francis report

PART B : FRANCIS TURBINE1.0 INTRODUCTION In Francis Turbine water flow is radial into the turbine and exits the Turbine axially. Water pressure decreases as it passes through the turbine imparting reaction on the turbine blades making the turbine rotate. Read more about design and working principle of Francis Turbine in this article.Francis Turbine is the first hydraulic turbine with radial inflow. It was designed by American scientist James Francis. Francis Turbine is a reaction turbine. Reaction Turbines have some primary features which differentiate them from Impulse Turbine. The major part of pressure drop takes place up to the entry point and the turbine passage is completely filled by the water flow during the operation.For power generation using Francis Turbine the turbine is supplied with high pressure water which enters the turbine with radial inflow and leaves the turbine axially through the draft tube. The energy from water flow is transferred to the shaft of the turbine in form of torque and rotation. The turbine shaft is coupled with dynamos or alternators for power generation. For quality power generation speed of turbine should be maintained constant despite the changing loads. To maintain the runner speed constant even in reduced load condition the water flow rate is reduced by changing the guide vanes angle.

2.0 OBJECTIVETo determine the relationship between the head, flow rate, velocity, power and efficiency of Francis Turbine.

3.0 LEARNING OUTCOMESAt the end of the course, students should be able to apply the knowledge and skills they have learned to: a. Understand the basic operating system of the Francis Turbine. b. Understand on the factors which influence the efficiency of turbine.4.0 THEORY

Hydraulic power can be obtain at the turbine inlet (usually known in watt unit) and can be calculated as, , and Q as the volume discharge that can be read from the measuring equipment (m3/s) and is a water weight per unit of volume (9820 N / m3). is a total head (m) which can be calculated (using theorem Bernoulli method) in a circuit section flow just before the turbine in a location of pressure head as a total of three parameter, which is Hman is a the differences of pressure head position which measured the pressure and the turbine shaft bar. As using the experimental table, (if the pressure decreasing at the turbine point out which cause by the mixer which count as an available head, Hman can be calculated as the height differences between the pressure gauging point position and the point of water level at the outflow of reservoir) is a kinetic parameter which cause by water velocity, vm (in m / s), at the pressure gauging location, where the pipe section is Sm (this value is not S and v value at the inlet section). , where as g is the gravity constant (9.81m/s2) and vm=Q/Sm. is a resultant parameter from water pressure, (in unit N / m2) as measured by pressure gauging. In calculation table, the readable gauging value which has been read by four gauging, all the readable value from the calculation can be added by the related values, vm, Hman, Hkin, Hpres, Htot, Phyd, which has been described first. This following relationship, has been used to calculated the value of mechanical power out, Pmec (usually in watt), , where as , is the turbine angular velocity (in rad / s), n is a turbine average velocity obtain by reader equipment (RPM) and c is a braking torque velocity at the turbine shaft as obtain by reader equipment (Nm). Lastly, the turbine overall efficiency, is calculated as the ratio of power at outlet point compared to the generated power. Last table arrangement for the experimental value and the calculation value (for each ) can be explained in a provided table.

5.0 EQUIPMENTSFrancis Turbine equipment

6.0 PROCEDURES

The delivery control shutter closed.The distributor leverage at the chosen value (50% and 100%) was fixed.The delivery control shutter was opened until desired flow rate.The reading of pressure (, speed (n), flow rate (Q), and voltage (v) for each distributor leverage value in Table 6.1 was recorded.Step 3 4 steps were repeated with different distributor leverage value.PRECAUTION : Must ensure the delivery control shutter should be closed before switch on the equipment. Carelessness may cause equipment damage.

7.0 8.0 SAMPLE DATA

Sm= 1256mm2Hmm=0.20m

Z Pressure Pm Flowrate QSpeed, nVoltage VCurrent Reading ICurrent Reading IAverage PmAverage QAverage nAverage VAverage I

(%) (bar)(m2/h)(PRM)(volt)(ampere)(%)

