EXPERIMENT NO 1 Anil Chejara

IMPACT OF JET ON VANE Roll No. 9003022

Group No. 4

Background:

The momentum equation based on Newton’s 2nd

law of motion states that the algebraic sum

of external forces applied to control volume of fluid in any direction equal to the rate of

change of momentum in that direction.

The external forces include the component of the weight of the fluid and of the forces exerted

externally upon the boundary surface of control volume.

If a vertical water jet moving with velocity ‘V’ made to strike a target (Vane) which is free,

to move in vertical direction, force will be exerted on the target by the impact of jet.

Applying momentum equation in z- direction, force exerted by the jet on the vane, FZ is given

by

FZ = ρQ (Vzout- VZ in)

For flat plate, Vz out= 0

Fz = ρQ(0-v)

FZ= ρQv

For hemispherical curved plate, vz out= -v, vz in= v

Fz = ρQ[v+(-v)]

FZ = 2 ρQv

Where Q= Discharge from the nozzle (Calculated by volumetric method)

V= Velocity of jet = (Q/A)

Objectives:

1. To determine the force produced by a water jet when it strikes a flat vane and

a hemispherical cup.

2. To compare the results measured with the theoretical values calculated from

the momentum flux in the jet.

Experimental Setup:

In experimental arrangement the bench supply is led to a vertical pipe terminating in a

tapered nozzle. This produces a jet of water which impinges on a vane in a form of a flat or a

hemispherical cup. The nozzle and the vane are contained within a transparent cylinder at the

base of the cylinder where there is an outlet by which the flow may be directed to the

weighing tank.

The vane is supported by a rod which carries a weight, and which is restrained by a spring.

The rod may be set to a balance position by placing the weight at its zero position. Any force

generated by the impact of the jet on the vane may now be measured by moving the rod

weight along the scale until the rod head reaches to equilibrium position.

Procedure:

1. Clean and fill the sump tank with water.

2. First of all note down the relevant dimensions as area of collecting tank and diameter

of nozzle.

3. When jet is not running, note down the position of upper disc or plate.

4. Admit water supply to the nozzle.

5. As the jet strikes the disc, the disc moves upward, now for given water supply to

nozzle and weight note the equilibrium position of disc on the scale.

6. At this position find out the discharge and note down the weights placed above the

disc.

7. The procedure is repeated for different values of flow rate by reducing the water

supply in steps.

Flow pattern of impact

Observations and Calculation:

a. Data used

Density of water ρ = 1000 kg/ m3

Area of cross section of nozzle A=d2/ 4 = 7.85 X 10

-5 m

2

Area of measuring tank =0.1 m 2

Diameter of nozzle d = 0.010 m

weight of Aluminium disc + rod = 14.5 gm.

Weight of Flat plate disc = 89.5 gm.

Weight of hemispherical vane = 97.5 gm.

Sample weights used = 98.5, 192, 481, 286 gm.

b. Sample calculation for Flat plate vane with 98.5 gm. Weight;

Given area of measuring tank = 0.25 x .40 m2 = 0.1 m

2

time = 10 sec

height rise in measuring tank = 6.5 cm = 0.065 m

Volume flow rate = Volume / time

Volume of fluid = Area x height

= 0.0065 m3

Volume flow rate = 0.0065 m3 / 10 sec

= 0.00065 m3/ sec

c. Constant weights: total = 104 gm. For Flat plate and 112 gm. For hemispherical vane

Theoretical force, using the formula

Flat Plate: Ft = ρ a v2

=1000 X 7.85 X 10-5

X8.28 X 8.28 / 9.8 N

= 0.55 N

Applying Newton’s second law in the direction of incident jet

Weight Flat Plate vane Hemispherical vane

(gm.)

Estimated

force (for

flat plate)

(N)

Estimated

force (for

hemispherical)

(N)

Q

(Volume

flow rate)

(m3/s)

Jet

velocity

(m/s)

Ft

(N)

Q (Volume

flow rate)

(m3/s)

Jet

velocity

(m/s)

Ft

(N)

0 1.02 1.10 0.0006 7.64 4.59 0.0005 6.37 6.37

98.5 1.98 2.06 0.00065 8.28 5.38 0.00056 7.13 7.99

192 2.90 2.98 0.00065 8.28 5.38 0.00062 7.9 9.79

286 3.82 3.90 0.00067 8.54 5.72 0.00065 8.28 10.76

481 5.73 5.81 0.00072 9.17 6.60 0.00068 8.66 11.79

Force = Mass x Acceleration

. = Mass flow rate x Change in velocity per second

In vertical direction

F = Ṁ x v F=

Ṁ x ( VcosƟ – V )

F = ṀV(1-cosƟ)

where Ṁ=AV

F=AV2(1-cosƟ)

For flat vane, Ɵ = 0 and for spherical cup Ɵ = 180.

d. Graphical Representation

0

1

2

3

4

5

6

7

0 2 4 6 8 10

Forc

e o

n p

late

(N

)

Jet velocity (m/s)

Force Vs Jet velocity for Flat Plate

Theretical Values

Experimental Values

0

2

4

6

8

10

12

14

0 2 4 6 8 10

Forc

e o

n t

he

van

e (

N)

Jet velocity (m/s)

Force Vs Jet velocity for Hemispherical vane

Theoretical value

Experimental value

Conclusion:

1. The force produced on each of the vanes is proportional to the momentum flow in the

jet as it strikes the vane.

2. As weight increases the required velocity of jet increases.

3. For the hemispherical vanes the force is more than that of the flat plate. It is almost

twice.

Precaution:

1. Clean the apparatus and make All Tanks free from dust.

2. Apparatus should be in levelled condition.

3. Reading must be taken in steady conditions.

4. Discharge must be varied very gradually from a higher to smaller value

Sources of Error:

1. Change in velocity due to height difference between nozzle and vanes.

2. All the losses like frictional, nozzle, etc. were neglected.

3. In the theoretical calculations the viscosity effects were neglected.

4. Non uniform flow rate due to fluctuation in the power supply of pump.

5. In the theoretical calculations the viscosity effects were neglected.

6. All the losses like frictional, nozzle, etc. were neglected.

7. The vane was never in equilibrium due some fluctuation so it will cause error.