Force exerted by a jet of fluid study material
A jet of fluid contains energy and will exert a force on anything which lies in its path. In water, steam and gas turbines, the jet impinges on the rotor blades causing them to rotate. Some of the
energy in the jet is thus changed into rotational kinetic energy.
Turbine output and efficiency calculations can be quite complex and will be left for study at a higher level. Here we will consider only the force exerted by a jet of fluid on stationary objects, which will lay the foundation for future work.
Force exerted by a jet of fluid on a stationary
flat plate
a parallel jet of fluid of density 𝛒 kg m⁻³ and cross sectional area A m² which strikes a stationary flat plate at right angles with a velocity of v ms⁻¹. On striking the plate, the jet is deflected at right angles in all directions so that its final velocity in the original x-direction is zero, as shown in Figure 1 .The mass of fluid striking the plate per second is given by equation
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| FIG 1 |
mass flow rate = density× volume flow rate
m=𝛒 A v
velocity is a vector quantity. It has both magnitude and direction, and if either of should change, a change
in velocity will occur.Here the jet direction is changed by the plate and as a result, its velocity changes from v ms⁻ ¹ to zero in the x-direction. Momentum is the product of mass and velocity, and so there will also have been a change in momentum in the x-direction. From Newton’s second law of motion, the force acting on the plate is given by
force = rate of change of momentum
force = (mass × change in velocity) / time taken
But the mass divided by the time taken is the mass flow rate and the change in velocity is from v ms⁻ ¹ to zero
F = m (v-0)
F =𝛒 A v ( v -0)
F =𝛒 A v²
it can be seen that the thrust on the plate is proportional to the density of the fluid, the cross-sectional area of the jet and the square of the jet velocity. This means that if you double the velocity of the jet, the force on the plate will increase by a factor of 4. Similarly, if you double the diameter of the jet, its cross-sectional area will increase by a factor of 4 and once again, there will be four times as much force on the plate.Force exerted by a jet of fluid on a stationary
hemispherical cup
the effect of replacing the flat plate by a hemispherical cup as shown in Figure 2 .The mass striking the cup per second is the same as for the flat plate
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| FIG 2 |
From Newton’s second law of motion, the force exerted on the cup is again given by
force= (mass / time taken ) × change in velocity
The difference now is that the jet is deflected through 180⁰and the change in velocity in the x direction is from v ms⁻ ¹ to - v ms⁻ ¹
F = m[v - (-v)]
F = 𝛒 A v [v - (-v)]
F = 2 𝛒 A v²
In reversing the velocity of the fluid, the change of velocity, and so also the change of momentum is double that for the flat plate. As a result there is twice as much force exerted by the jet.
Reaction of a convergent nozzle
You may have experienced the same backward force when holding a water hose pipe. Applying Newton’s second law of motion enables the force to be found.
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| FIG 3 |
The mass flow rate in the pipe and issuing from the nozzle is again given by
m=𝛒 A₁ v₁ =𝛒 A₂ v₂
From Newton’s second law of motion, reaction of the nozzle is given by
F = m ( v₂ – v₁ )
F = 𝛒 A₁ v₁ (v₂ – v₁)
OR
F = 𝛒 A₂ v₂ (v₂ – v₁)
Power of a jet
The power delivered by a jet of fluid is the rate at which kinetic energy emerges from the nozzle. For a jet of cross-sectional area A m², density 𝛒 kg m⁻³ and velocity v ms⁻ ¹ and mass flow rate m kg s⁻¹
POWER = 1/2 𝛒 A v³
the power of a jet of fluid increases as the cube of its velocity. In other words, if you double
the velocity of a jet you will have 2³=8 times more power available.
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