500.8713.5562423.20.0000.7614.724478.4015.881.36

0.8214.3559819.60.7225

0.7515.5472515.91.5950

0.7115.0371811.91.9575

0.6615.327278.82.52100

1000.9013.1598923.00.0000.8014.524345.2014.981.26

0.8614.1538018.80.6825

0.7914.9436014.61.4450

0.7414.8356411.51.8575

0.7115.724337.02.35100

Table 7.1 : Results of Francis Turbine Experiment

Sm= 1256mm2Hmm=0.20m

ZCerrent Reading I Electrical Power PelPipe Cross Section SmFlowrate Qx10-3 Water Speed VmDistance Between Turbine Shaft & Pressure Tap HmanKinetic HkinPressure HpresTotal Head HtotMechanical Power PmHydraulic Power PhydEfficiency Pm/Phyd

(%)(%)(w)(m2)(m3/s)(m/s)(m)(m)(m)(m)(watt)(watt)(watt)

50098.20.0012563.752.9860.200.4548.8599.514588.944350.3460.595

253.973.1630.518.359.06586.221353.4090.603

504.313.4280.5997.6378.436494.801356.6960.721

754.173.3170.5617.237.991389.348326.9680.84

1004.253.3840.5846.7217.505285.571313.2031.097

10003.642.8970.4289.1659.793627.167349.9340.558

253.923.1180.4968.7589.453561.088363.5880.648

504.143.2950.5538.0458.798456.578357.5960.783

754.113.2730.5467.5368.282373.221334.3410.896

1004.363.4720.6147.238.045254.783344.521.352

Table 7.2 : Results of Francis Turbine Experiment

9.0 ANALYSIS OF RESULTS, EQUATIONS USED

A) DATA at Z = 50%

Pressure, Pm (N/m2)* 1 bar = 100kPa = 100000 Pa = 100000N/m2

At current reading, I = 0= 0.87 bar x 100000 N/m2= 87000 N/m2





Flow rate, Q (m3/s)At current reading, I = 0,13.5 m3 x 1 h = 3.75 x 10-3 m3/s h 3600s

At current reading, I = 25,14.3 m3 x 1 h = 3.97 x 10-3 m3/s h 3600s

At current reading, I = 50,15.5 m3 x 1 h = 4.31 x 10-3 m3/s h 3600 s


At current Reading, I =100,15.3 m3 x 1 h = 4.25 x 10-3 m3/s h 3600 s

Water Speed, Vm (m/s)At current reading, I = 0,Vm = Q / Sm = 3.75 x 10-3/ 0.001256 = 2.986 m/s

At current reading, I = 25,Vm = Q / Sm = 3.75 x 10-3/ 0.001256 = 2.986 m/s

At current reading, I = 50,Vm = Q / Sm = 3.75 x 10-3/ 0.001256 = 2.986 m/s

At current reading, I = 75,Vm = Q / Sm = 3.75 x 10-3 / 0.001256 = 2.986 m/s

At current Reading, I =100,Vm = Q / Sm = 3.75 x 10-3/ 0.001256 = 2.986 m/s

Kinetic, Hkin (m)At current reading, I = 0,Hkin = Vm2 / 2g = (2.986) 2 / (2 x 9.81) = 0.454 m

At current reading, I = 25,Hkin = Vm2 / 2g = (3.163) 2 / (2 x 9.81) = 0.510 m

At current reading, I = 50,Hkin = Vm2/ 2g = (3.428) 2 / (2 x 9.81) = 0.599 m

At current reading, I = 75,Hkin = Vm2 / 2g = (3.317) 2 / (2 x 9.81) = 0.561 m

At current Reading, I =100,Hkin = Vm2 / 2g = (3.384) 2 / (2 x 9.81) = 0.584 m

Pressure, Hpres (m)At current reading, I = 0,Hpres = Pm / = 87000 = 8.859 m 9820

At current reading, I = 25,Hpres = Pm / = 82000 = 8.350 m 9820



At current Reading, I =100,Hpres = Pm / = 66000 = 6.721 m 9820

Total Head, Htot (m)At current reading, I = 0,Htot = Hman + Hkit + Hpres = 0.20 + 0.454 + 8.859 = 9.514 m

At current reading, I = 25,Htot = Hman + Hkit + Hpres = 0.20 + 0.510 + 8.350 = 9.060 m



At current Reading, I =100,Htot = Hman + Hkit + Hpres = 0.20 + 0.584 + 6.721 = 7.505 m

Mechanical Power, Pm (watt)At current reading, I = 0,Pm = 2n / 60 = 2(5624) = 588.944 watt

