationships between these factors. These results can then be used to find the coefficient of discharge. n Procedure: 1. Attach the rectangular weir to the channel. 2. Close the pump flow control valve and start the pump. 3. Regulate the flow to maintain a water level in the flow channel so that the weir is filled to the top of the machined section. Be careful to avoid flooding above this level. 4. Allow a steady flow to develop throughout the entire circuit. Measure the height of the water level above the top of the weir as an H to start the analysis. 5. Measure the flow rate with a stopwatch and graduated measuring cylinder, and measure the water level in the approach channel using the gage at a location about halfway along the approach channel; the first reading of the gage should be used as a calibration. For each set of measurements, measure the flow rate at least three times and take the average. 6. Reduce the approach channel water level in about 2 or 3 even steps, each time recording the water level differential in the channel with the gage. Also record the flow rate. 7. Attach the triangular (90°) weir (V-notch), where θ = 0.5 (90°) = 45° 8. Repeat Steps 2 through 6; however, this time regulate the flow to maintain a level in the approach channel so that the weir is filled only to the top of the triangular section. n Questions: (interpretation of results) 1. By plotting a graph of the logarithm of the flow rate vs. the logarithm of the depth, compare the theoretical power law and coefficient with those obtained from the graph. Comment on your results. See Notes 1 and 2, and Appendix A-1. 2. Calculate the coefficient of discharge. See Eq. (8-6), (8-13) and Appendix 3. Notes: 1. Consider the flow through a rectangular notch or sharp-crested weir as shown in Figure 8-1. A horizontal differential element is taken at a depth y below the free surface. The area of the element is given by, dA = B dy (8-1) The velocity through the element is given by, v = 2 gy

(8-2)

Therefore, the theoretical discharge through the element is, dQ = B 2 gydy Integrating Eq. 8-3 yields the theoretical discharge,

(8-3)

H

Qt = B 2 g ∫ y 1/ 2 dy

(8-4)

0

or, Qt =

2 B 2 g H 3/ 2 3

(8-5)

The actual discharge is given by, 2 B 2 g H 3/ 2 3 2 3 where Cd = the coefficient of discharge, K = B 2 g , N = and B = 3 cm. 3 2 Qa = Cd

(8-6)

2. Consider the flow through the triangular notched weir shown in Figure 8-2. Consider an element at depth y. The breadth of the element is given by, B = 2 (H - y) tan θ (8-7) and the area of the differential element is then given by, dA = 2 (H - y) tan θ dy (8-8) while the velocity through the element is given by, v = 2 gy (8-9) The discharge through the element is, dQ = 2( H − y ) 2 gy tan θdy (8-10) and the total theoretical discharge is obtained by integrating Eq. 8-10, H

Qt = 2 tan θ 2 g ∫ ( Hy1 / 2 − y 3 / 2 )dy

(8-11)

0

which yields, Qt =

8 tan θ 2 g H 5/ 2 15

(8-12)

The actual discharge is given by, 8 tan θ 2 g H 5 / 2 15 in which Cd = the coefficient of discharge. θ = 1/2 of the machined angle = 45° N = 5/2 (triangle), and 8 K= 2 g tan θ 15 Qa = C d

(8-13)

Figure 8-1 Flow Over a Rectangular-Notched, Sharp-Crested Weir

Figure 8-2 Flow Over a Triangular V-Notched Weir n Data: Water temperature=

Rectangular Weir: Set Q H (cm) (mL/sec) 1 2 3 4 5

Triangular Weir (V-notch): Set Q H (cm) (mL/sec) 1 2 3 4 5

Experiment 9 Design of an Outfall Diffuser ________________________________________________________________________ Purpose: The objective of this experiment is to design a manifold diffuser to distribute flow uniformly. Problem Description: In general, it is possible to specify the desired flow distribution in a multi-port distribution system and to solve for the required flow areas to achieve this distribution provided that the upstream energy in the system is known. However, for a diffuser, a range of discharges may be experienced and the upstream energy level is likely to be a variable as well. In addition, the construction of different sized orifices at each discharge point is generally not feasible from an economic point of view. Therefore, it is generally better to specify a given diameter for all the discharge orifices or at least a combination of only a few different orifice diameters and then to compute the flow distribution from the proposed system.

Qo

< 1.5

S

Figure 1 Schematic of a Diffuser Procedure: Prepare a computer program or spreadsheet, which calculates the distribution of flow in a diffuser pipeline. The following design parameters will be used in the analysis:

design discharge: Qo # of orifices: N Diffuser pipe friction: F

Diffuser diameter: Spacing between Orif.:

Discharge coefficient for orifice: cd = 0.63 − 0.58

v2 2 gE

DIA S

in which v is the velocity (in the pipe) just upstream from the orifice and E is the difference between the total energy inside the pipeline and the static head outside. The analysis should begin with an assumed energy head at the upstream end of the diffuser and proceed downstream with repeated applications of orifice and energy equations. The repeated calculations will include the following steps: 1. Calculation of the orifice cd based on local conditions. 2. Calculation of orifice flow, Qi = cd Ai 2 gE . 3. Calculation of velocity in the pipe downstream from the orifice. 4. Calculation of the friction loss to the next orifice. 5. Calculation of the velocity head and energy at the next downstream orifice. The repeated calculations will yield a total orifice discharge associated with the assumed upstream energy. Adjustments in this energy will then be necessary until the computed discharge agrees with the design discharge, while satisfying the following constraints: 1. The flow rates from the orifices must be within 7% of each other. 2. At the design discharge, the available head at the upstream end of the manifold cannot exceed a specific value. Questions: Prepare a brief description of the computations including a description of the input and output. Details of the computations must be submitted with the attached output to show your solution. The output must include the orifice diameters and the flow distribution from the orifices. Also, provide a listing of the required energy head at the upstream orifice to develop this flow condition. It is also useful to print out the maximum and minimum orifice discharges. Repeat the analysis for a flow rate of 0.5 m3/s to see how changing the rate affects the flow distribution. Comment on all relevant results. Data: Q0 = 5.0 m3/s; DIA = 2.0 m; F = 0.02; S = 3.0 m; N = 40; allowable upstream head difference = 1.5 m.

Appendix

1. Verification of power laws. If a theoretical relation can be expressed as, Q = k Hn

(A-1)

then by plotting log(Q) vs. log(H), k and n can be easily found. Taking logarithms of both sides of Eq. A-1, log(Q) = log(k) + log(Hn) or,

log(Q) = log(k) + n log(H)

which is the equation of a straight line in slope-intercept form if the variables are log(Q) and log(H). The slope of the line gives n while the y-axis intercept is the value of log(k). To find k, simply raise 10 to the log(k) power. 2. Pressure recovery, in percent, %

(A-2)

3. Discharge coefficient Cd = Actual discharge/Theoretical discharge

(A-3)