Lab Experiment No.: 4Date Performed: February 21, 2012Title: Head Loss Across A Gate ValveDate Submitted:

Group No.: 4 4CECGroup Members: MEDINA, ROUENE JEAN P.OLPOT, NEZELLE MAY T.NOLASCO, LUIS PATRICK V.ORBE, MELIZA P.OLALO, LEONARDO T.OSMEA, SALVIO JR. E.IntroductionIn addition to skin friction losses in pipe flow, there are additional minor losses due to fittings, valves, gradual and sudden expansions or contractions, and entrance and exit effects. Since the flow patterns in valves and fittings is quite complex and varies from one manufacturer to another, there is no accepted theoretical means of determining these minor losses. The losses are usually measured experimentally and correlated with the pipe flow parameters. The measured minor loss is typically expressed as a ratio of the head loss hm=p/g through the fitting or valve to the velocity head V2/2g of the attached piping system. The resulting ratio is called the loss coefficient K for the fitting or valve

Thus, the minor losses may also be expressed in terms of the velocity head once the loss coefficient has been determined

Objective To determine the loss factors for flow through a range of pipe fittings including a gate valve To determine the minor loss coefficient and associated loss in head due to gate valve

Apparatus and Supplies

Procedure Equipment Set UpSet up the losses apparatus on the hydraulic bench so that its base is horizontal (this is necessary for accurate height measurements from the manometers). Connect the test rig inlet to the bench flow supply and run the outlet extension tube to the volumetric tank and secure it in place.

Open the bench valve, the gate valve and the flow control valve and start the pump to fill the test rig with water. In order to bleed air from pressure tapping points and the manometers, close both the bench valve and the test rig flow control valve and open the air bleed screw and remove the cap from the adjacent air valve. Connect a length of small-bore tubing from the air valve to the volumetric tank. Now, open the bench valve and allow flow through the manometers to purge all air from them; then, tighten the air bleed screw and partly open both the bench valve and the test rig flow control valve. Next, open the air bleed screw slightly to allow air to enter the top of the manometers, re-lighten the screw when the manometer levels reach a convenient height.

Check that all the manometer levels are on scale at the maximum volume flow rate required (approximately 17 liters/minute). These levels can be adjusted further by using the air bleed screw and the hand pump supplied. The air bleed screw controls the air flow through the air valve, so when using the hand pump, the bleed screw must be open. To retain the hand pump, pressure in the system, the screw must be closed after pumping.

Taking a Set of ResultsExercise B measures the loss across a gate valve only. Clamp off the connecting tubes to the mitre bend pressure tapping (to prevent air being drawn into the system). Start with the gate valve closed and open fully both the bench valve and the test rig flow control valve. Now open the gate valve by approximately 50% of one turn (after taking up any backlash). For each of at least 5 flow rates, measure pressure drop across the valve from the pressure gauge; adjust the flow rate by the use of the test rig flow control valve. Once measurements have started, do not adjust the gate valve. Determine the volume flow rate by timed collection.

Repeat this procedure for the gate valve opened approximately 70% of one turn and then approximately 80% of one turn.

Data and Results

# of turnsPz(m)HL (hg)(m)Vol.(L)t(s)Q(m3/s)V(m/s)Kg

BarMeter

1.250.22.040.152.19310.42.830 x 10-4

0.93848.836

310.8

310.6

1.000.44.080.154.23310.42.876 x 10-4

0.95391.381

310.4

310.5

0.750.66.120.156.27311.12.695 x 10-4

0.893154.264

310.8

311.5

ComputationsConversion factor:1 bar = 10.2 m of water

Formula:hg = P + zQ = Vol.V = QKg = 2g (hg) t AV2

P1 = 0.2 bar x 10. 2 m = 2.04 mhg1 = 2.04 + 0.15 = 2.19 m 1 barP2 = 0.4 bar x 10. 2 m = 4.08 mhg1 = 4.08 + 0.15 = 4.23 m 1 barP3 = 0.6 bar x 10. 2 m = 6.12 mhg1 = 6.12 + 0.15 = 6.27 m 1 bar

tave = 10.4 + 10.8 + 10.6 = 10.6 sQ1 = 3 x 10-3 = 2.830 x 10-4 m3/s3 10.6tave = 10.4 + 10.4 + 10.5 = 10.43 sQ2 = 3 x 10-3 = 2.876 x 10-4 m3/s3 10.43tave = 11.1 + 10.8 + 11.5 = 11.13 sQ3 = 3 x 10-3 = 2.695 x 10-4 m3/s3 11.13

V1 = 2.830 x 10-4 = 0.938 m/sKg1 = 2(9.81)(2.19) = 48.836 /4 (0.0196)20.9382V2 = 2.876 x 10-4 = 0.953 m/sKg2 = 2(9.81)(4.23) = 91.381 /4 (0.0196)20.9532V3 = 2.695x 10-4 = 0.893 m/sKg3 = 2(9.81)(6.27) = 154.264 /4 (0.0196)20.8932

Drawings/Figures

Apparatus Determining the volume flow rate by timed collection Reading the pressure dropGate valve and pressure gauge for 1.0 turnRemarksReferencesApplication of theory