At current reading, I = 25,Pm = 2n / 60 = 2(5598) = 586.221 watt



At current reading, I =100, Pm = 2n / 60 = 2(2727) = 285.571 watt

Hydraulic Power, Phyd (watt)At current reading, I = 0,Phyd = HtotQ = (9820)(9.514)(3.75 x 10-3) = 350.346 watt

At current reading, I = 25,Phyd = HtotQ = (9820)(9.060)(3.75 x 10-3) = 353.409 watt



At current reading, I =100, Phyd = HtotQ = (9820)(7.991)(3.75 x 10-3) = 313.203 watt

Efficiency, Pm / Phyd (watt)At current reading, I = 0,(Pm / Phyd) x 100 = (588.944 / 350.346) x 100 = 0.595 watt

At current reading, I = 25,(Pm / Phyd) x 100 = (586.221 / 353.409) x 100 = 0.603 watt



At current reading, I =100, (Pm / Phyd) x 100 = (285.571 / 313.203) x 100 = 1.097 watt

Average Data at Z = 50%i) Average of Pressure, Pm (N/m2)= (87000 + 82000 + 75000 + 71000 + 66000) / 5= 76200 N/m2

ii) Average of Flowrate, Q (m3/s)= (3.75 + 3.97 + 4.31 + 4.17 + 4.25) /5 x (10-3 m3/s)= 4.09 x 10-3 m3/s

iii) Average of Speed, n (RPM)= (5624 + 5598 + 4725 + 3718 + 2727) /5= 4478.4

iv) Average of Voltage, V (volt)= (23.20 + 19.60 + 15.90 + 11.90 + 8.80) / 5= 15.88 volt

v) Average of Current reading, I (ampere)= (0 + 0.72 + 1.59 + 1.95 + 2.52) / 5 = 1.356 Ampere

B) DATA at Z = 100%Pressure, Pm (N/m2)* 1 bar = 100kPa = 100000 Pa = 100000N/m2






Flow rate, Q (m3/s)At current reading, I = 0,13.10 m3 x 1 h = 3.64 x 10-3 m3/s h 3600 s




At current Reading, I =100,15.7 m3 x 1 h = 4.36 x 10-3 m3/s h 3600 s

Water Speed, Vm (m/s)At current reading, I = 0,Vm = Q / Sm = 3.64 x 10 -3 / 0.001256 = 2.897 m/s

At current reading, I = 25,Vm = Q / Sm = 3.92 x 10 -3 / 0.001256 = 3.118 m/s



At current Reading, I =100,Vm = Q / Sm = 4.36 x 10 -3 / 0.001256 = 3.472 m/s

Kinetic, Hkin (m)At current reading, I = 0,Hkin = Vm2 / 2g = (2.897)2 / (2 x 9.81) = 0.428 m

At current reading, I = 25,Hkin = Vm2 / 2g = (3.118)2 / (2 x 9.81) = 0.496 m



At current Reading, I =100,Hkin = Vm2 / 2g = (3.472)2 / (2 x 9.81) = 0.614 m

Pressure, Hpres (m)At current reading, I = 0,Hpres = Pm / = 90000 = 9.793 m 9820


At current reading, I = 50,Hpres = Pm / = 79000 = 8.798m 9820


At current Reading, I =100,Hpres = Pm / = 71000 = 8.045 m 9820

Total Head, Htot (m)At current reading, I = 0,Htot = Hman + Hkin + Hpres = 0.20 + 0.428 + 9.165 = 9.793 m

At current reading, I = 25,Htot = Hman + Hkin + Hpres = 0.20 + 0.496 + 8.758 = 9.453 m



At current Reading, I =100,Htot = Hman + Hkin + Hpres = 0.20 + 0.614 + 7.230 = 8.045 m

Mechanical Power, Pm (watt)At current reading, I = 0,Pm = 2n / 60 = 2(5989) = 627.167 watt




At current reading, I =100, Pm = 2n / 60 = 2(2433) = 254.783 watt

Hydraulic Power, Phyd (watt)At current reading, I = 0,Phyd = HtotQ = (9820)(9.793)(2.897 x 10-3) = 349.934 watt

At current reading, I = 25,Phyd = HtotQ = (9820)(9.453)(3.118 x 10-3) = 363.588watt



