1.Impact of a Jet - Ahmed E. Mansour -...

ATOOMUS مؤمن توفيق احمد العتوم

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A H M 531 Impact of a Jet By: Mu'men Al-Otoom

The Civil Engineering Center 1 Visit www.AHM531.com for more lecture notes and E-books!

Object: To study the relation between the force produce and the change of momentum when a jet strikes a vane . Introduction : One way of producing mechanical work from fluid under pressure is to use the pressure to accelerate the fluid to a high velocity in a jet .the jet is directed on the vanes of a turbine wheel which is rotated by the force generated on the place as the jet strikes the vanes . Theory : If a flow Q m3/s with a velocity Uo m/s strikes a vane perpendicular to it , the flow is deflected by an angle ß so that the fluid leaves the vanes with velocity U1 inclined at an angle ß to the original axis of flow .the force acting on the fluid being equal to the rate of change of momentum that is :

F = ρ Q (U1 cos ß - Uo ) Where : ρ =density in Kg/m3 The force R on the vane is equal & opposite to F so that :

R = ρ Q ( Uo - U1 cos ß) Case one : In the case of flat plate ß = 90o and the force acting on the plate :

R = ρ Q ( Uo) Case two : In the case of a hemispheric cup we may assume that ß = 180o hence :

R = ρ Q ( Uo + U1) Since the changes in peizometric pressure and elevation are neglected the maximum value of U1 will be Uo thus max. force on the hemispheric cup is ( 2 ρ Q Uo ) i.e. twice the force on the flat plate .


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Apparatus : The bench supply is led to a vertical pipe terminating in a tapered nozzole ( d = 10 mm). This produces a jet of water which impinges on a vane in the form of a flat plate or a hemispheric cup. The nozzle and the vane are contained within a transparent cylinder at the base of which there is an outlet that leads the flow to the weighing tank .The vane is supported by a lever which carries a jockey weight ( w = 600gm ) and restrained by a spring. The lever may be set to a balanced position by placing the jockey weight at its zero position and then adjusting the knurled nut above the spring. Any force generated by impact of the jet on the vane may be measured by moving the jockey weight along the lever a distance Y until the tally shows that the lever has been restored to its original balanced position .Taking moments about the pivot of the lever which is 150mm away from the center of the vane gives the force exerted on the vane R2 thus @ F = 0 , T(x) – mg(0.15) = 0 --------------- ( a ) @ F ≠ 0 , F(0.15) +T(x) –mg(0.15 +Y ) = 0 , F(0.15) +T(x) –mg(0.15) + mgY = 0 , put ( a ) equation …….. → F( 0.15 ) + mgY = 0 , remember that R = - F ………. → -R(0.15) = - mgY , R = (mgY/0.15) = ( 0.6 * 9.81*Y /0.15 ) = ( 9.81*4*Y ) …….. R2 = ( 9.81*4*Y ) = practice measured force on flat plate & hemispheric cup.


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Procedure : - Level the apparatus and set the lever to a balanced position

with the jockey weight at its zero position . - Admit the water at max. Flow rate through the supply valve . - Note the position of the jockey which restates the lever to the

balanced position - To determine the flow rate ,take reading of volume and time

using the volumetric tank . - A series of reading with equally y ( spaced position ) of the

jockey . - The experiment may be then repeated using the hemispheric

cup .

Calculation : ( for flat plate ) :

- Q = ( ( volume of used water ) /( the time for this volume) ) . - U nozzle = the velocity at exit from the nozzle = ( Q / Area of nozzle ) , where dnozzle = 10 mm . - Uo

2 = U nozzle 2 – ( 2 g S ) , where S = 35 mm = the height of vane

above tip of nozzle . - R1 = ρ Q ( Uo) = theoretical value ( predicted value ) . - R2 = 9.81 * 4 * Y = practice value .

Calculation : ( for hemispherical cup ) :

Similar with flat plate but : - R1 = 2 ρ Q ( Uo) = theoretical value ( predicted value ) .

FOR FLAT PLATE CALCULATION : Y

mm Q

Liter Time sec.

Q x10-4 m3/sec.

U nozzle m/sec.

Uo m/sec.

R1 prediction

R2 practice

10 5 59 0.847 1.06 0.689 0.0584 0.3924 20 5 30 1.66 2.1 1.92 0.3187 0.785 30 5 19 2.53 3.18 3.07 0.7767 1.177 40 5 15 3.33 4.18 4.09 1.362 1.5696 50 5 14 3.57 4.5 4.42 1.578 1.962 60 5 12 4.16 5.2 5.13 2.134 2.3544 68 =Ymax 5 11 4.54 5.71 5.54 2.5151 2.668


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FOR HEMISPHERIC CUP CALCULATION : Y

mm Q

Liter Time sec.

Q x10-4 m3/sec.

U nozzle m/sec.

Uo m/sec

R1 prediction

R2 practice

30 5 25 2 2.545 2.41 0.964 1.18 60 5 19 2.53 3.182 3.07 1.553 2.35 90 5 14 3.57 4.5 4.42 3.156 3.532 120 5 12 4.16 5.27 5.2 4.33 4.71 137 = Ymax

5 11 4.5 5.66 5.6 5.04 5.38

Sample of calculation : For flat plate : Y = 10 mm T = 59 sec. Q Vol. = 5 L Q = ( Vol. / T ) = (5x10-3)/(59) = 8.47x10-5 m3/sec. A nozzle = ( π x (0.01)2/4) = 7.95x10-5 m2

U nozzle = (Q/A nozzle ) =1.06 m/sec. Uo = ( U nozzle 2 – 2 x 9.81 x S )½ = 0.689 m/sec. R1 = ρ Q ( Uo) = 0.0584 N . R2 = 9.81 * 4 * Y = 0.3924 N . There is a large amount of error . For hemispheric cup : Same steps but R2 = 2 ρ Q ( Uo) .

Discussion & Conclusion :

From the results above , we can conclude that the force on the hemispheric cup is double the force on the flat plate ( take Ymax as example 2x2.668 ~= 5.38 ) .We can see the relation between Ymax in each flat plate & hemispheric cup : Ymax for hemispheric = 2 Ymax for flat plate . We can see the different between the predicted value for force (R1) & measured value for force ( R2 ) .This different come from many errors :

1. when we calibrating the machine . 2. personal errors in taking the volume reading .

☺ In this experiment we know how can we generated energy from water force . ☺ If ß = 60o , then R1 = 0.5 ρ Q ( Uo) .


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Note : 2R Flat = R Hemispherical 2Y Flat = Y Hemispherical З% = ( Rexp. /Rtheo. )x100 Rexp. = (mass / arm length) x 9.81Y = (weight / arm length) x Y


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