At current reading, I =100, Phyd = HtotQ = (9820)(8.045)(3.472 x 10-3) = 344.520 watt

Efficiency, Pm / Phyd (watt)At current reading, I = 0,(Pm / Phyd) x 100 = (627.167 / 349.934) x 100 = 0.558watt




At current reading, I =100, (Pm / Phyd) x 100 = (254.783 / 344.520) x 100 = 1.352 watt

Average Data at Z = 100%i) Average of Pressure, Pm (N/m2)= (90000 + 86000 + 79000 + 74000 + 71000) / 5= 80000 N/m2

ii) Average of Flowrate, Q (m3/s)= (3.64 + 3.92 + 4.14 + 4.11 + 4.36) /5 x (10-3 m3/s)= 4.034 x 10-3 m3/s

iii) Average of Speed, n (RPM)= (5989 + 5358 + 4360 + 3564 + 2433) /5= 4340.8

iv) Average of Voltage, V (volt)= (23.0 + 18.8 + 14.6+ 11.5 + 7.0) / 5= 14.98 volt

v) Average of Current reading, I (ampere)= (0 + 0.68 + 1.44 + 1.85 + 2.35) / 5 = 1.264 Ampere

10.0 QUESTIONS1. .Plot a graph of :

a. Turbine velocity versus flow rate

b. Turbine velocity versus output torque

c. Turbine velocity versus hydraulic power

d. Turbine velocity versus mechanical power

e. Turbine velocity versus efficiency.

2. Give your comment(s) from the graph obtained.From the turbine velocity against flow rate, the flow rate increased accordingly to turbine velocity. For the 2nd, the output torque is increased when the velocity reduced and this happened to graph velocity against hydraulic power also graph velocity against mechanical power as well. Graph velocity against efficiency shown that with the increased of velocity, the efficiency (Pm/Phyd) has increased accordingly.3. State five (5) safety factors that have been taken in the experiment Make sure that the power is turn on before any work is carried out. Make sure all the equipment is in good condition before work and after work done. Make sure that the distributor leverage is lift up or down correctly. Read the constant value of data carefully. Checks the sample data taken for accuracy.

11.0 DISCUSSSION

From these experiment, we can see the relationship between the head (m), flow rate (Q), velocity (v), speed (n), power efficiency of a Francis Turbine. In Francis Turbine, water flow is radial into the turbine and exits the Turbine axially. Water pressure decreases as it passes through the turbine imparting reaction on the turbine blades making the turbine rotate. The function of turbines is to get electricity converted from natural hydraulic power. Francis turbine may also be used for pumped storage, where a reservoir is filled by the turbine (acting as a pump) during low power demand and then reserved and used to generate power during peak demand.Francis Turbines are generally installed with their axis vertical. Water with high head (pressure) enters the turbine through the spiral casing surrounding the guide vanes. The water loses a part of its pressure in the volute (spiral casing) to maintain its speed. Then water passes through guide vanes where it is directed to strike the blades on the runner at optimum angles. As the water flows through the runner its pressure and angular momentum reduces. This reduction imparts reaction on the runner and power is transferred to the turbine shaft.If the turbine is operating at the design conditions the water leaves the runner in axial direction. Water exits the turbine through the draft tube, which acts as a diffuser and reduces the exit velocity of the flow to recover maximum energy from the flowing water.

12.0 CONCLUSION

The experiment objective is achieve. In producing the highest power through turbines, all factors must be taken. The same flow rate produces a different velocity. While the velocity of water will produce a different power and efficiency according to the different capabilities of the turbine. Factor such as the flow rate, velocity, power and efficiency of this interaction and should be taken into account. Results of experiment showed the relationship between head, flow rate, velocity, power and efficiency were produce the desired energy. Aperture-opening in these experiments show differences and advantages between an aperture so that we can to reflect the actual operating situation of a Francis Turbine.

13.0 REFERENCE

1. Mifflin, Boston, MA.White, F.M. (1994). Fluid Mechanics, 3rd edition, McGraw-Hill,Inc., New York, NY.2. R. E. Featherstone, C. Naluri. (1995.) Civil Engineering Hydraulics. Bodmin, Cornwall:Blackwell Science3. http://www.brighthubengineering.com/fluid-mechanics-hydraulics/27407-hydraulic-turbines-francis-turbine/

francis report

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